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R
R n

R
R n

2 × 2
SO(3)

2 × 2
SO(3)

R
a, b ∈ R a < b
]a, b[:= {x ∈ R : a < x < b}
[a, b] := {x ∈ R : a ≤ x ≤ b}
]a, b] := {x ∈ R : a < x ≤ b}
[a, b[:= {x ∈ R : a ≤ x < b}
[a, a] := {a}
]a, a[:= ∅
]a, +∞[:= {x ∈ R : x > a}
] − ∞, a[:= {x ∈ R : x < a}
[a, +∞[:= {x ∈ R : x ≥ a}
] − ∞, a] := {x ∈ R : x ≤ a}
] − ∞, +∞[:= R
R ]a, b[ ]a, +∞[ ] − ∞, a[ ∅ R
A ⊆ R R
B ⊆ R
R \ B
R
τ := {A ⊆ R : A R}
A A A
A
R
∅ R
Q R
Z R
r ≥ 0 a ∈ R
r a
r a
B(a, r)
B(a, r[:= {x ∈ R : |x − a| < r} =]a − r, a + r[;
B(a, r] := {x ∈ R : |x − a| ≤ r} = [a − r, a + r].
E ⊆ R R > 0 E ⊆ B(0, R]
(a, b) (a, +∞)
(a, b) ]a, b[ (a, b) ∈ R 2

R
A R a ∈ A δa > 0
B(a, δa[⊆ A
R
∅ R
{Aλ}λ∈Λ
R A :=
Aλ R
λ∈Λ
A1, ..., Am R
A = A1 ∩ ... ∩ Am
R
∅ R
{Cλ}λ∈Λ
R C :=
Cλ R
λ∈Λ
C1, ..., Cm R
C = C1 ∪ ... ∪ Cm
{An = B(0, 1 + 1/n[}n∈N
{An = B(0, 1 − 1/n]}n∈N
x ∈ R V ⊆ R V x A R
x ∈ A A ⊆ V x x
V x V ⊆ U U x
x x x x
x x
V ⊆ R V ∪ {+∞} +∞ a ∈ R
V ⊇]a, +∞[ V ∪ {−∞} −∞ b ∈ R
V ⊇] − ∞, b[
E ⊆ R E
R E R E
E R E
Ē clR(E)
Ē := {C ⊆ R : C ⊇ E, C }.
E ⊆ R x ∈ R Ē
U x R U ∩ E = ∅
x /∈ Ē x R E

R
F G R F G ¯ F ⊇ G
F G G F
E ⊆ R p ∈ R E R
p R E p q ∈ E E
E
E R E
E
R
E R c ∈ R
c E
{xj}j∈N E c c
c E
c ∈ R E ⊆ R c ∈ Ē {xj}j∈N
E c C R
{cj}j∈N C c ∈ R c ∈ C
K R
{xj}j∈N K {xjn}n∈N
x ∈ K
R
E R E
intR(E) := {x ∈ R : E x}.
E
E
E R p ∈ R E p
E p E
R \ E p E
E frR(E) ∂E bdry(E)
D R f : D → R c ∈ D f
c V f(c) U c f(U ∩ D) ⊂ V
R

R n
X τ X τ
X
∅ ∈ τ X ∈ τ
{Aλ}λ∈Λ τ A :=
Aλ ∈ τ
A1, ..., Am τ A := A1 ∩ ... ∩ Am ∈ τ
τ (X, τ)
(X, τ1) (X, τ2) X
τ1 τ2 τ1 ⊇ τ2
τ1 = τ2
X τ1 = {∅, X} τ2 = {A : A ⊆ X}
X
X
R
(X, τ) B ⊆ τ
τ τ B
B X
τ X B
X B X B
X A, B ∈ B x ∈ A ∩ B C ∈ B x ∈ C
C ⊆ A ∩ B X ∅ B
X, Y f : X → Y a ∈ X
V f(a) f −1 (V ) := {x ∈ X : f(x) ∈ V } a
X f X f X
U f : X → Y f(U)
X = R n
R
R n
λ∈Λ

R n
x, y ∈ Rn x = (x1, ..., xn) y = (y1, ..., yn)
d(x, y) := x − y = n
(xi − yi) 2 .
d(x, y) = d(y, x) d(x, y) ≥ 0 d(x, y) = 0 x = y x, y, z ∈ R n
d(x, y) ≤ d(x, z) + d(z, y) a ∈ R n r > 0
r a B(a, r[:= {x ∈ R n : x − a < r}
r a B(a, r] := {x ∈ R n : x − a ≤ r}
R n
B(x1, r1) B(x2, r2) x ∈ B(x1, r1) ∩ B(x2, r2)
δx > 0 B(x, δx) ⊆ B(x1, r1) ∩ B(x2, r2) z ∈ B(x, δx)
d(z, x1) ≤ d(z, x) + d(x, x1) = δx + d(x, x1) d(z, x2) ≤ d(z, x) + d(x, x2) = δx + d(x, x2)
z ∈ B(x1, r1)∩B(x2, r2) d(z, x1) < r1 d(z, x2) < r2
δx < min{r1 − d(x, x1), r2 − d(x, x2)} r1 > d(x, x1) r2 > d(x, x2) δx > 0
{xk}k∈N R n x ∈ R n
lim
k→+∞ xk − x = 0.
R
Rn Rn
R n
R n
x, y ∈ Rn ℓ∞ x = (x1, ..., xn) y = (y1, ..., yn)
dℓ∞(x, y) := x − yℓ∞ = max{|xi − yi|}.
dℓ∞(x, y) = dℓ∞(y, x) dℓ∞(x, y) ≥ 0 dℓ∞(x, y) = 0 x = y
x, y, z ∈ Rn dℓ∞(x, y) ≤ dℓ∞(x, z) + dℓ∞(z, y) a ∈ Rn r > 0
ℓ∞ r a
Bℓ ∞(a, r[:= {x ∈ Rn : x − aℓ ∞ < r} =]x1 − r, x1 + r[× · · · ×]xn − r, xn + r[;
ℓ ∞ r a
Bℓ ∞(a, r] := {x ∈ Rn : x − aℓ ∞ ≤ r} = [x1 − r, x1 + r] × · · · × [xn − r, xn + r].
n = 2 n = 3 2r x
ℓ ∞ R n
ℓ ∞
ℓ ∞ R n
R n
A x
x A x
A x ∈ A
x A A ℓ ∞
R n
i=1

R n
{xk}k∈N R n x = (x (1) , ..., x (n) ) ∈ R n
lim
k→+∞ xk − xℓ∞ = 0 lim
k→+∞ xk − x = 0
j = 1, ..., n lim
k→+∞ x(j)
k = x(j)
Rn
R

(X, τ) D ⊆ X X
τ |D = {A ∩ D : A ∈ τ}
(D, τ |D) τ |D X D τ |D
X D B X B |D = {B ∩D : B ∈ B}
(X, τ)
T0 x, y ∈ X x y
y x
T1 x, y ∈ X U V x ∈ U y /∈ U
y ∈ V x /∈ V
T2 x, y ∈ X U V
x ∈ U y ∈ V
R n
R ∅ R {x ∈ R : x > d}
d ∈ R T0 T1
R ∅ R R
T1 T2
V ∞ R n R n \ V
D ⊆ R n x0 ∈ R n ∪ {∞} D
f : D → R m ℓ ∈ R m ∪ ∞ x x0 f ℓ
V ℓ f −1 (V ) := {x ∈ D : f(x) ∈ V } x0 D
R n
lim
x→x0
x∈D
f(x) = ℓ
x0 ∈ R n ℓ ∈ R m ε > 0
δ > 0 x − x0 < δ x ∈ D f(x) − ℓ < ε
lim
lim
x→x0
x∈D
x−x0→0 +
x∈D
f(x) = ℓ,
f(x) − ℓ = 0.
dℓ∞ d ℓ = (ℓ1,
(f1, ..., fm)
..., ℓm) f =
lim f(x) = ℓ lim fj(x) = ℓj ∀j = 1...m.
x→x0
x∈D
x→x0
x∈D
x0 = ∞ ℓ ∈ Rm ε > 0 M > 0 x > M x ∈ D
f(x) − ℓ < ε
lim f(x) = ℓ,
x→∞
x∈D

lim
x→+∞
x∈D
f(x) − ℓ = 0.
x0 ∈ Rn ℓ = ∞ M > 0 δ > 0 x − x0 < δ x ∈ D
f(x) > M
lim f(x) = ∞,
x→x0
x∈D
lim
x−x0→0 +
x∈D
f(x) = +∞.
x0 = ∞ ℓ = ∞ M > 0 N > 0 x > N x ∈ D
f(x) > M
lim f(x) = ∞,
x→∞
x∈D
lim
x→+∞
x∈D
f(x) = +∞.
R n
x → x0 f :
D → R m m fj : D → R
f : R n → R m = 1
f : R 2 → R (x0, y0) ∈ R 2
lim f(x, y),
(x,y)→(x0,y0)
lim
x→x0
y→y0
f(x, y).
lim lim f(x, y)
x→x0 y→y0
y x
x x
lim lim f(x, y)
y→y0 x→x0
x y
y y
lim lim f(x, y) = lim lim f(x, y)
y→y0 x→x0
x→x0 y→y0
f(x, y) = x 2 /(x 2 + y 2 ) R 2 \ {(0, 0)}
lim
y→0
y=0
lim
x→0
x=0
⎛
⎝ lim
x→x0
x=0
⎛
⎝ lim
y→y0
y=0
x 2
x 2 + y 2
x 2
x 2 + y 2
⎞
⎠ = lim 0 = 0.
⎞
y→0
y=0
⎠ = lim 1 = 1.
y→0
y=0

f : D → R
D ⊆ Rn γ : [a, b] → D a, b ∈ R
a < b γ(a) γ(b) x0 f ◦ γ : [a, b] → R
γ(b) = x0 lim
x→x0
x∈D
γ(a) = x0 lim
x→x0
x∈D
lim
t→b
f(x) lim f ◦ γ(t),
t→b− f(x) lim f ◦ γ(t),
t→a +
t→a
f ◦ γ(t), lim f ◦ γ(t),
− +
γ
f : D → R D ⊆ Rn c D
lim f(x, y) ℓ
x→c
x∈D
{xn}n∈N ⊆ D xn → c lim
n→∞ f(xn) = ℓ
γ : [a, b[→ D lim
t→b
t→b
γ(t) = c lim f(t) = ℓ
− −
f : D → R D ⊆ R2 (x0, y0)
D ε > 0 ] − ε, ε[×{y0} ∪ {x0}×] − ε, ε[⊂ D lim f(x, y)
lim
y→y0
lim f(x, y)
x→x0
= lim
x→x0
lim f(x, y)
y→y0
= lim
f(x, y) =
(x,y)→(x0,y0)
(x,y)∈D
(x,y)→(x0,y0)
(x,y)∈D
f(x, y),
x 2
x 2 + y 2 D = R2 \ {(0, 0)} m ∈ R x > 0
γm : [0, x] → D γm(t) = (t, mt) γm
[0, x] R t = 0 γ(t) ∈ D
lim f ◦ γ(t) = lim f(t, mt) = lim
t→a + t→0 + t→0 +
t2 t2 + m2 1
= .
t2 1 + m2 m γm
γm m m f
γm
f(x, y) = xy2
x 2 + y 4 D = R \ {(0, 0)} γm(t) = (t, mt)
lim
t→0
f ◦ γ(t) = lim f(t, mt) = lim
t→0 t→0
m2t3 t2 + m4 = lim
t4 t→0
m2t 1 + m2 = 0,
t2 m γ+(t) = (t 2 , t)
γ−(t) = (−t 2 , t)
lim
t→0 f ◦ γ+(t) = lim f(t
t→0 2 t
, t) = lim
t→0
4
t4 1
=
+ t4 2 ,
lim
t→0 f ◦ γ−(t) = lim f(−t
t→0 2 −t
, t) = lim
t→0
4
t4 = −1
+ t4 2 ,
f
(x, y) → (0, 0)

f : D → Rm D ⊆ Rn c D f
c
lim f(x) − f(c) = 0,
x→c
x∈D
j = 1, ..., m
f = (f1, ..., fm)
lim fj(x) = fj(c),
x→c
x∈D
R 2
(x, y) ∈ R x = ρ cos θ y = ρ sin θ
R 2 \ {0} ρ = x 2 + y 2 |(x, y)| = ρ f : D → R D ⊆ R 2
(0, 0) D
lim
(x,y)→(0,0)
(x,y)∈D
f(x, y) = lim
ρ→0 +
(ρ cos θ,ρ sin θ)∈D
f(ρ cos θ, ρ sin θ),
θ
θ
θ
(x, y) → (0, 0) f(x, y) = x 4 arctan y
f(ρ cos θ, ρ sin θ) = ρ 4 cos θ arctan(ρ sin θ)
| cos θ arctan(ρ sin θ)| ≤ π/2
0 ≤ |f(ρ cos θ, ρ sin θ)| ≤ π
2 ρ4 ,
ρ → 0 + 0 θ
0
R2
sin
f(x, y) =
arctan y
x , x = 0;
0, x = 0.
(x, y) x = 0 (0, 0)
lim sin arctan y
= lim sin (arctan tan θ) = sin θ
x ρ→0 +
(x,y)→(0,0)
x=0
θ f (0, 0) (0, ¯y) ¯y > 0
lim
(x,y)→(0,¯y)
x>0,y=¯y
lim
(x,y)→(0,¯y)
x=0
f(x, y) = 0
f(x, y) = lim f(x, ¯y) = 1,
x→0 +
(0, ¯y) ¯y > 0 (0, ¯y) ¯y < 0
lim
(x,y)→(0,¯y)
x>0,y=¯y
f(x, y) = lim f(x, ¯y) = −1,
x→0 +
(0, ¯y) ¯y < 0
f (x, y) x = 0

A =]0, +∞[×]0, +∞[ f : A → R
f(x, y) = x2 − y 2 + 2xy
y 2 + 3xy + x .
lim
(x,y)→(0,0)
(x,y)∈A
f(x, y).
x = y
2x
f(x, x) =
2
x2 + 3x2 2x
=
+ x 4x + 1 .
0 x → 0 0 y = mx
x = my
0
y = mx x = my
x = my 2 m > 0
f(my 2 , y) = m2 y 4 − y 2 + 2my 3
y 2 + 3my 3 + my 2
= m2y2 − 1 + 2my
1 + 3my + m
lim
y→0 + f(my2 , y) = − 1
1 + m
m > 0
lim
(x,y)→(0,0)
x,y∈A
f(x, y)

α > 0
⎧
| sin(xy) − xy|
⎪⎨
f(x, y) =
⎪⎩
α
(x2 + y2 ) 3 (x, y) = (0, 0),
0 (x, y) = (0, 0).
α f (0, 0)
α ≤ 0
α ≤ 0 (x, y) → 0
| sin(xy) − xy| α
(x 2 + y 2 ) 3
≥
1
(x 2 + y 2 ) 3
sin(xy) − xy β > 0
sin(s) − s
lim
s→0 sβ
sin(s) − s
lim
s→0 sβ cos(s) − 1
= lim
s→0 βsβ−1 = lim
s→0
− sin(s)
,
β(β − 1)sβ−2 β − 2 = 1 β = 3
α > 0
lim
(x,y)→(0,0)
(x,y)=(0,0)
f(x, y) = lim
(x,y)→(0,0)
(x,y)=(0,0)
sin(s) − s
lim
s→0 s3 = − 1
6 .
| sin(xy) − xy|
|xy| 3
α
|xy| 3α
(x2 + y2 1
=
) 3 6α |xy| ≤ 1
2 (x2 + y 2 )
0 ≤ |xy|α
x2 1
≤
+ y2 2α (x2 + y 2 ) α−1 .
α > 1
lim
(x,y)→(0,0)
(x,y)=(0,0)
|xy| α
x2 = 0, lim
+ y2 (x,y)→(0,0)
(x,y)=(0,0)
⎛
⎜
⎝ lim
(x,y)→(0,0)
(x,y)=(0,0)
f(x, y) = 0 = f(0, 0),
|xy| α
x2 + y2 α > 1 f α ≤ 1 y = mx
|xy| α
x2 |m|α
=
+ y2 m2 + 1
|x| 2α
,
x2 α < 1 x → 0 +∞ α = 1 |m|/(m 2 + 1) m
f f α > 1
⎞
⎟
⎠
3

lim
(x,y)→(0,0)
(x,y)=(0,0)
|xy| α
x2 =
+ y2 lim
ρ→0 +
x=ρ cos θ
y=ρ sin θ
|ρ2 sin θ cos θ| α
ρ 2 = lim
ρ→0 +
x=ρ cos θ
y=ρ sin θ
1
2 α ρ2α−2 | sin 2θ| α .
α > 1 θ α = 1 +∞
0 < α < 1 θ /∈ {0, π/2, π, 3π/2}
lim
(x,y)→(0,0)
1 − cos(xy)
log(1 + x 2 + y 2 )
lim
x→0
y→0
1 − cos(xy)
(xy) 2
(x2 + y2 )
log(1 + x2 + y2 ) (x2 + y2 )
(xy) 2
lim
(x,y)→(0,0)
1
=
2 lim
x→0
y→0
x3y2 x4 .
+ y6 (xy) 2
(x2 + y2 1
=
) 2 lim
ρ→0 +
ρ4 cos2 θ sin2 θ
ρ2 z = x 2 v = y 3 |x| = z 1/2 y = v 1/3
lim
(x,y)→(0,0)
x3y2 x4 + y6
= lim
(x,y)→(0,0)
z = ρ cos θ v = ρ sin θ
lim
x
3y2
(x,y)→(0,0)
x 4 + y 6
|x| 3y2 x4 + y6 = lim
ρ→0 + ρ3/2+2/3 | cos θ|3/2 sin 2/3 θ
ρ 2
= lim
z→0 +
v→0
|z| 3/2v2/3 z2 = lim
+ v2 (z,v)→(0,0)
|z| 3/2v2/3 z2 .
+ v2 = 0.
= lim
ρ→0 + ρ1/6 | cos θ| 3/2 sin 2/3 θ ≤ lim
ρ→0 + ρ1/6 = 0.
lim
(x,y)→(1,0)
y 2 log x
(x − 1) 2 + y 2 lim
(x,y)→(0,0)
sin(x 2 + y 2 )
x 2 + y 2
sin(x
lim
(x,y)→(0,0)
y=0
2 + y2 )
x2y2 + y4 x
lim
(x,y)→(0,0)
3 + x sin2 (y)
x2 + y2 xy(x
lim y arctan(y/x) lim
(x,y)→(0,0) (x,y)→(0,0)
2 − y2 )
x2 + y2 xy
lim
(x,y)→(0,0) x2 + xy + y2 xyz
lim
(x,y,z)→(0,0,0) x
(x,y)=(0,0)
2 + y2 (x
lim
|(x,y,z)|→+∞
2 + y2 ) 2
x2 + z2 lim
|(x,y,z)|→+∞
lim
|(x,y,z)|→+∞ x4 + y 2 + z 2 − x + 3y − z lim
|(x,y,z)|→+∞ x4 + y 2 + z 2 − x 3 + xyz − x + 4
(0, 0) 0
1
1
xz

y 2
y = 0 +∞
0
arctan α ≤ π/2 0
0
θ
0
x 2 + y 2 ≥ 2|xy|
|z|/2 0
(t, 0, 0) (0, 0, t)
(t, t, t) (t −1 , t, t −1 )
+∞ ±∞ |(x, y, z)| → +∞
+∞ +∞
(t, 0, 0) (t, t 2 , t 2 ) t → ±∞
n ∈ N \ {0}
lim
(x,y)→(0,0)
x|y| 1/n
x2 + y2 = 0
+ |y|
|y|
|x||y| 1/n−1 n = 1 n ≥ 2 γ(t) = (t, t 2 )
0 1/2 n = 2 +∞ n > 2
α ∈ R
|y|
lim
(x,y)→(0,0)
α
x e−y2 /x2 = 0.
|f(x, y)| = ρ α−1 | sin θ| α−1 | | tan θ| e − tan2 θ ≤ Mρ α−1 ,
M = maxt∈R{|t|e−t2} t ↦→ |t|e−t2
α > 1 γ(t) = (t, t) e−1
α = 1 ∞ t < 1
E ⊆ R 2 E :=
{(x, y) ∈ R 2 : x 2 + cos(y) > 1}
f(x, y) = x 2 + cos(y) E = f −1 (]1, +∞[)
f E
f −1 ([1, +∞[)
E f −1 ([1, +∞[) ⊇ E E

E E E
f −1 ([1, +∞[) ⊆ E (¯x, ¯y) ∈ f −1 ([1, +∞[)
{(xn, yn)}n∈N E (xn, yn) → (¯x, ¯y) f(¯x, ¯y) ≥ 1
¯x 2 + cos(¯y) ≥ 1
¯x ≥ 0 xn = ¯x + 1/n yn = ¯y (xn, yn) → (¯x, ¯y) x 2 n > ¯x 2
f(xn, yn) > f(¯x, ¯y) ≥ 1 f(xn, yn) > 1 {(xn, yn)}n∈N ⊆ E
E (¯x, ¯y) (¯x, ¯y) ∈ E
¯x < 0 xn = ¯x − 1/n yn = ¯y (xn, yn) → (¯x, ¯y) x 2 n > ¯x 2
f(xn, yn) > f(¯x, ¯y) ≥ 1
E = f −1 ([1, +∞[) = {(x, y) ∈ R 2 : x 2 + cos(y) ≥ 1}.
E
E
R 2 \ E = R 2 \ f −1 (]1, +∞[) = f −1 (] − ∞, 1]),
E E
∂E = E ∩ R 2 \ E = E ∩ (R 2 \ E) = f −1 (] − ∞, 1]) ∩ f −1 ([1, +∞[) = f −1 (1),
∂E = {(x, y) ∈ R 2 : f(x, y) = 1}
(X, d) {xn}n∈N X x ∈ X
E = {xn : n ∈ N} ∪ {x}
E x
(X, d) E ⊆ X Ē = {x ∈ X : inf{d(x, y) : y ∈
E} = 0}
dE(x) = inf{d(x, y) : y ∈ E} x ∈ Ē dE(x) >
0 x X \ E x /∈ Ē
dE(x) = 0 x /∈ Ē x X \E
δ > 0 x E dE(x) ≥ δ > 0
dE(x) = 0

D ⊆ RN {fn}n∈N fn : D → RM
f : D → RM
{fn}n∈N f f {fn}n∈N
x ∈ D lim
n→∞ fn(x) = f(x) lim
n→∞ fn(x) − f(x) = 0
{fn}n∈N f f {fn}n∈N
lim
n→∞ fn − f∞ = 0 lim
n→∞ sup fn(x) − f(x) = 0
R
R
{an}n∈N an → 0 |fn(x) − f(x)| ≤ an x ∈ D
D
D D ′ ⊂ D
fn, f C 1
sup
|fn − f|
e
fn : R → R fn(x) =
1
−xt
dt
1 + t2 fn
fn
x, n lim |fn(y) − fn(x)| = 0 y = x + h
y→x
n
e
|fn(y) − fn(x)| = |fn(x + h) − fn(x)| =
−(x+h)t n e
dt −
1 + t2 xt
dt
1 + t2
=
=
n
1
n
1
e −xt (e −ht − 1)
1 + t 2
e −xt |e −ht − 1|
1 + t 2
x∈D
dt
≤
dt =
1
1
n
e
1
−xt (e−ht − 1)
1 + t2 n e−xt |1 − e−ht |
1 + t2 dt.
h > 0 |1 − e −ht | = 1 − e −ht t > 0 h > 0 e −ht ≤ 1
|fn(y) − fn(x)| ≤
n
1
1
e −xt (1 − e −ht )
1 + t 2
dt
dt
n

s ↦→ 1 − e −s s ≥ 0 1 − e −s ≤ s
s ≥ 0 w(s) = (1 − e −s ) − s w(0) = 0 w ′ (s) = e −s − 1 < 0
s > 0 w w(s) < w(0) s > 0
1 − e −s ≤ s s ≥ 0 s = ht
|fn(y) − fn(x)| ≤
n
1
e −xt (1 − e −ht )
1 + t 2
dt ≤
n
1
e−xt n
ht
e
dt = h
1 + t2 1
−xtt dt
1 + t2 x, n t
[1, n] M = M(x) [1, n]
n e
|fn(y) − fn(x)| ≤ h
1
−xt n
t
dt ≤ h
1 + t2 1
M dt = hM(n − 1)
h → 0 +
h < 0 |e −ht − 1| = e −ht − 1 = e |h|t − 1 h < 0 t > 0
e −ht > 1
e
lim
|h|t→0
|h|t − 1
= 1 < 2,
|h|t
|h|t e |h|t − 1 < 2|h|t
|fn(y) − fn(x)| ≤
n
1
e −xt |e −ht − 1|
1 + t 2
dt ≤
n
1
e−xt n
2|h|t
e
dt = 2|h|
1 + t2 1
−xtt dt
1 + t2 h → 0 −
lim
h→0 |fn(x + h) − fn(x)| = 0 fn
x ∈ R
fn [1, n] [1, n + 1]
{fn(x)}n∈N x
x < 0 t ↦→ e−xt
1 + t 2 +∞ t → ∞ ¯t > 1
e −xt
1 + t 2 > 1 n > ¯t
|fn(x)| = fn(x) =
≥
n
e−xt dt =
1 + t2 1
¯t e
1
−xt n
dt +
1 + t2 ¯t
1 dt =
¯t e
1
−xt n e
dt +
1 + t2 ¯t
−xt
dt
1 + t2 ¯t e
1
−xt
1 + t2 dt + (n − ¯t).
+∞ n → +∞ fn(x)
x < 0
x < 0 e−xt ≤ 1
fn(x) ≤
n
1
1
π
dt = arctan n −
1 + t2 4
π π
≤ −
2 4
= π
4 ,
fn(x)

x ∈ [0, +∞[
fn
f(x) =
∞
1
e−xt dt
1 + t2 R [0, +∞[
[0, +∞[
|f(x, y) − fn(x, y)| =
∞
1
e−xt dt −
1 + t2 n
1
e−xt
dt
1 + t2 =
∞
n
e−xt π
dt ≤ − arctan n,
1 + t2 2
fn(x) ≤ f(x) x
eα < 1 α < 0
sup |f(x) − fn(x)| ≤
x∈[0,+∞[
π
− arctan n,
2
[0, +∞[
fn :
R 2 → R fn(x, y) =
2 n (x + y)
1 + n2 n (x 2 + y 2 )
fn f(x, y) = 0
R 2 fn(0, 0) = 0 limn→∞ fn(0, 0) = 0 (x, y) = (0, 0)
|fn(x) − f(x)| = |fn(x)| ≤ 1
n
|x + y|
(x 2 + y 2 )
n → ∞ 1 +
n2 n (x 2 + y 2 ) > n2 n (x 2 + y 2 )
1
1 + n2 n (x 2 + y 2 ) <
1
n2 n (x 2 + y 2 ) .
fn
f fn
sup
(x,y)∈R2 |fn(x, y) − f(x, y)| = sup
(x,y)∈R2 |fn(x, y)| = sup |fn(ρ cos θ, ρ sin θ)|
ρ≥0
θ∈[0,2π]
= sup
ρ≥0
θ∈[0,2π]
2nρ 1 + n2n | cos θ + sin θ|
ρ2 | cos θ + sin θ| ≤ √ 2 θ ∈ [0, 2π]
θ1 = π/4 θ2 = 5π/4
sup |fn(ρ cos θ, ρ sin θ)| = sup
ρ≥0
ρ≥0
θ∈[0,2π]
2 n√ 2ρ
1 + n2 n ρ 2 = √ 2 sup
ρ≥0
2 n ρ
1 + n2 n ρ 2
g(θ) := cos θ + sin θ [0, 2π]
0 = − sin θ + cos θ θ = π/2, 3/2π tan θ = 1 θ1 = π/4 θ2 = 5π/4
|g(θ1)| = |g(θ2)| = √ 2 > 1 = |g(π/2)| = |g(3π/2)| = |g(0)| = |g(2π)|.

Fn : [0, +∞[→ R Fn(ρ) = 2n ρ
1+n2 n ρ 2 Fn(0) = 0 lim
ρ→+∞ Fn(ρ) = 0
F ′ n(ρ) = 2n − 4nnρ2 (2nnρ2 2 ,
+ 1)
ρn = 1/ √ n2 n
Fn
sup
ρ≥0
θ∈[0,2π]
|fn(ρ cos θ, ρ sin θ)| = √ 2Fn (ρn) = √ 2
2 n
2 √ .
n2n +∞ n → +∞ R 2
{(ρn cos θ1, ρn sin θ1), (ρn cos θ2, ρn sin θ2)}
(0, 0)
(0, 0)
(0, 0) ¯ρ > 0
sup |fn(x, y) − f(x, y)| = sup |fn(x, y)| = sup |fn(ρ cos θ, ρ sin θ)|
(x,y)∈R\B((0,0),¯ρ)
(x,y)∈R\B((0,0),¯ρ)
ρ≥¯ρ
θ∈[0,2π]
= √ 2 sup
ρ≥¯ρ
2 n√ 2ρ
1 + n2 n ρ 2 = √ 2 sup
ρ≥¯ρ
Fn [ρn, +∞[ ρn
n ρn < ¯ρ Fn
[¯ρ, +∞[ Fn(¯ρ) ≥ Fn(ρ) ρ ≥ ¯ρ
F (ρ)
sup |fn(x, y) − f(x, y)| =
(x,y)∈R\B((0,0),¯ρ)
√ 2
2 sup
ρ≥¯ρ
n√2ρ 1 + n2nρ2 = √ 2F (¯ρ) = √ 2
2 n ¯ρ
1 + n2 n ¯ρ 2
|fn(¯ρ cos θ1, ¯ρ sin θ1)|,
R
fn(x) = nxe −n2x2 fn : R → R
fn(x) =
nx
1 + n 2 x 2 fn : R → R
fn(x) = nx
1 + nx fn : [0, 1] → R
fn(x) = (x 2 − x) n fn : [0, 1] → R
x ∈ R |fn(x)| n → ∞
f(x) = 0 R fn

f ′ n(x) = ne −n2 x 2
(1 − 2n 2 x 2 ) x + n = 1/n √ 2
x − n = −1/n √ 2
sup
x∈R
|fn(x) − f(x)| = sup |fn(x)| ≥ |f(x
x∈R
± n )| = 1
√ e
2 −2 .
n → +∞
R 2 fn(x) 0
0 {x : |x| ≥ ε} ε > 0 |fn(x)| = |fn(−x)|
sup |fn(x) − f(x)| = sup |fn(x)|.
|x|≥ε
|x|≥ε
n x + n , x − n ∈] − ε, ε[ |f(ε)| > |f(x)| |x| ≥ ε
|F |
x + n , x − n ] − ∞, x − n [ ]x + n , +∞[
sup |fn(x) − f(x)| = |fn(ε)|,
|x|≥ε
0
R
fn(0) = 0 x = 0 |fn(x)| n → ∞
f(x) = 0 R2 fn
x + n = 1/n x− n = −1/2
|fn(x ± n ) − f(x)| = 1/2
0
0
fn(0) = 0 x = 0 fn(x) 1 n → +∞ fn
[0, 1] f f(0) = 0
f(x) = 1 x ∈]0, 1] [0, 1]
f
[ε, 1] ε > 0
sup |fn(x) − f(x)| = sup |fn(x) − 1| = sup
1
[ε,1]
[ε,1]
[ε,1] 1 + nx
=
1
1 + nε
0 n → +∞
x2 − x ≤ 1/2 x ∈ [0, 1] |fn(x)| ≤ 1/2n → 0 x ∈ [0, 1]

I =]a, b[ R {fn}n∈N fn :
I → R {sn}n∈N
n
sn(x) = fj(x).
{sn}n∈N
∞
fj(x)
j=0
j=0
∞
fj(x) s : I → R
s s x ∈ I
lim
n→∞ sn(x) = s(x) lim
n→∞ sn(x) − s(x) = 0
n
x ∈ I lim fj(x) = s(x)
n→∞
j=0
s s lim
n→∞ sn − s∞ = 0
s
∞
{an}n∈N |fn(x)| ≤ an n ∈ N x ∈ I an < +∞
{fn}n∈N I
R R s s
I
{fn}n∈N I R
R s
∞
s(x) dx = fn(x) dx,
I
n=0
∞ e
n=0
−nx
]0, +∞[
n + x
]c, +∞[ c > 0
]0, +∞[
x < 0 n > |x|
e−nx en2
≥
n + x 2n
I
→ +∞.
n=0
j=0

x = 0 1/n
x > 0
e −(n+1)x
n + 1 + x
e −nx
−x n + x
= e
n + x + 1
= e−x
1
1 + 1
n+x
n + x
x > 0
s(·) {sN}N∈N
c > 0 sup
< 1,
fn(x) = e−nx
n + x , f ′ n(x) = −e −nx 1 + n2 + nx
(n + x) 2 .
f ′ n(x) = 0 x = −(1 + n 2 )/n < 0 fn(x) [c, +∞[
∞
sup |fn(x)| = fn(c),
x∈]c,+∞[
sup
n=0 x∈]c,+∞[
|fn(x)| =
∞
fn(c),
]c, +∞[
]0, +∞[ M > N + 1
∞
e
|s(x) − sN(x)| = sup
−nx
n + x −
N e−nx
∞
e
= sup
n + x
−nx
M
e
≥ sup
n + x
−nx
n + x
sup
x>0
x>0
n=0
n=0
n=0
x>0
n=N+1
{xj}j∈N xj → 0 +
∞
n=N+1
sup
1
n
x>0
M
n=N+1
e−nx
≥ lim
n + x
j→∞
M
n=N+1
e−nxj
n + xj =
+∞ M > 0
∞ e
sup
−nx
n + x −
x>0
n=0
N e−nx
> N
n + x
n=0
M
n=N+1
M
n=N+1
1
n
.
x>0
n=N+1
1
> N
n
N → ∞ +∞ ]0, +∞[
∞ n
n
n=1
3
+ x
[0, +∞[
1/n2 [0, +∞[
∞
sup
n
n3
+ x
≤
+∞ 1
< +∞,
n2 n=1
x≥0
∞ n + x
n
n=1
3 [0, +∞[
+ x
[0, +∞[ [0, +∞[
n=1

K [0, +∞[ R > 0 B(0, R) ⊇ K
∞
sup
n + x
n3
+ x
≤
+∞ n + R
n3 =
+∞
∞ 1 1
+ R < +∞,
n2 n3 n=1
x∈K
n=1
[0, +∞[ [0, +∞[
[0, +∞[
∞
n + x
sup
n3 + x −
N n + x
n3
≥ sup
+ x
2N n + x
n3
+ x
x≥0
n=1
n=1
n=1
x≥0
n=1
n=N+1
sup xj
2N
n + x
sup
n3
+ x
≥
2N
1 = N − 1,
x≥0
n=N+1
N → +∞
∞
0
n=1
n=1
n=N+1
e−n√x n2 [0, +∞[
+ 1
∞
e
sup
−n√x n2
+ 1
≤
n=1
x≥0
n=1
∞
n=1
1
< +∞,
n2
∞
log 1 +
n=1
x
n2
[0, +∞[
[0, +∞[
1 ∞
log 1 + x
n2 ∞
= (1 + n 2
) log 1 + 1
n2
− 1 .
s > 0 log(1 + s) < s g(s) =
log(1 + s) − s g(0) = 0 g ′ (s) = 1
1+s − 1 < 0 s > 0 g(s) < g(0) ≤ 0 s > 0
K
+∞
sup log 1 + x
n2 ∞
1
≤ sup |x| < +∞,
x∈K n2 n=1
x∈K
K
[0, 1]
1
log 1 +
0
x
n2
dx = n 2
1+1/n2 log y dy = n
1
2 [y log y − y] y=1+1/n2
y=1
= n 2
1 + 1
n2
log 1 + 1
n2
− 1 − 1
+ 1 ,
n2
n=1

a ∈ N a ≥ 2
∞ an−1zn +∞
, anz
n!
an−a .
n=0
∞ an−1zn n=0
n!
r = ∞
n=0
= 1
a
∞ anzn n=0
n!
n=1
= 1
a
∞
n=0
(az) n
n!
∞
anz an−a ∞
= a n(z a ) n−1
n=0
= eaz
a
w = za
∞
anz
n=0
an−a ∞
= a nw
n=0
n−1 ∞ d
= a
dw
n=0
wn = a d
∞
w
dw
n=0
n
= a d
1 a a
= =
dw 1 − w (1 − w) 2 (1 − za ) 2
∞ n=0 (za ) n |za | < 1
r = 1
f : R → R 2π f(t) = π
− t
2
t ∈ [0, π]
f
t = 0
f(t) = a0
2 +
∞
an cos nt n ∈ N n = 0
n=1
a0 = 1
π
f(t) dt =
π −π
2
π
f(t) dt = 0
π 0
an = 1
π
f(t) cos nt dt =
π −π
2
π
π
− t cos nt dt =
π 0 2 2
π π π
sin nt sin nt
− t − −
π 2 n 0 0 n dt
= 2
π cos nt
− =
nπ n 0
2 (1 − (−1)
π
n )
n2
f = 2
∞ (1 − (−1)
π
n )
n2 cos nt.
n=1

an = 0 n = 2k an = 4/(πn2 ) n = 2k + 1
f = 4
∞ cos ((2k + 1)t)
π (2k + 1) 2 .
k=0
L2
S(t) = 2
∞ (1 − (−1)
π
n )
n2 cos nt,
n=1
t = kπ k ∈ Z f
S(t) = f(t)
t = kπ k ∈ Z f
S(t) = f(t)
S(0) = f(0) = π/2 π
2
∞
k=0
1 π2
=
(2k + 1) 2 8
= 4
π
∞
k=0
1
(2k + 1) 2
f : R → R 2π f(t) = 3(π + t)
t ∈ [−π, 0] f
∞ 1
(2n + 1) 4
L2 (0, 2π)
π
|f(t)|
−π
2 0
dt = 2 9(π + t)
−π
2
(π + t) 3 0
= 18
= 6π
3 −π
3 .
f
S(t) = a0
2 +
∞
ak cos(kt),
k=1
a0 = 1
π
f(t) dt =
π −π
2
0
f(t) dt =
π −π
2
0
3(π + t) dt =
π −π
6
(π + t) 2 0
= 3π
π 2
ak = 1
π
= 6
π
π
f(t) cos(kt) dt =
−π
2
π
(π + t) sin(kt)
0 −
k
6
π
−π
0
−π
0
−π
f(t) cos(kt) dt = 2
π
sin(kt)
k
dt = − 6
kπ
0
−π
−π
3(π + t) cos(kt) dt
− cos(kt)
k
0
−π
n=0
= 6
πk 2 (1 − (−1)k )
k ≥ 1 ak = 0 k = 2n ak = 12/(πk2 ) k = 2n + 1
S(t) = 3
∞ 12 cos((2n + 1)t)
π +
2 π (2n + 1) 2 .
1
2π
π
−π
n=0
|f(t)| 2 dt =
a0 2
4
+ 1
2
∞
k=1
a 2 k ,

π 4 /96
3π 2 = 9
4 π2 + 72
π2 ∞
n=0
1
.
(2n + 1) 4
f : R → R 2π
2 t ∈ [−π, 0[
f(t) =
−1 t ∈ [0, π[
∞
n=0
1
.
(2n + 1) 2
f ∈ L 2 (−π, π)
f 2 L2 π
=
−π
|f(t)| 2 dt =
0
−π
|f(t)| 2 dt +
π
0
|f(t)| 2 dt = 4π + π = 5π < +∞.
f ∈ L2 (−π, π)
a0 = 1
π
f(t) dt =
π −π
1
0
f(t) dt +
π −π
1
π
f(t) dt = 2 − 1 = 1
π 0
ak = 1
π
f(t) cos kt dt =
π −π
1
0
f(t) cos kt dt +
π −π
1
π
f(t) cos kt dt
π 0
= 2
0
t=0 sin kt
−
π
k
1
t=π sin kt
= 0
π k
bk = 1
π
= 2
π
cos kt dt −
−π
1
π
cos kt dt =
π 0
2
π
π
0
f(t) sin kt dt = 1
π
−π
−π
0
sin kt dt −
−π
1
π
sin kt dt =
π 0
2
π
f(t) sin kt dt + 1
π
− cos kt
k
t=−π
π
0
t=0 t=−π
= − 2
1
(1/k − cos(−kπ)) + (cos(kπ)/k − 1/k)
π π
= 1
3
(−2 + 2 cos(kπ) + cos(kπ) − 1) = (cos kπ − 1).
πk πk
bk = 0 k bk = −6/(πk) k
f(t) sin kt dt
− 1
π
t=0
− cos kt
g(t) = f(t) − 1/2 3/2 t ∈ [−π, 0[ −3/2
t ∈ [0, π[ ak = 0 k > 0
π
π
g(t) cos kt dt = f(t) − 1
π
cos kt dt = f(t) cos kt dt,
2
−π
f(t) = a0
2 +
∞
ak cos kt + bk sin kt = 1 6
−
2 π
k=0
1
2π
π
−π
−π
|f(t)| 2 dt = a2 0
4
∞
k=0
−π
1
2
+ ak + b
2
2 k ,
k
sin ((2k + 1)t)
.
2k + 1
t=π
t=0

5 1 18
= +
2 4 π2 ∞
n=0
∞
n=0
1
,
(2n + 1) 2
1 π2
=
(2n + 1) 2 8 .
u : R → R 2π
−t − π ≤ t < 0,
u(t) =
π 0 ≤ t < π.
u
∞
k=0
1
.
(2k + 1) 2
u |u(t)| [−π, π]
a0 = 1
π
π
−π
π
f(t) dt = 1
π
t=−π
0
−π
an = 1
f(t) cos(nt) dt =
π −π
1
π
= 1
t=0 −t sin(nt)
+
π n
1
nπ
−t dt + 1
π
0
−π
0
−π
π
0
π dt = π 3π
+ π =
2 2 .
−t cos(nt) dt + 1
π
π
sin(nt) dt + 1
[sin nt]t=π t=0
n
0
π cos(nt) dt
= − 1
n2 1 − (−1)n
[cos nt]t=0 t=−π = −
π n2 .
π
bn = 1
π
f(t) sin(nt) dt =
π −π
1
0
−t sin(nt) dt +
π −π
1
π
π sin(nt) dt
π 0
= 1
t=0 t cos(nt)
−
π n
1
0
cos(nt) dt −
nπ
1
[cos nt]t=π t=0
n
t=−π
= (−1)n
n − (−1)n − 1
=
n
1
n .
f(t) = a0
2 +
∞
n=1
−π
an cos(nx) + bn sin(nx) = 3π
4 +
∞
n=1
(−1) n − 1
n 2 π
cos(nx) +
sin nx
n .
u C ∞ u
u
xk = 2kπ k ∈ Z
π 0 (2k + 1)π k ∈ Z
0 u π/2
π
2
= 3π
4 −
∞ 1 − (−1) n
πn2 .
n=1

∞ π 2 1
= ,
4 π (2n + 1) 2
π 2 /8
n=0

g(x) := x(π−x) x ∈ [0, π] u
g [−π, π] R 2π
u
an u
bn = 1
π
u(x) sin nx dx =
π −π
2
π
u(x) sin nx dx =
π 0
2
π
g(x) sin nx dx
π 0
= 2
π
x(π − x) sin nx dx =
π 0
2
π
(−x
π 0
2 + πx) sin nx dx
= 2
cos nx
−
π n (−x2 π + πx) +
0
2
π
cos nx(−2x + π) dx =
πn 0
2
π
cos nx(−2x + π) dx
πn 0
= 2
π sin nx
(−2x + π) +
πn n 0
4
πn2 π
sin nx dx =
0
4
πn2 π
sin nx dx
0
= 4
πn2
cos nx
π − =
n 0
4
πn3 (1 − (−1)n )
b2k = 0 b2k+1 = 8/(π(2k + 1) 3 ) k ∈ N
u(x) = 8
∞ 1
π (2k + 1) 3 sin (2k + 1)x .
x ∈ R
1
(2k
+ 1) 3 sin (2k + 1)x
1
≤ ,
(2k + 1) 3
∞
sup
1
(2k
+ 1) 3 sin (2k + 1)x ∞
1
≤
< +∞,
(2k + 1) 3
k=0
k=0
x∈R
k=0
(2k + 1) −3 < 2−3k−3 < 1/(8k2 )
u(x)
u g(x) = π
2 − x − π
2 [0, π]
[−π, π] 2π R
u
an bn
bn = 1
π
u(x) sin nx dx =
π −π
2
π
u(x) sin nx dx =
π 0
2
π
g(x) sin nx dx
π 0
= 2
π
π
π 0 2 −
x − π
sin nx dx
2
= 2
π/2
x sin nx dx +
π
2
π
(π − x) sin nx dx =
π
4
sin n
πn2 π
.
2
0
π/2

u(x) = 4
π
∞
sup
n2 n=1
x∈R
sin n π
2
∞
n=1
sin n π
2
n2 sin(nx).
sin(nx)
≤
∞
n=1
1
< ∞,
n2 u g(x) = x [0, π]
[−π, π] 2π R u
S1 :=
n=1
∞
k=0
1
(2k + 1) 2 , S2 :=
a0 = 2
π
x = π
π 0
an = 2
π
x cos(nx) dx =
π 0
2
x=π x sin(nx)
π n x=0
x = π
∞ 2 1 − (−1)
−
2 π
n
n2 cos(nx) = π 4
−
2 π
∞
n=1
1
.
n2 − 2
π
sin(nx) dx = −
nπ 0
2
n2π (1 − (−1)n ),
∞
k=0
cos(2k + 1)x
(2k + 1) 2 .
1/n2
x = 0
0 = π
∞ 4 1
− .
2 π (2k + 1) 2
S2 =
∞
n=1
1
=
n2 ∞
k=1
∞
k=0
1
+
(2k + 1) 2
k=0
1 π2
=
(2k + 1) 2 8 .
∞
k=1
1
(2k) 2 = S1 + 1
4
S2 = 4
3 · S1 = π2
6 .
∞
k=1
1
k 2 = S1 + S2
4 ,

X Y
T : X → Y T ℓ > 0
T xY ≤ ℓxX.
ℓ
T L
ℓ = sup{T xY : xX ≤ 1} = sup{T xY : xX = 1}.
L(X, Y ) X Y
(L(X, Y ), · L)
X Y R D X
f : D → Y u ∈ X X uX = 1 p ∈ D
f(p + tu) − f(p)
lim
=: v ∈ Y.
t→0,t=0 t
v f p u
v = ∂f
∂u (p) = Duf(p) = ∂uf(p).
X = R n u = ei Dei f(p) = Dif(p)
f p
X Y R D X p ∈ D
f : D → Y T : X → Y f p
f p T
f(x) − f(p) − T (x − p)Y
lim
= 0.
x→p x − pX
f p T = f ′ (p) = Df(p) f
p p Df(p)u = ∂uf(p) ∈ Y
X = R n Y = R m p R n
R m
X, Y R n R m
Matn×m(R)
f : Rn → Rm p n × m
f = (f1, ..., fm)
⎛
∂x1
⎜
Jac f(p) := ⎝
f1(p)
. . . ∂xnf1(p)
⎞
⎟
⎠ .
∂xnfm(p) . . . ∂xnfm(p)

v = (v1, ..., vm)
df(p)(v) = Jac f(p)(v) =
⎛
⎜
⎝
∂x1f1(p)
. . . ∂xnf1(p)
∂xnfm(p) . . . ∂xnfm(p)
⎞ ⎛
⎟ ⎜
⎠ ⎝
f : R n → R f Jac f(p) =
(∂x1 f(p), . . . ∂xnf(p)) R n n
∇f(p) grad f(p) f p
f p f p
Df(p)u = ∂uf(p)
X = Rn Y = R p Rn R h = (h1, ..., hn) Rn
h = n j=1 hjej
⎛ ⎞
n
n
n
df(p)(h) = df(p) ⎝ ⎠ = df(p)(ej) · hj = ∂xjf(p) hj ∈ R.
j=1
hjej
df(p) =
j=1
n
j=1
∂xjf(p) dxj,
df (p) h = (h1, ..., hn)
n
∂xjf(p) hj.
j=1
D R n p ∈ D f : D → Y
f p f p
D(αf + βg) = αDf + βDg.
D(f ◦ g)(p) = Df(g(p)) ◦ Dg(p) ◦
C 1 f : D → Y D R n
C 1 (D, Y ) D f
X, Y D ⊆ X f : D → Y
u ∈ X x ∈ D ∂uf(x) ∂uf : D → Y
x Y ∂uf(x) v ∈ X
∂v(∂uf)(x) f D f ′ : D → L(X, Y )
f ′ p f p
f ′′ (p) D 2 f(p) f ′′ (p) ∈ L(X, L(X, Y )) L 2 (X ×X, Y )
L : X × X → Y
K R C
E a, b ∈ R f : E × [a, b] → K
F (x) :=
F : E → K
b
a
f(x, t) dt
j=1
v1
vm
⎞
⎟
⎠ .

X K E X a, b ∈ R f : E × [a, b] → K
u X x ∈ E t ∈ [a, b] ∂uf(x, t)
E × [a, b] ∂uf(x, t) E
∂uF (x) =
b
a
∂uf(x, t) dt
X E X I R f : E × I → Y Y
Φ : E × I × I → Y
Ψ(x, α, β) =
β
α
f(x, t) dt
Ψ
Ψ α, β
∂βΨ(x, α, β) = f(x, β), ∂αΨ(x, α, β) = −f(x, α);
X ≈ Kn ∂if(x, t) i = 1...n
Ψ(x, α, β) α, β
Ψ ′ n
β
(x, α, β)(h, △α, △β) = ∂jf(x, t) dt hj + f(x, β)△β − f(x, α)△α,
j=1
α
h = (h1, ..., hn) ∈ Kn x ↦→ α(x) x → β(x) R I α, β : E → I ⊆ R
∂kG(x) =
β(x)
α(x)
G(x) =
β(x)
α(x)
f(x, t) dt
∂kf(x, t) dt + f(x, β(x))∂kβ(x) − f(x, α(x))∂kα(x).
f : R 2 → R
f(x, y) = x 2 sin y
f(x, y) = |x|
f(x, y) = |xy|
f(x, y) = |x| + |y|
f(x, y) = |xy|
f(x, y) = sign(2 − x 2 − y 2 ) |2 − x 2 − y 2 |
∂xf(x, y) = 2x sin y ∂yf(x, y) = x 2 cos y R 2
R 2 Df(x, y) = 2x sin y dx + x 2 cos y dy.
∂xf(x, y) = sign(x)
2 |x| x = 0 ∂yf(x, y) = 0 R 2 \ ({0} × R)
Df(x, y) = sign(x)
2 |x| dx
∂xf(x, y) = |y| sign(x) ∂yf(x, y) = |x| sign(y)
xy = 0 Df(x, y) = |y|sign(x) dx + |x|sign(y) dy
∂xf(x, y) = sign(x) ∂yf(x, y) = sign(y) xy = 0
Df(x, y) = sign(x) dx + sign(y) dy

y
∂xf(x, y) =
2 |xy| sign(xy) ∂yf(x,
x
y) =
2 sign(xy)
|xy|
y
xy = 0 Df(x, y) =
2 sign(xy) dx +
|xy|
x
2 sign(xy) dy.
|xy|
x
∂xf(x, y) = −√
∂yf(x, y) = −√
y
|2−x2−y2 | |2−x2−y2 |
R 2 x 2 + y 2 = 2
x
y
Df(x, y) = −
dx − dy.
|2 − x2 − y2 | |2 − x2 − y2 |
v = (1/ √ 2, 1/ √ 2, 0) v (0, 0, 0)
f(x, y, z) = (2x − 3y + 4z) cos(xyz)
f
∂xf(x, y, z) = 2 cos(xyz) − yz(2x − 3y + 4z) sin(xyz)
∂yf(x, y, z) = −3 cos(xyz) − xz(2x − 3y + 4z) sin(xyz)
∂zf(x, y, z) = 4 cos(xyz) − xy(2x − 3y + 4z) sin(xyz).
R 3 R 3
∂f
∂u (0, 0, 0) = Df(0, 0, 0)u = ∂xf(0,
√
2
0, 0)ux + ∂yf(0, 0, 0)uy + ∂zf(0, 0, 0)uz = −
2 .
f(x, y) = x3y x6 (x, y) = 0 f(0, 0) = 0
+ y2 f(x, y) = log(1 + 3y3 )
x2 + y2 (x, y) = 0 f(0, 0) = 0
sin y +
f(x, y) =
|x| log(1 + y2 )
(x, y) = 0 f(0, 0) = 0
x 2 + y 2
f(x, y) = arctan(x2 + y 2 )
x 2 + y 2
(x, y) = 0 f(0, 0) = 0
γ(t) = (t, t 3 ) (0, 0) t → 0 +
lim f(γ(t)) = lim
t→0 t→0
t6 1
= = 0 = f(0, 0).
2t6 2
(0, 0)
v = (vx, vy)
vx = 0 vy = 0
f((0, 0) + t(vx, vy)) − f(0, 0) f(t(vx, vy)) 1
∂vf(0, 0) = lim
= lim
= lim
t→0
t
t→0 t
t→0 t
= lim
t→0
tv 3 xvy
t 4 v 6 x + v 2 y
= 0.
t 4 v 3 xvy
t 6 v 6 x + t 2 v 2 y
(0, 0)
(0, 0)

|f(x, y)| ≤
log(1 + 3y 3 )
y 2
=
log(1 + 3y 3 )
3y 3
|3y| → 0.
(0, 0) y = 0
∂xf(0, 0) = 0
f(0, 0 + h) − f(0, 0)
∂yf(0, 0) = =
h
log(1 + 3h3 )
h3 → 3,
∂yf(0, 0) = 3 L(x, y) = 3y
f(x,
y) − f(0, 0) − L(x, y)
x2 + y2 =
log(1+3y
3 )
y2 +x2
− 3y
x2 + y2
≤
log(1 + 3y
3 ) − 3y(x2 + y2 )
(x2 + y2 ) 3/2
=
log(1 + 3y
3 ) − 3y3 3yx
−
2
(x 2 + y 2 ) 3/2
γ(t) = (t, t)
f(t, t) − f(0, 0) − L(t, t)
√
t2 + t2 =
23/2t3 = 1
23/2
t3
log(1 + 3t 3 ) − 3t 3
(x 2 + y 2 ) 3/2
23/2t3 log(1 + 3t3 ) − 3t3 − 3t3
− 3
3
→ √ = 0
8
v = (1, 1)
f(0 + t, 0 + t) − f(0, 0)
lim
= lim
t→0 tv
t→0
1
2 √ 2
log(1 + 3t 3 )
t 3 = 3
√ 8 .
L L(0, 1) = ∂yf(0, 0)
L(1, 1) = ∂vf(0, 0) L(1, 0) = ∂xf(0, 0) (0, 1) (1, 1)
L(0, 1) = 0 L(1, 1) = 0 L(vx, vy) = 0 vx = vy = 0
L(1, 0) = 0
|f(x, y)| ≤
sin y +
|x|
log(1 + y 2 )
y 2
1 f(x, y)
(0, 0) y = 0 ∂xf(0, 0) = 0
f(0, y) = sin y log(1 + y2 )
y2 .
f(0, y) − f(0, 0)
=
y
sin y log(1 + y
y
2 )
y2 → 1.
∂yf(0, 0) = 1 (1, 1)
f(t, t) − f(0, 0)
tv
= 1 sin(t +
√
2
|t|) log(1 + t
t
2 )
t2 = 1 sin(t +
√
2
|t|)
t + t +
|t|
|t| log(1 + t
t
2 )
t2 → ∞.

|f(ρ cos θ, ρ sin θ)| =
arctan ρ
2
ρ2
ρ → 0,
f(x, y) = f(y, x)
lim
t→0 +
f(t, 0) − f(0, 0)
=
t
arctan t2
t2 ∂xf(0, 0) = ∂yf(0, 0) = 1 L L(x, y) = x+y
f(x,
y) − f(0, 0) − L(x, y)
x2 + y2 =
arctan(ρ
2
)
ρ − ρ(cos θ + sin θ)
ρ
=
− (cos θ + sin θ)
θ = π/4
lim
f(x,
y) − f(0, 0) − L(x, y)
= lim
x2 + y2
(x,y)→(0,0)
y=x
(0, 0)
arctan(ρ 2 )
ρ 2
ρ→0 +
= 1,
arctan(ρ 2 )
ρ 2
− √
2
→ √ 2 − 1 = 0.
f(x, y) = y 2/3 (y + x 2 − 1) ∂yf
f(x, y) = 3 x 2 (y − 1)+1 (0, 1)
Dvf(0, 1) v
R 2
f(x, y) =
x 2 y
0
arctan t
t
y = 0 ∂yf(x, y) = 2
3 3√ y (y + x2 − 1) + y 2/3 y = 0
dt.
f(x, y) − f(x, 0)
∂yf(x, 0) = lim
= lim
y→0 y
y→0
y + x2 − 1
√ .
3 y
x 2 − 1 = 0 x = ±1
∂yf(±1, 0) = 0
(0, 1) x = ρ cos θ y = ρ sin θ + 1
f(ρ cos θ, ρ sin θ + 1) = 3 ρ 3 cos θ sin θ = ρ 3√ cos θ sin θ.
(0, 1) ρ = 0 f(0, 1) = 1 v v = (cos θ, sin θ)
∂vf(0, 1) = lim
t→0 +
f((0, 1) + tv) − f(0, 1)
=
t
3√ cos θ sin θ.
v ↦→ ∂vf(0, 1)

(x, y) ∈ R 2
xy = 0
∂xf(x, y) = 2xy arctan(x2 y)
x 2 y
∂yf(x, y) = x 2 arctan(x2 y)
x 2 y
= arctan(x2y) ,
x
= arctan(x2y) .
y
xy = 0 f(x + h, y) =
f(x, y + h) = 0
R 2 \ {(x, y) : xy = 0}
Σ := {(x, y) : xy = 0}
Σ
lim
(x,y)→(¯x,¯y)
xy=0
lim
(x,y)→(¯x,¯y)
xy=0
arctan(x
∂xf(x, y) = lim
(x,y)→(¯x,¯y)
xy=0
2y) = lim
x
(x,y)→(¯x,¯y)
xy=0
arctan(x
∂xf(x, y) = lim
(x,y)→(¯x,¯y)
xy=0
2y) = lim
y
(x,y)→(¯x,¯y)
xy=0
arctan(x2y) x2 xy
y
arctan(x2y) x2 x
y
2
arctan s
lim =
s→0 s
d
arctan(0) = 1
ds
| arctan s/s| ≤ 1
lim
(x,y)→(¯x,¯y)
xy=0
lim
(x,y)→(¯x,¯y)
xy=0
∂xf(x, y) = 0 = ∂xf(¯x, ¯y)
∂yf(x, y) = ¯x 2
Σ \ {(0, 0)}
v = (vx, vy) 0 (¯x, ¯y) ∈ Σ \ {(0, 0)}
f(¯x + svx, ¯y + svy) − f(¯x, ¯y)
s
¯x = 0 ¯y = 0
f(¯x + svx, ¯y + svy) − f(¯x, ¯y)
s
= 1
|s|
= 1
s
(¯x+svx) 2 (¯y+svy)
0
s 2 v 2 x(¯y+svy)
0
arctan t
t
arctan t
t
dt
≤
dt
s 2 v 2 x(¯y + svy)
s
s → 0 ¯x = 0
¯x = 0 ¯y = 0
f(¯x + svx, ¯y + svy) − f(¯x, ¯y)
s
= 1
|s|
s(¯x+svx) 2 vy
0
arctan t
t
s → 0 s
1/2
vx = vy = 1
f(¯x + svx, ¯y + svy) − f(¯x, ¯y)
s
= 1
|s|
s(¯x+s) 2
0
1
2 dt
≥
s(¯x + s)
2
2s
dt
= |¯x|
2
= 0.
.

¯x = 0 y = 0

X, Y D ⊆ X f : D → Y u ∈ X
x ∈ D ∂uf(x) ∂uf : D → Y x
Y ∂uf(x) v ∈ X
∂v(∂uf)(x) = ∂ 2 vuf(x)
X, Y D ⊆ X f : D → Y
D df : D → L(X, Y ) p ↦→ df(p)
L(X, Y )
df p
f p f ′′ (p) D 2 f(p) D 2 f(p) ∈ L(X, L(X, Y ))
X, Y, Z K K = R C
B : X × Y → Z x, x1, x2 ∈ X y, y1, y2 ∈ Y α, β ∈ K
B(αx1 + βx2, y) = αB(x1, y) + βB(x2, y)
B(x, αy1 + βy2) = αB(x, y1) + βB(x, y2),
B
X × Y X Y
α(x1, y1) + β(x2, y2) = (αx1 + βx2, αy1 + βy2).
R2 (x, y) ∈ X × Y
p ≥ 1
(x, y)p|X×Y = x p 1/p X + yp Y , (x, y)∞|X×Y = max{xX, yY },
X
L 2 (X × X, Y ) := {B : X × X → R },
X × X
X, Y K
L(X, L(X, Y )) L 2 (X × X, Y ).
X = R n Y = R D X f : D → R
R n L(R n , R) L(R n , R)
R n f R n
R n R n R n n × n
p ∈ D H ∈ Matn×n(R)
df(p)(h) (k) = 〈H h, k〉,

H
h, k ∈ R n H(h, k) f
p Hf(p) D 2 f(p) ∇ 2 f(p) Hess f(p)
Hess f(p) :=
⎛
⎜
⎝
∂x1x1f(p) . . . ∂2 x1xnf(p)
∂2 xnx1f(p) . . . ∂2 xnxnf(p) X f : X → R
a ∈ X f f(y) ≥ f(a) y ∈ X
f(y) > f(a) y ∈ X y = a
a ∈ X f f(y) ≤ f(a) y ∈ X
f(y) < f(a) y ∈ X y = a
X f : X → R
a ∈ X f U a f(y) ≥ f(a)
y ∈ U f(y) > f(a) y ∈ U y = a
a ∈ X f U a f(y) ≤ f(a)
y ∈ U f(y) < f(a) y ∈ U y = a
X D ⊂ X f : D → R a ∈ D a f
u ∈ X Duf(a) Duf(a) = 0 f a
Df(a)
X D ⊂ X f : D → R D a ∈ D
a f Df(a) = 0
X D X f : D → R p ∈ D
∂uf(x) ∂vf(x) ∂u∂vf(x) p p ∂v∂uf(p)
∂v∂uf(p) = ∂u∂vf(p).
X = R n f
H(p) = (∂i∂jf(p))ij
f p Hess f(p)
D ⊆ R n f ∈ C 2 (D, R) a f
f ′′ (a)(h, h) f
a f
f ′′ (a)(h, h) f
a f
a f f ′′ (a)(h, h)
p p
H = D 2 f(p) f p
D 2 f(p)
p
D 2 f(p)
p
H
⎞
⎟
⎠ .

H
A 2 × 2
λ 2 − tr(A) λ + det A = 0,
tr(A) A A
f : Ω → R Ω R 2 (x0, y0) ∈ Ω
∇f(x0, y0) = 0
· ∂ 2 xxf(x0, y0) > 0 det D 2 (x0, y0) > 0 (x0, y0)
· ∂ 2 xxf(x0, y0) < 0 det D 2 (x0, y0) > 0 (x0, y0)
· det D 2 (x0, y0) < 0 (x0, y0)
· det D 2 (x0, y0) = 0
R
f(x) = |x| 0
f(x, y) = 2x 3 + y 3 − 3x 2 − 3y
f(x, y) = x 3 + y 3 − (1 + x + y) 3
f(x, y) = cos x sin y
f(x, y) = x 4 + x 2 y + y 2 + 3
f(x, y) = x 4 + y 4 − 2(x 2 + y 4 ) + 4xy
f
∂xf(x, y) = 6x 2 − 6x = 0 =⇒ x ∈ {0, 1},
∂yf(x, y) = 3y 2 − 3 = 0 =⇒ y ∈ {1, −1}.
(0, ±1) (1, ±1)
f
D 2
∂2 f(x, y) =: xxf(p) ∂2 xyf(p)
∂2 yxf(p) ∂2
,
yyf(p)
∂ 2 xxf(x, y) = 12x − 6, ∂ 2 yxf(x, y) = ∂ 2 xyf(x, y) = 0, ∂ 2 yyf(x, y) = 6y.
D 2 f(x, y) =:
12x − 6 0
0 6y
D 2 f(x, y) λ1(x, y) = 12x − 6 λ2(x, y) = 6y
λ1(0, 1) = −6 λ2(0, 1) = 6 (0, 1)
λ1(0, −1) = −6 λ2(0, −1) = −6 (0, −1)
λ1(1, 1) = 6 λ2(1, 1) = 6 (1, 1)
λ1(1, −1) = 18 λ2(1, −1) = −6 (1, −1)
.

f(x, y) = f(y, x)
f
∂xf(x, y) = 3x 2 − 3(1 + x + y) 2 = 0
∂yf(x, y) = 3y 2 − 3(1 + x + y) 2 = 0.
x = ±y x = −y 3y 2 − 3 = 0 y = ±1
(−1, 1) (1, −1) x = y 3y 2 − 3(1 + 4y 2 + 2y) = 0 y = −1
y = −1/3 (−1, −1) (−1/3, −1/3)
(x, y) ↦→ (y, x)
∂ 2 xxf(x, y) = 6x − 6(1 + x + y), ∂ 2 yyf(x, y) = 6y − 6(1 + x + y), ∂ 2 xyf(x, y) = ∂ 2 yxf(x, y) = −6(1 + x + y).
D 2
−12 −6
f(−1, 1) =:
, D
−6 0
2
0 −6
f(1, −1) =:
,
−6 −12
D 2
0 6
f(−1, −1) =: , D
6 0
2
−4 −2
f(−1/3, −1/3) =:
−2 −4
(−1, 1) λ 2 + 12λ − 36 = 0 λ1 =
−6 + 6 √ 2 λ1 = −6 − 6 √ 2
(1, −1) (−1, 1)
(−1, −1) λ 2 − 36 = 0 λ = ±6
(−1/3, −1/3) λ 2 + 8λ + 12 = 0
λ1 = −6 λ2 = −2
2π
[0, 2π[×[0, 2π[
∂xf(x, y) = − sin x sin y ∂yf(x, y) = cos x cos y
∂xf(x, y) = 0 x ∈ {0, π} y ∈ {0, π}
∂yf(x, y) = 0 x ∈ {π/2, 3π/2} y ∈ {π/2, 3π/2}
(0, π/2) (0, 3π/2) (π, π/2) (π, 3π/2) (π/2, 0) (3π/2, 0) (π/2, π) (3π/2, π)
∂ 2 xxf(x, y) = − cos x sin y, ∂ 2 yyf(x, y) = − cos x sin y, ∂ 2 xyf(x, y) = − sin x cos y.
D 2
−1 0
f(0, π/2) =:
, D
0 −1
2
1 0
f(0, 3π/2) =: ,
0 1
D 2
1 0
f(π, π/2) =: , D
0 1
2
−1 0
f(π, 3π/2) =:
,
0 −1
D 2
0 −1
f(π/2, 0) =:
, D
−1 0
2
0 1
f(3π/2, 0) =: ,
1 0
D 2
0 1
f(π/2, π) =: , D
1 0
2
0 −1
f(3π/2, π) =:
.
−1 0

(2kπ, π/2+2hπ) (π+2kπ, 3π/2+2hπ) h, k ∈ Z
1
(2kπ, 3π/2+2hπ) (π +2kπ, π/2+2hπ) h, k ∈ Z
−1
(π/2 + kπ, hπ) h, k ∈ Z
∂xf(x, y) = 4x 3 + 2xy = x(4x 2 + y), ∂yf(x, y) = x 2 + 2y.
(0, 0)
∂ 2 xxf(x, y) = 12x 2 + 2y ∂ 2 yyf(x, y) = 2 ∂ 2 xyf(x, y) = 2x
D 2 f(0, 0) =:
0 0
0 2
.
(0, 0)
f(x, y) = g(x 2 , y) g(v, w) = v 2 + vw + w 2 + 3
v 2 + vw + w 2 v > 0 v = x 2 > 0 x = 0 v > 0
v 2 + vw + w 2 = 0 w
v 2 − 4v 2 < 0 v 2 + vw + w 2 v > 0
w → ±∞ v > 0
f(x, y) = g(x 2 , y) > 3 = f(0, 0) x = 0 (0, 0)
f(x, y) = x 4 + y 4 − 2(x 2 + y 4 ) + 4xy
∂xf(x, y) = 4x 3 − 4x + 4y, ∂yf(x, y) = −4y 3 + 4x.
x = y 3 4y 9 − 4y 3 + 4y = 0 x = y 3
y(y 8 −y 2 +1) = 0 (0, 0)
y 8 −y 2 +1 = 0 y = 0 0 < |y| ≤ 1 1−y 2 ≥ 0 y 8 −y 2 +1 ≥ y 8 > 0
|y| > 1 y 8 > y 2 y 8 − y 2 + 1 > 1 > 0
∂ 2 xxf(x, y) = 12x 2 − 4 ∂ 2 yyf(x, y) = −12y 2 ∂ 2 xyf(x, y) = 4
D 2 f(0, 0) =:
−4 4
4 0
,
λ 2 + 4λ − 16 = 0 λ = −2 ± 2 √ 5

λ, µ ∈ R
f(x, y) = x 3 + xy + λx + µy
P = (1/ √ 3, 0) f(x, y)
λ µ
λ, µ ∈ R C 2
∂xf(x, y) = 3x 2 + y + λ,
∂yf(x, y) = x + µ.
(1/ √ 3, 0) µ = −1/ √ 3 λ = −1
f(x, y) = x 3 + xy 2 − x − 1
√ 3 y,
∂xf(x, y) = 3x 2 + y − 1 ∂yf(x, y) = x − 1/ √ 3
P
∂ 2 xxf(x, y) = 6x, ∂ 2 yyf(x, y) = 0, ∂ 2 xyf(x, y) = 1,
D 2 f(P ) =
2 √ 3 1
1 0
λ 2 − 2 √ 3λ − 1 = 0 √ 3 ± 2
α ∈ R
.
f(x, y, z) = cos 2 x + y 2 − 2y + 1 + αz 2 .
f
C 2 R 3 γ(t) = (0, t, 0)
lim f ◦ γ(t) = +∞
t→+∞
α < 0 γ(t) = (0, 0, t) lim f ◦ γ(t) = −∞ α < 0
t→+∞
α ≥ 0
f(x, y, z) ≥ y 2
π
− 2y + 1 = f + kπ, y, 0 , k ∈ Z,
2
y ↦→ y2 − 2y + 1 y = 1 ( π
2 + kπ, 1, 0) k ∈ Z
f f( π
2 + kπ, 1, 0) = 0 k ∈ Z

∂xf(x, y, z) = −2 cos x sin x = − sin(2x)
∂yf(x, y, z) = 2y − 2
∂zf(x, y, z) = 2αz.
∂ 2 xxf(x, y, z) = −2 cos(2x)
∂ 2 yyf(x, y, z) = 2
∂ 2 zzf(x, y, z) = 2α
∂ 2 xyf(x, y, z) = ∂ 2 xzf(x, y, z) = ∂ 2 zyf(x, y, z) = 0.
α = 0 Pk = (kπ/2, 1, 0) k ∈ Z ∂ 2 xxf(Pk) =
2(−1) k+1 ∂ 2 yyf(Pk) = 2 ∂ 2 zzf(Pk) = 2α
D 2 ⎛
f(Pk) = ⎝ 2(−1)k+1 0
0
2
0
0
⎞
⎠ .
0 0 2α
α > 0 Pk k k Pk
f(Pk) = 0 α < 0 Pk
α = 0 Pkz = (kπ/2, 1, z) k ∈ Z z ∈ R
∂ 2 xxf(Pk) = 2(−1) k+1 ∂ 2 yyf(Pk) = 2
D 2 f(Pkz) =
⎛
⎝ 2(−1)k+1 0 0
0 2 0
0 0 0
z f(x, y, z) = f(x, y, 0) z
g(x, y) = f(x, y, 0) = f(x, y, z) (x, y, z) f
g g Qk = (kπ/2, 1)
D 2 g(Qk) =
2(−1) k+1 0
0 2
Pkz = (Qk, z) Pkz k k
Pkz f(Pkz) = 0
α > 0 Pk = (kπ/2, 1, 0)
k ∈ Z k k
f 0
α < 0
Pk = (kπ/2, 1, 0) k ∈ Z
α = 0
⎞
.
⎠ .
Pkz = (kπ/2, 1, z) k ∈ Z, z ∈ R,
k k f 0
n ∈ N\{0}
fn : R 2 → R
fn(x, y) = (x 2 + 3xy 2 + 2y 4 ) n .

fn g : R 2 → R
g(x, y) = x 2 + 3xy 2 + 2y 4 h(s) = s n h : R → R
f(x, y) = h(g(x, y)).
n n
h n n
f(x1, y1) = h(g(x1, y1)) > h(g(x2, y2)) = f(x2, y2) ⇐⇒ g(x1, y1) > g(x2, y2),
f
g n
n g γ(t) = (0, t)
lim g ◦ γ(t) = +∞ g
t→∞
f
g(x, y) = x + 3
2 y2
2 − 1
4 y4 .
γ(t) = (−3/2t2 , t) lim g◦γ(t) = lim
t→∞
g f
g
∂xg(x, y) = 2x + 3y 2 , ∂yg(x, y) = 6xy + 8y 3 = 2y(3x + 4y 2 )
t→∞ −t 4 /4 = −∞
y y = 0
x x = 0 y = 0 y x =
−4y 2 /3 x −8y 2 /3 + 3y 2 = 0
(−8/3+3)y 2 = 0
g(0, 0) = 0 V γ(t) = (−3/2t 2 , t)
t > 0 γ(t) ∈ V
ε > 0 B((0, 0), ε) ⊆ V
|γ(t)| = 9/4t 4 + t 2 t → 0 δ > 0 |t| < δ
|γ(t)| < ε γ(t) ∈ B((0, 0), ε) ⊆ V
g ◦ γ(t) = −t 4 /4 < 0 = f(0, 0) t ∈]0, δ[ 0
g g(0, 0) γ2(t) = (t, 0)
t γ(t) V
g ◦ γ(t) = t 2 > 0 = g(0, 0) t = 0 (0, 0)
g g(0, 0) g g(0, 0) (0, 0)
g f
(0, 0)
f h 0 (x, y) g(x, y) = 0
f f
g g (0, 0)
n h(s)
s = 0 x 2 + 3xy 2 + y 4 = 0 f
0
h [0, +∞[ ] − ∞, 0]
G + := {(x, y) : x 2 + 3xy 2 + y 4 > 0} g
f g
g(G + ) ⊆]0, +∞[ h g
(0, 0) /∈ G +
g f G − := {(x, y) : x 2 + 3xy 2 + y 4 < 0} f
x 2 + 3xy 2 + y 4 = 0 f 0

(0, 0) f : R 2 → R
f(x, y) = log(1 + x 2 ) − x 2 + xy 2 + y 3 + 2.
f(0, 0) = 2 (0, 0) ∂xf(0, 0) =
∂yf(0, 0) = 0 γ(t) = (0, t) t = 0 f ◦ γ(t) = t3 + 2
|t| γ1(t)
lim γ(t) = (0, 0) t > 0 f ◦ γ(t) > 2 = f(0, 0) f ◦ γ(t) < 2 = f(0, 0) t < 0
t→0
(0, 0)
f, g : R 3 → R
f(x, y, z) = x 2 (y − 1) 3 (z + 2) 2 , g(x, y, z) = 1/x + 1/y + 1/z + xyz.
γ(t) = (1, t, 1) f ◦ γ1(t) = 9(t − 1) 3
t → ±∞ f ◦ γ1(t) ±∞ f
f
f
∂xf(x, y, z) = 2x(y − 1) 3 (z + 2) 2
∂yf(x, y, z) = 3x 2 (y − 1) 2 (z + 2) 2
∂zf(x, y, z) = 2x 2 (y − 1) 3 (z + 2).
x = 0 y = 1 z = −2
(0, y, z) (x, 1, z) (x, y, −2) x, y, z ∈ R 0
(0, y, z) (x, y, −2) y > 1 (0, y, z) (x, y, −2)
y ′ y > 1 (x ′ ) 2 (y ′ − 1) 3 (z + 2) 2 > 0
y < 1 (0, y, z) (x, y, −2) y = 1
(0, 1, z) (x, 1, −2)
(0, y, z) (x, y, −2) y > 1 y < 1
(0, 1, z) (x, 1, −2) (x, 1, z)
(0, 1, z) (x, 1, −2)
γ(t) = (1, t, 1) g ◦ γ1 = 2 + 1/t + t t → ±∞
g ◦ γ1(t) ±∞ g
g(x, y, z) = −g(−x, −y, −z)
∇g(x, y, z) = yz − 1 1 1
, xz − , xy −
x2 y2 z2
.
x 2 yz = xy 2 z =
xyz 2 = 1 xyz x = y = z
x 2 yz = 1 x 4 = 1 x = ±1 (1, 1, 1) (−1, −1, −1)
⎛
Hess g(1, 1, 1) = ⎝
2 1 1
1 2 1
1 1 2
⎛
Hess g(x, y, z) = ⎝
⎞
2
x3 z y
2 z y3 x
2 y x z3 ⎞
⎠ ,
⎛
⎠ , Hess g(−1, −1, −1) = ⎝
−2 −1 −1
−1 −2 −1
−1 −1 −2
Hess g(1, 1, 1) i
i i = 1, 2, 3 2 3 4
⎞
⎠ .

(1, 1, 1) g(1, 1, 1) = 4 g(−x, −y, −z) =
−g(x, y, z) (−1, −1, −1) g(−1, −1, −1) = −4
α ∈ R
f : R2 → R
(x
f(x, y) =
2 + y2 ) α log(x2 + y2 ) (x, y) = (0, 0),
0 (x, y) = (0, 0).
f(ρ cos θ, ρ sin θ) = 2ρ 2α log ρ =: gα(ρ).
α = 0 lim
ρ→0 + g0(ρ) = −∞ lim
ρ→+∞ g0(ρ) = +∞
g0 g0(0) = 0 0
g0
α < 0 lim
ρ→0 + g0(ρ) = −∞ lim
ρ→+∞ gα(ρ) = 0 =
gα(1) gα
gα(0) = 0 0 gα
g ′ α(ρ) = 4αρ 2α−1 log ρ + 2ρ 2α−1 = 2ρ 2α−1 (2α log ρ + 1),
ρ = e −1/2α x 2 + y 2 = e −1/α
α > 0 f lim
ρ→0 + gα(ρ) = 0 = g(1) gα(ρ) ≤ 0 0
lim
ρ→+∞ g0(ρ) = +∞
gα(0) = gα(1) = 0 [0, 1] gα
g ′ α(ρ) = 4αρ 2α−1 log ρ + 2ρ 2α−1 = 2ρ 2α−1 (2α log ρ + 1),
ρ = e −1/2α < 1 α > 0
x 2 + y 2 = e −1/α

α ∈ R (0, 0)
f(x, y) = 2 + αx 2 + 4xy + (α − 3)y 2 + (2x + y) 4 .
f ∈ C 2 f(0, 0) = 2 f
∂xf(x, y) = 2αx + 4y + 8(2x + y) 3
∂yf(x, y) = 4x + 2(α − 3)y + 4(2x + y) 3
∂xxf(x, y) = 2α + 48(2x + y) 2
∂xyf(x, y) = 4 + 24(2x + y) 2
(0, 0)
D 2
2α 4
f(0, 0) =
4 2(α − 3)
∂yyf(x, y) = 2(α − 3) + 12(2x + y) 2 .
, detD 2 f(0, 0) = (α − 4)(α + 1).
λ1, λ2 ∈ R −1 <
α < 4 detD 2 f(0, 0) < 0
detD 2 f(0, 0) > 0 1, 1
α > 4 detD 2 f(0, 0)
α < −1
λ 2 − 2(2α − 3)λ + 4(α 2 − 3α − 4) = 0
λ = 2α − 3 ± 4α 2 + 9 − 12α − 4α 2 + 12α + 16 = 2α − 3 ± 5,
λ1 = 2(α + 1) λ2 = 2(α − 4) λ1 > λ2 λ2 > 0 α > 4
(0, 0) λ1 < 0 α < −1
(0, 0) −1 < α < 4 λ2 < 0 λ1 > 0
α ∈ {−1, 4}
α = 4 (x, y) = (0, 0)
f(x, y) = 2 + 4x 2 + 4xy + y 2 + (2x + y) 4 = 2 + (2x + y) 2 + (2x + y) 4 > 2,
f(x, −2x) = 2
x
α = −1
f(x, y) = 2 − x 2 + 4xy − 4y 2 + (2x + y) 4 = 2 − (x − 2y) 2 + (2x + y) 4 .
γ1(t) = (2t, t) t > 0 f ◦ γ1(t) = 2 + 5 4 t 4 > 2
γ2(t) = (t, −2t) t > 0 f ◦ γ2(t) = 2 − 25t 2 < 2 t → 0
γ1(t) → (0, 0) γ2(t) → (0, 0) (0, 0) γ1(t) f
f(0, 0) γ2(t) f f(0, 0)
(0, 0)

f(x, y) = x 3 − 6xy + 3y 2 + 3x
f
γ1(t) = (t, 0) limt→±∞ f ◦ γ1(t) = ±∞
∂xf(x, y) = 3x 2 − 6y + 3, ∂yf(x, y) = −6x + 6y.
(1, 1)
∂ 2 xxf(x, y) = 6x ∂ 2 yyf(x, y) = 6 ∂ 2 xyf(x, y) = −6
D 2 f(1, 1) =:
6 −6
−6 6
λ 2 − 12λ = 0 λ1 = 0 λ2 = 12 > 0
f(1, 1) = 1 v = (v1, v2)
kerD 2 f(1, 1) (v1, v2) = (1, 1) γ(t) = (1, 1) + tv
f ◦ γ(t) = (1 + t) 3 − 6(1 + t) 2 + 3(1 + t) 2 + 3(1 + t) = (1 + t)((1 + t) 2 − 3(1 + t) + 3)
= (1 + t)(−5 − 5t 2 − 10t + 3 + 3t + 3) = (1 + t)(1 + t 2 + 2t − 3 − 3t + 3)
= (1 + t)(t 2 − t + 1) = t 3 + 1.
t → 0 γ(t) → (1, 1) t > 0 f ◦ γ(t) > 1 t < 0 f ◦ γ(t) < 1
(1, 1) f f(1, 1) = 1 f
f(1, 1) = 1 (1, 1)
α ∈ R (0, 0, 0)
gα(x, y, z) = 5 + αx 2 + 2xy + 4αxz − 6y 2 − 3z 2 .
gα(0, 0, 0) = 5
x√6
gα(x, y, z) = 5 − − √ 2 6 y + x2
6 + αx2 + 4αxz − 3z 2
x√6
= 5 − − √ 2 6 y + x2
6 −
√3z 2
2α
− √3x
x√6
= 5 − − √ 2 6 y −
√3z − 2α
√3 x
= 5 − A(x, y) − Bα(x, z) + Cαx 2
.
2
+ 4α2
3 x2 + αx 2
+ 1
6 (8α2 + 6α + 1)x 2 .
Cα < 0 8α 2 + 6α + 1 < 0 −1/2 < α < −1/4 (x, y, z) =
(0, 0, 0) g(x, y, z) < 5 A(x, y) ≥ 0 Bα(x, y) ≥ 0 Cαx 2 < 0 x = 0
g
gα(x, y, z) = gα(0, 0, 0) x = A(x, y) = Bα(x, z) = 0
x = y = z = 0 α = 0
Cα = 0 α ∈ {−1/2, −1/4} (x, y, z) = (0, 0, 0) gα(x, y, z) ≤ 5
A(x, y) = Bα(x, z) =
0 γ(t) = (6t, t, 4αt) limt→0 γ(t) = (0, 0, 0) U
gα 5
γ(t) t = 0 |t| U

Cα > 0 α /∈ [−1/2, −1/4] γ
f ◦ γ(t) = 5 + 6Ct 2 5 t = 0
γ2(t) = (0, t, 0) f ◦ γ2(t) = 5 − 6t 2 5 t = 0
t → 0 0
|t|
gα gα(0, 0, 0) = 5 gα
gα(0, 0, 0) = 5
gα
∇gα(x, y, z) = (2αx + 2y + 4αz, 2x − 12y, 4αx − 6z) ,
α
Hα := D 2 ⎛
⎞
2α 2 4α
g(0, 0, 0) = ⎝ 2 −12 0 ⎠ , det D
4α 0 −6
2 g(0, 0, 0) = 192α 2 + 144α + 24.
α ∈ {−1/2, −1/4} α ∈] − 1/2, −1/4[
α /∈ [−1/2, −1/4] 3
2α −(6α+1) α ∈]−1/2, −1/4[
α ∈] − ∞, −1/2[∪] − 1/4, 0[
α ∈ [0, +∞[
gα
gα
gα(v) = gα(0, 0, 0) + 〈Hαv, v〉,
α ∈ {−1/2, −1/4}
α /∈] − 1/2, −1/4[ α ∈ [−1/2, −1/4]

f, ϕ : Ω → R ¯x
f ϕ = 0 ϕ(¯x) = 0 ¯x f |ϕ=0
f ϕ = 0 f
ϕ = 0
Ω ⊆ R n Γ ⊆ Ω f : Ω → R
g : V → Γ V ⊆ R m m ≤ n C 1
f Γ g f ◦ g
V V int V V ∩ ∂V
Ω ⊆ R n f :
R n → R C 1 (Ω) ϕ : Ω → R C 1 ¯x ∈ Ω f
ϕ = 0 Dϕ(¯x) = 0 λ ∈ R
Df(¯x) + λDϕ(¯x) = 0.
¯x ∈ Ω f ϕ = 0 ¯x
Γ := {x ∈ R n : ϕ(x) = 0}, Cx0 := {x ∈ Rn : f(x) = f(x0)}.
f(x, y) = x 3 + 4xy 2 − 4x
x 2 + y 2 − 1 = 0
g(x, y) = x 2 + y 2 − 1 ∇g(x, y) = (0, 0)
x = y = 0 g(x, y) = 0 g(0, 0) = 0
L(x, y, λ) := f(x, y) + λg(x, y) = x 3 + 4xy 2 − 4x + λ(x 2 + y 2 − 1).
∂xL(x, y, λ) = 3x 2 + 4y 2 − 4 + 2λx
∂yL(x, y, λ) = 8xy + 2λy = 2y(λ + 4x).
∂xL(x, y) = ∂yL(x, y) = 0 y = 0 4x = −λ
y = 0 x = ±1
y = 0 x = −λ/4 y 2 = 1 − λ 2 /16
0 ≤ λ ≤ 4
3 λ2
+ 4
16
1 − λ2
16
− 4 + 2λ
− λ
= 0,
4
λ = 0 x = 0 y = ±1 (±1, 0) (0, ±1)
f(1, 0) = −3 f(−1, 0) = 3 f(0, ±1) = 0 (−1, 0)
(1, 0)

x = cos θ y = sin θ θ ∈ [−π, π]
h(θ) := f(cos θ, sin θ) = cos 3 θ + 4 cos θ sin 2 θ − 4 cos θ.
h ′ (θ) = −3 cos 2 θ sin θ + 8 cos 2 θ sin θ − 4 sin 3 θ + 4 sin θ = sin θ(5 cos 2 θ − 4 sin 2 θ + 4) = 9 cos 2 θ sin θ,
θ = 0, ±π/2, ±, π
h ′′ (θ) = −18 cos θ sin 2 θ + 9 cos 3 θ = 9 cos θ(−2 sin 2 θ + cos 2 θ).
h ′′ (±π/2) = 0 h ′′ (0) = 9 h ′′ (±π) = −9 h(0) =
f(1, 0) = −3 h(±π) = f(−1, 0) = 3 h(±π/2) = 0 ±π/2
h ′′′ (θ) = − 9
sin(θ)(9 cos(2θ) + 5),
2
h ′′′ (±π) = ±18 (1, 0) (−1, 0)
(0, ±1)
f(x, y) = x + y
x 2 + 4y 2 − 1 = 0
g(x, y) = x 2 + 4y 2 − 1 ∇g(x, y) = (0, 0)
x = y = 0 (0, 0) g(x, y) = 0
L(x, y, λ) := f(x, y) + λg(x, y) = x + y + λ(x 2 + 4y 2 − 1).
∂xL(x, y, λ) = 1 + 2λx
∂yL(x, y, λ) = 1 + 8λy.
∂xL(x, y) = ∂yL(x, y) = 0 λ = 0
λ(x − 4y) = 0 x = 4y
20y 2 = 1 ±(2/ √ 5, 1/(2 √ 5)) f(2/ √ 5, 1/(2 √ 5)) = √ 5/2
f(−2/ √ 5, −1/(2 √ 5)) = − √ 5/2
x 2 + (2y) 2 − 1 = 0
x = cos θ 2y = sin θ
sin θ
sin θ
h(θ) := f cos θ, = cos θ +
2
2 .
h ′ (θ) = − sin θ + cos θ/2 θ tan θ = 1/2
ξ 2 + ζ 2 = 1
ζ = ξ/2
5ζ 2 = 1 ζ = ±1/ √ 5 = sin θ ξ = ±2/ √ 5 = cos θ
(ξ, 2ζ) = ±(2/ √ 5, 1/(2 √ 5)) h ′′ (θ) = − cos θ − sin θ/2 =
− cos θ(1 + tan θ/2) (2/ √ 5, 1/(2 √ 5)) −(2/ √ 5, 1/(2 √ 5))
f(x, y) = sin x + sin y
cos x − cos y + 1 = 0
2π
g(x, y) = cos x − cos y + 1
∇g(x, y) = (− sin x, sin y) Zhk = (kπ, hπ) h, k ∈ Z
g(Zhk) = (−1) k + (−1) h + 1 h, k ∈ Z h, k g(Zhk) = −1 = 0
g(Zhk) = 3 = 0 g(Zhk) = 1 = 0

Zhk h, k ∈ Z g(x, y) = 0
L(x, y, λ) := f(x, y) + λg(x, y) = sin x + sin y + λ(cos x − cos y + 1).
∂xL(x, y, λ) = cos x − λ sin x
∂yL(x, y, λ) = cos y + λ sin y.
∂xL(x, y) = ∂yL(x, y) = 0 sin x = 0 sin y = 0 cot x = cot(−y)
x = −y + kπ k ∈ Z cos x − cos(−x + kπ) + 1 = 0
cos x − cos(x − kπ) + 1 = 0 cos x − (−1) k cos x + 1 = 0 (1 − (−1) k ) cos x = −1
k = 2j k = 2j + 1 cos x = −1/2
x1 = 2/3π + 2hπ x2 = 4/3π + 2hπ h ∈ Z y1 = −2/3π + 2hπ + (2j + 1)π =
π/3 + 2(j − h)π y2 = −4/3π − 2hπ + (2j + 1)π = −1/3π + 2(j − h)π
m = j − h ∈ Z
Phm =
2 π
4
π + 2hπ, + 2mπ , Qhm = π + 2hπ, −2 π + 2mπ .
3 3 3 3
h, m ∈ Z f(Phm) = √ 3/2 + √ 3/2 = √ 3 f(Qhm) =
− √ 3/2 − √ 3/2 = − √ 3
f(x, y) = 3x 2 +4y 2 −6x+3
x 2 + y 2 = 4
g(x, y) = x 2 + y 2 − 4 ∇g(x, y) = (0, 0) x = y = 0
(0, 0) g(x, y) = 0 g(0, 0) = 0
L(x, y, λ) := f(x, y) + λg(x, y) = 3x 2 + 4y 2 − 6x + 3 + λ(x 2 + y 2 − 4).
∂xL(x, y, λ) = 6x − 6 − 2λx
∂yL(x, y, λ) = 8y − 2λy = 2y(4 − λ).
∂xL(x, y) = ∂yL(x, y) = 0 y = 0 λ = 4
y = 0 x = ±2 λ = 4
x = −3 9 + y 2 = 4
(2, 0) (−2, 0) f(2, 0) = 3 f(−2, 0) = 27
g
x = 2 cos θ y = 2 sin θ θ ∈ [0, 2π[
h(θ) = f(2 cos θ, 2 sin θ) = 12 cos 2 θ + 16 sin 2 θ − 12 cos θ + 3 = 4 sin 2 θ − 12 cos θ + 15.
h ′ (θ) = 8 sin θ cos θ + 12 sin θ = 4 sin θ(2 cos θ + 3),
h ′′ (θ) = 4 cos θ(2 cos θ + 3) + 4 sin θ(−2 sin θ).
h ′ (θ) = 0 θ = 0, π h ′′ (0) = 20 > 0 h ′′ (π) = −4 < 0 (2, 0) (−2, 0)
f(x, y) = x(x 2 + y 2 )
xy = 1

γ(t) = (t, 1/t) t → ±∞
L(x, y, λ) = x(x 2 + y 2 ) + λxy ∂xL(x, y, λ) = 3x 2 + y 2 + λy ∂yL(x, y, λ) =
2yx + λx = x(2y + λ) x = 0
y = −λ/2 x = 0 y = −λ/2 λ = 0
x = −2/λ
3 4 λ2 λ2
+ −
λ2 4 2
12 λ2
= 0 ⇐⇒ =
λ2 4 ,
λ = ±2 4√ 3 ±(1/ 4√ 3, 4√ 3)
f(1/ 4√ 3, 4√ 3) = 4/3 3/4 f(−1/ 4√ 3, − 4√ 3) = −4/3 3/4
x = 0
y = 1/x h(x) = f(x, 1/x) = x 3 + 1/x h ′ (x) = 3x 2 − 1/x 2
3x 4 = 1 x = ±1/ 4√ 3 h(1/ 4√ 3) = 4/3 3/4 h(−1/ 4√ 3) = −4/3 3/4
f(x, y) = e x + e y
x + y = 2
g(x, y) = x + y − 2 S = {(x, y) : g(x, y) = 0}
γ(t) = (t, 2 − t)
t → +∞ (xn, y2)
+∞ −∞
fS (x, y) → ∞ (x, y) ∈ S +∞ f L(x, y, λ) =
e x + e y + λ(x + y − 2) ∂xL(x, y, λ) = e x + λ ∂yL(x, y, λ) = e y + λ
e x = e y = −λ
x = y x = y = 1
f(1, 1) = 2e
x = 2 − y
h(y) = f(2 − y, y) = e 2−y + e y .
h ′ (y) = −e 2−y + e y −e 2 + e 2y = 0 y = 1
x = 1 h ′′ (y) = e 2−y + e y h ′′ (1) = 2e > 0 (1, 1)
E a, b, c > 0
g(x, y, z) := x2
− 1 = 0.
a b c2 P [−x, x]×[−y, y]×[−z, z] V (x, y, z) =
8xyz x, y, z ≥ 0 P E (x, y, z) ∈ E
V E ∇g(x, y, z) = (0, 0, 0) x = y = z = 0
(0, 0, 0) /∈ E g(0, 0, 0) = 0
x2 y2 z2
L(x, y, z, λ) = 8xyz + λ + + − 1 .
a2 b2 c2 x, y, z V = 0
xyz > 0
⎧
⎪⎨ ∂xL(x, y, z, λ) = 8yz + 2λx/a
⎪⎩
2
∂yL(x, y, z, λ) = 8xz + 2λy/b2 ∂xL(x, y, z, λ) = 8xy + 2λz/c2 y2 z2
+ + 2 2

λ = 0 yz = 0
x = 0 y = 0 z = 0
x 2 /a 2 = y 2 /b 2 = z 2 /c 2 x = a/ √ 3 y = b/ √ 3
z = c/ √ 3 V = 8abc/(3 √ 3) r > 0
a 2 = b 2 = c 2 = r 2 x = y = z = r/ √ 3
f(x, y) = x + y − 1
V = {(x, y) ∈ R 2 : x 2 + y 2 − 2x = 0}
V = {(x, y) ∈ R 2 : (x − 1) 2 + y 2 = 1},
V 1 (1, 0)
R 2 f
V
(1, 0)
ϕ(θ) = (cos θ + 1, sin θ), θ ∈ [0, 2π].
f ◦ ϕ(θ) = cos θ + 1 + sin θ − 1 = cos θ + sin θ,
d
f ◦ ϕ(θ) = − sin θ + cos θ,
dθ
d2 f ◦ ϕ(θ) = − cos θ − sin θ.
dθ2 cos θ = sin θ
cos θ = 0 sin θ = ±1
θ cos θ = 0 cos θ tan θ = 1
θ = π/4, 5/4π d2
dθ2 f ◦ ϕ(π/4) = − √ 2
d2
dθ 2 f ◦ ϕ(5π/4) = + √ 2
ϕ(π/4) = (1 + √ 2/2, √ 2/2) f ◦ ϕ(π/4) = √ 2
ϕ(5π/4) = (1 − √ 2/2, − √ 2/2) f ◦ ϕ(5π/4) = − √ 2
f V f x + y − 1 = c
y = c + 1 − x
V d = c + 1 y = d − x V
x 2 −2x+d 2 +x 2 −2dx = 0 2x 2 −2x(d+1)+d 2 = 0
(d + 1) 2 − 2d 2 = 0
−d 2 + 2d + 1 = 0 d = 1 ± √ 2 x = (d + 1)/2
x1 = 1 + √ 2/2 y1 = √ 2/2 x2 = 1 − √ 2/2 y2 = − √ 2/2
f(x1, y1) = √ 2 f(x2, y2) = − √ 2
(x1, y1) (x2, y2)
g(x, y) = x 2 + y 2 − 2x V = g −1 (0) L(x, y, λ) =
f(x, y) + λg(x, y) ∇g(x, y) = (2x − 2, 2y) (1, 0)
(1, 0) /∈ V g(1, 0) = 0
∇L(x, y, λ) = 0
⎧
⎪⎨ ∂xf(x, y) + λ∂xg(x, y) = 0,
∂yf(x, y) + λ∂yg(x, y) = 0,
⎪⎩
g(x, y) = 0

⎧
⎪⎨ 1 + λ(2x − 2) = 0,
1 + 2λy = 0,
⎪⎩
x 2 + y 2 − 2x = 0.
y = 0
λ = −1/(2y) x = 1
λ = −1/(2x − 2) y = x − 1
x 2 +(x−1) 2 −2x = 0 2x 2 −4x+1 = 0 x1 = 1+ √ 2/2
y1 = √ 2/2 x2 = 1 − √ 2/2 y2 = − √ 2/2
f(x1, y1) = √ 2 f(x2, y2) = − √ 2 (x1, y1)
(x2, y2)
V = {(x, y) ∈ R 2 :
x 2 /9 + y 2 = 1}
f f(x, y) := x 2 + y 2 ≥ 0 f
V
3 1
f
F (x, y) = f 2 (x, y) := x 2 + y 2
3 1
ϕ(θ) = (3 cos θ, sin θ) θ ∈
[0, 2π]
F (ϕ(θ)) = 9 cos 2 θ + sin 2 θ = 8 cos 2 θ + 1
d
F (ϕ(θ)) = 16 cos θ sin θ = 8 sin 2θ
dθ
d2 F (ϕ(θ)) = 16 cos 2θ.
dθ2 sin 2θ = 0 θ = 0, π/2, π, 3π/2
θ = 0, π θ = π/2, 3π/2
(±3, 0) 3 (0, ±1)
1
x 2 + y 2 = c 2
x 2 /9 + y 2 = 1
y 2 = c 2 − x 2
x 2 /9 + c 2 − x 2 = 1 8x 2 − 9(c 2 − 1) = 0 c 2 − 1 = 0
x = 0 y = ±1
x 2 = c 2 − y 2 8/9(c 2 − y 2 ) + y 2 = 1 y 2 − (9 − 8c 2 ) = 0
9 − 8c 2 = 0 y = 0 x = ±3
(±3, 0) 3 (0, ±1) 1
(±3, 0) (0, ±1)
g(x, y) = x 2 /9+y 2 −1 ∇g(x, y) = 0 x = y = 0 V = g −1 (0)
(0, 0) /∈ V L(x, y, λ) = F (x, y) + λg(x, y)
∇L = 0 ⎧
⎪⎨ ∂xF (x, y) + λ∂xg(x, y) = 0,
∂yF (x, y) + λ∂yg(x, y) = 0,
⎪⎩
g(x, y) = 0

⎧
⎪⎨ 2x + λ2x/9 = 0,
2y + 2λy = 0,
⎪⎩
x 2 /9 + y 2 = 1
y = 0 λ = −1 y = 0 x = ±3
λ = 1 x = 0 y = ±1
(±3, 0) 3 (0, ±1) 1
(±3, 0) (0, ±1)
f(x, y) = (x−1) 2 −y 2
V = {(x, y) ∈ R 2 : x 2 + y 2 = 1}
1 f
V
ϕ(θ) = (cos θ, sin θ) θ ∈ [0, 2π]
f ◦ ϕ(θ) = (cos θ − 1) 2 − sin 2 θ
d
f ◦ ϕ(θ) = 2(cos θ − 1) sin θ − 2 sin θ cos θ = 2 sin θ(1 − 2 cos θ)
dθ
d2 dθ2 f ◦ ϕ(θ) = 4 sin2 θ + 2(1 − 2 cos θ) cos θ.
θ = 0, π, π/3, 5/3π ϕ(0) = (1, 0) f(1, 0) = 0
d2 dθ2 f ◦ ϕ(0) = −2 ϕ(π) = (−1, 0)
f(−1, 0) = 4 d2
dθ2 f ◦ ϕ(π) = −6
ϕ(π/3) = (1/2, √ 3/2) f(1/2, √ 3/2) = −1/2 d2
dθ2 f ◦ ϕ(π/3) = 3
ϕ(5π/3) = (1/2, − √ 3/2) f(1/2, − √ 3/2) = −1/2
d2 dθ2 f ◦ ϕ(5π/3) = 3 (−1, 0)
(1/2, ± √ 3/2)
(x − 1) 2 − y2 = c
x2 + y2 = 1 (x0, y0)
(x0 − 1)(x − x0) − y0(y − y0) = 0
(x0, y0) x0(x − x0) + y0(y − y0) = 0
y0(x0 − 1) + x0y0 = 0 y0 = 0 x0 = ±1
x0 = 1/2 y0 = ± √ 3/2 f(1, 0) = 0 f(−1, 0) = 4
f(1/2, ± √ 3/2) = −1/2 (−1, 0) (1/2, ± √ 3/2)
g(x, y) = x2 + y2 − 1 V = g−1 (0) L(x, y, λ) =
f(x, y) + λg(x, y) ∇L = 0
⎧
⎪⎨ ∂xf(x, y) + λ∂xg(x, y) = 0,
∂yf(x, y) + λ∂yg(x, y) = 0,
⎪⎩
g(x, y) = 0.
⎧
⎪⎨ 2(x − 1) + λ2x = 0,
−2y + 2λy = 0,
⎪⎩
x 2 + y 2 = 1
y = 0 λ = 1 y = 0 x = ±1 f(1, 0) = 0
f(−1, 0) = 4 λ = 1 x = 1/2 y = ± √ 3/2
f(1/2, ± √ 3/2) = −1/2 (−1, 0) (1/2, ± √ 3/2)

f(x, y) = x 2 + y 2
V = {(x, y) ∈ R 2 : x 2 + y 2 + x 2 y 2 = 1}
V g(x, y) = x 2 + y 2 + x 2 y 2 − 1
1 (x, y) ∈ V 0 ≤ x 2 + y 2 =
1 − x 2 y 2 ≤ 1 f
L(x, y, λ) = f(x, y) + λg(x, y) ∇L = 0
⎧
⎪⎨ ∂xf(x, y) + λ∂xg(x, y) = 0,
∂yf(x, y) + λ∂yg(x, y) = 0,
⎪⎩
g(x, y) = 0
⎧
⎪⎨ 2x + λ(2x + 2xy
⎪⎩
2 ) = 0,
2y + λ(2y + 2yx2 ) = 0,
x 2 + y 2 + x 2 y 2 = 1
x = 0 λ = −1/(1 + y 2 ) x = 0 y = ±1
y = 0 λ = −1/(1 + x 2 ) y = 0 x = ±1
x = 0 y = 0 x = ±y 2x 2 + x 4 = 1 x = ± √ 2 − 1
y = ± √ 2 − 1 f(0, ±1) = f(±1, 0) = 1 f(± √ 2 − 1, √ 2 − 1) =
f(± √ 2 − 1, − √ 2 − 1) = 2( √ 2 − 1) 2( √ 2 − 1) < 1 √ 2 − 1 <
1/2 √ 2 < 3/2 2 < 9/4 (0, ±1) (±1, 0)
x 2 =
1 − y2 f , y2 =
1 + y2 1 − y2
1 − y2
,
1 + y2 1 + y 2 + y2 .
0 ≤ t ≤ 1
1 − t
k(t) = + t.
1 + t
k ′ (t) = t2 +2t−1
(t+1) 2 t = √ 2 − 1
t = √ 2 − 1
t = 0 t = 1 t = 0 y = 0
x = ±1 t = 1 y = ±1 x = 0 t = √ 2 − 1 |y| = √ √
2 − 1 |x| = 2 − 1
f(0, ±1) = f(±1, 0) = 1 f(± √ √ √ √ √
2 − 1, 2 − 1) = f(± 2 − 1, − 2 − 1) = 2( 2 − 1)
(0, ±1) (±1, 0)
R 2
V V = {(x, y) ∈ R 2 : x 2 + y 2 + x 2 y 2 = 1}
f(x, y) = x 2 + y 2
f(x, y, z) = x 2 − x + y 2 +
y(z + x − 1) V = {(x, y, z) ∈ R 3 : x 2 + y 2 = 1 ∧ x + y + z = 1}

g1(x, y, z) = x 2 +y 2 −1 g2(x, y, z) = x+y+z−1 =
0 g −1
1 (0)∩g−1 2 (0)
∇g1(x, y, z) = 0 x = y = 0 g1(x, y, z) = 0 ∇g2(x, y, z) = 0
x 2 +y 2 = 1 |x| ≤ 1 |y| ≤ 1
x + y + z = 1 z = 1 − x − y |z| ≤ 3
f
L(x, y, z, λ, µ) = f(x, y, z) + λg1(x, y, z) + µg2(x, y, z) ∇L = 0
⎧
∂xf(x, y, z) + λ∂xg1(x, y, z) + µ∂xg2(x, y, z) = 0
⎪⎨ ∂yf(x, y, z) + λ∂yg1(x, y, z) + µ∂yg2(x, y, z) = 0
∂zf(x, y, z) + λ∂zg1(x, y, z) + µ∂zg2(x, y, z) = 0
g1(x, y, z) = 0
⎪⎩
g2(x, y, z) = 0
⎧
−1 + 2x + y + 2λx + µ = 0
⎪⎨ −1 + x + 2y + z + 2λy + µ = 0
y + µ = 0
x 2 + y 2 = 1
⎪⎩
x + y + z = 1
y = −µ −1+2x+2λx = 0 2x(1+λ) = 1
λ = −1 x = 1
2(1+λ)
λy = 0 y = 0 λ = 0 λ = 0 x = 1/2 y = ± √ 3/2
y = 0 x = ±1 (1, 0, 0)
(−1, 0, 2) (1/2, √ 3/2, 1/2 − √ 3/2) (1/2, − √ 3/2, 1/2 + √ 3/2) f(1, 0, 0) = 0 f(−1, 0, 2) = 2
f(1/2, √ 3/2, 1/2 − √ 3/2) = f(1/2, − √ 3/2, 1/2 + √ 3/2) = −1/4 (−1, 0, 2)
(1/2, √ 3/2, 1/2 − √ 3/2) (1/2, − √ 3/2, 1/2 + √ 3/2)
R 3
V V = {(x, y, z) : x 2 + y 2 + xy − z 2 = 1 ∧ x 2 + y 2 = 1}
g1(x, y, z) = x2 + y2 + xy − z2 − 1 g2(x, y, z) =
x2 + y2 − 1 = 0
g −1
1
(0) ∩ g−1
2 (0) ∇g1(x, y) = (2x + y, 2y + x, −2z) x = y = z = 0
g1(x, y, z) = 0 ∇g2(x, y) = 0 x = y = 0 g2(x, y) = 0
x 2 +y 2 = 1 |x| ≤ 1
|y| ≤ 1 x 2 +y 2 +xy−z 2 = 1 z 2 = 1+x 2 +y 2 +xy |z| ≤ 2
f(x, y, z) = x 2 +y 2 +z 2
L(x, y, z, λ, µ) = f(x, y, z) + λg1(x, y, z) + µg2(x, y, z)
∇L = 0
⎧
∂xf(x, y, z) + λ∂xg1(x, y, z) + µ∂xg2(x, y, z) = 0
⎪⎨ ∂yf(x, y, z) + λ∂yg1(x, y, z) + µ∂yg2(x, y, z) = 0
∂zf(x, y, z) + λ∂zg1(x, y, z) + µ∂zg2(x, y, z) = 0
g1(x, y, z) = 0
⎪⎩
g2(x, y, z) = 0

⎧⎪
2x + λ(2x + y) + 2µx = 0
⎨2y
+ λ(2y + x) + 2µy = 0
2z − 2λz = 0
⎪⎩
x2 + y2 + xy − z2 = 1
x2 + y2 = 1
z 2 = xy z = 0 λ = 1
z = 0 xy = 0 x = 0 y = 0 x =
y = 0 x = 0 y = ±1 y = 0 x = ±1
(±1, 0, 0) (0, ±1, 0) λ = 1
2(x − y) + (x − y) + 2µ(x − y) = 0 (3 + 2µ)(x − y) = 0 x = y
µ = −3/2 x = y z = ±x x = y = ± √ 2/2
( √ 2/2, √ 2/2, ± √ 2/2) (− √ 2/2, − √ 2/2, ± √ 2/2) µ = −3/2 λ = 1
2(x + y) + 3(x + y) − 3x = 0 2x = −5y x, y
z 2 = xy
(±1, 0, 0) (0, ±1, 0) ( √ 2/2, √ 2/2, ± √ 2/2) (− √ 2/2, − √ 2/2, ± √ 2/2)
f(±1, 0, 0) = f(0, ±1, 0) = 1 f( √ 2/2, √ 2/2, ± √ 2/2) = f(− √ 2/2, − √ 2/2, ± √ 2/2) = 3/2
(±1, 0, 0) (0, ±1, 0)
f(x, y) = √ ye−x2−y2 x2 + (y − 1) 2 ≤ 1}
D := {(x, y) ∈ R 2 :
f D
f D f(x, y) ≥ 0 f(x, y) = 0 y = 0
D y = 0 O(0, 0)
f(x, y) = √ ye−x2e−y2 ≤ √ ye−y2 = f(0, y)
g(y) = √ ye−y2 0 ≤ y ≤ 2
g ′ (y) = e −y2
1
2 √
− 2y3/2 =
y e−y2
2 √ y (1 − 4y2 ),
]0, 2[ y = 1/2
g(0) = 0 < g(2) = √ 2e −4 √
2
< g(1/2) =
2 e−1/4 ,
y = 1/2 g [0, 2] (0, 1/2) f D
D f
∇f(x, y) =
−2x √ ye −x2 −y 2
, e−x2 −y 2
2 √ y − 2y3/2 e −x2 −y 2
D y > 0 x = 0 y = 1/2
f
⎛
⎞
Hess f(x, y) =
⎝ 4e−x2 −y2 x2√y − 2e−x2−y2√ y 4e−x2−y2 xy3/2 − e−x2 −y 2 √ x
y
4e−x2−y2 xy3/2 − e−x2 −y 2 √ x
y 4e−x2−y2 y5/2 − 4e−x2−y2√ y − e−x2 −y 2
4y3/2 Hess f(0, 1/2) =
√
2
− 4√ 0
e
0 − 2√2 4√
e
.
.
⎠ ,

√
2
(0, 1/2) f f(0, 1/2) =
2 e−1/4 D x = cos t y = sin t + 1 t ∈ [0, 2π[
h(t) := f(cos t, sin t + 1) = e −2(sin(t)+1) sin(t) + 1.
h ′ (t) = − e−2(sin(t)+1) (4 sin(t) + 3) cos(t)
2 .
sin(t) + 1
cos t = 0 t = π/2, 3π/2 sin t = −3/4
t (0, 0) (0, 2) (± √ 7/4, 1/4) f(0, 0) = 0 f(0, 2) = √ 2e−4 f(± √ 7/4, 1/4) =
e−1/2 √
2
/2 f(0, 1/2) =
2 e−1/4 (0, 0)
(0, 1/2)

R 2 D ⊆ R 2 f : D → R (x0, y0) ∈ D
f(x0, y0) = 0 (x0, y0) ∂yf(x, y) ∂yf(x, y) = 0
U V R x0 y0 ϕ : U → V
{(x, y) ∈ D : f(x, y) = 0} ∩ (U × V ) = {(x, ϕ(x)) : x ∈ U}, ϕ(x0) = y0.
ϕ f x x0
f (x0, y0)
ϕ x0
ϕ ′ (x0) = − ∂xf(x0, y0)
ϕyf(x0, y0) .
f (x0, y0)
ϕ ′ (x) = − ∂xf(x, ϕ(x))
, ∀x ∈ U.
∂yf(x, ϕ(x))
X, Y, Z D X × Y f : D → Z
(x0, y0) ∈ D f(x0, y0) = 0 (x0, y0)
∂Y f(x, y) ∂Y f(x0, y0) Y Z U ⊂ X V ⊂ Y
x0 y0 ϕ : U → V
{(x, y) ∈ D : f(x, y) = 0} ∩ (U × V ) = {(x, ϕ(x)) : x ∈ U}, ϕ(x0) = y0.
ϕ f x x0 f
(x0, y0)
ϕ ′ −1 (x0) = − ∂Y f(x0, y0) ◦ ∂Xf(x0, y0).
y = ϕ(x)
x = 2
x 2 + y 3 − 2xy − 1 = 0,
ϕ(2) = 1
f(x, y) = x 2 + y 3 − 2xy − 1 P = (2, 1) f(P ) = 0
∂yf(x, y) = 3y 2 − 2x ∂yf(P ) = −1 = 0
ϕ 2 F (x, ϕ(x)) = 0 ϕ(2) = 1 f ∈ C 1
ϕ C 1
ϕ ′ (2) = 2
d 2
d
dx ϕ(x) = −∂xf(x, ϕ(x)) 2x − 2ϕ(x)
= −
∂yf(x, ϕ(x)) 3ϕ2 (x) − 2x ,
dx2 ϕ(x) = −(2 − 2ϕ′ (x))(3ϕ2 (x) − 2x) − (2x − 2ϕ(x))(6ϕ(x)ϕ ′ (x) − 2)
(3ϕ2 (x) − 2x) 2
x = 2 ϕ(2) = 1 ϕ ′ (2) = 2 ϕ ′′ (2) = 1/16
,

df(x, y) = ∂xf(x, y) dx + ∂yf(x, y) dy = 2(x − y) dx + 2(3y 2 − x) dy.
∇f(x, y) = 2(x − y, 3y 2 − x)
S = {(x, y) ∈ R 2 : f(x, y) = 0} (x, y) ∈ S C ∈ R
C + df(x, y)(h, k) = 0 (h, k) (x, y) (2, 1)
C + 2h − k = 0 (2, 1) h = 2 k = 1
C = −3 (2, 1) k = 2h − 3 ϕ ′ (2) = 2
xe y − y + 1 = 0,
x = 0 y = ϕ(x)
x = 0 1
f(x, y) = xe y − y + 1 f ∈ C ∞ (R 2 ) f(0, 1) = 0
∂xf(x, y) = e y ∂yf(x, y) = xe y − 1 ∂yf(0, 1) = −1 = 0
ϕ f(x, y) = 0
(0, 1) ϕ(0) = 1 C 1 f ∈ C 1
ϕ ′ (0) = e
ϕ ′ (x) = − ∂xf(x, ϕ(x))
xe y − 1
= − eϕ(x)
xe ϕ(x) − 1 .
ϕ ′′ (x) = − eϕ(x) ϕ ′ (x)(xe ϕ(x) − 1) − e ϕ(x) (e ϕ(x) + xe ϕ(x) ϕ ′ (x))
(xe ϕ(x) − 1) 2
ϕ ′′ (0) = e 2
x 2 y − z 3 + 4xy 3 z + x 3 z + 30 = 0
= e2ϕ(x) (1 − xϕ ′ (x))
(xeϕ(x) − 1) 2
,
(1, 2) z = ϕ(x, y)
x = 1 y = 2 −1 ϕ
f(x, y, z) = x 2 y − z 3 + 4xy 3 z + x 3 z + 30 f(1, 2, −1) =
2 + 1 − 32 − 1 + 30 = 0 f
∂xf(x, y, z) = 2xy + 4y 3 z + 3x 2 z,
∂yf(x, y, z) = x 2 + 12xy 2 z,
∂zf(x, y, z) = −3z 2 + 4xy 3 + x 3 .
∂zf(1, 2, −1) = −3+32+1 = 30 = 0
z = ϕ(x, y) (1, 2) ϕ(1, 2) = −1 C 1
f ∈ C 1
∂xϕ(x, y) = − ∂xf(x, y, ϕ(x, y))
∂zf(x, y, ϕ(x, y)) = −2xy + 4y3ϕ(x, y) + 3x2ϕ(x, y)
−3ϕ2 (x, y) + 4xy3 + x3 ∂yϕ(x, y) = − ∂xf(x, y, ϕ(x, y))
∂zf(x, y, ϕ(x, y)) = − x2 + 12xy 2 ϕ(x, y)
−3ϕ 2 (x, y) + 4xy 3 + x 3
∂xϕ(1, 2) = 31/30 ∂yϕ(1, 2) = 47/30
f1(x, y, z) = x 2 + y 2 − z 2 + 2 = 0,
f2(x, y, z) = xy + 2yz − xz + 7 = 0,

x = 1 y = −1 z = 2 y z
x x = 1 y(x) z(x) x = 1 −1 2
y ′ (1) z ′ (1)
F (x, y, z) := (f1(x, y, z), f2(x, y, z)).
F : R 3 → R 2 F
Jac(F )(x, y, z) =
2x 2y −2z
y − z x + 2z 2y − x
y z x
y z (1, −1, 2)
Jac(F ) f1, f2 y
z
Jac(F )(1, −1, 2) =
2 −2 −4
−3 5 −3
.
−2 −4
, Jacy,z(F )(1, −1, 2) =
5 −3
(1, −1, 2) det Jacy,z(F )(1, −1, 2) = 26 = 0
x = 1
y z y = y(x) z = z(y) y(1) = −1 z(1) = 2
ϕ(x) = (y(x), z(x))
Jacy,z(F )(1, −1, 2)
[Jacy,z(F )(1, −1, 2)] −1 = 1
26
−3 4
−5 −2
x F (1, −1, 2)
F Jacx(F )(1, −1, 2) =
y ′ (1) = 9/13 z ′ (1) = 2/13
2
−3
.
ϕ ′
y ′ (x)
(1) =
z ′
= −[Jacy,z(F )(1, −1, 2)]
(x)
−1 Jacx(F )(1, −1, 2)
= − 1
−3 4 2
= −
26 −5 −2 −3
1
−18
26 −4
1
p1 > 0 p2 > 0
m3 m1 m2
1/3 1/2
m3 = m 1/3
1 m1/2
2 .
p1 > 0 p2 > 0 (m1, m2)
p1 p2 (m1, m2)
m3
1
pi mi i = 1, 2
I(m1, m2, m3, p1, p2) = m3 − p1m1 − p2m2,
.

I g(m1, m2, m3) = 0 g(m1, m2, m3) = m3 − m 1/3
1 m1/2 2
m1, m2, m3 ≥ 0 p1 p2
h(m1, m2) := I(m1, m2, m 1/3
1 m1/2
2 , p1, p2) = m 1/3
1 m1/2
2 − p1m1 − m2p2, m1, m2 ≥ 0.
mi i = 1, 2 m 1/3
1 ≤ p1m1/2 m 1/2
2 ≤ p2m2/2
|(m1, m2)| h(m1, m2) ≤ −p1m1/2−m2p2/2 |(m1, m2)| → +∞
h(m1, m2) → −∞ R > 0 |(m1, m2)| > R h(m1, m2) ≤ h(0, 0) = 0
(m1, m2) /∈ B(0, R) h (m1, m2) h
(1, 1) ∈ B(0, R) B(0, R) ∩ {m1 ≥ 0, m2 ≥ 0}
h
h(0, m2) = −p2m2 h(m1, 0) = −p1m1
h(0, 0) = 0
( ¯m1, ¯m2) ( ˜m1, ˜m2) {m1, m2 > 0}
(1−λ)( ¯m1, ¯m2)+λ( ˜m1, ˜m2) λ[−ε, 1+ε]
ε > 0 {m1, m2 > 0}
ψ(λ) := h((1 − λ)( ¯m1, ¯m2) + λ( ˜m1, ˜m2)),
d
dλ ψ(λ) = ∇h((1 − λ)( ¯m1, ¯m2) + λ( ˜m1, ˜m2)) · ( ˜m1 − ¯m1, ˜m2 − ¯m2),
d 2
dλ 2 ψ(λ) = 〈D2 h((1 − λ)( ¯m1, ¯m2) + λ( ˜m1, ˜m2))( ˜m1 − ¯m1, ˜m2 − ¯m2), ( ˜m1 − ¯m1, ˜m2 − ¯m2)〉
d
d
dλψ(0) = dλψ(1) = 0 ( ¯m1, ¯m2) ( ˜m1, ˜m2)
h(m1, m2)
D 2 ⎛
−
⎜
h(m1, m2) = ⎜
⎝
2
9 m−5/3 1 m1/2
1
2 6 m−2/3 1 m−1/2 2
1
6 m−2/3 1 m−1/2 2 − 1
4 m1/3 1 m−3/2
⎞
⎟
⎠
2 .
.
m −4/3
1 /(36m2) > 0
d2
dλ2 ψ(λ) < 0 ψ
( ¯m1, ¯m2) = ( ˜m1, ˜m2)
∇h(m1, m2) = (0, 0)
⎧
1
⎪⎨ 3 m−2/3 1 m1/2 2 − p1 = 0,
⎪⎩
1
2 m1/3 1 m−1/2 2 − p2 = 0,
p1 = 1
3 m−2/3 1 m1/2 2 p2 = 1
2 m1/3 1 m−1/2 2
I m1, m2, m 1/3
1 m1/2 2 , 1
3 m−2/3 1 m1/2 2 , 1
2 m1/3 1 m−1/2
2 = 1
6 m1/3 1 m1/2 2 > 0.
{m1, m2 ≥ 0}

⎛
1
⎜ 3
F (m1, m2, p1, p2) = ⎜
⎝
m−2/3 1 m1/2 2 − p1,
1
2 m1/3 1 m−1/2
⎞
⎟
⎠
2 − p2,
.
F (m1, m2, p1, p2) = 0.
Dm1,m2 F F (m1, m2) h
m1 = m1(p1, p2)
m2 = m2(p1, p2)
m1 m2 p1, p2
s(p1, p2) = ((p1, p2), m2(p1, p2))
Jac s(p1, p2) = −[Dm1,m2 F (m1, m2, p1, p2)] −1
|(m1,m2)=s(p1,p2) ◦ D[p1, p2]F (m1, m2, p1, p2) |(m1,m2)=s(p1,p2)
⎛
−
⎜
= − ⎜
⎝
2
9 m−5/3 1
1
6 m−2/3 1
⎛
−
⎜
= ⎜
⎝
2
9 m−5/3 1
1
6 m−2/3 1 2
⎛
=
⎜
⎝
m1/2 2
m−1/2 2
m1/2 2
m−1/2
−9 m 5/3
1 m−1/2 2
−6 m 2/3
1 m1/2 2
1
6 m−2/3 1
m−1/2 2
− 1
4 m1/3 1 m−3/2 2
1
6 m−2/3 1
m−1/2 2
− 1
4 m1/3 1 m−3/2 2
−6 m 2/3
1 m1/2 2
−8 m −1/3
1 m3/2 2
∂
m1(p1, p2) = −9 m
∂p1
5/3
∂
∂p2
∂
∂p1
∂
∂p2
⎞
⎟
⎠
⎞
⎟
⎠
⎞
⎟
⎠
−1
|(m1,m2)=s(p1,p2)
−1
|(m1,m2)=s(p1,p2)
|(m1,m2)=s(p1,p2)
1 (p1, p2)m −1/2
2
(p1, p2) < 0,
m1(p1, p2) = −6 m 2/3
1 (p1, p2)m 1/2
2 (p1, p2) < 0,
m2(p1, p2) = −6 m 2/3
1 (p1, p2)m 1/2
2 (p1, p2) < 0,
m2(p1, p2) = −8 m −1/3
1 (p1, p2)m 3/2
2 (p1, p2) < 0.
−1 0
·
0 −1
mi = mi(p1, p2) i = 1, 2

Γ = {(x, y) ∈ R 2 : x 3 − 3xy + y 3 = 0},
y = ϕ(x) C 1 x 3 − 3xy + y 3 = 0
f(x, y) = x 3 − 3xy + y 3 f(x, y) = f(y, x) Γ
∇f(x, y) = 3(x 2 − y, y 2 − x) (0, 0) (1, 1)
(0, 0) f(0, 0) = 0 f(1, 1) = 0 (0, 0)
f
Γ x = k k ∈ R f(k, y) = 0 k 3 − 3ky + y 3 = 0
f
y = ϕ(x)
pk(y) := k 3 − 3ky + y 3
k
pk(y) y x = k Γ
x Γ
pk(y) y x = k Γ
x Γ
y ↦→ pk(y) p ′ k (y) = 3(y2 − k) p ′′
k
(y) = 6y
k < 0 y ↦→ pk(y) p ′ k (y) > 0
yk pk(0) = k3 < 0 y3 k > 0
x < 0 y = y(x) ∈]0, +∞[ (x, y(x)) ∈ Γ
k = 0 p0(y) = y3 0 (0, y) ∈ Γ y = 0
k > 0 pk y± = ± √ k pk(·)
√ k p ′ k (√ k) = 0 p ′′
k (√ k) = 6 √ k > 0 − √ k
pk(·) y < − √ k |y| < k
y > k
pk(− √ k) = k 3 + 3k 3/2 − k 3/2 > pk( √ k) = k 3/2 (k 3/2 − 2).
lim
y→−∞ pk(y) = −∞ pk(− √ k) > 0 pk ] − ∞, − √ k[
pk(·) x > 0
y ∈ R (x, y) ∈ Γ y ≤ √ x
pk( √ k)
pk(·)
k 3/2 − 2 > 0 k > 3√ 4
0 < k < 3√ 4

k = 3√ 4 ξ ζ
p 3 √ 4 (y) = C(y − ξ)2 (y − ζ),
p ′ 3 √ 4 (y) = 2C(y − ξ)(y − ζ) + C(y − ξ)2 = C(y − ξ)(3y − 2ζ − ξ) = 3C(y − ξ)
y − 2
ξ
ζ − .
3 3
p ′ k (y) = 3(y2 − k) k = 3√ 4 p ′ k (y) = 3(y − 3√ 2)(y + 3√ 2)
C = 1 ξ = 3√ 2
ξ = − 3√ 2 p 3 √ 4 ( 3√ 2) = 0 p 3 √ 4 (− 3√ 2) = 0 ξ = 3√ 2
− 2
3 ζ − 3√ 2
3 = 3√ 2 ζ = −2 3√ 2 x = 3√ 4 (x, y) ∈ Γ
y ∈ {−2 3√ 2, 3√ 2}
x < 0 Γ y = ϕ1(x)
ϕ1(x) > 0 ∂yf(x, ϕ1(x)) > 0
ϕ1 ∈ C 1 (] − ∞, 0[; R)
Γ y = x x < 0
y > 0 (x, y) ∈ Γ
x > 0 y < 0 (x, y) ∈ Γ
ϕ2 :]0, +∞[→] − ∞, 0[ ϕ1
0 < x < 3√ 4 ϕ3, ϕ4 :]0, 3√ 4[→ ]0, +∞[ ϕ3(x) >
ϕ4(x) x
f(x, ϕi(x)) = 0 f
0 = lim sup
x→0 −
f(x, ϕ1(x)) = f(0, lim sup
x→0− ϕi(x)),
f(0, y) = 0 (0, y) ∈ Γ y = 0 lim sup ϕi(x))
lim inf
x→0− ϕi(x)) lim ϕi(x))
x→0−
x→0 −
lim
x→0− ϕ1(x) = lim
x→0 + ϕi(x) = 0, i = 2, 3, 4,
ϕ2(x) ≤ 0 x > 0 lim sup
x→ 3√ 4
0 = lim sup
x→ 3√ 4
ϕ2(x) ≤ 0
f(x, ϕ2(x)) = f( 3√ 4, lim sup
x→ 3√ ϕ2(x)),
4
f(0, y) = 0 (0, y) ∈ Γ y ∈ {−2 3√ 2, 3√ 2}
lim sup
x→ 3√ ϕ2(x) ∈ {−2
4
3√ 2, 3√ 2}.
lim sup
x→ 3√ ϕ2(x) = −2
4
3√ 2,
lim inf lim
x→ 3√ ϕ2(x) = −2
4
3√ 2
ϕ3, ϕ4 x
ϕi(x) ≤ 0 i = 3, 4 ϕ3 ϕ4
lim sup
x→ 3√ 4
( 3√ 4, 3√ 2)

Γ ∂yf(x, y) = 0
f(x, y) = 0 y 3 − 3y 3 + y 6 = 0 x = 0 y = 3√ 2 y = 3√ 2
Γ
Γ ϕ1 C 1
Γ
0 3√ 4 ϕ3 ϕ4 C1
3√
4 ϕ2
( 3√ 4, 3√ 2)
Γ ϕ3 ϕ4 Γ
( 3√ 2, 3√ 4) Γ
ϕ3
ϕ4 ∂yf ∂xf ϕ4
ϕ3 0 < x < 3√ 2
3√ 2 < x < 3√ 4 x = 3√ 2
γ1 ∂yf ∂xf
0 x → 0 −
γ2
(x, mx) ∈ Γ x < 0
(m 3 + 1)x 3 − 3mx 2 = 0 x = 3m/(m 3 + 1) x → −∞ m → −1 +
q = lim ϕ(x) + x = lim
x→−∞ m→−1− ϕ(x)
lim = lim
x→−∞ x m→−1 +
3m 2
m 3 + 1
3m
m 3 + 1
3m2 m3 3m
+
+ 1 m3 = lim
+ 1 m→−1− = −1.
3m(m + 1)
(m + 1)(m2 = −1
− m + 1)
ϕ1 y = −x − 1 ϕ1
x → 0 − m → 0 −
lim
x→0− ϕ1(x) − ϕ1(0)
= lim
x − 0 x→0− ϕ1(x)
= lim
x m→0− 3m 2
m 3 + 1
3m
m 3 + 1
limx→0− ϕ ′ 1 (x) = 0
x = −y − 1 γ2 limx→0 + ϕ ′ 2 (x) = −∞
f(r cos t, r sin t) = 0 r = 0
r =
3 sin(t) cos(t)
sin 3 (t) + cos 3 (t) .
t → −π/4 t → 3/4π r → +∞ tan(−π/4) =
tan(3/4π) = −1
ϕ5 :] − ∞, 3√ 4[→ [0, +∞[ ϕ5(x) =
ϕ1(x) x ≤ 0 ϕ5(x) = ϕ3(x) 0 ≤ x < 3√ 4 C1 ϕ ′ 5 (0) = 0 0
0
Γ = {(x, y) ∈ R 2 : x 4 − 4xy + y 4 = 0}.
f(x, y) = x 4 − 4xy + y 4 f(x, y) = f(y, x) = f(−x, −y)
Γ y = x (0, 0) ∈ Γ
= 0

x = ρ cos θ y = ρ sin θ ρ 2 (ρ 2 cos 4 θ − 4 cos θ sin θ + ρ 2 sin 4 θ) = 0
ρ = 0
ρ 2 =
4 cos θ sin θ
cos 4 θ + sin 4 θ .
cos θ sin θ > 0 Γ
ρ(θ) =
4 cos θ sin θ
cos 4 θ + sin 4 θ
cos 4 θ + sin 4 θ x 4 + y 4
x 2 + y 2 = 1 x 4 + y 4
ρ
Γ
ρ 0 (0, 0) ∈ Γ lim θ→0 + ρ(θ) = 0
lim
θ→0
θ→π/2
θ→π
θ→3/2π
ρ(θ) = lim ρ(θ) = lim ρ(θ) = lim ρ(θ) = 0.
+ − + −
y 2 Γ ρ 2 (θ) sin 2 θ θ =
0, π/2, π, 3/2π y = 0 0 < θ < π/2
f1(θ) = ρ 2 (θ) sin 2 θ = 4 cos θ sin3 θ
cos 4 θ + sin 4 θ = 4 tan3 θ
1 + tan 4 θ .
g(s) = 4s 3 /(1 + s 4 ) s > 0
g ′ (s) = 4 3s2 (1 + s 4 ) − 4s 6
(1 + s 4 ) 2
= 4s2 (3 − s4 )
(1 + s4 .
) 2
s = 4√ 3
tan :]0, π/2[→]0, +∞[
f1 = g ◦ tan θm = arctan 4√ 3 f1 [0, π/2]

x 4 − 4xy + y 4 = 0
cos θm sin θm tan θm = 4√ 3 0 < θm < π
cos 2 θm + sin 2 θm = 1,
sin θm = 4√ 3 cos θm.
cos θm = (1+ √ 3) −1/2 sin θm = 4√ 3 (1+ √ 3) −1/2
f1 g 3√
4 ymax = g( 4√ 3) = 33/8 x∗ ymax
x ∗
= ρ(θm) cos θm =
4 cos3 θm sin θm
=
4 tan θm
8√
3
= 2 √ =
4 8√ 3.
cos 4 θm + sin 4 θm
1 + tan 4 θm
Γ [−3 3/8 , 3 3/8 ] × [−3 3/8 , 3 3/8 ]
Γ P1 = ( 8√ 3, 3 3/8 )
P2 = (− 8√ 3, −3 3/8 ) P3 = (3 3/8 , 8√ 3) P4 = (−3 3/8 , − 8√ 3)
α z = z(x, y)
e z+x2
+ αx + y 2 − z 2 = 1, z(0, 0) = 0
(0, 0) α

f(x, y, z) = ez+x2 f
+ αx + y 2 − z 2 − 1 f(0, 0, 0) = 0 α
∂xf(x, y, z) = 2xe z+x2
+ α
∂yf(x, y, z) = 2y
∂zf(x, y, z) = e z+x2
− 2z.
∂zf(0, 0, 0) = 1 = 0 f = 0
z = z(x, y) (0, 0) z(0, 0) = 0 f ∈ C 1
C 1
∂xz(x, y) = − ∂xf(x, y, z(x, y)) 2xez(x,y)+x2 + α
= −
∂zf(x, y, z(x, y)) ez(x,y)+x2 − 2z(x, y) ,
∂yz(x, y) = − ∂yf(x, y, z(x, y))
2y
= −
∂zf(x, y, z(x, y)) ez(x,y)+x2 − 2z(x, y) .
∂xz(0, 0) = α ∂yz(0, 0) = 0 (0, 0)
z ∈ C 1 α = 0
α z(x, y) (0, 0)
∂ 2 xxz(x, y) = ∂xf(x, y, z(x, y)) ∂xz(x, y)∂2 zzf(x, y, z(x, y)) + ∂2 xzf(x, y, z(x, y))
[∂zf(x, y, z(x, y))] 2
+
− ∂zf(x, y, z(x, y)) ∂xz(x, y)∂2 xzf(x, y, z(x, y)) + ∂2 xxf(x, y, z(x, y))
[∂zf(x, y, z(x, y))] 2
∂ 2
∂yz(x, y)∂
xyz(x, y) =
2 zzf(x, y, z(x, y)) + ∂2 yzf(x, y, z(x, y)) ∂xf(x, y, z(x, y))
[∂zf(x, y, z(x, y))] 2
+
− ∂zf(x, y, z(x, y)) ∂yz(x, y)∂2 xzf(x, y, z(x, y)) + ∂2 xyf(x, y, z(x, y))
[∂zf(x, y, z(x, y))] 2
∂ 2 yyz(x, y) = ∂yf(x, y, z(x, y)) ∂yz(x, y)∂2 zzf(x, y, z(x, y)) + ∂2 yzf(x, y, z(x, y))
[∂zf(x, y, z(x, y))] 2
+
− ∂zf(x, y, z(x, y)) ∂yz(x, y)∂2 yzf(x, y, z(x, y)) + ∂2 yyf(x, y, z(x, y))
[∂zf(x, y, z(x, y))] 2
f
∂xxf(x, y, z) = 2e z+x2
∂yyf(x, y, z) = 2, ∂yyf(0, 0, 0) = 2
+ 4x 2 e z+x2
, ∂xxf(0, 0, 0) = 2
∂zzf(x, y, z) = e z+x2
− 2, ∂zzf(0, 0, 0) = −1
∂xyf(x, y, z) = ∂zyf(x, y, z) = 0, ∂xyf(0, 0, 0) = ∂zyf(0, 0, 0) = 0,
∂xzf(x, y, z) = 2xe z+x2
, ∂xzf(0, 0, 0) = 0.

∂xz(0, 0) = ∂yz(0, 0) = 0 ∂xf(0, 0, 0) = ∂yf(0, 0, 0) = 0
∂zf(0, 0, 0) = 1
∂ 2 xxz(0, 0) = ∂xf(0, 0, 0) ∂xz(0, 0)∂2 zzf(0, 0, 0) + ∂2 xzf(0, 0, 0)
[∂zf(0, 0, 0)] 2
+
− ∂zf(0, 0, 0) ∂xz(0, 0)∂2 xzf(0, 0, 0) + ∂2 xxf(0, 0, 0)
[∂zf(0, 0, 0)] 2
= −2,
∂ 2
∂yz(0, 0)∂
xyz(0, 0) =
2 zzf(0, 0, 0) + ∂2 yzf(0, 0, 0) ∂xf(0, 0, 0)
[∂zf(0, 0, 0)] 2
+
− ∂zf(0, 0, 0) ∂yz(0, 0)∂2 xzf(0, 0, 0) + ∂2 xyf(0, 0, 0)
[∂zf(0, 0, 0)] 2
= 0,
∂ 2 yyz(0, 0) = ∂yf(0, 0, 0) ∂yz(0, 0)∂2 zzf(0, 0, 0) + ∂2 yzf(0, 0, 0)
[∂zf(0, 0, 0)] 2
+
− ∂zf(0, 0, 0) ∂yz(0, 0)∂2 yzf(0, 0, 0) + ∂2 yyf(0, 0, 0)
[∂zf(0, 0, 0)] 2
= −2.
z (0, 0)
(0, 0) z(x, y)
H(z)(0, 0) =
−2 0
0 −2
,
Γ (x, y) ∈ R 2
y 3 − xy 2 + x 2 y = x − x 3 .
Γ ϕ : R → R
ϕ
ϕ ϕ
x → ±∞
f(x, y) = y 3 − xy 2 + x 2 y − x + x 3 Γ = {(x, y) : f(x, y) = 0}
f(−x, −y) = −f(x, y) Γ
f(0, y) = 0 y = 0 f(x, 0) = 0 −x + x 3 = 0 x ∈ {0, ±1}
f f ∈ C ∞
∂xf(x, y) = −y 2 + 2xy − 1 + 3x 2 = 4x 2 − (y − x) 2 − 1
∂yf(x, y) = 3y 2 − 2xy + x 2 = 2y 2 + (x − y) 2
∂yf(x, y) x = y = 0 x = 0
x γ C 1 x = 0
f(0, y) = 0 y = 0 ϕ(0) = 0
ϕ {(xn, yn)}n∈N Γ xn → 0
0 = lim inf
n→∞ f(xn, yn) = lim inf
n→∞ y3 n,
0 = lim sup
n→∞
f(xn, yn) = lim sup y
n→∞
3 n,
yn → 0 = limn→∞ ϕ(xn) ϕ ϕ(0) = 0 ϕ(±1) = 0
ϕ ∈ C 1 R \ 0 f (0, 0)
df(0, 0) = − dx C ∈ R C + x = 0 γ
(0, 0) x = 0 ϕ 0 x = 0
ϕ ′ (x) = − ∂xf(x, ϕ(x))
∂yf(x, ϕ(x)) = −4x2 − (ϕ(x) − x) 2 − 1
2ϕ2 ,
(x) + (x − ϕ(x)) 2

ϕ(1) = 0 ϕ ′ (1) = − ∂xf(1,0)
∂yf(1,0) = −2 < 0 ϕ(x)
1 x > 1
1 0 < x < 1
0 1 f(x, 0) = 0 x ∈ {0, ±1}
ϕ(x) > 0 0 < x < 1 x < −1 ϕ(x) = 0 x ∈ {0, ±1}
ϕ(x) < 0 −1 < x < 0 x > 1
y = mx f(x, mx) = 0 (m3 − m2 + m + 1)x3 − x = 0
x((m3 − m2 + m + 1)x2 − 1) = 0 x = 0
(m3 − m2 + m + 1) > 0
x = ±(m3 − m2 + m + 1) −1/2
y = ±
h(m) =
m
√ m 3 − m 2 + m + 1 .
m
√ m 3 − m 2 + m + 1 ,
m3 − m2 + m + 1 − m
h ′ (m) =
3m 2 − 2m + 1
2 √ m 3 − m 2 + m + 1
(m 3 − m 2 + m + 1)
= 2m3 − 2m 2 + 2m + 2 − m(3m 2 − 2m + 1)
2(m 3 − m 2 + m + 1) 3/2
=
−m 3 + m + 2
2(m 3 − m 2 + m + 1) 3/2
m 3 −m−2 = 0
v(m) = m 3 − m − 2 v v ′ (m) = 3m 2 − 1
¯m ± = ± 1/3 | ¯m ± | < 1 | ¯m ± | 3 < 1 v( ¯m ± ) < 0
v m 3 −m−2 = 0
v(0) = v(1) = −2 limm→∞ v(m) = +∞
v(2) = 4 > 0
2 1 < ¯m < 2 h ′ ( ¯m) = 0
ϕ ±( ¯m 3 − ¯m 2 + ¯m + 1) −1/2
x + = −x − = ( ¯m 3 − ¯m 2 + ¯m + 1) −1/2 ¯m 3 − ¯m − 2 = 0 x + = −x − =
(3 − ¯m 2 + 2 ¯m) −1/2
0, 1 W (t) =
3 − t 2 + 2t x + = −x − = (W ( ¯m)) −1/2 ¯m ∈ [1, 2]
W ]1, 2[
W ′ (t) = −2t + 2 < 0 W (2) = 3 < W (t) < W (1) = 4 0 < 4 −1/2 <
x + < 3 −1/2 < 1 x − = −x + 0 < x + < 1 −1 < x − < 0 x +
x −
x ≥ 0 x < 0
x = (m 3 − m 2 + m + 1) −1/2 ϕ(V (m)) =
m(m 3 −m 2 +m+1) −1/2 V (m) = m 3 −m 2 +m+1 V ′ (m) = 3m 2 −2m+1
V ′ > 0 V
m ∗ V (0) = 1 V (−1) = −2 < 0
−1 < m ∗ < 0 A(m) = V (m)/(m − m ∗ ) A(m)
ϕ(x)
lim = lim
x→∞ x m→m∗+ ϕ(V (m))
(m3 − m2 + m + 1) −1/2 = m(m3 − m2 + m + 1) −1/2
(m3 − m2 + m + 1) −1/2 = m∗ .

q = lim
x→∞ ϕ(x) − m∗ x
= lim
m→m ∗ m(m3 − m 2 + m + 1) −1/2 − m ∗ (m 3 − m 2 + m + 1) −1/2
= lim
m→m ∗
= lim
m→m ∗
m − m ∗
(m3 − m2 = lim
+ m + 1) −1/2 m→m∗
m − m ∗
A(m)
= 0
m − m ∗
(m − m ∗ ) 1/2 A(m) 1/2
y = m∗x x → +∞
y = m∗x x → −∞
¯m m∗ x ± , y ±
¯m = 1
3
m ∗ = 1
3
3
27 − 3 √ 78 + 3
3 9 + √
78
1.52,
2
1 − √ +
3 33 − 17 3
3 √
33 − 17 −0.54,
3
x + = −x − = (3 − ¯m 2 + 2 ¯m) −1/2 0.51,
y + = −y − = h( ¯m) 0.79.

y 3 − xy 2 + x 2 y = x − x 3

f(x1, ..., xn) dx1 ... dxn,
D
D ⊆ R n
gi(x1, ..., xn) ≤ 0 i = 1, ..., n gi : R n → R C 1 f : D → R
D
D = D1 ∪ ... ∪ DM D
Dj
D f D
Dn ⊂ D D f
f Dn In n → ∞
In Dn → D
f Dn
f : X1 × X2 → R X1, X2
R
f(x1, x2) dx1 dx2 = f(x1, x2) dx2 dx1 = f(x1, x2) dx1 dx2.
X1
X2
X, Y R n ϕ : X → Y
C 1 A X B = ϕ(A) f : B → R
B A
B f B
x ↦→ f(ϕ(x))| det Jac(ϕ)(x)|
A
f(y) dy = f(ϕ(x))| det Jac(ϕ)(x)| dx,
B
Jac(ϕ)(x) ϕ x
A
X2
X1

X = {(ρ, θ) ∈ R 2 : ρ > 0, 0 < θ < 2π}
f : R 2 → R R 2 (ρ, θ) ↦→ f(ρ cos θ, ρ sin θ)ρ X
R 2
f(x, y) dx dy =
X
f(ρ cos θ, ρ sin θ)ρ dρ dθ.
X = {(ρ, θ, ϕ) ∈ R 2 : ρ > 0, 0 < θ < 2π, 0 < ϕ <
π} f : R 3 → R R 3
X
f(x, y, z) dx dy dz =
R 3
(ρ, θ, ϕ) ↦→ f(ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ)ρ 2 sin ϕ
I = xy dx dy,
X
f(ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ)ρ 2 sin ϕ.
D y 2 = 4x x 2 = 4y
D
y 4 = 64y y = 0 y = 4
(0, 0) (4, 4) x ≥ 0 y = 2 ± √ x y = 2 ± √ x
y = x 2 /4 0 < x < 4 y = x 2 /4
(2 ± √ x) 2 = 4x > (x 2 /4) 2 = x 4 /16
I =
4
0
2x3 =
3
D = {(x, y) : 0 ≤ x ≤ 4, x 2 /4 ≤ y ≤ 2 √ x},
√
2 x
xy dy dx =
x 2 /4
− x6
192
x=4
x=0
= 64
3 .
4
0
xy 2
2
y=2 √ x
y=x 2 /4
I = 1 − y2 dx dy,
D = B((1, 0), 1)
D
=
4
0
2x 2 − x5
dx
32
D x = 1 − 1 − y2 x = 1 + 1 − y2 −1 < y < 1
1
√
1+ 1−y2 I =
−1 1− √ 1−y2
1
1 − y2 dx dy = 4 (1 − y
0
2 ) dy = 8
3 .
D xy dx dy D D
A(0, 4) B(1, 1) C(4, 0)
P1 = (x1, y1) P2 = (x2, y2)
(y − y1)(x1 − x2) = (y1 − y2)(x − x1).
x = x1 y = y1 P1 x = x2
y = y2 P2

y = 4 − x A C y = 4 − 3x A B y = (4 − x)/3
B C
I =
=
1 4−x
0
1
0
1
4−3x
xy dy dx +
4 4−x
1
(4−x)/3
4
x (4 − x)2 − (4 − 3x) 2
dx +
2
4
xy dy dx
= x
0
2 (8 − 4x) dx + 8
x(4 − x)
9 1
2 dx = 5 4
+
3 9
= 5
4
+ 8x
3 9
2 − 8
3 x3 + x4
x=4 4 x=1
= 5
4 8 · 64
8 1
+ 8 · 16 − + 64 − 8 − −
3 9
3 3 4
= 5
4 64 67
+ − =
3 9 3 12
5 26
+ 7 =
3 3 .
I = e x2
4 +y2
x
2
4 + y2
− 1
dx dy,
D
1
x
2
(4 − x) 2
2
4 − x
−
dx
3
D x ≥ 0 y ≥ 0 x 2 + 4y 2 ≤ 16
D
D = (x, y) ∈ R 2 : x ≥ 0, y ≥ 0, x2
4 + y2
≤ 4
= (x, y) ∈ R 2
x
2 : x ≥ 0, y ≥ 0, + y
2
2
≤ 4
= (2x, y) ∈ R 2 : x ≥ 0, y ≥ 0, x 2 + y 2 ≤ 4
= {(2ρ cos θ, ρ sin θ) : θ ∈ [0, π/2], 0 ≤ ρ ≤ 2}
4
(16x − 8x
1
2 + x 3 ) dx
ϕ(x, y) = (2ρ cos θ, ρ sin θ)
Jac(ϕ)(ρ, θ) =
2 cos θ −2ρ sin θ
sin θ ρ cos θ
2ρ
I =
π/2 2
0
= π
2
0
e ρ2
|ρ 2 − 1| 2ρ dρ dθ = π
[e s (1 − s)] s=1
s=0 +
1
0
2
4
e s
ds + π
2
e s |s − 1| ds = π
2
[e s (s − 1)] s=4
s=1 −
= π
π
(−1 + e − 1) +
2 2 (3e4 − e 4 + e) = π(e 4 + e − 1).
0
1
0
4
y
I = arcsin dx dy,
x2 + y2 D 1 ≤ x 2 + y 2 ≤ 4 y ≥ 0
D
1
.
e s (1 − s) ds + π
2
e s
ds
D = {(ρ cos θ, ρ sin θ) : 1 ≤ ρ ≤ 2, 0 ≤ θ ≤ π},
4
1
e s (s − 1) ds

arcsin sin θ = θ θ ∈] − π/2, π/2[
I =
= 2
2 π
1 0
2 π/2
1
0
arcsin(sin θ) ρ dρ dθ = 2
θ ρ dρ dθ = 2
2
1
ρ dρ ·
2 π/2
1 0
π/2
0
arcsin(sin θ) ρ dρ dθ
θ dθ = 3
8 π2 .

V = π
b
a
f 2 (x)
z = f(x, y) D
A = 1 + (∂xf) 2 + (∂yf) 2 dxdy
D
S
x γ z = 0 x ≥ 0
γ x = x(t) y = y(t) a ≤ t ≤ b
AS = 2π
b
a
y(t) x ′ (t) 2 + y ′ (t) 2 dt
γ y = y(x) a ≤ x ≤ b
AS = 2π
b
a
y(x) 1 + y ′ (x) 2 dx
D ⊆ R2
x dx dy
y dx dy
xB = D
, yB = D
.
dx dy
dx dy
D
D
D ⊆ R3
x dx dy dz
y dx dy dz
z dx dy dz
xB = D , yB = D , zB = D
dx dy dz
dx dy dz
dx dy dz
D ⊆ R2
I = δ 2 (x, y) dx dy
D
D
δ 2 (x, y) (x, y)
T
θ z E x ≥ 0 (x, z)
λ3(T ) = θrGλ2(E),
rG E
D
D

I R r : I → Rd
γ µ : r(I) → [0, +∞[
massa(γ) := µ(r(t)) |r ′ (t)| dt = 0,
r µ
r(t)µ(r(t)) |r
G I :=
′ (t)| dt
.
massa(γ)
d = 2 r(t) = (x(t), y(t))
⎛
⎜
x(t)µ(x(t), y(t))
G := ⎜ I
⎝
˙x 2 (t) + ˙y 2
(t) dt y(t)µ(x(t), y(t))
I ,
massa(γ)
˙x 2 (t) + ˙y 2 ⎞
(t) dt
⎟
massa(γ)
⎠ .
d = 2 µ = cost
massa(γ) = µ |r ′ (t)| dt = µ lunghezza(γ),
I
I
G
1
:=
x(t)
lunghezza(γ) I
˙x 2 (t) + ˙y 2
(t) dt, y(t)
I
˙x 2 (t) + ˙y 2
(t) dt .
γ
y ≥ 0 µ > 0
γ r(t) = (x(t), y(t)) = (cos t, sin t) θ ∈ I := [0, π]
µ lunghezza(γ) = π
x(t)
I
˙x 2 (t) + ˙y 2 π
(t) dt = cos t sin
0
2 t + cos2 π
t dt = cos t dt = 0.
0
y(t) ˙x 2 (t) + ˙y 2 π
(t) dt = sin t sin2 t + cos2 π
t dt = sin t dt = 2.
G = (0, 2/π)
I
0
γ r(t) = (t 2 , 2t)
t ∈ I := [0, 1] µ > 0
x(t) = t2 y(t) = 2t
γ
|r ′ 1
1
(t)| dt = (2t) 2 + 22 dt = 2 1 + t2 dt.
I
α ∈ R
α
t
+ t2 dt = t α + t2 −
2 dt
= t α + t2 α
− + t2 dt + α
0
√ α + t 2 = t α + t 2 −
0
0
dt
√ α + t 2
α + t 2 − α dt
√ α + t 2
√ α + t2 = −t + w t = w2 − α
2w dt = w2 + α
2w2
dt
√
α + t2 =
2w
w2 w
+ α
2
+ α 1
dw =
2w2 w = log |w| + c = log |t + α + t2 | + c,

α + t 2 dt = t α + t 2 −
α 1
+ t2 dt =
2
α = 1
lunghezza(γ) = 2
1
0
α + t 2 dt + α log |t + α + t 2 | + c,
t α + t2 + α log |t + α + t2
| + c .
1
0
1 + t 2 dt = √ 2 + log(1 + √ 2).
s = t2 t = √ s dt = ds
2 √ s
t
I
2 1 + t2 dt = 1
1
1 1 1
s(1 + s) ds = (s2 + s + 1/4) − 1/4 ds =
2 0
2 0
2
= 1
1
3 1
(2s + 1) 2 − 1 ds = w2 − 1 dw,
4
8
0
1
1
0
(s + 1/2) 2 − 1/4 ds
w = 2s + 1 = 2t2 + 1 α = −1
t
I
2 1 + t2 dt = 1
3
1 1
w2 − 1 dw = w
8 1
8 2
w2 − 1 − log(w + w2
− 1)
w=3 w=1
= 1
16 (3√8 − log(3 + √ 8)) = 1
3
8
√ 2 − 1
2 log(1 + √ 2) 2
= 1
3
8
√ 2 − log(1 + √
2) ,
(1 + √ 2) 2 = 3 + √ 8 s = t2
2t
I
1 + t2 1
dt = (1 + s)
0
1/2 2
ds = v
1
1/2 dv = 2
3 [v3/2 ] 2 v=1 = 2
3 (2√2 − 1).
G :=
=
=
1
lunghezza(γ)
2
√ 2 + log(1 + √ 2)
1
√ 2 + log(1 + √ 2)
x(t) ˙x 2 (t) + ˙y 2
(t) dt,
I
1
0
1
4
t 2 1 + t 2 dt,
1
3 √ 2 − log(1 + √ 2)
0
I
y(t) ˙x 2 (t) + ˙y 2
(t) dt
2t 1 + t2
dt
, 4
3 (2√
2 − 1)
γ ρ(θ) = 3θ θ ∈ I := [0, 5π]
γ
µ : R 2 → [0, +∞[
µ(x, y) = 6(x 2 + y 2 ) γ
x(θ) = ρ(θ) cos θ = 3θ cos θ,
y(θ) = ρ(θ) sin θ = 3θ sin θ,
˙x 2 (θ) + ˙y 2 (θ) = ( ˙ρ(θ) cos θ − ρ(θ) sin θ) 2 + ( ˙ρ(θ) sin θ + ρ(θ) cos θ) 2
= ˙ρ 2 (θ) cos2 θ + ρ2 (θ) sin2 θ − 2 ˙ρ(θ)ρ(θ) sin θ cos θ + ˙ρ 2 (θ) sin2 θ + ρ2 (θ) cos2 θ + 2 ˙ρ(θ)ρ(θ) sin θ cos θ
= ˙ρ 2 (θ) + ρ 2 (θ).
˙x 2 (θ) + ˙y 2 (θ) = 3 1 + θ 2 .

γ
lunghezza(γ) = |r ′ (θ)| dθ = 3
I
5π
0
3
1 + θ2 dθ = 5π
2
1 + 25π2 + log(5π + 1 + 25π2
) ,
√ 1 + t 2
µ(x(θ), y(θ)) = µ(ρ(θ) cos θ, ρ(θ) sin θ) = ρ 2 (θ) = 9θ 2 .
massa(γ) = µ(x(θ), y(θ)) ˙x 2 (θ) + ˙y 2 (θ) dθ = 27
I
5π
0
θ 2 1 + θ 2 dθ.
w = 2θ 2 + 1 √ w 2 − 1
5π
50 2 π+1
massa(γ) = 27 θ
0
2 1 + θ2 dθ = 27
w2 − 1 dw
8 1
= 27
1
2
+ 50π (1 + 50π
16
2 ) 2
− 1 − log 1 + 50π 2
+ (1 + 50π2 ) 2
− 1 .
T r
z a a > r
C r zy (0, a, 0)
C := {(0, s cos φ + a, s sin φ) : 0 ≤ s ≤ r, φ ∈ [0, 2π]}
θ ∈ [0, 2π] (x, y, z) z θ
( x 2 + y 2 cos θ, x 2 + y 2 sin θ, z)
T := {(|s cos φ + a| cos θ, |s cos φ + a| sin θ, s sin φ) : ρ, φ ∈ [0, 2π], 0 ≤ s ≤ r}
a > r s cos φ + a > −s + a > −r + a > 0
T := {((s cos φ + a) cos θ, (s cos φ + a) sin θ, s sin φ) : ρ, φ ∈ [0, 2π], 0 ≤ s ≤ r},
T
ϕ(s, θ, ϕ) = ((s cos φ + a) cos θ, (s cos φ + a) sin θ, s sin φ).
T
V =
T
dx dy dz.
⎛
cos φ cos θ −(s cos φ + a) sin θ
Jac(ϕ)(s, θ, ϕ) = ⎝ cos φ sin θ (s cos φ + a) cos θ
⎞
−s sin φ cos θ
−s sin φ sin θ ⎠ .
sin φ 0 s cos φ
det Jac(ϕ)(s, θ, ϕ) = sin φ(s(s cos φ + a) sin 2 θ sin φ + s sin φ cos 2 θ(s cos φ + a))+
+ s cos φ(cos φ cos 2 θ(s cos φ + a) + (s cos φ + a) sin 2 θ cos φ)
= s sin 2 φ(s cos φ + a) + s cos 2 φ(s cos φ + a)
= s(s cos φ + a) > 0

V =
2π 2π r
s(s cos φ + a) dr dφ dθ
0 0 0
r 2π
= 2πa (s
0 0
2 cos φ + as) dφ ds = 4π 2 r
a s dr = 2π
0
2 ar 2 .
C πr2
2πa V = 2π2ar2
I = cos(x + y)e x−y dxdy
D
D = {(x, y) ∈ R 2 : |x + y| ≤ π
, |x − y| ≤ 1}.
2
u = x+y v = x−y x = (u+v)/2 y = (u−v)/2
−1/2 1/2
I = 1
1 π/2
cos u e
2 −1 −π/2
v du dv = e − e −1 = 2 sinh(1).
1 + y
I =
x dx dy,
D 1 − y
D
y 4 + x 2 − 2x = 0
1 − y 4 = (x − 1) 2 (1 − y)(1 + y)(1 + y 2 ) = (x − 1) 2
−1 ≤ y ≤ 1 1 − 1 − y4 < x < 1 + 1 − y4
1
√
1+ 1−y4 1
√
1+ 1−y4 1 + y
I =
x dx dy =
1 − y
= 1
2
= 2
= 4
−1 1− √ 1−y4 1
−1
1
−1
1
0
−1
1− √ 1−y 4
1 + y
x dx dy
1 − y
1 + y
1 − y ((1 + 1 − y4 ) 2 − (1 − 1 − y4 ) 2
1 1 + y
) dy = 2
1 − y4 dy
−1 1 − y
(1 + y) 1 + y2 1
dy = 2 1 + y2 dy − 2 y 1 + y2 dy
−1
1 + y 2 dy = 2( √ 2 + arc sinh(1)) = 2( √ 2 + log(1 + √ 2))
|(x, y)| −p B = B(0, 1) ⊆ R 2
p < 2 R 2 \ B(0, 1) p > 2 |(x, y, z)| −p
B(0, 1) ⊆ R 3 p < 3 R 3 \ B(0, 1) p > 3
p = 2
|(x, y)|
B
−p 2π 1 1
dx dy =
ρ dρ dθ = 2π
0 0 ρp p < 2 p = 2
2π
0
1
ρ = 2π[log ρ]10 = +∞,
B(0, 1) ⊆ R2 p < 2
1
1
0
ρ 1−p dρ = 2π
(2 − p) [ρ2−p ] ρ=1
ρ=0 ,

p = 2
|(x, y)| −p dx dy =
R 2 \B
2π +∞
p > 2 p = 2
∞ 1
2π
1 ρ = 2π[log ρ]∞1 = +∞,
R2 \ B p > 2
0
1
1
2π
ρ dρ dθ =
ρp (2 − p) [ρ2−p ] ρ=∞
ρ=1
x = ρ sin φ cos θ y = ρ sin φ sin θ z = ρ cos φ
ρ2 sin φ p = 3
π 2π 1
0 0 0
ρ 2−p sin φ dρ dθ dφ = 4π
3 − p
p < 3 p = 3
1
π/2
4π
0
1
= +∞
ρ
B(0, 1) ⊆ R3 p < 3
p = 3
R 3 \B(0,1)
p > 3 p = 3
0
sin φ dφ · [ρ 3−p ] ρ=1 4π
ρ=0 =
3 − p [ρ3−p ] ρ=1
ρ=0
|(x, y, z)| −p dx dy dz = 4π
3 − p [ρ3−p ] ρ=+∞
ρ=1
R3 |(x, y, z)|
\B(0,1)
−p dx dy dz = 4π[log ρ] ∞ 1 = +∞.
R 3 \ B(0, 1) p > 3
I := dx dy
D ρ = 1 + cos θ θ ∈ [0, π]
x
D = {(s cos θ, s sin θ) : θ ∈ [0, π], s ∈ [0, 1 + cos θ]}
π 1+cos θ
I :=
ρ dρ dθ =
0 0
1
(1 + cos θ)
2 0
2 dθ
= 1
π
dθ +
2 0
1
π
2 cos θ dθ +
2 0
1
π
cos
2 0
2 θ dθ
= π π 3
+ =
2 4 4 π.
D := {(x, y) ∈ R 2 : (x + 1) 2 + y 2 ≥ 1, (x − 1) 2 + y 2 ≥ 1, x 2 + y 2 ≤ 2}.
f(x, y) =
I =
D
D
π
|y|
x 2 + y 2 ,
f(x, y) dx dy.

|y|
x 2 + y 2
D √ 2
(±1, 0) 1
D f
D ∩ {x ≥ 0, y ≥ 0} 4 x 2 + y 2 = 2
(x − 1) 2 + y 2 = 1 x 2 − (x − 1) 2 = 1 x 2 − x 2 + 2x − 1 − 1 = 0 x = 1 y = ±1
(1, 1) x
≤ 1,
D ∩ {x ≥ 0, y ≥ 0} = {(x, y) : 0 ≤ x ≤ 1, 1 − (x − 1) 2 ≤ y ≤ 2 − x 2 }
I = 4
= 4
= 4
= 4
1 √ 2−x2 √ 1−(x−1) 2
0
1 √ 2−x2 0
1
0
1
0
y
dy dx = 4
x2 + y2 1 √ 2−x2 ∂
√ x2 + y2 dy dx
1−(x−1) 2 ∂y
x2 + 2 − x2 − x2 + 1 − (x − 1) 2 dx
√ √
2 − 2x dx = 4√2 3 .
I = dx dy dz,
T
0
2y
√
1−(x−1) 2 2 x2 dy dx
+ y2 T z = a 2 x 2 + b 2 y 2 z = k 2 k = 0
T := {(x, y, z) : (ax) 2 + (by) 2 ≤ z ≤ k 2 }
2 2 =
(x, y, z) :
x
√z/a
+
y
√z/b
≤ 1, 0 < z ≤ k 2
x
√z/a = s cos θ
√ √
zs cos θ/a y(s, θ, z) = zs sin θ/b z(s, θ, z) = z
⎛
a
⎜
Jac(ψ)(s, θ, z) = ⎝
√ z cos θ − √ zs sin θ/a s cos θ
2a √ z
⎞
⎟
⎠ ,
zs/(ab)
y
√ z/b = s sin θ z = z 0 ≤ s ≤ 1 0 < z ≤ k 2 x(s, θ, z) =
I =
b √ z sin θ √ zs cos θ/b s sin θ
2b √ z
0 0 1
2π 1 k2 0
0
0
zs
πk4
dz ds dθ =
ab 2ab .
a > 0
x(x 2 + y 2 ) = a(x 2 − y 2 )

ρ 3 cos θ = aρ 2 (cos 2 θ − sin 2 θ)
cos 2θ
ρ = a
cos θ
θ ∈ [−π/4, π/4] ∪ [π/2, 3/4π] ∪ [5/4π, 3/2π] x(θ) = ρ cos θ = a cos 2θ
lim ρ = +∞ x = −a ρ(θ) = 0 θ =
θ→±π/2∓ ±π/4 −π/4 < θ < π/4 0 < x < a |y| ≤
x (a − x)/(a + x) C C
C
C
dx dy =
x dx dy =
π/4 a cos(2θ)/ cos θ
= a2
2
= a2
2
= a2
2
= a2
2
−π/4
0
π/4
2
2
4 cos
−π/4
2 θ + 1
π/4
−π/4
π/4
−π/4
(4 − π)
π/4 a cos(2θ)/ cos θ
= a3
3
= a3
3
= a3
3
= a3
3
= a3
3
−π/4 0
π/4
s ds dθ = a2
2
cos2 − 4 dθ
θ
π/4
−π/4
(cos 2θ + 1) dθ + [tan θ] π/4
cos 2θ dθ + 2 − π
−π/4
π/4
cos 2 (2θ)
cos 2 θ dθ
− 2π
s 2 cos θ ds dθ = a3
3 −π/4
2 cos θ −
−π/4
1
3 cos θ dθ
cos θ
π/4
8 cos
−π/4
4 θ − 12 cos 2 θ − 1
cos2
+ 6 dθ
θ
π/4
2(cos 2θ + 1)
−π/4
2 dθ − 3(2 + π) − 2 + 3π
π/4
cos 3 (2θ)
cos 2 θ dθ
2(cos
−π/4
2 2θ + 2 cos 2θ + 1) dθ − 3(2 + π) − 2 + 3π
π/4
(cos 4θ + 1 + 4 cos 2θ) dθ + π − 8
−π/4
= a3
π/4
(cos 4θ) dθ +
3 −π/4
3
π − 4
2
= a3
3
π − 4 =
3 2 a3
(3π − 8)
6
x dx dy
C
xB = =
dx dy
a 3π − 8
3 4 − π , yB = 0.
C

x
y = sin x 0 ≤ x ≤ π
A = 2π
π
0
sin x 1 + cos 2 x dx = 4π
π/2
= 2π( √ 2 + arc sinh(1)) = 2π( √ 2 + log(1 + √ 2)).
0
sin x 1 + cos 2 x dx = 4π
S
z = 2
3 · (x3/2 + y 3/2 )
D x = 0 y = 0 x + y = 3
∂xz(x, y) = √ x ∂yz(x, y) = √ y
3 4 √ 2
A = 1 + x + y dx dy = t dt dx =
3
D
= 16 − 2
3
4
1
0
s 3/2 ds = 16 − 2
3
62
5
1+x
= 116
15 .
3
0
1
0
1 + t 2 dt
(8 − (1 + x) 3/2 ) dx

Γ := {(x, y) ∈ R 2 : (x 2 + y 2 )(y 2 + x(x + 1)) = 4xy 2 }
cos(3θ) = cos θ(1 − 4 sin2 θ)
Γ Γ
2π
3
Γ P1 = (−1, 0) P2 = (1/2, √ 3/2)
P3 = (1/2, − √ 3/2)
Γ P0 y = ϕ1(x) ϕ1(−1) = 0
ϕ ′ 1 (0)
Γ P1 y = ϕ2(x) ϕ2(1/2) =
√ 3/2 ϕ ′ 2 (1/2)
Γ P2 y = ϕ3(x) ϕ3(1/2) =
− √ 3/2 ϕ ′ 3 (1/2)
h(x, y) = x 2 + y 2 Γ Γ
Γ
x y ↦→ −y
f(x, y) = (x 2 + y 2 )(y 2 + x(x + 1)) − 4xy 2 .
cos 3θ = cos(θ + 2θ) = cos θ cos 2θ − 2 sin 2 θ cos θ
= cos θ(1 − 2 sin 2 θ) − 2 sin 2 θ cos θ = cos θ(1 − 4 sin 2 θ)
Γ x = ρ cos θ y = ρ sin θ
f(ρ cos θ, ρ sin θ) = ρ 2 (ρ 2 + ρ cos θ) − 4ρ 3 cos θ sin 2 θ = ρ 3 (ρ + cos θ(1 − 4 sin 2 θ))
ρ = 0 ρ + cos θ(1 − 4 sin 2 θ) = 0
θ = π/2
Γ = {(ρ cos θ, ρ sin θ) : ρ = − cos 3θ, ρ ≥ 0, θ ∈ [0, 2π]}.
cos 3θ = cos(3(θ + 2π/3))
f
df(x, y) = (2x(y 2 + x(x + 1)) + (x 2 + y 2 )(2x + 1) − 4y 2 ) dx+
+ (2y(y 2 + x(x + 1) + x 2 + y 2 ) − 8xy) dy
= (4x 3 + 3x 2 + 4xy 2 − 3y 2 ) dx + 2y(2y 2 + 2x 2 − 3x) dy

P1 P2 P3
√
1 3
df(−1, 0) = − dx, df , ±
2 2
= 1
√
3
dx ±
2 2 dy.
P1 r1 : x = q1 q1
P1 x = −1
P2 1
P2 r2 : 1
2 x +
2 x+
√
3
√ 3
2
P3 1
P3 r3 : 1
2
x −
2 y = q2 q2
y = 1
2 x−
y = 1
√ 3
2
√ 3
2 y = q3 q3
r2 ∩ r3 = {(2, 0)} r1 ∩ r2 = {(−1, √ 3)} r1 ∩ r2 = {(−1, − √ 3)}
√ 12 = 2 √ 3
∂yf(P1) = 0
Γ ∂yf(P2) = − √ 3/2 = 0 ∂yf(P3) =
√ 3/2 = 0
y = ϕ2(x) y = ϕ3(x) x = 1/2 C 1 ϕ2(1/2) = √ 3/2
ϕ3(1/2) = − √ 3/2 ϕ2 ϕ3 1/2
ϕ ′ 2 (1/2) = −√ 3/3 ϕ ′ 3 (1/2) = √ 3/3
ρ 2 ρ = − cos 3θ
w(θ) = cos 2 (3θ) w
w ′ (θ) = −6 cos(3θ) sin(3θ) = −3 sin(6θ)
w ′′ (θ) = 18 cos(6θ)
θ = kπ/6 k = 0, 1, ..., 5
k = 2, 3, 4 k = 0, 1, 5
k = 0, 1, 5 Γ ρ = 0
(0, 0) ρ 2 = 0 k = 2, 3, 4
(−1, 0) (1/2, √ 3/2) (1/2, − √ 3/2)
ρ 2 1
f ρ ≤ 1 Γ
Γ
Γ
1 1 P1 P2 P3
Γ \ {(0, 0)}
∂yf(x, y) = 0 f(x, y) = 0 y = 0 y 2 =
(3x − 2x 2 )/2 x = 0 x = −1 x 2 (16x − 9)/4 =
0 x = 9/16 y = ±3 √ 15/16
P1 = Q1 = (−1, 0), Q2 =
9
16 , 3√
15
9
, Q3 =
16
16 , −3√
15
,
16
Q3 x Q2 Γ
q(x, y) = x Γ
∇f ∇q Γ
∇f = (0, 0)
q(Q1) < q(0, 0) = 0 < q(Q2) = q(Q3) Q1

x Γ Q2 Q3
x
Γ \ {(0, 0)}
∂xf(x, y) = 0 f(x, y) = 0 y 2 = (−4x 3 − 3x 2 )(4x − 3)
2x 3 32x 2 + 6x − 9
(3 − 4x) 2
x = 0 32x2 + 6x − 9 = 0 x = 3(−1 ± √ 33)/32
y
3
S1 =
32 (−1 − √ 33), − 1
3
69 + 11
16 2
√
33
3
S2 =
32 (−1 − √ 33), 1
3
69 − 11
16 2
√
33
3
S3 =
32 (−1 + √ 33), − 1
3
69 + 11
16 2
√
33
3
S4 =
32 (−1 + √ 33), 1
3
69 + 11
16 2
√
33
,
S2 S4 S1 S3 x
s(x, y) = y Γ
S3 y Γ S3
Γ s(S3) = −s(S4) = s(0, 0) s(S1) = −s(S2) |s(S3)| 2 > |s(S1)| 2
Γ − := Γ \ {(x, y) : x ≤ 0} S1
s(x, y) Γ − S2 s(x, y) Γ −
s(S1) = −s(S2) = s(0, 0)
= 0,
y = mx
x = 0
0 = f(x, mx) = x 2 (1 + m 2 )(m 2 x 2 + x(x + 1)) − 4x 3 m 2
= x 3 (1 + m 2 )(m 2 x + x + 1) − 4x 3 m 2 ,
⎧
⎪⎨
x(m) = 3m2 − 1
(1 + m 2 ) 2
⎪⎩ y(m) = m(3m2 − 1)
(1 + m 2 ) 2
x = 0 f(0, y) = 0 y = 0 f(0, y) = y 4
m ∈ R m Γ (0, 0)
(x(m), y(m)) m → ±∞ (0, 0)
(x(m), y(m)) = (0, 0) m = ±1/ √ 3 m > 1/ √ 3
−1/ √ 3 < m < 1 √ 3 m < −1/ √ 3
−1/ √ 3 < m < 1 √ 3 x < 0
Γ
2π/3 Γ
Γ ∩ {−1 < x < 0, y ≥ 0} ±2π/3

f ∈ C ∞ C 1
−1 < k < 0 Γ x = k (k 2 + y 2 )(y 2 +
k(k + 1)) = 4ky 2 y 4 + (2k 2 − 3k)y 2 + k 4 + k 3 = 0 t = y 2
pk(t) := t 2 + (2k 2 − 3k)t + (k 4 + k 3 ) = 0
y t
limt→±∞ pk(t) = +∞ pk(0) = k 4 + k 3 < 0
t − < 0 t + > 0
t− y± = ± √ t +
t+ = −(2k2 − 3k) + k2 (9 − 16k)
.
2
Γ
y = ϕ + (x) y = ϕ− (x)
C1 y = y(m) y ′ (m) = 0
0 = (9m 2 − 1)(1 + m 2 ) 2 − 4m 2 (3m 2 − 1)(1 + m 2 )
= (1 + m 2 )((9m 2 − 1)(1 + m 2 ) − 4m 2 (3m 2 − 1))
= (1 + m 2 )(−1 + 12m 2 − 3m 4 ),
m m = ± (6 ± √ 33)/3
y(m) m = ± (6 ± √ 33)/3
Si i = 1, . . . , 4 −1 < x < 0
y2 x = −1 x = 0
y = 0 ( 3
32 (1+√ 33) 2 −1 < x < 0
y = ϕ + (x)
(−1, 0) y P2
y = ϕ + (−) ±2π/3
x = x(m) x ′ (m) = 0
0 = 6m(1 + m 2 ) 2 − 4m(3m 2 − 1)(1 + m 2 )
= 2m(1 + m 2 )(3(1 + m 2 ) − 2(3m 2 − 1)) = 2m(1 + m 2 )(5 − 3m 2 ),
m = 0 m = ± 5/3 Qi
i = 1, 2, 3
Si Qj i = 1, ..., 4 j = 1, 2, 3 Γ
P1, P2, P3 Γ
r, R > 0 R > r
D R r = {(x, y) ∈ R 2 : x ≥ 0, y ≥ 0, r 2 ≤ x 2 + y 2 ≤ R 2 },
lim
r→0 +
DR r
√
ye x2 +y2 x2 dxdy.
+ y2

(x 2 + y 2 )(y 2 + x(x + 1)) = 4xy 2
f
˜D R r = {(ρ, θ) : r ≤ ρ ≤ R, θ ∈ [0, π/2]}.
f(ρ cos θ, ρ sin θ) = sin θe ρ /ρ,
ρ
√
ye x2 +y2 x2 π/2 R
dxdy = sin θe
+ y2 ρ π/2 R
dρ dθ = sin θ dθ · e ρ dρ = e R − e r .
D R r
e R − 1
0
r
[0, 1]
∞
nxe −nx2
− (n + 1)xe −(n+1)x2
.
n=1
[0, 1] f
0
r

[0, 1]
fn = nxe−nx2 N
N
N
N
sN = fn(x) − fn+1(x) = fn(x) − fn+1(x) = f1(x) − fN+1(x),
n=1
n=1
sN(x) = xe−x2 −(N +1)xe−(N+1)x2 x ∈ [0, 1]
limn→∞(n + 1)xe−(n+1)x2 = 0 f(x) = xe−x2
n=1
sup |xe
x∈[0,1]
−x2
− sn| = sup |(N + 1)xe
x∈[0,1]
−(N+1)x2
| = sup |fn+1(x)|
x∈[0,1]
fN+1(0) = 0 fn+1(1) = (N + 1)e −(N+1)
f ′ N+1(x) = (N + 1)e −(N+1)x2
− 2(N + 1) 2 x 2 e −(N+1)x2
= (N + 1)e −(N+1)x2
(1 − 2(N + 1)x 2 ),
[0, 1] x = 1/ 2(N + 1)
1
fN+1
=
2(N + 1)
(N + 1)
e
2(N + 1) −1/2 → +∞
N → +∞
Γ := {(x, y) ∈ R 2 : 4(x 2 + y 2 − x) 3 = 27(x 2 + y 2 ) 2 }.
Γ
Pi = (xi, yi) i = 1, 2, 3, 4 P1
Γ P2 P3 P4
i = 1, 2, 3, 4 Γ y = ϕi(x)
C 1 xi ϕi(xi) = yi
h(x, y) = x 2 + y 2 Γ
Γ
Γ
f(x, y) = 4(x 2 + y 2 − x) 3 − 27(x 2 + y 2 ) 2 f(x, −y) = f(x, y) Γ
f(ρ cos θ, ρ sin θ) = 4ρ 3 (ρ − cos θ) 3 − 27ρ 4 = ρ 3 (4(ρ − cos θ) 3 − 27ρ).
ρ > 0 4(ρ − cos θ) 3 = 27ρ
3√
3 Γ = {(ρ cos θ, ρ sin θ) : 4ρ − 3 √ ρ = 3√ 4 cos θ, ρ ≥ 0, θ ∈ [0, 2π[} ∪ {(0, 0)}.
f(0, 0) = 0 Γ
f(0, y) = 4y 6 − 27y 4 = y 4 (4y 2 − 27),
y = 0 y = ±3 √ 3/2 P3 = (0, 3 √ 3/2) P4 = (0, −3 √ 3/3)
f(x, 0) = 4x 3 (x − 1) 3 = 27x 4 ,
x = 0 4(x−1) 3 −27x = 0 4x 3 −12x 2 −15x−4 = 0

4
±1, ±2, ±4 ±1, ±2 4
x − 4
4x 3 −12x 2 −15x −4 x − 4
−4x 3 +16x 2 4x 2 + 4x + 1
4x 2 −15x
−4x 2 +16x
x −4
−x 4
0
4x 3 − 12x 2 − 15x − 4 = (x − 4)(4x 2 + 4x + 1) = (x − 4)(2x + 1) 2 ,
x = 4 x = −1/2 P1 = (−1/2, 0) P2 = (4, 0)
f
∂xf(x, y) = 12(x 2 + y 2 − x) 2 (2x − 1) − 108x(x 2 + y 2 )
= 12((x 2 + y 2 − x) 2 (2x − 1) + 9x(x 2 + y 2 ))
∂yf(x, y) = 12y(2(x 2 + y 2 − x) 2 − 9(x 2 + y 2 )).
P2 P3 P4
df(4, 0) = 12(144 · 7 − 36 · 4) dx = 144 · 72 dx (4, 0)
x = 0 x = 4
df(0, 3 √ 3/2) = 729
16 (− dx + √ 3 dy) (0, 3 √ 3/2)
−x + √ 3y = q q = 9/2 −x +
√ 3y = 9/2.
(0, −3 √ 3/2) x + √ 3y = −9/2
P3 P4 ∂yf(P3) = 0 ∂yf(P4) = 0
∂yf(P1) = ∂yf(P2) = 0
ρ ≥ 0 − 3√ 4 ≤ 3√ 4ρ − 3 3√ ρ ≤ 3√ 4
z(ρ) = 3√ 4ρ − 3 3√ ρ z(0) = 0
˙z(ρ) = 3√ 4 − ρ −2/3 > 0
z(ρ) ρ = 1/2 ¨z(ρ) > 0
z(1/2) = − 3√ 4 ρ > 1/2
3√ 4 ρ ρmax
z(ρmax) = 3√ 4 θ = 0 (4, 0)
ρ 1/2
ρ 4 ρ 2 16 f Γ
ρ Γ
H(ρ, θ) := z(ρ) − 3√ 4 cos θ = 0 ρ
θ ˙z(ρ) = 0
ρ = 1/2 ρ = ρ(θ) C 1
dρ
dθ (θ) = −∂θH(ρ(θ), θ)
∂ρH(ρ(θ), θ) = − 3√ sin θ
4
˙z(ρ(θ) ,
θ = 0, π Γ
ρ = 1/2
(4, 0) (0, 0) (−1/2, 0) (0, 0) ρ
(4, 0) ρ

x = k |k| > 4
x = 4 (4, 0)
f
f = 0
v = ρ 2 = x 2 + y 2 0 ≤ v ≤ 16
px(v) = 4(v − x) 3 − 27v 2 = 0,
v v ≥ x 2
x x = v − 3v 2/3 / 3√ 4 v ≥ x 2
h(v) = v − 3v 2/3 / 3√ 4 x = h(v) 0 ≤ v ≤ 16
v ≥ h 2 (v)
h ′ (v) = 1 − 2 1/3 /v 1/3 v = 2
h
0 < v < 2 2 < v < 16
x = h(v) 0 ≤ v ≤ 16 v ≥ x 2 − √ v ≤ x ≤ √ v
h(v) = 0 v = 0 v = 27/4 0 < 2 < 27/4 < 16 h
]0, 27/4[ ]27/4, 16[
v h(v) h(16) = 4
− √ 16 ≤ 4 ≤ √ 16 x
f(x, y) = 0 4 f(4, y) = 0
y = v − h 2 (v) v = 16 y = 0
x = h(v) f(x, y)
h v = 2 h(2) = −1 − √ 2 < −1
− √ v v > 2 h(v)
2 < v < 16 − √ v ≤ h(v) = x
(x, y) ∈ Γ x ≥ −1 −1
x Γ
f(−1, y) = 0 5 − 6y 2 − 3y 4 + 4y 6 = 0 t = y 2 5 − 6t −
3t 2 + 4t 3 = 0 5 ±1, ±5 1
5 − 6t − 3t 2 + 4t 3 x − 1 4t 2 + t − 5
1 −5/4
f(−1, y) = 0 t = y 2 = 1 y = ±1 ∂xf(1, ±1) = 0
x y
h(v) − √ v 2 < v < 16
2
4 2 < v < 16 h(v)
0 < x < 4 v h(v) = x x0 ≥ 0
f(x0, y) = 0 y
y = ± 1 − h 2 (v)
h 2
[0, 2] [2, 16] −1 < x < 0 v
y f(x, y) = 0
y 2 Γ y 2 = v − h 2 (v) 1 −
2h(v)h ′ (v) = 5 3√ 2v 2/3 −2v−3·2 2/3 3√ v+1 y 2
x = −1/2 0 v = 1/4 1 − 2h(v)h ′ (v)
v0 = 1/4
p(v) = 5 3√ 2v 2/3 − 2v − 3 · 2 2/3 3√ v + 1 p(1/4) = 0 v = 2t 3
p(t) = −4t 3 + 10t 2 − 6t + 1 1/4 = 2t 3
t = 1/2 p(t) t − 1/2

−2 + 8t − 4t 2 t = 1/2(2 ± √ 2)
v1 = 1/4(2 − √ 2) 3 v2 = 1/4(2 + √ 2) 3
x0 = h(v0) = −1/2 x1 = h(v1) = 1/2 − 1/ √ 2 x2 = h(v2) = 1/2 + 1/ √ 2
x2 > 0 y > 0 f(x2, ±y) = 0 y + 2 = v2 − h 2 (v2) =
17/4 + 3 √ 2 y − 2 = −y+ 2 y2 Γ
v1 y + 1 = v − h 2 (v) =
Γ
Q := [−1, 4] × [− 17/4 + 3 √
2, 17/4 + 3 √ 2].
17/4 − 3 √ 2 y − 1 = −y+ 1
0 < x < 4
(0, 3 √ 3/2) 1/2 + 1/ √
2
17/4 + 3 √ 2 (4, 0)
(4, 0)
−1/2 ≤ x ≤ 0
(−1/2, 0)
(0, 0) x 1/2 − 1/ √
2
17/4 − 3 √ 2
−1 < x < −1/2
(−1, 1)
(−1/2, 0)
−1/2 < x < 0
−x + √ 3y = 9/2 x + √ 3y = −9/2 Γ
(0, 3 √ 3/2) (0, −3 √ 3/2)
x = −1 Γ (−1, ±1) Γ
x = −1/2 (−1/2, 0) Γ
x = 1/2 − 1/ √ 2 Γ
−1/2 < x < 0 Γ (1/2 −
1/ √
2, ± 17/4 − 3 √ 2)
x = 0 Γ (0, ±3 √ 3/2)
x = 1/2 + 1/ √ 2 Γ
0 < x < 4 Γ (1/2 + 1/ √
2, ± 17/4 + 3 √ 2)
Γ Γ
17/4 + 3 √ 2
x = 4 Γ (4, 0) Γ
y = 17/4 + 3 √ 2 Γ (1/2 + 1/ √
2, 17/4 + 3 √ 2)
Γ
y = 3 √ 3/2 Γ
y = −1 Γ (−1, 1)
Γ

4(x 2 + y 2 − x) 3 = 27(x 2 + y 2 ) 2
y = 17/4 − 3 √ 2 Γ (1/2 + 1/ √
2, ± 17/4 − 3 √ 2)
y = 0 Γ (−1/2, 0) (0, 0) (4, 0)

a ∈ R R 2
(x 2 + y 2 ) 3 = 4a 2 x 2 y 2 .
a = 0 a = 0 f(x, y) = (x 2 + y 2 ) 3 −
4a 2 x 2 y 2
f(ρ cos θ, ρ sin θ) = ρ 6 − 4a 2 ρ 4 cos θ sin θ = ρ 4 (ρ 2 − a 2 sin 2 2θ).
(0, 0) ρ 2 = a 2 sin 2 2θ
ρ ≥ 0 θ = π ρ = 0 ρ(θ) = |a|| sin 2θ|
ρ θ = π/4+kπ/2 k = 1, 2, 3, 4
|a| |a|
(|a| √ 2/2, |a| √ 2/2)
| sin(2(θ + π/2))| = | sin(2θ + π)| = | sin(2θ)|
π/2 0 < θ < π/4
x(θ) = ρ(θ) cos θ 0 < θ < π/4
x(θ) = |a| sin 2θ cos θ = 2|a| sin θ cos 2 θ = 2|a|(sin θ − sin 3 θ)
˙x(θ) = 2|a| cos θ(1 − 3 sin 2 θ).
0 < θ < π/4 sin θ = 1/ √ 3 cos θ = 2/3
x θ x ∗ = ρ ∗ cos θ ∗ = ρ(θ ∗ ) cos θ ∗ = |a|4/(3 √ 3)
ρ ∗ sin θ ∗ = |a|(2/3) 3/2 P = (|a|4/(3 √ 3), |a|(2/3) 3/2 )
(±|a|4/(3 √ 3), ±|a|(2/3) 3/2 )
(±|a|(2/3) 3/2 , ±|a|4/(3 √ 3)) 0 < θ < θ ∗
y(θ) x(θ)
Γ (x(θ), y(θ)) 0 < θ < θ ∗ y = ϕ(x)
θ ∗ < θ < π/4
x(θ) y(θ)
P a = 1

(x 2 + y 2 ) 3 = 4x 2 y 2

(x 2 + y 2 ) 3 = 4x 2 y 2

(x 2 + y 2 ) 3 = 4x 2 y 2

a > 0
x = at
at2
, y = .
1 + t3 1 + t3 F : R \ {−1} → R2
at at2
F (t) = (F1(t), F2(t)) := ,
1 + t3 1 + t3
.
F1 F2
F1(0) = 0, lim
|t|→∞ F1(t) = 0, lim
t→−1 ± F1(t) = ∓∞, F ′ 1(t) = a(1 − 2t2 )
(1 + t3 > 0 |t| <
) 2
F2(0) = 0, lim
|t|→∞ F2(t) = 0, lim
t→−1 ± F2(t) = ±∞, F ′ 2(t) = at(2 − t3 )
(1 + t 3 ) 2 > 0 0 < t < 3√ 2.
t < −1
t > −1 t1, t2 ∈ R ∪ {+∞} t1 < t2
F (t1) = F (t2) =: (¯x, ¯y)
at1
1 + t 3 1
= at2
1 + t3 ,
2
at 2 1
1 + t 3 1
= at22 1 + t3 .
2
0 < t1 < √ 2/2 t2 > 3√ 2
at 2 1
1 + t 3 2
= at1t2
1 + t3 ,
1
(t1 − t2) · at1
1 + t 3 1
t1 = 0 (¯x, ¯y) = (0, 0) t2 = 0
(0, 0) t → +∞ t2 = +∞
γ = {F (t) : t ≥ 0}
= 0.
C
A = dx dy.
(P (x, y) dx + Q(x, y) dy) =
γ
C
C
∂Q
∂x
∂P
− dxdy,
∂y
P : R 2 → R Q : R 2 → R
C
√ 2
2 ,

P Q A
Q(x, y) = x P (x, y) = 0
+∞
A = x dy = F1(t)F ′ +∞ at
2(t) dt =
1 + t3 · at(2 − t3 )
(1 + t3 +∞ t
dt = a2
) 2 2 (2 − t3 )
(1 + t3 dt
) 3
= a2
3
γ
0
0
+∞ 2 − t
0
3
(1 + t3 ) 3 · 3t2 dt = a2
+∞
3 0
A = a 2 /6
2 − s a2
ds =
(1 + s) 3 3
+∞
1
0
3 − u a2
du =
u3 6 .
a > 0
ρ 2 (θ) = 2a 2 cos 2θ.
θ [−π/2, 3/2π] 2π
cos 2θ ≥ 0 θ ∈ [−π/4, π/4]∪[3/4π, 5/4π] =:
D ρ(−π/4) = ρ(π/4) = ρ(3/4π) = ρ(5/4π) = 0
γ C
A = dx dy.
(P (x, y) dx + Q(x, y) dy) =
γ
C
C
∂Q
∂x
∂P
− dxdy,
∂y
P : R 2 → R Q : R 2 → R
C
P Q A
Q(x, y) = x P (x, y) = 0 x(θ) = ρ(θ) cos θ
y(θ) = ρ(θ) sin θ
A = ρ(θ) cos(θ) · ρ
D
′
(θ) sin θ + ρ(θ) cos θ dθ
= ρ(θ) cos(θ) · ρ
D
′
(θ) sin θ + ρ(θ) cos θ dθ
= 1
ρ(θ) · ρ
2 D
′ (θ) sin 2θ dθ + 2a 2
cos 2θ cos
D
2 θ dθ
= 1
d
4 D dθ [ρ2 (θ)] sin 2θ dθ + a 2
cos 2θ · (cos 2θ + 1) dθ
D
= −a 2
sin 2 2θ dθ + a 2
cos 2 2θ dθ + a 2
cos 2θ dθ
= −a 2
+
−a 2
D
π/4
D
sin
D
−π/4
2 2θ dθ + a 2
π/4
cos
−π/4
2 2θ dθ + a 2
π/4
−π/4
5/4π
sin
3/4π
2 2θ dθ + a 2
5/4π
cos
3/4π
2 2θ dθ + a 2
5/4π
3/4π
cos 2θ dθ+
cos 2θ dθ
π
[3/4π, 5/4π] [−π/4, π/4]
A = 2
−a 2
= 2a 2
π/4
−π/4
π/4
sin
−π/4
2 2θ dθ + a 2
π/4
cos
−π/4
2 2θ dθ + a 2
π/4
−π/4
cos 2θ dθ = 2a 2 ,
cos 2θ dθ
.

cos(π − α) = − cos(α)
π/4
−π/4
sin 2 2θ dθ = 2
= 2
π/4
0
π/4
2a 2
0
sin 2 2θ dθ = 2
π/4
cos 2 (π − 2θ) dθ = 2
0
cos 2 (π − 2θ) dθ
π/4
0
cos 2 (2θ) dθ =
π/4
−π/4
cos 2 2θ dθ.
a > 0
ρ(θ) = a(1 + cos θ).
ρ(0) = ρ(2π) γ
C
A = dx dy.
(P (x, y) dx + Q(x, y) dy) =
γ
C
C
∂Q
∂x
∂P
− dxdy,
∂y
P : R 2 → R Q : R 2 → R
C
P Q A
Q(x, y) = x P (x, y) = 0 x(θ) = ρ(θ) cos θ
y(θ) = ρ(θ) sin θ
2π
A = x dy = ρ(θ) cos θ ·
= a 2
= a 2
= a 2
= a 2
= a2
8
= a2
4
γ
2π
0
2π
0
(1 + cos θ) cos θ ·
(1 + cos θ) 2 cos 2 θ dθ − a 2
2π
ρ ′
(θ) sin θ + ρ(θ) cos θ dθ
− sin 2
θ + (1 + cos θ) cos θ dθ
sin 2 θ(1 + cos θ) cos θ dθ
0
2π
0
cos
0
2 θ + cos 4 θ + 2 cos 3 θ − (cos θ − cos 3 θ − cos 4 θ + cos 2 θ) dθ
2π
4
eiθ + e−iθ 2 cos 4 θ + 3 cos 3 θ − cos θ dθ = 2a 2
2π
0
2π
(e
0
2iθ + e −2iθ + 2) 2 dθ = a2
8
2π
0
(cos 4θ + 4 cos 2θ + 3) dθ = 3πa2
2 ,
cos 4 θ dθ = 2a 2
2π
0
0
2π
(e
0
4iθ + e −4iθ + 4 + 2 + 4e 2iθ + 4e −2iθ ) dθ
(a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc 3πa 2 /2
2
dθ

I := x 2 dxdy.
D = {(x, y) ∈ R 2 : 1 ≤ x 2 + y 2 ≤ 2}
D
D γ1 γ2
1 √ 2 Br := {(x, y) ∈ R 2 : x 2 + y 2 < r} D = B2 \ B1
D
x 2
dxdy =
B2
x 2
dxdy −
(P (x, y) dx + Q(x, y) dy) =
γi
Bi
∂Q
∂x
B1
x 2 dxdy.
∂P
− dxdy,
∂y
P : R 2 → R Q : R 2 → R
Bi
P Q
x 2 Q(x, y) = x 3 /3 P (x, y) = 0
I := 1
3
γ2
x 3 dy − 1
3
γ1
x 3 dy.
γi
I := 1
3
=
= 1
16
2π
( √ 2 cos θ) 3 · ( √ 2 cos θ) dθ − 1
3
2π
cos 3 θ · (cos θ) dθ =
0
0
2π 4
eiθ + e−iθ dθ =
0 2
1
2π
(e
16 0
2iθ + e −2iθ + 2) 2 dθ
2π
(e
0
4iθ + e −4iθ + 4 + 2 + 4e 2iθ + 4e −2iθ ) dθ
2π
= 1
8
0
(cos 4θ + 4 cos 2θ + 3) dθ = 3
4 π.
3π/4
R 3
S = {x 2 + y 2 + z 2 = 1, z > 0}
F (x, y, z) =
D = {x 2 + y 2 ≤ 1, z = 0}
γ = {x 2 + y 2 = 1, z = 0}
y
1 + z2 , x5z 100 − y, z + x 2
2π
0
cos 4 θ dθ

I := F · ˆn dσ
ˆn Σ = S ∪ D
ω1 F 1 F
S
γ +
ω 1 F
γ + γ z = 0
F S
C = {(x, y, z) ∈ R 3 : x 2 + y 2 + z 2 ≤ 1, z ≥ 0}
Σ F = (F1, F2, F3)
F · ˆn dσ = F · ˆn dσ + F · ˆn dσ =
Σ
S
D
divF (x, y, z) = 0.
C
divF dxdydz
F · ˆn dσ = − F · ˆn dσ
S
D
D C ˆn = (0, 0, −1) F = (y, −y, x2 )
F · ˆn dσ = − F · ˆn dσ = x 2 dxdy
γ
S
∂Q ∂P
(P (x, y) dx + Q(x, y) dy) =
− dxdy,
∂x ∂y
D
P : R 2 → R Q : R 2 → R
D
D
P Q
x 2 Q(x, y) = x 3 /3 P (x, y) = 0
I := 1
3
γ
x 3 dy.
γ
I := 1
3
= 1
3
2π
= 1
3 · 16
= 1
3 · 8
cos 3 θ · cos θ dθ = 1
3
2π
D
cos 4 θ dθ
0
0
2π 4
eiθ + e−iθ dθ =
0 2
1
2π
(e
3 · 16 0
2iθ + e −2iθ + 2) 2 dθ
2π
(e
0
4iθ + e −4iθ + 4 + 2 + 4e 2iθ + 4e −2iθ ) dθ
2π
0
(cos 4θ + 4 cos 2θ + 3) dθ = π
4 .
I = π
4

ω 1 F
1 F
ω 1 F (x) = F1(x, y, z) dx + F2(x, y, z) dz + F3(x, y, z) dz.
ω
+γ
1 F (x) =
=
+γ
+γ
F1(x, y, z) dx + F2(x, y, z) dz + F3(x, y, z) dz
y dx − y dy
γ x = cos θ y = sin θ
ω 1
F (x) =
2π
y (dx − dy) = sin θ · (− sin θ + cos θ) dθ
+γ
= −
+γ
2π
0
sin 2 θ = −
0
2π
0
1 − cos 2θ
2
dθ = −π.
∂F3 ∂F2
rot F := −
∂y ∂z ,
∂F1 ∂F3
−
∂z ∂x ,
∂F2 ∂F1
− ,
∂x ∂y
= −100x 5 z 99 , −2yz
(1 + z2 ) 2 − 2x, 5x4z 100 − 1
1 + z2
0 = div(rotF ) dxdydz = rot F dσ = rot F · ˆn dσ + rot F · ˆn dσ,
C
Σ
rot F · ˆn dσ = − rot F · ˆn dσ,
S
D
D C ˆn = (0, 0, −1) rotF = (0, −2x, −1)
rot F · ˆn dσ = − rot F · ˆn dσ = (−1) dxdy
S
D
D
D −π
ω 1
F (x) = −π = rot F · ˆn dσ,
+γ
S z = xy/2 x 2 +y 2 ≤ 12
S F = (x, y, 1) S
S z = f(x, y) f : D → R f(x, y) = xy/2
D := {(x, y) ∈ R 2 : x 2 + y 2 ≤ 12} D √ 12 = 2 √ 3
dσ S
dσ = 1 + |∇f| 2
= 1 + y2 x2 1
+ dx dy = 4 + y2 + x2 dx dy.
4 4 2
S
S
D

S ψ(x, y) = (ψ1, ψ2, ψ3) = (x, y, f(x, y)) = (x, y, xy/2)
dσ =
⎛
Jac ψ(x, y) = ⎝
1 0
0 1
y/2 x/2
2
2
det 2
1 0
0 1
+ det 2
1 0
y/2 x/2
⎞
⎠
+ det 2
0 1
y/2 x/2
dx dy = 1 + x2 y2
+ dx dy.
4 4
Jac ψ
dσ = ∂xψ ∧ ∂yψ = − y
, −x , 1 dx dy = 1 +
2 2 x2
4
+ y2
4
Area(S) =
S
dσ =
D
1 + x2
4
+ y2
4
dx dy.
1
dxdy = 4 + y2 + x2 dxdy.
2 D
x = ρ cos θ y = ρ sin θ
ρ
Area(S) =
= π
2
2π
0
16
4
2 √ 3
0
√ t dt = π
2
1
4 + ρ2 ρ dρ dθ = π
2
t 3/2
3/2
t=16
t=4
2 √ 3
0
= π
56π
(64 − 8) =
3 3 .
4 + ρ 2 ρ dρ = π
2
12
0
√ 4 + w dw
S G(x, y, z) = z−f(x, y) = z−xy/2
G(x, y, z) = 0
± ∇G (−y/2, −x/2, 1) (−y, −x, 2)
= ± = ± ,
|∇G| |(−y/2, −x/2, 1)| x2 + y2 + 4
ˆn(x, y, z) = (n1, n2, n3) =
(−y, −x, 2)
x 2 + y 2 + 4 ,

⎛
det ⎝ n1
n2
n3
∂xψ1
∂xψ2
∂xψ3
∂yψ1
∂yψ2
∂yψ3
⎛ −y
⎞
√ 1 0
x2 +y2 +4
⎞ ⎜
⎟
⎜
⎟
⎜ −x
⎠ = det ⎜ √
⎟
0 1 ⎟
⎜ x2 +y2 +4
⎟
⎜
⎟
⎝
⎠
√ 2 y/2 x/2
x2 +y2 +4
⎛
⎞
−y 1 0
1
= det ⎝ −x 0 1 ⎠ =
x2 + y2 + 4 2 y/2 x/2
2 + y2 /2 + x2 /2
x2 + y2 + 4
= 1
x2 + y2 + 4 > 0.
2
F = (F1, F2, F3)
Φ(S, ⎛
F ) = det ⎝
D
F1
⎞
◦ ψ ∂xψ1 ∂yψ1
F2 ◦ ψ ∂xψ2 ∂yψ2 ⎠ dxdy =
⎛
x
det ⎝ y
1
0
0
1
⎞
⎠ dxdy
F3 ◦ ψ ∂xψ3 ∂yψ3
= (1 − xy) dxdy = Area(D) −
D 1 y/2 x/2
2π √
2 3
xy dxdy = 12π −
r
D
D
0 0
2
cos θ sin θ ρ dρ dθ
√
2 3
= 12π − r
0
3 2π
dr · cos θ sin θ = 12π.
12π
0
λ ∈ R λ < min{f(x, y) : x, y ∈ D} λ
f D λ < 0
div F (x, y, z) = ∂F1
∂x
+ ∂F2
∂y
+ ∂F3
∂z
C S S − := D × {λ}
C (0, 0, −1) C {(x, y, λ) : x, y, ∈ D}
S
= 2.
L = {(x, y, z) : x 2 + y 2 = 12, λ ≤ z ≤ f(x, y)}.
(x, y, 0)/ x2 + y2
div
F dxdydz = F · ˆn dσ.
C
∂C
2Volume(C) = Φ(S,
F ) +
2
f(x,y)
dz dxdy = Φ(S,
D λ
F ) −
S −
D
F · ˆn dσ +
dσ +
L
L
F · ˆn dσ
x 2 + y 2
x 2 + y 2 dσ
2Volume(C) = Φ(S, F ) − Area(D) + √ 12 Area(L)

C = {(x, y, z) : (x, y) ∈ D, λ < z < f(x, y)}
Area(D) = 12π
2π
2Volume(C) := 2
Φ(S, F ) = 12π − √ 12 Area(L) + 2Volume(C).
= 2
= 2
= 2
= 2
√ 12 f(r cos θ,r sin θ)
0 0 0
2π √ 12 f(r cos θ,r sin θ)
0 0
2π √ 12
0 0
√ 12 2π
0 0
√ 12 2π
0
0
λ
r dθdr
r dz dθ dr
r(f(r cos θ, r sin θ) − λ) dθ dr
(r 3 cos θ sin θ − rλ) dθ dr
r 3 cos θ sin θ dθ dr − 2λπ
√ 12
0
r dr = −24λπ,
λ < 0 L ψ1(θ, z) = ( √ 12 cos θ, √ 12 sin θ, z)
⎛
Jac ψ1(θ, z) = ⎝ −√ ⎞
12 sin θ
√
0
0 12 cos θ ⎠ .
0 1
ω2 =
=
det 2
− √ 12 sin θ 0
0 1
12 sin 2 θ + 12 cos 2 θ = √ 12.
+ det 2
√
12 cos θ 0
0 1
+ det 2
√
−
√
12 sin θ 0
12 cos θ 0

√ √
2π
12 Area(L) := 12
:= √ 12
:= √ 12
f( √ 12 cos θ, √ 12 sin θ)
0 λ
2π 6 cos θ sin θ
0
2π
0
λ
dz dθ
√ 12dz dθ
6 cos θ sin θ − λ dθ = 24πλ.
12π
I1 := F · ˆn dσ,
S1
F := (xz, xy, yz) S1 = ∂C C = {(x, y, z) ∈ R 3 : x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 1}
C
F · ˆn dσ = divF dxdydz = (x + y + z) dxdydz.
S1
1 1−z 1−z−y
(x + y + z) dxdydz =
(x + y + z)dx dy dz
C
I1 = 1/8
=
=
C
0 0 0
1 1−z 1−z−y
0 0
1 1−z
= 1
2
= 1
2
= 1
2
= 1
6
0 0
1 1−z
C
x2 x=1−z−y
+ yx + zx dy dz
0 2 x=0
(1 − z − y) 2
+ (y + z)(1 − z − y) dy dz
2
(1 − z − y)(1 + y + z) dy dz
0 0
1 1−z
(1 − (z + y)
0 0
2 ) dy dz w=y+z
= 1
1 1
2 0 z
1 w=1 1
0
1
0
w − w3
3
w=z
(2 − 3z + z 3 ) dz = 1
6
2 z3
− z +
3 3
dz = 1
2 0
2 − 3
1
+
2 4
= 1
8 .
(1 − w 2 ) dw dz
F : R 3 → R 3 F (x, y, z) = (y 2 , 0, x − y)
F S z = 1 − x 2 − y 2
(x, y) ∈ D := {(x, y) ∈ R 2 : x ≥ 0, y ≤ 0, x 2 + y 2 ≤ 1}
S
φ(x, y) = (φ1(x, y), φ2(x, y), φ3(x, y)) = (x, y, 1 − x 2 − y 2 ).
S G(x, y) = x 2 + y 2 + z − 1 = 0
∇G(x, y) = (2x, 2y, 1).
dz

ˆn = ∇G = ((∇G)1, (∇G)2, (∇G)3) ˆn = −∇G
⎛
⎞
∂φ1 ∂φ2 ∂φ3
⎜ ∂x ∂x ∂x ⎟ ⎛
⎞
⎜
⎟ 1 0 −2x
⎜
det ⎜ ∂φ1 ∂φ2 ∂φ3
⎟ = det ⎝ 0 1 −2y ⎠ = 4y
⎜ ∂y ∂y ∂y
⎟ 2x 2y 1
⎝
⎠
2 + 4x 2 > 0
(∇G)1 (∇G)2 (∇G)3
ˆn = ∇G(x, y) = (2x, 2y, 1) F
⎛
rot F = det ⎝ e1
e2
∂x
∂y
F1
F2
⎞ ⎛
⎠ = det ⎝ e1 ∂x y2 e2 ∂y 0
⎞
⎠ = (−1, −1, −2y)
e3 ∂z x − y
e3 ∂z F3
rot F · ˆn dσ = (−1, −1, −2y) · (2x, 2y, 1) dxdy = −
S
D
x = r cos θ y = r sin θ
1 π/2
1
rot F · ˆn dσ = − 2r(cos θ + 2 sin θ) r dθ dr = − 2r 2 dr ·
S
0
0
rot F · ˆn dσ =
S
+∂S
0
F dγ.
D
π/2
0
2x + 4y dxdy
(cos θ − 2 sin θ) dθ = 2
3 .
D γ1(θ) = (cos θ, sin θ) −π/2 <
θ < 0 γ2(t) = (1 − t, 0) 0 < t < 1 γ3(s) = (0, −s) 0 < s < 1 S
γ1 γ2 γ3 φ S
φ◦γ1
φ◦γ2
φ◦γ3
F dγ =
φ ◦ γ1(θ) = (cos θ, sin θ, 1 − cos 2 θ − sin 2 θ) = (cos θ, sin θ, 0);
φ ◦ γ2(t) = (1 − t, 0, 1 − (1 − t) 2 ) = (1 − t, 0, 2t − t 2 );
φ ◦ γ3(s) = (0, −s, 1 − s 2 ).
=
F dγ =
F dγ =
0
+∂S
−π/2
π/2
0
w=cos θ
= 1 −
1
0
1
0
F dγ =
φ◦γ1
F dγ + F dγ + F dγ
φ◦γ2
φ◦γ3
(sin 2 0
θ, 0, cos θ − sin θ) · (− sin θ, cos θ, 0) dθ = −
sin 3 θ dθ =
0
1
π/2
0
−w 2 dw = 2
3
(0, 0, 1 − t) · (−1, 0, 2 − 2t) dt =
(−s 2 , 0, s) · (0, −1, −2s) dt =
sin θ(1 − cos 2 θ) dθ = 1 −
1
0
1
0
π/2
0
2(1 − t) 2 dt = 2
3 .
−2s 2 dt = − 2
3 .
−π/2
sin 3 θ dθ
cos 2 θ sin θ dθ

S
rotF · ˆn dσ =
+∂S
F dγ = 2 2 2 2
+ − =
3 3 3 3 ,
S
ϕ(θ, y) = ( y 2 + 1 cos θ, y, y 2 + 1 sin θ), θ ∈ [0; 2π], |y| < 1
F (x, y, z) = (x 2 , y/2, x) S
(1, 0, 0) e1 = (1, 0, 0)
ϕ(θ, y) = (ϕ1, ϕ2, ϕ3) F = (F1, F2, F3)
Jac ϕ(θ, y) =
⎛
⎜
⎝
− y 2 + 1 sin θ
y cos θ
√ y 2 +1
0 1
y2 + 1 cos θ
√y sin θ
y2 +1
ˆn = (n1, n2, n3)
(1, 0, 0)
⎛
n1 −
⎜
det ⎜
⎝
y2 + 1 sin θ
n2
0 1
y2 + 1 cos θ
n3
y cos θ
√ y 2 +1
√y sin θ
y2 +1
⎞
⎟
⎠
(θ,y)=(0,0)
⎛
= det ⎝
⎞
⎟
⎠ .
1 0 0
0 0 1
0 1 0
⎞
⎠ = −1 < 0.
Φ(S, ⎛
2π F1 ◦ ϕ −
1 ⎜
F ) = − det ⎜
⎝
0 −1
y2 y cos θ
+ 1 sin θ √
y2 +1
F2 ◦ ϕ 0 1
F3 ◦ ϕ y2 ⎞
⎟
⎠ dy dθ
y sin θ
+ 1 cos θ √
y2 +1
⎛
2π (y
1 ⎜
= − det ⎜
⎝
0 −1
2 + 1) cos2 θ − y2 ⎞
y cos θ
+ 1 sin θ √
y2 +1 ⎟
y/2 0 1 ⎟
⎠ dy dθ
y2 + 1 cos θ y2 y sin θ
+ 1 cos θ √
y2 +1
⎛
2π 1
= − (−y/2)det ⎝
0 −1
−y2 ⎞
y cos θ
+ 1 sin θ √
y2 +1 ⎠ dy dθ+
y2 + 1 cos θ
2π
−
= −
1
0 −1
2π 1
0
= − 2
3 π.
−1
y sin θ
√ y 2 +1
(y2 + 1) cos2 θ − y2 + 1 sin θ
(−1)det dy dθ
y2 + 1 cos θ y2 + 1 cos θ
y 2 1 2π
/2 dy dθ + (y
−1 0
2 + 1) 3/2 cos 3 θ + (y 2
+ 1) sin θ cos θ dθ dy

2π
0
cos θ sin θ dθ = 1
2
2π
0
cos 3 θ dθ =
=
=
2π
0
3π/2
−π/2
π/2
−π/2
1
−1
sin(2θ) dθ = 1
4
cos 3 θ dθ =
4π
0
3π/2
−π/2
(1 − sin 2 θ) cos θ dθ +
(1 − w 2 ) dw +
−1
1
sin w dw = 0.
(1 − sin 2 θ) cos θ dθ
3/2π
π/2
(1 − w 2 ) dw = 0.
Σ ⊂ R 3
(1 − sin 2 θ) cos θ dθ
ϕ(θ, z) = ( 1 + 2z 2 cos θ, 1 + 2z 2 sin θ, z), |θ| ≤ π/4, |z| ≤ 1,
F (x, y, z) = (1/ 1 + 2z 2 , 1/ 1 + 2z 2 , x 2 + y 2 )
Σ (1, 0, 0) (−1, 0, 0)
F = (F1, F2, F3)
div F (x, y, z) = ∂F1
∂x
+ ∂F2
∂y
+ ∂F3
∂z
F
S S ∪ Σ
C
θ ∈ [−π/4, π/4] z γ x = √ 1 + 2z 2
y = 0
Σ : ϕ(θ, z) = ( √ 1 + 2z 2 cos θ, √ 1 + 2z 2 sin θ, z) γ
x = √ 1 + 2z 2
S + θ ∈ [−π/4, π/4]
(0, 0, 1) ( √ 2, 0, 1) y = 0 z
= 0.
S + = {(r cos θ, r sin θ, 1) ∈ R 3 : 0 ≤ r ≤ √ 2, |θ| ≤ π/4}
C (0, 0, 1)

S − θ ∈ [−π/4, π/4]
(0, 0, −1) ( √ 2, 0, −1) y = 0 z
C (0, 0, 1)
S − = {(r cos θ, r sin θ, 1) ∈ R 3 : 0 ≤ r ≤ √ 2, |θ| ≤ π/4}
L + L −
L + = {(r cos θ, r sin θ, z) : θ = π/4, 0 ≤ r ≤ 1 + 2z 2 , |z| ≤ 1},
L − = {(r cos θ, r sin θ, z) : θ = π/4, 0 ≤ r ≤ 1 + 2z2 , |z| ≤ 1}.
(− √ 2/2, √ 2/2, 0) (− √ 2/2, − √ 2/2, 0)
Σ C (1, 0, 0) (1, 0, 0)
Σ
− F · ˆn dσ =
F · ˆn dσ,
Σ
S + ∪S − ∪L + ∪L −
S + ∪S − ∪L + ∪L −
F · ˆn dσ,
C (x, y, z) ∈ L + F · ˆn = 0
L + F (x, y, 1) = F (x, y, −1) (x, y, 1) ∈ S +
(x, y, −1) ∈ S− ˆn C F ·ˆn(x, y, 1) = − F ·ˆn(x, y, −1)
(x, y, 1) ∈ S + (x, y, −1) ∈ S−
S +
F · ˆn dσ +
F · ˆn dσ = F3(x, y, z) dσ =
S +
S +
S −
F · ˆn dσ = 0.
S +
(x 2 + y 2
) dσ =
√ 2 π/2
0 −π/2
F · ˆn dσ.
(x, y, z) ∈ L −
L −
ψ(r, z) =
L −
√
F
2
· ˆn = −√
1 + 2z2 .
√
2
2 r,
√
2
r, z , z ∈ [−1, 1], r ∈ [0,
2 1 + 2z2 ].
⎛
Jac ψ(r, z) = ⎝
√ 2/2 0
√ 2/2 0
0 1
⎞
⎠
r 2 r drdθ = π
2 .
2
2
ω2(∂rψ, ∂zψ) = det 2
√
√
2/2 0
+ det
2/2 0
2
√
2/2 0
+ det
0 1
2
√
2/2 0
= 1.
0 1

L −
F · ˆn dσ = −
L− √ 1
2
√ dσ = −
1 + 2z2 −1
√ 1+2z2 0
−2 √ 2
ϕ = (ϕ1, ϕ2, ϕ3)
Jac ϕ(θ, z) =
⎛
⎝ ∂θϕ1 ∂zϕ1
∂θϕ2 ∂zϕ2
∂θϕ3 ∂zϕ3
⎛
−
⎞ ⎜
⎠ ⎜
= ⎜
⎝
√ 1 + 2z2 sin θ
√
1 + 2z2 cos θ
√
2
√
1 + 2z2 drdz = −√ 1
2 dz = −2
−1
√ 2.
2z cos θ
√ 1 + 2z 2
2z sin θ
√ 1 + 2z 2
0 1
Σ Ω =] −
π/4, π/4[×] − 1, 1[
Φ(Σ,
F ) :=
Σ
F · ˆn dσ =
Ω
ω3( F ◦ ϕ, ∂1ϕ, ∂2ϕ, ∂3ϕ)(x) dx
⎛
F1 ◦ ϕ −
⎜
⎜
= det ⎜
Ω ⎜
⎝
√ 1 + 2z2 sin θ
F2 ◦ ϕ √ 1 + 2z2 cos θ
−1
1
(1 + 2z
−π/4
2 )z dθ
π/4
2z cos θ
√ 1 + 2z 2
2z sin θ
√ 1 + 2z 2
⎞
⎟ (x) dx
⎟
⎠
⎞
⎟ .
⎟
⎠
F3 ◦ ϕ 0 1
= F3 ◦ ϕ(θ, z) (−2z sin
Ω
2 θ − 2z cos 2 θ) dθdz+
+
F1 ◦ ϕ(θ, z)
Ω
1 + 2z2 cos θ + F2 ◦ ϕ(θ, z) 1 + 2z2
sin θ dθdz
= −2 (ϕ
Ω
2 1(θ, z) + ϕ 2 2(θ, z))z dθdz+
1 1
+ √ 1 + 2z2 cos θ + √ 1 + 2z2 sin θ dθdz
Ω 1 + 2z2 1 + 2z2
= −2 (1 + 2z
Ω
2
)z dθdz + (cos θ + sin θ) dθdz
1
π/4
Ω
1
π/4
= −2
dz +
(cos θ + sin θ) dθ dz
−1 −π/4
= −π
= 2
−1
π/4
−π/4
(1 + 2z 2 )z dz + 2
cos θ dθ = 2 √ 2.
−π/4
(cos θ + sin θ) dθ

(1, 0, 0) ϕ(0, 0)
⎛
det ⎝ n1
⎛
−1 −
⎞
⎜
∂θϕ1 ∂zϕ1
⎜
n2 ∂θϕ2 ∂zϕ2 ⎠
⎜
= det ⎜
n3 ∂θϕ3 ∂zϕ3 (θ,z)=(0,0) ⎜
⎝
√ 1 + 2z2 2z cos θ
sin θ √
1 + 2z2 0
√ 1 + 2z2 ⎞
⎟
2z sin θ
⎟
cos θ √
⎟
1 + 2z2 ⎟
⎠
0
⎛
−1
= det ⎝ 0
0
1
0
⎞
0
0 ⎠ = −1 < 0,
1
(θ,z)=(0,0)
0 0 1
−2 √ 2

K ⊆ R n χK : R n → {0, 1} K
χK(x) = 0 x /∈ K χK(x) = 1 x ∈ K
D ⊂ R d
{fn : D → [0, +∞[}n∈N f : D → R
fi(x) ≥ fj(x) i ≥ j x ∈ D i, j ∈ N
f x ∈ D lim
n→∞ fn(x) = f(x)
f
D
f(x) dx = lim fn(x) dx.
n→∞
D
D ⊂ R d {fn : D → [0, +∞[}n∈N
x ∈ D
f
D
f(x) := lim inf
n→∞ fn(x),
f(x) dx ≤ lim inf
n→∞
D
fn(x) dx.
D ⊂ R d
{fn : D → R}n∈N f : D → R
x ∈ D
f(x) = lim
n→∞ fn(x).
g ∈ L 1 (D) |f(x)| ≤ g(x) x ∈ D f
g
D
f(x) dx = lim fn(x) dx.
n→∞
D

a ∈ R f(x) = e−x /x [a, +∞[
f(x) = e−x − 1
[0, 1] [0, +∞[
x
0
0 < a ≤ 1 f [a, 1]
M ∈ N M > 1 x ∈ [1, +∞[ f(x) ≤ e −x
[1, +∞[ e −x {χ [1,M](x)e −x } M∈N\{0}
χ [1,+∞[(x)e −x
+∞
e −x
dx =
1
R
χ [1,+∞[(x)e −x dx = lim
M→∞
χ [1,M](x)e
R
−x dx
1
= lim
M→∞ e − e−M = 1
e .
χ [1,+∞[(x)e−x χ [1,M](x)f(x)
χ [1,+∞[(x)e−x
χ [1,+∞[(x)f(x) χ [1,+∞[(x)f(x)
f [a, +∞[ a > 0
[0, r] r > 0 lim
x→0 + e−x = 1
0 < ε < 1 f(x) > 1
0 < x < ε n ∈ N
2x
0 < 1/n < ε f [1/n, ε]
ε
1/n
f(x) dx ≥ 1
2
χ ]1/n,r](x) 1
x
χ ]0,r](x) 1
x
χ ]0,ε](x)
R
1
1
dx = lim
2x n→+∞ 2
R
χ ]1/n,ε](x) 1
x
1
dx = (log ε − log(1/n)).
2
n → +∞
R
χ ]1/n,ε](x) 1
log ε − log(1/n)
dx = lim
= +∞.
x n→+∞ 2
f(x) > 1
]0, ε]
2x
r ε ε 1
f(x) dx ≥ f(x) dx ≥ dx = +∞.
0
0
0 2x
f [a, +∞[ a > 0
f(0) = −1 f [0, 1]
[1, +∞[
− 1
x = e−x − 1
−
x
e−x
x
[1, +∞[ e−x /x
[1, +∞[ x ↦→ 1/x [1, +∞[
1/x M → ∞
χ [1,M](x)1/x
∞
1
dx
x =
lim
R M→+∞
χ [1,M](x)
x
dx = lim
M→+∞
M
1
dx
x
= lim log M = +∞.
M→+∞

f [1, +∞[ [0, 1]
[0, +∞[ [1, +∞[
χ [0,+∞[f χ [0,1]f
[1, +∞[
+∞
fk(x) = e−x
lim
1 + kx k→∞ 0
fk(x) dx
χ [0,+∞[fk
|χ [0,+∞[(x)fk(x)| = χ [0,+∞[(x)fk(x) ≤ χ [0,+∞[(x) e−x
1 + x
χ [0,+∞[(x) e−x
1+x [0, 1] [0, 1]
e−x /x x ≥ 1 e−x /x [1, +∞[
χ [0,+∞[(x) e−x
1+x [1, +∞[ R
x < 0 |χ [0,+∞[(x)fk(x)|
+∞
lim
k→∞ 0
fk(x) dx = 0.
+∞
fk(x) = √ xe −kx lim fk(x) dx
k→∞ 0
χ [0,+∞[fk
χ [0,+∞[f1
lim
k→∞
+∞
0
fk(x) dx = 0.
M > 0 χ [0,+∞[f1
[0, M]
f1(x) x
lim = lim
x→+∞ 1/x2 x→+∞
5/2
= 0
ex
M > 0 f1(x) < 1/x2 x ≥ M 1/x2 [M, +∞[ f1 [M, +∞[
[0, +∞[ [0, M] χ [M,+∞[(x)/x2 N → +∞
χ [M,N[(x)/x2 1/M − 1/N
χ [M,+∞[(x)/x2 1/M f1 [0, +∞[
fk(x) = √ k + xe −kx
+∞
lim
k→+∞ 0
+∞
fk(x) dx =
0
lim
k→+∞ fk(x) dx.
fk ]0, +∞[
0
+∞
0
fk(x) =
k
0
= √ 2k
fk(x) dx +
1 − e−k2
k
≤ √ 1 − e−k2
2
k
+∞
k
+ √ 2
+ √ 2
fk(x) dx ≤
+∞
k
+∞
0
k
0
√ xe −kx dx
√ xe −kx dx.
√ k + ke −kx dx +
+∞
k
√ x + xe −kx dx

k → +∞
⌊x⌋ x
lim
k→0 +
+∞
1
(−1)
⌊x⌋ | sin πx|1/k
1/x2 x =
1/2 + k k ∈ N N := {x = 1/2 + k : k ∈ N}
lim
k→0 +
+∞
1
(−1)
x 2
⌊x⌋ | sin πx|1/k
x 2
dx,
dx, = 0.
gk(x) = k/π
1 + k2 k ∈ N \ {0}
x2
gk > 0 gk(x) dx = 1
R
ε > 0 gk {x : |x| > ε}
f
gk(x)f(x) dx = f(0).
R
lim
k→+∞
gk > 0 x ∈ R k ≥ 1
gk(x) dx = 1
k dx
π 1 + x2
1 dy
= = 1.
k2 π 1 + y2 R
R
R
ε > 0 0 < gk(x) = gk(−x) < gk(ε) x > ε gk(ε) → 0 + k → +∞
{x : |x| > ε}
gk(x)f(x) dx = 1
dy
f(y/k) dy
π 1 + y2 k f(y/k)/(1+y 2 ) f∞/(1+y 2 )
lim
k→∞
gk(x)f(x) dx = 1
π
dy
f(0)
lim f(y/k) dy =
k→∞ 1 + y2 π
R
y = kx
1
k
+∞
1/k
R
sin x 1
dx =
x2 k
+∞ 1
lim
k→∞ k 1/k
+∞
1
R
sin x
dx.
x2 R
sin(y/k)
y2 /k2 dy
k =
+∞ sin(y/k)
1 y2 dy
R
dy
= f(0).
1 + y2 1/y 2
α > 0 [0, +∞[
∞ 1
Fα(x) =
xα .
+ kα k=1

f α k =
1
x α + k α , sα n =
n
f α k .
f α k > 0 sαn
Fα
+∞
f
0
α k
+∞
0
Fα(x) dx = lim
n→∞
1
(x) dx =
kα +∞
0
+∞
0
sn(x) dx = lim
dx
(x/k) α 1
=
+ 1 kα +∞
0
k=1
n
+∞
f
n→∞
k=1
0
α k
(x) dx
k dy
yα 1
=
+ 1 kα−1 +∞
0
α > 1
n
+∞
lim f
n→∞
k=1
0
α k
(x) dx = lim
n 1
n→∞ k
k=1
α−1
+∞ dy
0 yα + 1 =
+∞
0
dy
y α + 1 lim
dy
y α + 1
n
n→∞
k=1
1
k α−1
α − 1 > 1 α > 2 Fα [0, +∞[
α > 2
k ∈ N \ {0} fk(x) = k 3 (x − k) 2 χ [k−1/k,k+1/k](x) fk
R
R
lim
k→∞ fk(x)
dx = lim
k→∞
R
fk(x) dx.
K R R > 0
|x| < R x ∈ K k > R + 1 K ∩ [k − 1/k, k + 1/k] = ∅ fk(x) = 0
x ∈ K K
fk
R
fk(x) dx =
k+1/k
k
k−1/k
3 (x − k) 2 1/k
dx =
−1/k
fk 2/3
k 3 y 2 1/k
dy = 2 k
0
3 y 2 dy = 2
3 ,
ϕ ∈ C∞ (R) S = {x ∈ R : ϕ(x) = 0}
− log |x|ϕ
R
′ (x) dx = lim
ε→0 +
ϕ(x)
R\[−ε,ε] x dx.
log |x|ϕ ′ (x) ≤ ϕ ′ ∞ log |x|χS(x) log |x|χS(x) L 1
0 /∈ S | log |x|| ¯ S L 1
1
0
| log |x|| dx = 1,

[−1, 0] log |x|χS(x) [−1, 1] ∪ S S
− log |x|ϕ
R
′
(x) dx = − log |x|ϕ
R
′
(x) lim
ε→0 + χ
R\[−ε,ε](x) dx
= − lim log |x|ϕ ′
(x)χR\[−ε,ε](x) dx = − lim log |x|ϕ ′ (x) dx
ε→0 +
= lim
ε→0 +
= lim
ε→0 +
R
−[log |x|ϕ(x)] −ε
−∞ +
−ε
ϕ(x)
x
−∞
+∞
= lim log(ε)(ϕ(ε) − ϕ(−ε)) +
ε→0 +
ε
+∞
= lim
ε→0 +
ε
R\[−ε,ε]
ϕ(x) − ϕ(−x)
dx
x
ϕ(x)
x dx.
ε → 0 +
ϕ
ε→0 +
|x|>ε
dx − [log |x|ϕ(x)]+∞
ε
ϕ(x) − ϕ(−x)
x
lim log(ε)(ϕ(ε) − ϕ(−ε)) = 0
ε→0 +
| log(ε)(ϕ(ε) − ϕ(−ε))| ≤ 2ϕ ′ ∞ε| log ε| → 0 +
ϕ(ε) − ϕ(−ε) =
ε
−ε
ϕ ′ (s) ds ≤ 2εϕ∞.
dx
+∞ ϕ(x)
+
ε x dx
n
un(t) :=
2 sin(nt) t ∈ − π
π
n , n ,
0 .
ϕ ∈ C∞ (R) {t ∈ R : ϕ(t) = 0}
un(t)ϕ(t) dt.
lim
n→∞
R
un(0) = 0 n ∈ N limn→∞ un(0) = 0 ¯t = 0
n > π/|¯t| |¯t| > π/n n > π/|¯t| un(¯t) = 0 lim
n→∞ un(¯t) = 0
un 0
R
un(t)ϕ(t) dt =
=
=
π/n
−π/n
π
−π
π
−π
n 2 sin(nt) ϕ(t) dt =
s sin s
s sin s
ϕ(s/n) − ϕ(0)
s/n
ϕ(s/n) − ϕ(0)
s/n
π
−π
ds +
n sin s ϕ(s/n) ds =
π
−π
ds + nϕ(0)
s sin s ϕ(0)
s/n ds
π
−π
sin s ds =
π
−π
π
s sin s ϕ(s/n)
s/n ds,
−π
s sin s
ϕ(s/n) − ϕ(0)
s/n
lim un(t)ϕ(t) dt = ϕ
n→∞
R
′ π
(0) s sin s ds
−π
= 2ϕ ′ π
(0) s sin s ds = 2ϕ
0
′
(0) [−s cos s] π π
0 + cos s ds
0
= 2πϕ ′ (0).
ds

D R X n
1 1 X C ℓ C ℓ ω : D → X ∗ D
X ∗ X X {dxi : i = 1...n}
1
ω(x) = ω1(x) dx1 + ... + ωn(x) dxn
ωj(x) ∈ C ℓ (D, R) ω
1 ω D D f : D → R
x ∈ D df(x) = ω(x) f ω
ω D ω
1
x ∈ D i, j = 1...n
∂kωj(x) = ∂jωk(x)
1
1 ω D D x ∈ D
U x D ω |U
D R n ω ∈ C 1 (D, (R n ) ∗ ) C 1
X R D X a, b ∈ R
D α : [a, b] → D C 1
α(a) α(b)
ω : D → X ∗ 1 C 0 α : [a, b] → D
ω α
ω :=
α
n
ω :=
α
j=1
[a,b]
[a,b]
ω(α(t))α ′ (t) dt
ωj(α(t))α ′ j(t) dt
ω : D → X ∗ 1 C 0
ω D
α, β D
γ D
ω = 0
γ
α
ω =
β ω

α, β : [a, b] → D α β D
h : [a, b] × [0, 1] → D
h(t, 0) = α(t) h(t, 1) = β(t) t ∈ [a, b]
h(a, λ) = h(b, λ) λ ∈ [0, 1]
ω 1 D α, β
D
ω = ω
α β
D X ω ∈ C 0 (D, X ∗ )
C 1
D X x0
x ∈ D {λx0 + (1 − λ)x : λ ∈ [0, 1]} D
1 C 1 D D
D R2 a1, ..., am ∈ D ω
1 D \ {a1, ..., am} ω D \ {a1, ..., am} γ1, ..., γm
aj ak
ω = 0 γj
j = 1, ..., m
S ⊆ R n S λ ≥ 0 x ∈ S λx ∈ S
α ∈ R f = f(x1, ..., xn) S R n
α (x1, ..., xn) ∈ S λ > 0 f(λx1, ..., λxn) =
λ α f(x1, ..., xn)
D ω = M(x, y) dx+N(x, y) dy
ω M, N α = −1
D ω D
f(x, y) = 1
[xM(x, y) + yN(x, y)]
α + 1
(t
γ(t) =
3 , 3t2 ) t ∈ [0, 1/2]
((1 − t2 )/6, (1 − t2 )) t ∈ [1/2, 1].
γ
ω1 = ydx + ydy, ω2 = ydx + xdy,
γ
ω1,
γ(t) γ1 : [0, 1/2] → R 2
γ1(t) = (t 3 , 3t 2 ) γ2 : [1/2, 1] → R 2 γ2(t) = ((1 − t 2 )/6, (1 − t 2 ))
γ1 γ1(t) = (x(t), y(t)) t = 3√ x
y = 3x 2/3 0 ≤ x ≤ 1/8
γ1(0) = (0, 0) γ2(1/2) = (1/8, 3/4)
γ2(t) 1 − t 2 = 6x y = 6x
ω2
γ

γ2(1/2) = (1/8, 3/4) γ2(1) = (0, 0) γ
3t2 −t/3
Jac(γ1) = , Jac(γ2) = ,
6t
−2t
1
1
ω1(t) =
√ 9t 4 + 36t 2 = 3t √ t 2 + 4 0 < t < 1/2,
t 2 /9 + 4t 2 = √ 37t/3 1/2 < t < 1,
1 1/2
lunghezza(γ) = ω1(t) dt = 3t
0
0
t2 √
1 37
+ 4 dt + t dt
1/2 3
= 3
√
1/4
√ 17/4
37 3
w + 4 dw + = z
2 0
8 2 4
1/2 √
37
dz +
8
= [z 3/2 ] z=5/4
z=17/4 +
√
37
8 = 17√ √
17 37
− 8 +
8 8
γ2
γ2 (3/4) 2 + (1/8) 2 = √ 37/8
γ1 y = 3x2/3 0 ≤ x ≤ 1/8
1/8
1/8
1 + ˙y 2 dx = 1 + 4x−2/3 1/2
dx = 1 + 4t−2 2
3t dt
0
0
1/2
=
0
3t t2 + 4 dt = 17√17 − 8.
8
ωi = ωx i dx + ωy
i dy i = 1, 2 ∂yωx 1 = 1
∂xω y
1 = 0 ω1
1/2
ydx + ydy = 3t
γ
0
2 · 3t 2 1/2
dt +
0
0
= 9 9
+
160 32 +
1/8
6x dx +
0
3t 2
· 6t dt +
0
1/8
= 9 9 3 9 3
+ − − =
160 32 64 32 320 .
6x · 6 dx
ω2 ω2 = df(x, y) f(x, y) = xy
γ ω2
x + By
ω =
x2 Cx + y
dx +
+ y2 x2 dy,
+ y2 BC R 2 \ {(0, 0)}
B C ω
ω Ω = {(x, y) : x > 0, y > 0}
B C ω
B C ω R 2 \ {(0, 0)}
γ2
ω1

ω = ωx dx + ωy dy
∂yωx = ∂xωy x 2 + y 2 = 0
∂
∂y
x + By
x 2 + y 2
B(x 2 + y 2 ) − (x + By)2y
(x 2 + y 2 ) 2
= ∂
∂x
Cx + y
x 2 + y 2
= C(x2 + y 2 ) − (Cx + y)2x
(x 2 + y 2 ) 2
Bx 2 + By 2 − 2xy − 2By 2 = Cx 2 + Cy 2 − 2Cx 2 − 2xy
(B − C)(x 2 + y 2 ) = 2By 2 − 2Cx 2
(x, y) ∈ R 2 \ {0, 0} x = 0 y = 1
B −C = 2B C = −B 2B(x 2 +y 2 ) = 2By 2 +2Bx 2
B ∈ R ω C = −B
C = −B ω Ω
Ω (1, 1)
u(x, y) = ω,
γ (x,y)
γ (x,y) Ω (x, y) ∈ Ω (1, 1) ∈ Ω
u(1, 1) = 0 γ (x,y)
C = −B
u(x, y) =
=
x
1
x
= 1
2
1
x
1
ωx(t, 1) dt +
y
y
1
ωy(x, s) ds
t + B
t2 + 1 +
Cx + s
1 x2 ds
+ s2 2t + 2B
t2 + 1 +
y
1
Cx + s
x2 ds
+ s2 = 1
2 [log(t2 + 1) + 2B arctan t] t=x
t=1 +
y
1
Cx + s
x2 ds
+ s2 = 1
2 log(x2 + 1) + B arctan x − log √ 2 − Bπ
4 +
= 1
2 log(x2 + 1) − C arctan x − log √ 2 + Cπ
4 +
= 1
2 log(x2 + 1) − C arctan x − log √ 2 + Cπ
4 +
+ C
y
1
1
s
1 +
x
2
ds
x
+ 1
2 [log(x2 + s 2 )] s=y
s=1
= 1
2 log(x2 + 1) − C arctan x − log √ 2 + Cπ
4 +
+ C
y/x
1/x
y
1 1
dz +
1 + z2 2 log(x2 + y 2 ) − 1
2 log(x2 + 1)
= −C arctan x − log √ 2 + Cπ
4
u(1, 1) = −C arctan 1 − log √ 2 + Cπ
4
Cx + s
x2 ds
+ s2 1
y Cx
1 x2 y s
ds +
+ s2 1 x2 ds
+ s2 + C arctan(y/x) − C arctan 1/x + 1
2 log(x2 + y 2 ).
1
+ C arctan(1) − C arctan(1) + log(2) = 0.
2

C = −B
∂yu(x, y) =
C/x
1 + y2 y
+
/x2 x2 Cx + y
=
+ y2 x2 + y2 ∂xu(x, y) = − C y/x2
+ B
1 + x2 1 + y2 −
/x2 C
1 + 1/x 2
− 1
x2
+
x
x2 x + By
=
+ y2 x2 + y2 γ(t) = (cos θ, sin θ) θ ∈ [0, 2π]
2π
2π
ω = (cos θ + B sin θ)(− sin θ) dθ + (−B cos θ + sin θ) cos θ dθ
γ
=
0
2π
0
− sin θ cos θ − B sin 2 θ − B cos 2
θ + sin θ cos θ dθ = −2πB
R 2 \ {(0, 0)}
1 ω
B = C = 0 ũ(x, y) = 1
2 log(x2 +y 2 ) R 2 \{(0, 0)}
α
y
ω = 2 + dx + 1 + α
x
x
dy
y
x > 0 y > 0 U(x, y) U(1, 1) = 2
ω =
y
dx +
x
0
α x
y dy + 2 dx + dy =: ω1 + df2(x, y)
f2(x, y) = 2x + y ω ω1 ω1
∂ y ∂
= α
∂y x ∂x
x
y
1
2 √ xy = √ α ∂
α
∂x
x
y
1
2 √ 1
= α
xy 2 √ xy ,
α = 1
α
y x
ω = dx +
x y dy + df2(x, y)
u(x, y) = f1(x, y)+f2(x, y)+c f1 ω1 c ∈ R
ω1 (1, 1)
f1(x, y) = ω1,
γ (x,y) (1, 1) (x, y) ω1 =
ωx 1 dx + ωy 1 dy
f1(x, y) =
x
ω
1
x y
1 (s, 1) ds +
1
γ (x,y)
ω y
1 (x, t) dt = 2(√ x − 1) + (2 √ x)( √ y − 1) = 2( √ xy − 1).
f2(1, 1) = 3 f1(1, 1) = 0 c = −1
U(x, y) = 2 √ xy + 2x + y − 3.

U(1, 1) = 2 ∂xU(x, y) = y/x+2 ∂yU(x, y) =
x/y + 1
1
ay
1
ω = cos
dx + cos
(x − 1)y − 1 ((x − 1)y − 1) 2 (x − 1)y − 1
Ω = {(x, y) : (x − 1)y < 1}
Ω
a ∈ R ω Ω
a (0, 0) sin 1
1 − x
dy
((x − 1)y − 1) 2
ω = ωx dx + ωy dy
Ω y = 1
x−1
(1, 0) Ω
Ω y Ω
ω Ω ∂yωx = ∂xωy v =
(x − 1)y − 1
∂yωx = ∂
1
ay
cos
∂y (x − 1)y − 1 ((x − 1)y − 1) 2
= ∂
1 ay
cos
∂v v v2
· ∂v
∂ 1 ay
+ cos
∂y ∂y v v2
= ay(x − 1) ∂
1 1
cos
∂v v v2
1 1
+ a cos
v v2 ∂xωy = ∂
1
1 − x
cos
∂x (x − 1)y − 1 ((x − 1)y − 1) 2
= ∂
1 1 − x
cos
∂v v v2
· ∂v
∂ 1 1 − x
+ cos
∂x ∂x v v2
= (1 − x)y ∂
1 1
cos
∂v v v2
1 1
− cos
v v2 a = −1
v(x, y) = (x − 1)y − 1
1
−y
1
1 − x
ω = cos
dx + cos
dy
(x − 1)y − 1 ((x − 1)y − 1) 2 (x − 1)y − 1 ((x − 1)y − 1) 2
1
1
= −
· cos
(ydx + (x − 1)dy)
((x − 1)y − 1) 2 (x − 1)y − 1
= d
sin(1/v) · dv(x, y)
dv
ω = df(x, y) f(x, y) = sin(1/v(x, y)) = sin(1/((x−1)y −1)) f(0, 0) = sin 1
f

dy N(x, y)
= −
dx M(x, y)
N M D R 2 R
y = y(x) I R J R
F : D → R
Γc := {(x, y) ∈ D : F (x, y) = c}, c ∈ R,
y = y(x) Γc
y = y(x)
dy
dx = −∂xF (x, y)
∂yF (x, y)
dy N(x, y)
= −
dx M(x, y) .
N, M : D → R D ⊆ R 2 F : D → R
C 1 λ : D → R \ {0} C 0 F ω = 0
ω(x, y) = N(x, y) dx + M(x, y) dy λ(x, y)
G(x, y) := (λ(x, y)N(x, y), λ(x, y)M(x, y))
F ∇F (x, y) = G
F (x, y) = c ω = 0
F λ(x, y)
∇F (x, y) = (∂xF (x, y), ∂yF (x, y)) = (λ(x, y)N(x, y), λ(x, y)M(x, y)) .
Γc := {(x, y) ∈ D : F (x, y) = c}, c ∈ R,
Γc y = y(x)
dy
dx = −∂xF (x, y) N(x, y)
= −
∂yF (x, y) M(x, y) ,
Γc
¯y(x)
F (x, y) = y − ¯y(x) λ(x, y) = M(x, y)
M(x, y) = 0 N(x, y) = 0
y = y(x)

x = x(y)
Γc
dx y)
= −M(x,
dy N(x, y) .
1 ω(x, y) = N(x, y) dx + M(x, y) dy
M N D R 2
ω(x, y) = 0
F (x, y) λ(x, y)
dF (x, y) = λ(x, y)ω(x, y) λ(x, y) = 0 D
F (x, y) = c c ∈ R
ω = 0 ω(x, y) = N(x, y) dx + M(x, y) dy
F (x, y) C1
∂F
∂F
(x, y) = M(x, y),
(x, y) = N(x, y),
∂x ∂y
ω G(x, y) =
(M(x, y), N(x, y))
D
∂M
∂y
λ(x, y) ≡ 1
∂N
(x, y) = (x, y).
∂x
ω = 0 γ C1
P (x0, y0) (x, y) ∈ D
F (x, y) = ω
D
P (x0, y) (x, y) P (x, y0) (x, y)
F (x, y) =
F (x, y) =
x
x0
x
x0
γ
M(t, y) dt +
y
y0
y
M(t, y0) dt +
y0
N(x0, s) ds.
N(x, s) ds
M N D α = −1
D
F (x, y) = 1
[x · M(x, y) + y · N(x, y)]
α + 1
D R n ω 1 C 1 A
ω λ ∈ C 1 (A, R) λω D
ω(x, y) = p(x, y)dx + q(x, y)dy = 0
f, q, p C 1
∂yp − ∂xq = f(x)q(x, y) − g(y)p(x, y)
x
h(x, y) = exp
x0
y
f(t)dt +
y0
g(t)dt
ω f ≡ 0 g ≡ 0

2xy dx + (x 2 + 1) dy = 0
(x 2 + y 2 − 2x) dx + 2xy dy = 0
y 2 dx + (xy − 1) dy = 0
R2
γ (0, 0) (x0, y0)
γ = γ1 ∪ γ2 γ1(x) = (x, 0) 0 ≤ x ≤ x0 x0 ≤ x ≤ 0 γ2(y) = (x0, y)
0 < y < y0 y0 ≤ y ≤ 0 ˙γ1(x) = (1, 0) ˙γ2(x) = (0, 1)
V (x0, y0) = ω = (2xy dx + (x 2
+ 1) dy) + (2xy dx + (x 2 + 1) dy)
γ γ1
γ2
= (2xy dx + (x
γ2
2 y0
+ 1) dy) = (x
0
2 0 + 1) dy = (x 2 0 + 1)y0.
V (x, y) = (x2 + 1)y (x2 + 1)y = c c ∈ R
1 + x2 = 0
y ′ (x) = dy 2xy
(x) = −
dx x2 + 1
y ′ 2x
= −
y x2 d
= −
+ 1 dx log(x2 + 1)
dy d
= −
y dx log(x2 + 1)
log |y| = − log(x 2 + 1) + d d ∈ R |y| = ed
x 2 +1
y = c
x 2 +1
c ∈ R c = ±e d
R 2
γ (0, 0) (x0, y0)
γ = γ1 ∪ γ2 γ1(x) = (x, 0) 0 ≤ x ≤ x0 x0 ≤ x ≤ 0 γ2(y) = (x0, y)
0 < y < y0 y0 ≤ y ≤ 0
V (x0, y0) = ω =
=
γ
x0
0
γ1
((x 2 + y 2 − 2x) dx + 2xy dy) +
(x 2 − 2x) dx +
y0
0
γ2
2x0y dy = x3 0
3 − x2 0 + x0y 2 0.
((x 2 + y 2 − 2x) dx + 2xy dy)
V (x, y) = x3
3 − x2 + xy 2 x(x 2 /3 − x + y 2 ) = c c ∈ R
R 2 \ {xy = 0} 2xy
y ′ (x) = dy
dx = −x2 + y2 − 2x
,
2xy
p(x, y) = y2 q(x, y) = xy − 1 ∂yp − ∂xq = 2y − y = y = 0 ω
∂yp − ∂xq = y = f(x)q(x, y) − g(y)p(x, y)
y f ≡ 0 g(y) = −1/y
(x0, y0) = (0, 1)
x y y
h(x, y) = exp f(t)dt + g(t)dt = exp −
1
1
t dt
= e log(1/y) = 1/y
x0
y0

R2 \ {y = 0}
h(x, y)ω(x, y) = y dx + x − 1
dy.
y
H + = {(x, y) : y > 0} H − = {(x, y) : y <
0}
V + V − H + H −
H + (0, 1) (x0, y0) H +
γ = γ1 ∪ γ2 γ1(x) = (x, 0)
0 ≤ x ≤ x0 x0 ≤ x ≤ 0 γ2(y) = (x0, y) 1 < y < y0 0 < y0 ≤ y ≤ 1
V +
x0
y0
(x0, y0) = h(x, y)ω(x, y) + h(x, y)ω(x, y) = dx + x0 − 1
dy
y
γ1
γ2
= x0 + [x0y − log |y|] y=y0
y=1 = x0 + x0y0 − log y0 − x0 = x0y0 − log y0
V + (x, y) = xy−log y y > 0 H + V + (x, y) =
c c ∈ R
V − H − (0, −1)
(x0, y0) H +
γ = γ1 ∪ γ2 γ1(x) = (x, 0) 0 ≤ x ≤ x0 x0 ≤ x ≤ 0 γ2(y) = (x0, y)
−1 < y < y0 < 0 y0 ≤ y ≤ −1
V −
(x0, y0) = h(x, y)ω(x, y) +
=
γ1
x0
0
γ2
y0
−1 dx + x0 −
−1
1
y
0
h(x, y)ω(x, y)
= −x0 + [x0y − log |y|] y=y0
y=−1 = x0 + x0y0 − log |y0| + x0
= x0y0 − log |y0| = x0y0 − log(−y0),
y0 < 0 V − (x, y) = xy − log y y < 0
H − V − (x, y) = c c ∈ R
V (x, y) = xy − log |y| R 2 \ {y = 0}
R 2 \ {y = 0} xy − log |y| = c c ∈ R
y ′ = y(1 − y)
y ′ = (x + y) 2 − (x + y) − 1
y ′ − y = e x√ y
y(x) = 0 y(x) = 1 y = 0, 1
dy
1
y(1 − y) dy − dx = 0.
0 < y < 1
1
x − dy = c, c ∈ R.
y(1 − y)
1
dy =
y(1 − y)
=
1
dy =
y(1 − y)
2dy
1 − (2y − 1) 2 =
1
dy =
y − y2 dt
√ 1 − t 2
1
dy
1/4 − (y − 1/2) 2
= arcsin(t) = arcsin(2y − 1).

x − c = arcsin(2y − 1) 0 < y < 1
−π/2 < x − c < π/2 y = (sin(x − c) + 1)/2 cos(x − c) ≥ 0
y ′ + 1 = d
dx (x + y) = (x + y)2 − (x + y).
v = y + x v ′ = v 2 − v
v = 0 v = 1 v = 0, 1
dv
= dx
v(v − 1)
1 A B
= +
v(v − 1) v v − 1
A = −B = −1
= Av − A + Bv
v(v − 1)
− log |v| + log |v − 1| = x + c
v = 0 v = 1
log
v − 1
v = x + c,
− 1
v = −1 ± ex+c
1
v = ,
1 ± ex+c v = −x v = 1 − x
1
y(x) = − x.
1 ± ex+c Ω := {(x, y) : y > 0}
˜ω(x, y) := ˜p(x, y) dx + ˜q(x, y) dy = (y + e x√ y) dx − dy = 0.
y ≡ 0 √ y
ω(x, y) := p(x, y) dx + q(x, y) dy = (y 3/2 + e x y) dx − √ y dy = 0.
∂yp(x, y) − ∂xq(x, y) = 3/2y 1/2 + e x = 1
y
h(x, y) = exp − dy
y +
−1/2 dx
h(x, y)ω(x, y) = e−x/2
y
1
p(x, y) − q(x, y),
2
= exp(−x/2 − log y) = e−x/2
.
y
(y 3/2 + e x y) dx − e−x/2 √ −x/2 x √ e
y dy = e (e + y) dx −
y
−x/2
√ dy
y
(0, 1) ∈ Ω (x0, y0) ∈ Ω
γ
x0
V (x0, y0) = h(x, y)ω(x, y) = e −x/2 (e x y0
+ 1) dx + − e−x0/2
√ dy
y
=
γ
0
1
x0
(e
0
x/2 + e −x/2 ) dx + 2(1 − √ y0 )e −x0/2 x0/2 √
= 2(e − y0e −x0/2
)

2e −x/2 (e x − √ y) = d, d ∈ R,
y = (e x + ce x/2 ) 2 , c ∈ R,
e x + ce x/2 ≥ 0 d ≥ −e x/2
y ′ = y
x +
1 − y2
x2 y ′ − 2x2 − 1
x(x 2 − 1) y = 2x2√ x 2 − 1
|y/x| ≤ 1
y
x +
1 − y2
x2
dx − dy = 0
0 x = ξ y = ξη
η + 1 − η2
dξ − (ξ dη − η dξ) = 0,
1 1
dξ − dη = 0.
ξ 1 − η2
V (x, y) = log |x| − arcsin(y/x)
log |x| − arcsin(y/x) = c c ∈ R
|x| ≥ 1
2x2 − 1
ω(x, y) := p(x, y) dx + q(x, y) dy =
x(x2 − 1) y + 2x2x2
− 1 dx − dy = 0.
f(x) = − 2x2 − 1
x(x2
A
= −
− 1) x
∂yp(x, y) − ∂xq(x, y) = 2x2 − 1
x(x2 = f(x)q(x, y),
− 1)
= − (A + B + C)x2 + (B − C)x − A
x(x 2 − 1)
B C
+ + = −
x − 1 x + 1
A(x2 − 1) + Bx(x + 1) + Cx(x − 1)
x(x2 − 1)
A = 1 B = C = 1/2 |x| ≥ 1
1 1 1 1 1
f(x) dx = − + + dx = − log |x| +
x 2 x − 1 2 x + 1
1
1
log |x − 1| + log |x + 1|
2 2
= − log |x| x2
− 1 .

h(x, y) = 1/(|x| √ x2 − 1)
2x
h(x, y)ω(x, y) =
2 − 1
x|x|(x2
1
y + 2|x| dx −
− 1) 3/2 |x| √ x2 − 1 dy
y
= d −
|x| √ x2
+ 2|x| dx
− 1
y
= d −
|x| √ x2
+ x|x|
− 1
x = 0 x sgnx = |x| sgn2x = 1 sgn(x)
R \ {0}
y
−
|x| √ x2 + x|x| = c, c ∈ R,
− 1
y(x) = (x 3 − c|x|) x 2 − 1, c ∈ R.

˙x = e t−x /x x(α) = 1 α ∈ R
ω(t, x) = xe x dx − e t dt = 0.
ω R 2
F (t0, x0) =
x0
0
xe x dx +
t0
0
−e t dt = [xe x ] x0
0 −
x0
0
e x dx − [e t ] t0
0 = (x0 − 1)e x0 − e t0 + 2.
F (t, x) = c c ∈ R
c 2
(x − 1)e x − e t = c
x(α) = 1 (α, 1) c = −e α
e t = (x − 1)e x + e α .
∂xF (t, x) = 0 x = 0 ∂tF (t, 0) = −1 = 0 F (t, x) = c
(t, 0) x = 0
x = x(t) x(α) = 1
x ≥ 0 t x t(x) = log((x−1)e x +e α )
(x − 1)ex + eα > 0 x > 0 g(x) = (x − 1)ex + eα g ′ (x) = xex
g(x) = +∞ x > 0 g
lim
x→+∞
0 e α − 1 ¯t x = x(t)
t > ¯t
lim t(x) − (log(x − 1) + x) = 0,
x→+∞
α x = x(t) t → +∞
et = (x − 1)ex α > 0 g eα − 1 > 0 g(x) > 0 x ≥ 0
t = log g(x) x ≥ 0
tα = lim
x→0 + log g(x) = log(eα − 1)
tα x(α) = 1
x = 0 tα < α t < tα
g t = log g(x)
x(t)
α = 0 g(x) > 0 x ≥ 0 g(0) = 0
tα = lim log g(x) = −∞
x→0 +
x(t) x = 0 t → −∞
g t = log g(x)
x(t)

˙x = e t−x /x x(α) = 1 α ∈ R
α < 0 g g
xα g(xα) = 0 (xα−1)e xα =
−e α 0 < xα < 1 e α > 0 g(x) > 0 x > xα
tα = lim log g(x) = −∞
x→(xα) +
t → −∞ xα
t ≥ α t → +∞ +∞ x = x(t)
t → +∞ e t = (x − 1)e x α > 0 ] log(e α − 1), +∞[
t → t + α = log(e α − 1) + 0 tα < α α = 0
R t → −∞ 0 α < 0 R t → −∞
0 < xα < 1 xα (xα − 1)e xα = −e α
˙x = x 2 /(1 − tx) x(0) = α α ∈ R
ω(t, x) = p(t, x) dx + q(t, x) dt = (1 − tx) dx − x 2 dt = 0.
∂tp(t, x) − ∂xq(t, x) = −x + 2x = x = 0 ω
∂tp(t, x) − ∂xq(t, x) = x = f(x)q(t, y) − g(t)p(t, y)

g(t) = 0 f(x) = −1/x x = 0 h(t, x) =
e f(x) dx = 1/|x| x > 0
1
h(t, x)ω(x, t) = − t dx − x dt =
x 1
dx − (t dx + x dt) = d(log(x) − tx)
x
x > 0 log(x) − tx = c1 c1 ∈ R x < 0
1
h(t, x)ω(x, t) = + t dx + x dt =
−x 1
dx + (t dx + x dt) = d(− log(−x) + tx)
−x
x < 0 − log(−x) + tx = c2 c2 ∈ R
F (t, x) := log(|x|) − tx = c c ∈ R x = 0
x(0) = α F (0, α) = c c = log |α|
(t, x) ↦→ (−t, −x) s = −t y(s) = −x(s)
dy
dt
(s) = −dx(−t)
ds dt ds = ˙x(−t) = x2 (−t)
1 − tx(−t) = x2 (s)
1 + sx(s) = y2 (s)
1 − sy(s) ,
α ≥ 0 α < 0
−α > 0
∂xF (t, x) = 1/x − t tx = 1
x(0) = α tx < 1
tx < 1
α = 0
α → 0 F (t, x) = log |α| F (t, x) ≡ −∞ c ≡ 0
x ′ > 0 xy < 1 0 < tα < +∞
tx = 1 −∞ ≤ t ≤ tα
F (t, x) = log |α| t t(x) = log(|x/α|)/x
x = 0
lim t(x) = ∓∞,
x→0 ±
α x = 0
x > 0 t ′ (x) = [1−log(|x/α|)]/x 2
x = αe t = 1/(αe) t ′′ (x) =
2 log(x/a)−3
x 3
t ′′ (αe) < 0 t :]0, αe[→] − ∞, 1/(αe)[
x = x(t)
x = 0 t → −∞
0 < x < 1 t < 0
g(t, x) = x2 x2 x2 x2
= > = = f(t, x)
1 − tx 1 + |t|x 1 + |t| 1 − t
ε > 0 ˙y = f(t, y) y(0) = α + ε
˙x = g(t, x) ] − ∞, 0]
t < 0
¯y
α
dy
=
y2 −∞
0
dt
= +∞
1 − t

˙x = x 2 /(1 − tx) x(0) = α α ∈ R
¯y = 0 t → −∞ 0 < x(t) ≤ y(t) → 0 x(t)
0 t → −∞
±α xt = 1
(t ∗ +, x ∗ +) = (1/(eα), αe) (t ∗ −, x ∗ −) = −(1/(αe), αe)

y ′ + y = sin x
y IV − 16y = 1 + cos 2x
y ′′′ − 6y ′′ + 11y ′ − 6y = x 2
y IV − y ′′ = x − 1
y ′ + y = 0 y0(x) = ce −x c ∈ R
sin x
A sin x + B cos x
A cos x − B sin x + A sin x + B cos x = sin x A + B = 0 A − B = 1
A = −B = 1/2 y(x) = ce −x + (sin x − cos x)/2 c ∈ R
y IV −16y = 0 λ 4 −16 = 0
{±2, ±2i} y0(x) = c1e 2x +c2e −2x +c3 cos(2x)+
c4 sin(2x) ci ∈ R i = 1, 2, 3, 4 1 + cos 2x
¯y1(x) y IV − 16y = 1 ¯y2(x)
y IV − 16y = cos 2x y IV − 16y = 1
0
0 ¯y1(x) =
−1/16
y IV −16y = cos 2x
2 1
¯y(x) = x(A cos(2x) + B sin(2x))
¯y ′ (x) = A cos(2x) + B sin(2x) + 2x(−A sin(2x) + B cos(2x))
¯y ′′ (x) = 2(−A sin(2x) + B cos(2x)) + 2(−A sin(2x) + B cos(2x)) + 4x(−A cos(2x) − B sin(2x))
= 4(−A sin(2x) + B cos(2x)) + 4x(−A cos(2x) − B sin(2x))
¯y ′′′ (x) = 8(−A cos(2x) − B sin(2x)) + 4(−A cos(2x) − B sin(2x)) + 4x(A sin(2x) − B cos(2x))
= 12(−A cos(2x) − B sin(2x)) + 8x(A sin(2x) − B cos(2x))
¯y IV (x) = 24(A sin(2x) − B cos(2x)) + 8(A sin(2x) − B cos(2x)) + 16x(A cos(2x) + B sin(2x))
= 32(A sin(2x) − B cos(2x)) + 16x(A cos(2x) + B sin(2x))
32(A sin(2x) − B cos(2x)) + 16x(A cos(2x) + B sin(2x)) − 16x(A cos(2x) + B sin(2x)) = cos 2x
A = 0 −32B cos(2x) = cos 2x B = −1/32
y(x) = c1e 2x + c2e −2x + c3 cos(2x) + c4 sin(2x) − 1 x
−
16 32 sin(2x),
ci ∈ R i = 1, 2, 3, 4

y ′′′ − 6y ′′ + 11y ′ − 6y = 0 λ 3 − 6λ 2 +
11λ − 6 = 0 −6
±1, ±2, ±3, ±6 λ1 = 1 λ2 = 2 λ3 = 3
x 2
0 ¯y(x)
2 ¯y(x) = ax 2 + bx + c
−12a + 11(2ax + b) − 6(ax 2 + bx + c) = x 2
a = −1/6 2 − 11x/3 + 11b − 6bx − 6c = 0 6b = −11/3 b = −11/18
6c = −85/18 c = −85/108
y(x) = c1e x + c2e 2x + c3e 3x − x2
6
11 85
− x −
18 108 ,
c1, c2, c3 ∈ R
y IV − y ′′ = 0 λ 4 − λ 2 = 0
0 2 ±1 x − 1
0 2
¯y(x) = x 2 (ax + b) = ax 3 + bx 2
−(6ax + 2b) = x − 1 a = −1/6 b = 1/2
y(x) = c1 + c2x + c3e x + c4e −x + x 2
1 x
− .
2 6
k ∈ N
− 1
k u′′ (x) + |u ′ (x)| = 1, x ∈ (0, 1),
u(0) = 0, u(1) = 2.
uk
{uk} [0, 1]
v(x) = u ′ (x) v ′ (x) = k(|v(x)| − 1)
v(x) ≡ 1 v(x) ≡ −1
v(x) < −1 v(x) = 1 −1 < v(x) < 1
v(x) = 1 v(x) > 1 x ∈ (0, 1)
u(x) = u(0) +
x
0
v(t) dt,
u(0) = 0 u(1) = 2
2 =
1
0
v(t) dt,
v(x) = u ′ (x) > 1
v ′ (x) = kv(x) − k
u(x)
u(x) = u(0) +
x
0
v(t) = ce kt + 1.
(ce kt + 1) dx = u(0) + c
k [ekt ] t=x
t=0 + x = u(0) + c
k (ekx − 1) + x.

u(0) = 0 2 = c
k (ek − 1) + 1 c = k/(ek − 1)
k ∈ N uk(x) = ekx +1
ek + x
−1
k → ∞ 0 ≤ x < 1 u∞(x) = x |u ′ (x)| = 1
u(0) = 0 uk(1) = 2 k u∞(1) = 2
[0, 1]
y ′′ + y = tan x
y ′′ + y = 1
sin x
y ′′ + 3y ′ + 2y = √ 1 + e x
y ′′′ − 3y ′′ + 3y ′ − y = ex
x
φ(x, c1, c2) = c1 cos x + c2 sin x
y(x) = c1(x) cos x + c2(x) sin x.
y ′ (x) = c ′ 1(x) cos x + c ′ 2(x) sin x − c1(x) sin x + c2(x) cos x.
c ′ 1 (x) cos x + c′ 2 (x) sin x = 0
y ′ (x) = −c1(x) sin x + c2(x) cos x,
y ′′ (x) = −c ′ 1(x) sin x + c ′ 2(x) cos x − c1(x) cos x − c2(x) sin x.
−c ′ 1(x) sin x + c ′ 2(x) cos x − c1(x) cos x − c2(x) sin x + c1(x) cos x + c2(x) sin x = tan x.
−c ′ 1 (x) sin x + c′ 2 (x) cos x = tan x
c ′ 1 c ′ 2
c ′ 1 (x) cos x + c′ 2 (x) sin x = 0,
−c ′ 1 (x) sin x + c′
cos x sin x
⇐⇒
2 (x) cos x = tan x, − sin x cos x
c ′ 1 (x)
c ′ 2 (x)
=
0
tan(x)
c ′ 1 (x) = − sin x tan x c′ 2 (x) = sin x c2(x) = − cos x
c1(x) cos x t = tan(x/2)
2
sin x 1 − cos2
x
1
c1(x) = − dx = −
dx = sin x −
cos x cos x
cos x
2dt
1 1
= sin x − = sin x + + dt = sin x + log
1 + t
1 − t2 1 − t 1 + t
1
− t
= sin x − log
cos(x/2) + sin(x/2)
cos(x/2)
− sin(x/2)
¯y(x) = cos x sin x − log
cos(x/2) + sin(x/2)
cos(x/2)
− sin(x/2) cos x − cos x sin x.
y(x) = c1 cos x + c2 sin x − log
cos(x/2) + sin(x/2)
cos(x/2)
− sin(x/2) cos(x).
t = tan(x/2) cos x = 1−t 2
1+t 2 sin x = 2t
1+t 2 dx = 2dt
1+t 2

c ′ 1 (x) cos x + c′ 2 (x) sin x = 0,
−c ′ 1 (x) sin x + c′ 2 (x) cos x = 1/ sin x,
cos x
⇐⇒
− sin x
sin x c ′
1 (x)
cos x c ′ 2 (x)
=
0
1/ sin x
c ′ 1 (x) = −1 c1 = −x c ′ 2 (x) = cos x/ sin x
c2(x) = log | sin x| ¯y(x) = log | sin x| sin x − x cos x
y(x) = c1 cos x + c2 sin x − x cos x + log | sin x| sin x.
y ′′ +3y ′ +2y = 0 λ 2 +3λ+2 = 0
λ = −1 λ = −2 Φ(x, c1, c2) = c1e −x + c2e −2x
¯y(x) = c1(x)e −x + c2(x)e −2x
¯y ′ (x) = c ′ 1(x)e −x + c ′ 2(x)e −2x − c1(x)e −x − 2c2(x)e −2x .
c ′ 1 (x)e−x + c ′ 2 (x)e−2x = 0 ¯y ′ (x) = −c1(x)e −x − 2c2(x)e −2x
¯y ′′ (x) = −c ′ 1(x)e −x − 2c ′ 2(x)e −2x + c1(x)e −x + 4c2(x)e −2x .
−c ′ 1(x)e −x −2c ′ 2(x)e −2x +c1(x)e −x +4c2(x)e −2x −3c1(x)e −x −6c2(x)e −2x +c1(x)e −x +c2(x)e −2x = √ 1 + e x
−c ′ 1 (x)e−x − 2c ′ 2 (x)e−2x = √ 1 + e x
c ′ 1 (x)e−x + c ′ 2 (x)e−2x = 0,
−c ′ 1 (x)e−x − 2c ′ 2 (x)e−2x = √ 1 + e x ,
c ′ 1 (x) = ex√ 1 + e x c ′ 2 (x) = −e2x√ 1 + e x
c1(x) = e x√ 1 + ex
√1
dx = + t dt = z 1/2 dz = 2
3 z3/2 = 2
3 (1 + t)3/2 = 2
c2(x) = − e 2x√ 1 + ex
dx = − t √ 1 + t dt = − 2
3 t(1 + t)3/2 + 2
(1 + t)
3
3/2 dt
= − 2
3 t(1 + t)3/2 + 4
15 (1 + t)5/2 = − 2
3 ex (1 + e x ) 3/2 + 4
15 (1 + ex ) 5/2
¯y(x) = 4
15 (1 + ex ) 5/2 e −2x ,
y(x) = c1e −x + c2e −2x + 4
15 (1 + ex ) 5/2 e −2x .
3 (1 + ex ) 3/2
y ′′′ − 3y ′′ + 3y ′ − y = 0 λ 3 − 3λ 2 +
3λ − 1 = 0 (λ − 1) 3 = 0
Φ(x, c1, c2, c3) = c1ex + c2xex + c3x2ex
⎛
⎞ ⎛
⎠ ⎝ c′ 1 (x)
c ′ 2 (x)
c ′ 3 (x)
⎞ ⎛
⎠ = ⎝ 0
0
ex /x
⎝ ex xe x x 2 e x
e x e x (x + 1) xe x (2 + x)
e x e x (2 + x) e x (2 + 4x + x 2 )
A
det(A) = e 3x ((x+1)(x 2 +4x+2)+x 2 (x+2)+x 2 (x+2)−x 2 (x+1)−x(x+2) 2 −x(x 2 +4x+2)) = 2e 3x .
⎞
⎠

A
A −1 ⎛ 1
2
= ⎝
e−x x2 + 2x + 2 −e−x 1
x(x + 1) 2e−xx2 −e−x (x + 1) e−x (2x + 1) −e−x ⎞
x ⎠ .
e−x 2 −e−x e−x
2
c ′ 1 (x) = x/2 c′ 2 (x) = −1 c′ 1
3 (x) =
2x c1(x) = x2 /4 c2(x) = −x c3(x) =
log |x|/2
¯y(t) = x 2 e x /4 − x 2 e x + x 2 e x log |x|/2
y(t) = c1e x + c2xe x + c3x 2 e x + x 2 e x /2(log |x| − 3/2)

y ′ + y 1 x
=
tan x sin x
y ′ − x
1 + x 2 y = e−x y 3
y ′ + y = x 2 y 2
sin x = 0
ω(x, y) = (x − y cos x) dx − sin x dy
V (x, y) = 1
2 x2 − y sin x.
V (x, y) = c
y(x) = x2 − 2c
2 sin x .
z = y 1−3 = y −2 y = 1/ √ z
z ′ = −2y −3 y ′ = −2y −3
e −x y 3 + x
y
1 + x2 = −2e −x − z
2x
.
1 + x2 z ′ 2x
+ z
1 + x2 = −2e−x .
ω(x, z) = p(x, z) dx + q(x, z) dz = 2e −x 2x
+ z
1 + x2
dx + dz = 0.
1 + x 2
(2e −x (1 + x 2 ) + 2xz) dx + (1 + x 2 ) dz = 0.
∂zp(x, z) − ∂xq(x, z) = 2x 2x
= q(x, z)
1 + x2 1 + x2 1 + x 2 > 0
2x dx
h(x, z) = e 1+x2 = (1 + x 2 ).

Ω
(0, 0) (x0, z0)
V (x0, z0) =
x0
0
2ex (1 + x2 dx +
)
z0
z
0
(1 + x 2 0) dz = 2(3 − e −x0 (3 + x0(2 + x0))) + z0(1 + x 2 0).
−2e −x (3 + x(2 + x)) + z(1 + x 2 ) = c, c ∈ R
z = c + e−x2(3 + x(2 + x))
1 + x2 y
y(x) = ±
1 + x2 c + e−x = ±ex/2
2(3 + x(2 + x))
1 + x 2
ce x + 2x 2 + 4x + 6 .
c ∈ R
z = y 1−2 = 1/y y = 1/z
z ′ = − y′
= z − x2
y2 z ′ − z = −x 2
ke x k ∈ R
ax 2 + bx + c
2ax+b−ax 2 −bx−c = −x 2 a = 1 b−c = 0 2a−b = 0
b = c = 2 z z(x) = ke x + x 2 + 2x + 2
y
k ∈ R
y(x) =
1
ke x + x 2 + 2x + 2 ,
˙x − 3x − 2y = cos(2t)
˙y − 4x − y = 0
A =
3 2
4 1
cos(2t)
, B(t) =
0
2 ˙y = ¨x − 3 ˙x + 2 sin(2t)
˙y
2(4x + y) = ¨x − 3 ˙x + 2 sin(2t).
¨x − 3 ˙x − 8x − 2y + 2 sin(2t) = 0
2y
.
¨x − 3 ˙x − 8x − ( ˙x − 3x − cos(2t)) + 2 sin(2t) = 0.
x
¨x − 3 ˙x − 8x − ˙x + 3x + cos(2t) + 2 sin(2t) = 0.
¨x − 4 ˙x − 5x = −2 sin(2t) − cos(2t).

λ 2 − 4λ − 5 = 0,
A det(A − λId) = 0
λ1 = 5 λ2 = −1
Φ(t, c1, c2) = c1e 5t + c2e −t
±2i ¨x−4 ˙x−
(t) = −2A sin(2t) +
(t) = −4A cos(2t) − 4B sin(2t)
5x = −2 sin(2t) ū1(t) = A cos(2t) + B sin(2t) ū ′ 1
2B cos(2t) ū ′′
1
−4A cos(2t) − 4B sin(2t) − 4(−2A sin(2t) + 2B cos(2t)) − 5(A cos(2t) + B sin(2t)) = −2 sin(2t)
−9A − 8B = 0 8A − 9B = −2 A = −16/145 B = 18/145
ū1(t) = − 16
18
cos(2t) +
145 145 sin(2t).
±2i ¨x−4 ˙x−5x =
− cos(2t) ū2(t) = A cos(2t) + B sin(2t) ū ′ 2
2B cos(2t) ū ′′
2
(t) = −4A cos(2t) − 4B sin(2t)
(t) = −2A sin(2t) +
−4A cos(2t) − 4B sin(2t) − 4(−2A sin(2t) + 2B cos(2t)) − 5(A cos(2t) + B sin(2t)) = − cos(2t)
−9A − 8B = −1 8A − 9B = 0 B = 8/145 A = 9/145
ū2(t) = 9
8
cos(2t) +
145 145 sin(2t).
x
x(t) = Φ(t, c1, c2) + ū1(t) + ū2(t)
= c1e 5t + c2e −t − 16
18
9
8
cos(2t) + sin(2t) + cos(2t) +
145 145 145 145 sin(2t)
= c1e 5t + c2e −t − 7
y
145
cos(2t) + 26
145 sin(2t).
y(t) = 1
( ˙x(t) − 3x(t) − cos 2t),
2
˙x(t) = 5c1e 5t − c2e −t + 14 52
sin(2t) +
145 145 cos(2t),
y(t) = c1e 5t − 2c2e −t − 32 36
sin(2t) −
145 145 cos(2t).
det(A) = 0
˙x + 2x − y = 4t 2
A =
−2 1
3 2
˙y − 3x − 2y = 0
4t2 , B(t) =
0

y = ˙x + 2x − 4t2 ˙y = 3x + 2y
˙y = ¨x + 2 ˙x − 8t
˙y
3x + 2y = ¨x + 2 ˙x − 8t.
¨x + 2 ˙x − 3x − 2y − 8t = 0
y
¨x + 2 ˙x − 3x − 2( ˙x + 2x − 4t 2 ) − 8t = 0.
x
¨x + 2 ˙x − 3x − 2 ˙x − 4x + 8t 2 − 8t = 0.
¨x − 7x = −8t 2 + 8t.
λ 2 − 7 = 0 λ1 = √ 7
λ2 = − √ 7 A
Φ(t, c1, c2) = c1e √ 7t + c2e −√ 7t
0
ū(t) = at 2 + bt + c
2a − 7at 2 − 7bt − 7c = −8t 2 + 8t
a = 8/7 b = −8/7 c = 16/49
ū(t) = 8
7 t2 − 8 16
t +
7 49 .
x
x(t) = Φ(t, c1, c2) + ū(t) = c1e
√ 7t + c2e −√ 7t + 8
7 t2 − 8 16
t +
7 49 .
y(t) = ˙x + 2x − 4t 2 = (2 + √ √ √
7t
7)c1e + (2 − 7)c2e −√7t 12t
− 2
7
− 24
49 .
det(A) = 0
A

F (x, y, ˙y) = 0,
F : A → R A R 3 F ∈ C 2 (A)
∂xF ≡ 0
Ā
∂pF (x, y, p) = 0
p = ϕ(x, y) F (x, y, y ′ ) = 0
y ′ = ϕ(x, y)
∂pF (x, y, p) = 0
F (x, y, p) = 0,
∂pF (x, y, p) = 0,
ψ(x, y) = 0 p
y = y(x)
y = y(x)
F (x, y(x), y ′ (x)) = 0 ∂pF (x, y, y ′ (x)) = 0 ∂pF F
y = y(x) F (x, y(x), ˙y(x)) = 0
(x, y(x), y ′ (x)) ∈ ∂A x
∂pF (x, y, p) = 0 ˙y = p p dx − dy = 0
(dx, dy, dp)
∂xF (x, y, p) dx + ∂yF (x, y, p) dy + ∂pF (x, y, p) dp = 0,
p dx − dy = 0.
∇F x dx = dy
p = 0
p
p = 0
∂xF (x, y, p)
+ ∂yF (x, y, p) dy + ∂pF (x, y, p) dp = 0,
p
F x
y, p V (y, p) = C C ∈ R
F (x, y, p) = 0,
V (y, p) = C.
p
y = f(p, C)
p dx − dy = 0 dy = ∂pf(p, C) dp

⎧
⎪⎨
y = f(p,
C),
∂pf(p, C)
x =
dp =: h(p, C),
⎪⎩
p
F (h(p, C), f(p, C), p) = 0.
p = 0 C ∈ R F (x, C, 0) = 0
y ≡ C
∇F y dy = p dx
(∂xF (x, y, p) + p∂yF (x, y, p)) dx + ∂pF (x, y, p) dp = 0,
F y
x, p V (x, p) = C C ∈ R
F (x, y, p) = 0,
V (x, p) = C.
p
x = f(p, C)
p dx − dy = 0 dx = ∂pf(p, C) dp
⎧
⎪⎨
x =
f(p, C),
y = p ∂pf(p, C) dp =: h(p, C),
⎪⎩
F (f(p, C), h(p, C), p) = 0.
˙y n (x) + Pn−1(x, y) ˙y n−1 (x) + · · · + P1(x, y) ˙y(x) + P0(x, y)y(x) = 0.
λ
Q(λ) := λ n + Pn−1(x, y)λ n−1 + · · · + P0(x, y)λ,
n
n−1
Q(λ) = (λ − Fk(x, y)).
k=0
fk(x, y, Ck) = 0 ˙y(x) = Fk(x, y) k = 0, . . . , n − 1
Ck ∈ R
n−1
k=0
fk(x, y, Ck) = 0.
x = y ′ + ey′
y ′ = p dy = p dx F (x, y, p) := x−p−e p = 0
F R 3 ∂pF = 0
∇F (x, y, p) = (1, 0, −1 − e p )
dx + (−1 − e p ) dp = 0
x(p) = (1 + e p ) dp = p + e p + c, c ∈ R.

F (x, y, p) = 0 c = 0 dy = p dx = p(1 + ep ) dp
y(p) = p(1 + e p ) dp = p2
2 +
pe p = p2
2 + (p − 1)ep + d, d ∈ R.
⎧
⎪⎨
x(p) = (1 + e p ) dp = p + e p ,
⎪⎩ y(p) =
p(1 + e p ) dp = p2
2 +
x = y ′ + log |y ′ |
pe p = p2
2 + (p − 1)ep + d, d ∈ R.
R 3 \{p = 0} y ′ = p dy = p dx
F (x, y, p) := x − p − log |p| = 0 F p = 0
∂pF (x, y, y ′ ) = 0 y ′ = −1 y(x) = −x + c
F (x, y(x), y ′ (x)) = 0
∇F (x, y, p) = (1, 0, −1 − 1/p) dx + (−1 − 1/p) dp = 0
x(p) = (1 + 1/p) dp = p + log |p| + c, c ∈ R.
F (x, y, p) = 0 c = 0 dy = p dx = p(−1 − 1/p) dp
y(p) = p(1 + 1/p) dp = p2
+ p + d, d ∈ R.
2
⎧
⎪⎨
x(p) = p + log |p|,
⎪⎩ y(p) = p2
+ p + d, d ∈ R.
2
p p = −1 ± 1 − 2(d − y)
x = −1 ±
1 − 2(d − y) + log −1 ±
1 − 2(d − y) , d ∈ R.
C = 1 − 2d ∈ R
x = −1 ±
C + 2y + log −1 ±
C + 2y
, C ∈ R.
y = ey′ (y ′ − 1)
F (x, y, p) = −y + e p (p − 1) F R 3
∂pF (x, y, p) = e p (p − 1) + e p = pe p p = 0
∂pF (x, y, y ′ ) = 0 y ′ = 0 y ≡ c c ∈ R F (x, c, 0) = −c − 1
c = −1 y(x) ≡ −1
∇F (x, y, p) = (0, −1, pep ) − dy + (pep ) dp = 0
y(p) = (pe p ) dp = e p (p − 1) + c, c ∈ R.

F (x, y, p) = 0 c = 0 dy = p dx − dy + (pep ) dp = 0
p = 0 p = 0 dx = ep dp x(p) = ep + d d ∈ R
⎧
⎪⎨ y(p) = ep (p − 1),
⎪⎩
x(p) = ep + d, d ∈ R.
p = log(x − d)
y(x) = (x − d)(log(x − d) − 1),
y(x) = −1
y = [y ′ ] 2 + 1 + [y ′ ] 2
F (x, y, p) = −y + p2 + 1 + p2 F R3
p
∂pF (x, y, p) = 2p + = p
1 + p2 p
2 +
1 + p2 p = 0 ∂pF (x, y, y ′ ) = 0 y ′ = 0 y ≡ c c ∈ R
F (x, c, 0) = −c + 1 c = 1 y(x) ≡ 1
∇F (x, y, p) =
p
0, −1, 2p +
1 + p2 ,
− dy +
2p +
p
dp = 0.
1 + p2
y(p) =
p
2p + dp = p
1 + p2 2 + 1 + p2 + c, c ∈ R.
F (x, y, p) = 0 c = 0 dy = p dx − dy + (pep
) dp = 0
p = 0 p = 0 dx =
1
2 +
1 + p2 dp
x(p) = 2p +
1
1 + p 2 dp
1 + p 2 = p + v 1 + p 2 = p 2 + 2pv + v 2
p =
1 − v2
, dp =
2v
−4v2 − 2 + 2v2 4v2 = v2 + 1
dv.
2v2
1
dp =
1 + p2 1
1−v2 v
2v + v
2
+ 1 1
dv =
2v2 v dv = log |v| = log |p − 1 + p2 | + d, dR.
⎧
⎪⎨ y(p) = p2 + 1 + p2 ,
y(x) = 1
⎪⎩
x(p) = 2p + log |p − 1 + p2 | + d, d ∈ R,

y = xy ′ + f(y ′ )
F (x, y, p) = xp − y + f(p) y ′ = p dy = p dx
∂pF (x, y, p) = x + f ′ (p) x = −f ′ (p)
⎧
⎪⎨ y(p) = −f ′ (p)p + f(p),
⎪⎩
x(p) = −f ′ (p),
x + f ′ (p) = 0 ∇F (x, y, p) = (p, −1, x + f ′ (p))
p dx − dy + (x + f ′ (p)) dp = 0.
∇F y dy = p dx (x + f ′ (p)) dp = 0 dp = 0
p = c c ∈ R y = cx + f(c) c ∈ R
y = xf(y ′ ) − g(y ′ ) f(p) = p
f(p) = p
f(p) = p F (x, y, p) = xf(p) − g(p) − y y ′ = p
dy = p dx ∂pF (x, y, p) = xf ′ (p) − g ′ (p)
f ′ (p) = 0 g ′ (p) = 0 f(p) = c1 g(p) = c2
c ∈ R y = c1x − c2 c1, c2 ∈ R f ′ (p) = 0
⎧⎪ ⎨x
=
⎪⎩
g′ (p)
f ′ (p) ,
y = g′ (p)
f ′ f(p) − g(p).
(p)
xf ′ (p)−g ′ (p) = 0 ∇F (x, y, p) = (f(p), −1, xf ′ (p)−g ′ (p))
f(p) dx − dy + (xf ′ (p) − g ′ (p)) dp = 0.
∇F y dy = p dx
(f(p) − p) dx + (xf ′ (p) − g ′ (p)) dp = 0.
f(p) = p
dx
dp + f ′ (p)
f(p) − p x = g′ (p)
f(p) − p ,
x = x(p) x(p, c) c ∈ R
c ∈ R
x = x(p, c),
y = x(p, c) f(p) − g(p).

α ∈ R x ≥ 0
y ′ (x) = 1 − x 2 y 2 (x),
y(0) = α.
α x → +∞
α ∗
y ∗ [0, +∞)
(−∞, 0] R
y ′ = (1 − xy)(1 + xy) y ′
xy = ±1
y ′ > 0
z(x) = −y(−x) z
y y(x) x ≥ 0
x ≤ 0 x ≥ 0
˙x = 1 ˙y = 1 − x 2 y 2
{|xy| > 1, x ≥ 0}
{|xy| > 1, x ≤ 0}
α > 0 x ≥ 0 α > 0 x > 0
(xα, yα) 1/xα = yα > α
x > xα yα
α > 0
x ≥ 0 x > xα
y ′ (x) = 1 − x 2 y(x) 2
x → +∞
y(x) → 0 + x → +∞
α > 0 x ≥ 0 α = 0 y ′ (0) = 1 0
ε > 0 y(ε) > 0 y ′ (ε) > 0
α > 0
α < 0 x ≥ 0
x0 > 0
x0
(xα, yα) xα > x0 yα > 0
0 y ′ < 1 x ≥ 0
y = x + α α < −1
α < −1

y → 0
−∞
−∞
ε = ε(α) > 0
|x| < ε x > ε y ′ < 1 − ε 2 y 2
˙z = 1 − ε 2 z 2
dz
= t + C
(1 − εz)(1 + εz)
1
2ε log
1 + εz
1 − εz = t + C
K ∈ R
1 + εz
= Ke2εt
1 − εz
z(t) = 1 Ke
ε
2εt − 1
Ke2εt + 1
K < 0 z(t) → −∞ t → tk,ε := log(|1/K|)/(2ε)
K < −1 y(x) < z(x) y(x) t → t −
k,ε
z(ε) < −1/ε
[0, +∞[ x > 0
xy = −1 y(α, x)
x y(α, 0) = α α1 < α2 y(α1, x) <
y(α2, x) x
A = {α ∈ R : xα > 0 : y(α, xα) = 1/xα},
[0, +∞[⊂ A α /∈ A α < −1
α + ∈ R α + = inf A
B = {α ∈ R : xα > 0 : y(α, xα) = −1/xα},
] − ∞, −1] ⊂ A α /∈ A α > 0
α − ∈ R α − = sup B
α + /∈ A α + ∈ A y(α + , x)
(x α +, 1/x α +) ¯x > x α +
¯y ¯x 1/¯x y ′ < 1
[0, ¯x] 1
(¯x, 1/¯x) y(α + , ¯x) y(α + , ¯x) > ¯y(¯x)
¯y(0) = ¯α < y(α + , 0) = α + ¯α ∈ A α +
α − /∈ B
α ∈ [α − , α + ] x ≥ 0
α1 < α2 α1, α2 ∈
[α − , α + ] y1(x) y2(x) y1(x) < y2(x) < 0
x
d
dx (y2(x) − y1(x)) = x 2 (y 2 1(x) − y 2 2(x)) > 0

˙y = 1 − x 2 y 2 y(0) = α α ∈ R
y2(x)−y1(x) α2 −α1 > 0
0 x → +∞
α + = α − = α ∗
x ≤ 0 x ≥ 0
R |α| ≤ −α ∗
R
⎧
⎪⎨
y ′ = y 4 −
⎪⎩
y(1) = 0,
x2 ,
(1 + |x|) 2
f(x, y) = y 4 x
−
2
.
(1 + |x|) 2
˙y = 0
|x|
y = ±
1 + |x| ,

x > 0
x
y = ±
1 + x
x < 0
−x
y = ±
1 − x .
z(x) = −y(−x)
z ′ (x) = y ′ (−x) = y 4 x
(x) −
2
(1 + |x|) 2 = z4 x
−
2
(1 + |x| 2 )
R2 \ {(x, y) : f(x, y) = 0}
R− R + Q + Q− R2 \ {(x, y) : F (x, y) = 0}
(−1, 0) (1, 0) (0, 1) (0, −1) R ±
R +
[1, +∞[
t > 1 R +
{(x, y) : y ≥ −1} ℓ +∞
t → +∞ −1
t > 1 t < 1 t = 1
0 < ¯t < 1 0 < M < 1
R− R−
R −∞
1 R−
−∞ 1
(0, 01303, 0, 113630)
(−0, 01303, 0, 113627) δy/δx
1, 1 · 10−4

y ′ = y 4 −
x2 y(1) = 0
(1 + |x|) 2

[0, π]
utt + 2ut − uxx = 0, ux(0, t) = ux(π, t) = 0,
u(x, 0) = 0 ut(x, 0) = x
u(x, t) = T (t)X(x)
¨T (t)X(x) + 2 ˙ T (t)X(x) − T (t) ¨ X(x) = 0
T (t)X(x)
¨T (t) + 2 ˙ T (t)
−
T (t)
¨ X(x)
= 0,
X(x)
¨X(x) − λX(x) = 0
¨T (t) + 2 ˙ T (t) − λT (t) = 0,
ux(0, t) = T (t) ˙ X(0) = 0 ux(π, t) = T (t) ˙ X(π) = 0 ˙ X(0) = ˙ X(π) = 0
¨X(x) − λX(x) = 0
˙X(0) = ˙ X(π) = 0,
λ ∈ R µ 2 = λ
λ > 0
√
λ x
X(x) = c1e + c2e −√λ x
, c1, c2 ∈ R
√ √ √
λ x −
λe − c2 λe √ λ x
, c1, c2 ∈ R
˙X(x) = c1
0 = ˙ X(0) = (c1 − c2) √ λ c1 = c2 0 = ˙ X(π) =
√ √
c1 λ(e λπ − e− √ λπ ) c1 = 0
λ = 0 X(x) = c1 + c2x c1, c2 ∈ R ˙ X(x) = c2 c2 = 0
X(x) = c1 ∈ R \ {0}
λ < 0 ω = |λ| X(x) = c1 cos(ωx) + c2 sin(ωx) c1, c2 ∈ R
˙ X(x) = −c1ω sin(ωx) + c2ω cos(ωx) 0 = ˙ X(0) = c2ω c2 = 0
0 = ˙ X(π) = −c1ω sin(ωx) ω ∈ Z λ = −n 2 n ∈ N n = 0
X(x) λ = −n 2 n ∈ N Xn(x) =
cn cos(nx) λ = n = 0 T (t)
¨T (t) + 2 ˙
T (t) + n 2 T (t) = 0,
T (0) = 0.

µ 2 + 2µ + n 2 = 0 ∆ = 4(1 − n 2 )
n ∈ N
n = 0 ∆ > 0 µ1 = 0 µ2 = −2 T (t) = d1+d2e −2t
d1, d2 ∈ R T (0) = 0 d1 = −d2
T0(t) = d1(1 − e −2t )
n = 1 ∆ = 0 µ = −1 T (t) =
d1e −t + d2te −t d1, d2 ∈ R T (0) = 0 d1 = 0
T1(t) = d2te t
n > 1 ∆ < 0 µ1 = −1 + i √ n 2 − 1 µ2 = −1 −
i √ n 2 − 1 T (t) = d1e −t cos( √ n 2 − 1 t) + d2e −t sin( √ n 2 − 1 t)
T (0) = 0 d1 = 0 Tn(t) = dne −t sin( √ n 2 − 1 t)
un(x, t) = Tn(t)Xn(x)
u0(x, t) = d0(1 − e −2t )c0 = a0(1 − e −2t )
u1(x, t) = d1te −t c1 cos x = a1te −t cos x
un(x, t) = dne −t sin( n 2 − 1 t) cn cos(nx) = ane −t sin( n 2 − 1 t) cos(nx).
t 0
u(x, t) =
x = ∂tu(x, 0) =
n=0
∂tu0(x, 0) = 2a0
∂tu1(x, 0) = a1 cos x
∂tun(x, 0) = an n2 − 1 cos(nx).
∞
un(x, t) t t = 0
n=0
∞
∞
∂tun(x, 0) = 2a0 + a1 cos x + an n2 − 1 cos(nx)
f(x) = x [0, π]
[−π, π] 2π R n > 1
1
2π
π
0
x dx = π
2
π 2
x cos(nx) dx =
π 0
2
π
|x| ≤ π
x = π 2
−
2 π
∞
n=1
(1 − (−1) n )
n 2
n=2
x sin(nx)
π −
n 0
2
π
sin(nx) dx = −
nπ 0
2(1 − (−1)n )
πn2 x = 2a0 + a1 cos x +
cos(nx) = π 4 2
− cos x −
2 π π
∞
n=2
k=1
∞ 1 − (−1) n
n2 cos(nx),
n=1
an n2 − 1 cos(nx).
a0 = π
4 a1 = − 4
π a2k = 0 a2k+1 = − 4 1
k ∈ N k ≤ 1
π 4k(1 + k)(2k + 1) 2
u(x, t) = π
4 (1 − e−2t ) − 4
π te−t cos x − 4
∞ e
π
−t sin( 4k(1 + k) t)
cos((2k + 1)x).
4k(1 + k)(2k + 1) 2

e
|uk(x, t)| = C
−t sin(
4k(1 + k) t)
cos((2k + 1)x)
4k(1 + k)(2k + 1) 2 ≤
C C
=
2k · 4k2 k3 u
x
e
|∂xuk(x, t)| = C1
−t sin(
4k(1 + k) t)
1
(− sin((2k + 1)x)) ≤ C1
4k(1 + k)(2k + 1) 2k · 2k
≤ C1
k 2
∂xu
x
∞
∂ 2 ∞ e
xxuk(x, t) =
−t sin( 4k(1 + k) t)
− cos((2k + 1)x).
4k(1 + k)
k=1
k=1
1/k L 2
t
e
|∂tuk(x, t)| ≤ |uk(t, x)| +
−t cos( 4k(1 + k) t)
(2k + 1) 2
cos((2k + 1)x)
≤
1 1
+
(2k + 1) 2 k2 ∂tu
t
∞ ∂2 2
∂t
e−t sin(
4k(1 + k) t)
cos((2k + 1)x).
4k(1 + k)(2k + 1) 2
k=1
bk(t)
|∂ 2
ttbk(t)| ≤ |bk(t)| +
∂t
e−t cos( 4k(1 + k) t)
(2k + 1) 2
cos((2k + 1)x)
1 1
e
≤ + +
(2k + 1) 2 k2
−t cos( 4k(1 + k) t)
(2k + 1) 2
cos((2k + 1)x)
+
e
+
−t cos( 4k(1 + k) t)
(2k + 1) 2
4k(1 + k) cos((2k + 1)x)]
2 1 4k(k + 1) 2 1
≤ + +
≤ + 2
(2k + 1) 2 k2 (2k + 1) 2 k2 k + 1
L 2
L 2
[0, π]
ut − uxx − 2ux − u = 0, x ∈ [0, π], t > 0
u(0, t) = u(π, t) = 0 ∀ t > 0
u(x, 0) = x(π − x)e −x 0 ≤ x ≤ π

u(t, x) = T (t)X(x)
˙
T (t)X(x) − T (t) ¨ X(x) − 2T (t) ˙ X(x) − T (t)X(x) = 0,
u(t, x) = T (t)X(x) = 0 (t, x)
T ˙ (t)
T (t) − ¨ X(x)
X(x) − 2 ˙ X(x)
− 1 = 0,
X(x)
T ˙ (t)
T (t) = ¨ X(x) + 2 ˙ X(x)
+ 1 = λ ∈ R,
X(x)
T ˙ (t) = λT (t)
¨X(x) + 2 ˙ X(x) + (1 − λ)X(x) = 0.
u(0, t) = u(π, t) = 0 t > 0 X(0) = X(π) = 0
λ ∈ R
¨X(x) + 2 ˙ X(x) + (1 − λ)X(x) = 0
X(0) = X(π) = 0.
µ 2 + 2µ + 1 − λ = 0
µ1 = −1 − 1 − (1 − λ) = −1 − √ λ, µ2 = −1 + √ λ.
λ > 0 c1, c2 ∈ R
X(x) = c1e µ1t + c2e µ2t .
X(0) = 0 c1 + c2 = 0 X(π) = 0
0 = c1(e µ1π − e µ2π ) µ1 = µ2 c1 = c2 = 0
λ > 0
λ = 0 µ1 = µ2 = −1 c1, c2 ∈ R
X(x) = c1e −t + c2te −t .
X(0) = 0 c1 = 0 X(π) = 0
c2πe −π = 0 c2 = 0
λ < 0 ω = |λ| λ < 0
µ1 = −1 − iω µ2 = −1 + iω c1, c2 ∈ C
d1, d2 ∈ R
X(x) = c1e −t e −iωx + c2e −x e iωx = e −x c1e −iωx + c2e iωx = e −x (d1 cos ωx + d2 sin ωx) .
X(0) = 0 d1 = 0 X(π) = 0
d2 sin πω = 0 d2 = 0
ω = n ∈ N \ {0}
λ = −n 2 n ∈ N \ {0}
¨Xn(x) + 2 ˙ Xn(x) + (1 − n 2 )Xn(x) = 0
Xn(0) = X(π) = 0.
Xn(x) = dne−x sin nx dn ∈ R ˙ Un(t) = −n2T (t)
Tn(t) = Tn(0)e−n2t un(t, x) = Tn(t)Xn(x) n
un(0, t) = un(π, t) = 0 t > 0
bn = Tn(0)dn ∈ R n ∈ N\{0} un(t, x) = bne−n2te−x sin nx
bn
∞
un(x, 0) = u(x, 0)
n=1

∞
n=1
bne −x sin nx = x(π−x)e −x x(π−x) =
k=0
∞
bn sin nx bn
x(π −x) [−π, π]
R 2π
bn = 2
π
x(π − x) sin nx dx =
π 0
2
π
(−x
π 0
2 + πx) sin nx dx
= 2
cos nx
−
π n (−x2 π + πx) +
0
2
π
cos nx(−2x + π) dx =
πn 0
2
π
cos nx(−2x + π) dx
πn 0
= 2
π sin nx
(−2x + π) +
πn n 0
4
πn2 π
sin nx dx =
0
4
πn2 π
sin nx dx
0
= 4
πn2
cos nx
π − =
n 0
4
πn3 (1 − (−1)n )
b2k = 0 b2k+1 = 8/(π(2k + 1) 3 ) k ∈ N
u(t, x) = 8
∞ 1
π (2k + 1) 3 e−(2k+1)2 t−x
sin (2k + 1)x .
t ≥ 0 x ∈ [0, π]
1
(2k
+ 1) 3 e−(2k+1)2 t−x
sin (2k + 1)x
1
≤
(2k + 1) 3
∞
1
(2k
+ 1) 3 e−(2k+1)2 t−x
sin (2k + 1)x
∞
≤
sup
t>0
k=0
x∈[0,π]
n=1
k=0
e−(2k+1)2t < +∞,
(2k + 1) 3
e −(2k+1)2 t < 1 t > 0
x t p(k)e −(2k+1)2 t−x sin (2k + 1)x
p(k) k k
|p(k)e −(2k+1)2 t−x | ≤ 1
k2

[0, π]
ut − 5uxx = 0, 0 ≤ x ≤ π , t > 0 ,
ux(0, t) = ux(π, t) = 0
u(x, 0) = 2x
u(t, x) = T (t)X(x)
˙ T (t)X(x) − 5T (t) ¨ X(x) = 0 5T (t)X(x)
T ˙ (t)
5T (t) = ¨ X(x)
= λ ∈ R.
X(x)
˙
T (t) = 5λT (t),
¨X(x) = λX(x),
ux(0, t) = T (t) ˙ X(0) = 0 ux(π, t) = T (t) ˙ X(π) = 0
˙X(0) = ˙ X(π) = 0 ¨X(x) − λX(x) = 0,
˙X(0) = ˙ X(π) = 0
µ 2−λ = 0
λ > 0 µ = ± √ λ
X(x) = c1e
c1, c2 ∈ R
√
˙X(x) = c1 λe
√ λ x + c2e −√ λ x ,
√ √
λ x −
− c2 λe √ λ x
,
0 c1 = c2
π ˙ X(π) = 0 c1 = c2 = 0
λ = 0 X(x) = c1 + c2x c1, c2 ∈ R
˙X(x) = c1 c2 = 0 X(x) = c1
c1 = 0
λ < 0 ω = |λ| X(x) = c1 cos(ωx) + c2 sin(ωx)
˙ X(x) = −ωc1 sin(ωx) + ωc2 cos(ωx) 0 π
c2 = 0 sin(ωπ) = 0 ω = |λ| ∈ Z
λ = −n 2 n ∈ N Xn
λ = −n 2 Xn(x) = c1 cos (nx)
n = 0

T ˙
Tn(t) = −5n 2 Tn(t) Tn(t) = Tn(0)e −5n2 t
n ∈ N
un(t, x) = Tn(t)Xn(x) = ane −5n2 t cos (nx) ,
an = T (0)c1 an ∈ R \ {0}
u(0, x) = 2x =
∞
∞
un(0, x) = a0 + cos (nx) ,
j=0
an
f(x) = 2x a0 0 f
f [−π, π] 2π R
a0 = 1
π
an = 2
π
π
0
π
0
2x dx = π
2x cos nx dx = 4
π
x
= 4
n2π [cos(nx)]π0 = 4 (−1)n − 1
n2π n=1
π sin nx
−
n 0
4
π
sin(nx) dx
nπ 0
a0 = π a2k = 0 a2k−1 = −8/(π(2k − 1) 2 ) k ∈ N k ≥ 1
u(t, x) = π − 8
∞ e
π
−5(2k−1)2
t cos (2k − 1)x
(2k − 1) 2 .
∞
e
sup
−5(2k−1)2
t cos (2k − 1)x
(2k − 1) 2
≤
∞ 1
< +∞,
(2k − 1) 2
k=1
k=1
1/k 2
⎧
−ut + 2uxx + 3ux + u = 0, t > 0, x ∈]0, π[,
⎪⎨
u(0, t) = u(π, t) = 0,
⎪⎩
3
−
u(x, 0) = e 4 x π
2 −
x − π
,
2
u(x, t) = T (t)X(x)
− ˙ T (t)X(x) + 2T (t) ¨ X(x) + 3T (t) ˙ X(x) + T (t)X(x) = 0
T (t)X(x)
k=1
− ˙ T (t) + T (t)
= −
T (t)
2 ¨ X(x) + 3 ˙ X(x)
X(x)
λ ∈ R
− ˙ T (t) + (1 − λ)T (t) = 0,
2 ¨ X(x) + 3 ˙ X(x) + λX(x) = 0.

λ ∈ R X(x)
λ ∈ R 2µ 2 + 3µ + λ = 0 ∆ = 9 − 8λ
u(0, t) = T (t)X(0) = 0 t u(π, t) = T (t)X(π) = 0 t X(0) = X(π) = 0
∆ > 0 λ1 λ2 X(x)
X(x) = c1e λ1x + c2e λ2x 0 = c1 + c2
X(x) = c1(e λ1x − e λ2x ). X(π) = 0 0 = c1(e λ1π − e λ2π )
λ1 = λ2 c1 = c2 = 0
∆ = 0 λ1 X(x)
X(x) = c1e λ1x + c2xe λ1x
c1 = 0 0 = c2πe λ1π c2 = 0
∆ < 0 λ1 = α + iω
λ2 = α − iω α = −3/4 ω = |∆|/4 = 0 X(x) =
e αx (c1 cos ωx + c2 sin ωx) c1 = 0 0 = c2e απ sin ωπ
ω ∈ Z \ {0} 4ω = |∆| ω = n ∈ N \ {0}
16n 2 = −∆ ∆ < 0 16n 2 = −9 + 8λn λn = (2n 2 + 9/8)
n ∈ N λn
Xn(x) = cne −3/4x sin nx.
T (t) − ˙
T (t) = (λ − 1)T (t) T (t) =
T (0)e −(λ−1)t λ λn Tn(t) =
Tn(0)e −(2n2 +1/8)t bn = Tn(0)cn
un(x, t) = bne −(2n2 +1/8)t e −3/4x sin nx.
f(x) =
∞
un(x, 0),
n=1
π
2 −
x − π
=
2
∞
bn sin nx.
bn
x ↦→ π
2 −
x − π
2
[0, π] [−π, π] 2π R
bn
bn = 2
π
π
π 2 −
x − π
sin nx dx
2
0
π/2
= 2
x sin nx dx + (π − x) sin nx dx
π 0
π/2
= 4
sin n
πn2 π
.
2
u(x, t) = 4
∞ sin
π
n π
2
n2 e −(2n2 +1/8)t −3/4x
e sin(nx).
n=1
∞
sin
sup
n π
2
n2 n=1 (x,t)∈[0,π]×[0,+∞[
n=1
π
e −(2n2 +1/8)t e −3/4x sin(nx)
≤
∞
n=1
1
< ∞,
n2

⎧
−utt + 3uxx = 0 ]0, π[×]0, +∞[
⎪⎨
ux(0, t) = ux(π, t) = 0
u(x, 0) = 0
⎪⎩
ut(x, 0) = x.
u(x, t) = T (t)X(x) ˙ X(0) = ˙ X(π) = 0
λ ∈ R
− ¨ T (t)X(x) + 3T (t) ¨ X(x) = 0,
T (t)X(x)
− ¨ T (t)
T (t) = −3 ¨ X(x)
= λ.
X(x)
− ¨ T (t) − λT (t) = 0,
3 ¨ X(x) + λX(x) = 0.
X(x) X(0) = X(π) = 0
3µ 2 + λ = 0 ∆ = −12λ ∆ > 0 λ < 0
X(x) = c1e µ1x +
c2e µ2x ˙ X(x) = c1µ1e µ1x + c2µ2e µ2x ˙ X(0) = 0 ˙ X(π) = 0
c1 c2
c1µ1 + c2µ2 = 0
c1µ1e µ1π + c2µ2e µ2π = 0
µ1µ2(e µ2π − e µ1π ) = 0 c1 = c2 = 0
∆ = 0 λ = 0 µ1 = 0
X(x) = c1 + c2x ˙ X(x) = c2
c2 = 0 X(x) = c1 c1 ∈ R \ {0}
∆ < 0 λ > 0 ±iω ω =
λ/3 X(x) = c1 cos(ωx)+c2 sin(ωx) ˙ X(x) = −ωc1 sin(ωx)+
ωc2 cos(ωx) ˙ X(0) = 0 c2 = 0 X(x) = c1 cos(ωx) ˙ X(π) = 0
ω ∈ N ∆ < 0 ω = 0 λ = 3n 2 n = 0
Xn(x) = cn cos(nx) n ∈ N ∆ ≤ 0
T λ = 3n 2 n ∈ N ¨ T (t) + 3n 2 T (t) = 0
T (0) = 0 n = 0 U(t) = c1 + c2t
U(t) = c2t n = 0 µ 2 + 3n 2 = 0
µ1 = in √ 3 µ2 = −in √ 3
T (t) = c1 cos(n √ 3t) + c2 sin(n √ 3t) T (0) = 0 c1 = 0 Tn(t) = dn sin(n √ 3t)
kn = cndn
u0(x, t) = U(t)X(x) = k0t
un(x, t) = U(t)X(x) = kn sin(n √ 3t) cos(nx)
∂tu0(x, 0) = k0
∂tun(x, 0) = n √ 3kn cos(nx)

a0 = 1
π
an = 2
π
π
0
π
0
x = π
2
x cos(nx) dx = 2
π
x sin(nx)
n
x = π 2
−
2 π
n=1
∂tu(x, 0) = k0 +
k0 = π/2 n √ 3kn = − 2
π
u(x, t) = π
2 t − 2√3 3π
= π
2 t − 4√3 3π
k=0
x=π
x=0
− 2
π
sin(nx) dx = −
nπ 0
2
n2π (1 − (−1)n ),
∞ 1 − (−1) n
n2 cos(nx).
1−(−1) n
n 2
∞
n √ 3kn cos(nx)
n=1
kn = − 2√ 3
3π
1−(−1) n
n3
∞ 1 − (−1)
n=1
n
n3 sin(n √ 3t) cos(nx)
∞ 1
(2k + 1) 3 sin((2k + 1)√3t) cos((2k + 1)x).
1/n 3
⎧
⎪⎨ −∂tu(t, x) + ∂xxu(t, x) + 2∂xu(t, x) − 3u = 0, (t, x) ∈]0, +∞[×[0, π],
u(t, 0) = u(0, π) = 0,
⎪⎩
u(0, x) = xe−x .
u(t, x) = T (t)X(x)
− ˙
T (t, x) + T (t) ¨ X(x) + 2T (t) ˙ X(x) − 3T (t)X(x) = 0
U(t)X(x) λ ∈ R
−T ˙ (t, x) − 3T (t)
− =
T (t)
¨ X(x) + 2 ˙ X(x)
=: λ
X(x)
λ ∈ R
¨X(x) + 2X(x) ˙ − λX(x) = 0
− ˙ T (t) + (−3 + λ) T (t) = 0.
X X(0) = X(π) = 0
µ 2 + 2µ − λ = 0 ∆ = 4(1 + λ)
λ ∈ R
∆ > 0 µ1 = −2−√∆ 2 µ2 = −2+√∆ 2
Φ(c1, c2, x) = c1e µ1x +c2e µ2x Φ(c1, c2, 0) =
X(0) = 0 c1 = −c2 Φ(c1, c2, π) = X(π) = 0
c1(e µ1π − e µ2π ) = 0 ∆ > 0 µ1 = µ2
c1 = c2 = 0

∆ = 0 λ = −1 µ1 = µ2 = −1
Φ(c1, c2, x) = c1e−x + c2xe−x
c1 = c2 = 0
√
|∆|
∆ < 0 α = −1 ω = 2 Φ(c1, c2, x) =
e−x (c1 cos ωx + c2 sin ωx)
c1 = 0 Φ(c1, c2, 0) =
X(0) = 0 Φ(c1, c2, π) = X(π) = 0 ω = n ∈ Z
ω ∆ ∆ < 0
−4n2 = 4(1 + λ) λ λn = −1 − n2
Xn(x) = cne −x sin(nx).
U λ
− ˙ T (t) + −3 − 1 − n 2 T (t) = 0.
Tn(t) = un(0)e −(4+n2 )t
X U λn
bn = cnun(0)
un(t, x) = bne −(4+n2 )t e −x sin(nx).
u(0, x) = xe −x ∞
= bne −x sin(nx) = e −x
∞
bn sin(nx),
n=1
bn x
x ∈ [0, π] [−π, π] 2π R
bn = 2
π
π
= 2
π
0
−π
x sin nx dx = 2
−
π
cos nπ
n
u(t, x) = e −x
x=π
x cos(nx)
n x=0
1
+ sin(nπ) =
n2 2(−1)n+1
n
∞
n=1
n=1
2(−1) n+1
e −(4+n2 )t
sin(nx).
n
+ 1
π
cos nx dx
n 0

f : R 2 → R
D := {(x, y) ∈ R 2 : y ≥ |x|}
f(x, y) =
x 2 − y 4
x 2 + sin 2 x + y .
lim f(x, y), lim
(x,y)→(0,0)
(x,y)→(0,0)
(x,y)∈D
f(x, y)
(t, 0) t → 0 ±
lim f(t, 0) = lim
t→0 ± t→0 ±
t2 t2 + sin2 1
=
t 2 .
(0, t) t → 0 ±
lim f(0, t) = lim
t→0 ± t→0 ±
−t4 t
(x, y) ∈ D f(x, y) ≤ 0
0
f(x, y) ≥
−y 4
x 2 + sin 2 x + y
= 0.
≥ −y3
F (x, y) = (x 2 + y 2 ) 2 − (x 2 − y 2 ),
¯ B = {(x, y) ∈ R 2 : x 2 + y 2 ≤ 1}
∂xF (x, y) = 4x x 2 + y 2 − 2x
∂yF (x, y) = 4y x 2 + y 2 + 2y.
⎧
⎪⎨ ∂xxF (x, y) = −2 + 12x2 + 4y2 ∂yyF (x, y) = 2 + 4x 2 + 12y 2
⎪⎩
∂xyF (x, y) = 8xy
−2 2
∂B = {(x, y) : x 2 + y 2 = 1}

x = cos θ y = sin θ F
F (cos θ, sin θ) = 1 − (cos 2 θ − sin 2 θ) = 1 − cos 2θ = 2 sin 2 θ
θ = π/2, 3π/2 2
(0, ±1)
θ = 0, π 0 (±1, 0)
⎧
⎪⎨ ∂x(F (x, y) − λ(x
⎪⎩
2 + y2 − 1)) = 4x x2 + y2 − 2x − 2λx = 0
∂y(F (x, y) − λ(x2 + y2 − 1)) = 4y x2 + y2 + 2y − 2λy = 0
x 2 + y 2 = 1.
⎧
⎪⎨ 0 = 2(1 − λ)x
0
⎪⎩
= 2(3 − λ)y
1 = x 2 + y 2 .
x = 0 y = ±1 y = 0 x = ±1 λ = 1 y = 0
x = ±1 λ = 3 x = 0 y = ±1 (0, ±1)
(±1, 0) F (0, ±1) = 2 F (±1, 0) = 0
α > 0
∞
n=1
fn(x, y, α) =
xy
n2 .
+ |xy| α
xy
n2 .
+ |xy| α
K R 2
1
|fn(x, y, α)| ≤ max |xy| ·
(x,y)∈K n2 |xy| K
R 2
R 2
s = |xy|
|fn(x, y, α)| = gn(s, α)
gn(s, α) =
s
n 2 + s α
g ′ n(s, α) = n2 + (1 − α)s α
(n 2 + s α ) 2
s > 0 ¯sn = n 2/α /(α − 1) 1/α
|gn(s, α)| ≤
n 2/α
(α−1) 1/α
n 2 + n2
α−1
= (α − 1)1−1/α
α
n2/α (α − 1)1−1/α
=
n2 α
1
n 2−2/α
α > 2
R 2 α > 2

α < 2 N n=1 fn(x,
y, α)
N∈N
R2 ε > 0 N = Nε N ′ , M ′ ≥ N
M
′
N
fn(x, y, α) −
′
fn(x, y, α)
n=1
n=1
N > Nε M = 2N {(xN, yN)}N∈N xN, yN > 0
xNyN = N
M
′
N
fn(x, y, α) −
′
2N
N
fn(x, y, α) ≥ fn(xN, yN, α) − fn(xN, yN, α)
n=1
n=1
∞
n=1
2N
≥
n=N+1
2N
≥
∞
≤ ε.
n=1
fn(xN, yN, α) ≥
n=N+1
2N
n=N
N
n 2 + N α
N
4N 2 N(N − 1)
=
+ N α 4N 2 + N α
1/4 α < 2 1/5 α = 2
ε = 1/6 > 0
R 2 α ≤ 2
R 4
x 4 + 5y 2 + tan z − z − 6 sin πt = 0
x 2 − 2y 4 + cos z − z − arctan t − t = 1
x = y = z = t = 0 (0, 0, 0, 0)
R 3 z = z(x, y) t = t(x, y) z(0, 0) = 0 t(0, 0) = 0
∇z(0, 0) ∇t(0, 0)
F (x, y, z, t) =
x 4 + 5y 2 + tan z − z − 6 sin πt
x 2 − 2y 4 + cos z − z − arctan t − t − 1
F (0, 0, 0, 0) = (0, 0) F (0, 0, 0, 0)
Jac(F ) =
(0, 0, 0, 0)
4x 3 10y −1 + 1/ cos 2 z −6π cos(πt)
2x −8y 3 −1 − sin z −1 − 1/(1 + t 2 )
Jac(F )(0, 0, 0, 0) =
0 0 0 −6π
0 0 −1 −2
F (z, t)
F (x, y)
∂z,tF (0, 0) =
0 −6π
0 − 1 −2
∇z(0, 0)
∇t(0, 0)
,
.
,
, [∂z,tF (0, 0)] −1
1/(3π) −1
=
−1/(6π) 0
= −[∂z,tF (0, 0)] −1 ∂x,yF (0, 0) =
0 0
0 0
.
.

I := arctan y dx − xy dy
γ
γ T (1, 0) (0, 1) (0, −1)
T
M := x 2 dx dy.
T
T = {(x, y) ∈ R 2 : −1 ≤ y ≤ 1, 0 ≤ x ≤ 1 − |y|}
γ = {(t, t − 1) : t ∈ [0, 1]} ∪ {(1 − t, t) : t ∈ [0, 1]} ∪ {(−t, 0) : t ∈ [1, −1]}
P (x, y) dx + Q(x, y) dy = arctan y dx − xy dy,
T
I = − (∂xQ − ∂yP ) dxdy = − y −
T
T
1
1 + y2
dx dy
1 1−|y|
= −
y − 1
1 + y2
dx dy
= −
= 2
−1
1
−1
1
0
0
y(1 − |y|) dy +
1 − y
1 + y 2
1
−1
1 − |y|
dy
1 + y2 = [2 arctan y − log(1 + y 2 )] y=1 π
y=0 =
M := x 2
dx dy =
Q(x, y) = x3 /3 P = 0
x
M = Q(x, y) dy =
γ
γ
3
dy =
3
= 1 1 1
+ =
12 12 6 .
M = x 2 dx dy =
= 1
3
= 2
3
T
T
1
−1
1
0
1
0
1
(1 − |y|) 3 dy
−1
T
t 3
3
(1 − y) 3 dy = 1
6 .
− log 2.
2
(∂xQ − ∂yP ) dxdy,
1 (1 − t)
dt +
0
3 1
dt − 0 dt
3
−1
1−|y|
x 2
dx dy
Γ = {(x, y) ∈ R 2 : (x 2 + y 2 + 12x + 9) 2 = 4(2x + 3) 3 },
0

cos(3θ) = cos θ(1 − 4 sin 2 θ) = cos θ(cos 2 θ − 3 sin 2 θ)
Γ Γ
2π
3
Γ P1 = (−1, 0) P2 = (1/2, √ 3/2)
P3 = (1/2, − √ 3/2)
Γ P0 y = ϕ1(x) ϕ1(−1) = 0
ϕ ′ 1 (0)
√ Γ P1 y = ϕ2(x) ϕ2(1/2) =
3/2 ϕ ′
2 (1/2)
Γ P2 y = ϕ3(x) ϕ3(1/2) =
− √ 3/2 ϕ ′ 3 (1/2)
h(x, y) = x2 + y2 Γ Γ
(3, 0) Γ
F (x, y) = (x 2 + y 2 + 12x + 9) 2 − 4(2x + 3) 3 .
F (x, y) =
F (x, −y)
cos 2 θ + sin 2 θ = 1
cos 3θ = cos(θ + 2θ) = cos θ cos 2θ − 2 sin 2 θ cos θ
= cos θ(1 − 2 sin 2 θ) − 2 sin 2 θ cos θ = cos θ(1 − 4 sin 2 θ)
F (ρ cos θ, ρ sin θ) = ρ 4 (sin 4 (θ) + cos 4 (θ) + 2 sin 2 (θ) cos 2 (θ))+
+ ρ 3 (−8 cos 3 (θ) + 24 sin 2 (θ) cos(θ)) + 18ρ 2 − 27
= ρ 4 (sin 2 (θ) + cos 2 (θ)) 2 − 8ρ 3 cos θ(cos 2 (θ) − 3 sin 2 (θ)) + 18ρ 2 − 27
= ρ 4 − 8ρ 3 cos(3θ) + 18ρ 2 − 27
θ 2π/3
2π/3 (0, 0) /∈ Γ
f(ρ) := ρ4 + 18ρ 2 − 27
8ρ 3
= cos(3θ)
Γ
Γ (ρn, θn) ρn → +∞
lim sup
+∞
n→∞
2π/3
∂yF (0, −1) = 0 ∂xF (0, −1) = 0
(0, −1) x = −1
α = ±2π/3
− 1
x ±
2
√
3
y = −1
2
√ 3 P3 − √ 3 P2

f(ρ)
f ′ (ρ) = 4ρ3 + 36ρ
8ρ 3
− 3 ρ 4 + 18ρ 2 − 27
8ρ 4
= (−9 + ρ2 ) 2
8r 4
ρ = 3 f ρ
cos 3θ = −1 θ1 = π/3 θ2 = 5π/3 θ3 = π cos 3θ = 1
θ = 0 θ = 2/3π θ = 4π/3 ρ f(ρ) = −1
(ρmin, π)
ρ f(ρ) = 1
(ρmax, 0) f(ρ) = −1
ρ 4 + 8ρ 3 + 18ρ 2 − 27 = 0
27 ±1, ±3, ±9
1
ρ 4 + 8ρ 3 + 18ρ 2 − 27 = (ρ − 1)(ρ 3 + bρ 2 + cρ + d) = ρ 4 + (b − 1)ρ 3 + (c − b)ρ 2 + (d − c)ρ − d)
b = 9 c = 27 d = 27
≥ 0,
ρ 4 + 8ρ 3 + 18ρ 2 − 27 = (ρ − 1)(ρ 3 + 9ρ 2 + 27ρ + 27) = (ρ − 1)(ρ + 3) 3
ρ = 1 ρ
f(ρ) = +1
ρ 4 − 8ρ 3 + 18ρ 2 − 27 = 0
27 ±1, ±3, ±9
−1
ρ 4 − 8ρ 3 + 18ρ 2 − 27 = (ρ + 1)(ρ 3 + bρ 2 + cρ + d) = ρ 4 + (b + 1)ρ 3 + (c + b)ρ 2 + (d + c)ρ + d)
b = −9 c = 27 d = −27
ρ 4 + 8ρ 3 + 18ρ 2 − 27 = (ρ + 1)(ρ 3 − 9ρ 2 + 27ρ − 27) = (ρ − 1)(ρ − 3) 3 ,
ρ = 3 ρ
ρ 2 9
(3, 0) ∂xF (3, 0) =
∂yF (3, 0) = 0 (3, 0) 2π/3
0
R 3 S
ϕ(θ, y) = ( y 2 + 1 cos θ, y, y 2 + 1 sin θ), θ ∈ [0, 2π], |y| < 1,
F : R 3 → R 3 F (x, y, z) = (x 2 , y/2, x)
F F
F
√ 2 (0, 1, 0) y = 1
γ(θ) = ( √ 2 cos θ, 1, √ 2 sin θ), θ ∈ [0, 2π].
ϕ 2
ϕ
(1, 0, 0)
F S
ϕ(θ, y) = (ϕ1, ϕ2, ϕ3) F = (F1, F2, F3)

div F (x, y, z) = ∂xF1 + ∂yF2 + ∂zF3 = 2x + 1/2,
⎛
⎞
rot F = det ⎝ e1 ∂x x2 e2 ∂y y/2 ⎠ = (0, −1, 0).
e3 ∂z x
D =
{(x, 1, z) : x2 + z2 ≤ 2} (0, −1, 0) (0, −1, 0) D
γ
rot
F · ˆn dσ = dσ = Area(D) = 2π.
D
2π
F dγ = F ( √ 2 cos θ, 1, √ 2 sin θ) · (− √ 2 sin θ, 0, √ 2 cos θ) dθ
γ
=
0
2π
0
D
− 2 3/2 cos 2 θ sin θ + 2 cos 2
θ dθ = 2π
F

Jac ϕ(θ, y) =
⎛
⎜
⎝
− y 2 + 1 sin θ
y cos θ
√ y 2 +1
0 1
y2 + 1 cos θ
√y sin θ
y2 +1
ω2 = det 2 B1 + det 2 B2 + det 2 B3
B1 =
− y 2 + 1 sin θ
⎛
y cos θ
√ y 2 +1
0 1
B2 = ⎝ −y2 + 1 sin θ
y2 + 1 cos θ
B3 =
0 1
y2 + 1 cos θ
√y cos θ
y2 +1
√y sin θ
y2 +1
√y sin θ
y2 +1
⎞
⎞
⎟
⎠ .
, det 2 B1 = (y 2 + 1) sin 2 θ.
⎠ , det 2 B2 = y 2 ,
, det 2 B3 = (y 2 + 1) cos 2 θ.
ω2 = 2y2 + 1
ϕ
(1, 0, 0) = ϕ(0, 0) (0, 1, 0) (0, 0, 1)
(±1, 0, 0)
⎛
det ⎝
±1 0 0
0 0 1
0 1 0
⎞
= ⎠ = ∓1.
(1, 0, 0) (−1, 0, 0)
Φ(S, ⎛
2π F1 ◦ ϕ −
1 ⎜
F ) = det ⎜
⎝
0 −1
y2 y cos θ
+ 1 sin θ √
y2 +1
F2 ◦ ϕ 0 1
F3 ◦ ϕ y2 ⎞
⎟
⎠ dy dθ
y sin θ
+ 1 cos θ √
y2 +1
⎛
2π (y
1 ⎜
= det ⎜
⎝
0 −1
2 + 1) cos2 θ − y2 ⎞
y cos θ
+ 1 sin θ √
y2 +1 ⎟
y/2 0 1 ⎟
⎠ dy dθ
y2 + 1 cos θ y2 y sin θ
+ 1 cos θ
=
=
2π 1
0
2π
+
−1
1
0 −1
2π 1
0
= 2
3 π.
−1
⎛
(−y/2)det ⎝ −y2 + 1 sin θ
y2 + 1 cos θ
√y cos θ
y2 +1
√y sin θ
y2 +1
⎞
√ y 2 +1
⎠ dy dθ+
(y2 + 1) cos2 θ − y2 + 1 sin θ
(−1)det
y 2 /2 dy dθ +
y 2 + 1 cos θ y 2 + 1 cos θ
1 2π
−1
0
dy dθ
(y 2 + 1) 3/2 cos 3 θ + (y 2
+ 1) sin θ cos θ dθ dy

2π
0
cos θ sin θ dθ = 1
2
2π
0
cos 3 θ dθ =
=
=
2π
0
3π/2
−π/2
π/2
−π/2
1
−1
sin(2θ) dθ = 1
4
cos 3 θ dθ =
4π
0
3π/2
−π/2
(1 − sin 2 θ) cos θ dθ +
(1 − w 2 ) dw +
−1
1
sin w dw = 0.
(1 − sin 2 θ) cos θ dθ
3/2π
π/2
(1 − w 2 ) dw = 0.
y ′ + 1 1
y =
sin x y
(1 − sin 2 θ) cos θ dθ
R 2 \ {y = 0}
z = y 1−(−1) = y 2 z ′ = 2yy ′ = 2 − 2z/ sin x
ω(x, z) = p(x, z) dx + q(x, z) dz =
2 − 2z
dx − dz = 0
sin x
∂zp(x, z) − ∂xq(x, z) = − 2 2
=
sin x sin x q(x).
2/ sin x sin x t =
tan(x/2)
dx 1 + t2 2 dt
2 = 2
= 2 log |tan(x/2)|
sin x 2t 1 + t2 h(x, z) = tan2 (x/2)
2 tan 2 z tan(x/2)
(x/2) −
cos2
dx − tan
(x/2)
2 (x/2)dz = 0
V (x, z) = 4 tan(x/2) − 2x − z tan 2 (x/2),
V (x, z) = c c ∈ R
z sgn(tan(x/2))
c + 2x − 4 tan(x/2)
z(x) = −
tan2 (x/2)
y
c + 2x − 4 tan(x/2)
y(x) = ±
.
| tan(x/2)|
˙x + 2x + 3y = 3e −2t ,
˙y + 5x + y = 0.
t = tan(x/2) cos x = 1−t 2
1+t 2 sin x = 2t
1+t 2 dx = 2dt
1+t 2

z = (x, y) ˙z = Az + B(t)
−2
A =
−5
−3
,
−1
3e−2 B(t) =
0
.
T = tr(A) = −3 D = det(A) = −13 λ2 − T λ + D = 0
λ2 + 3λ − 13 = 0
λ1 = 1
−3 −
2
√
61 , λ2 = 1
−3 +
2
√
61 .
D = 0 (0, 0)
−3y = ˙x + 2x − 3e
−2t
˙y = −5x − y
−3 ˙y = ¨x + 2 ˙x + 6e−2t ˙y
−3(−5x − y) = ¨x + 2 ˙x + 6e −2t .
¨x + 2 ˙x − 15x − 3y + 6e −2t = 0
y
¨x + 2 ˙x − 15x + ( ˙x + 2x − 3e −2t ) + 6e −2t = 0.
x
¨x + 2 ˙x − 15x + ˙x + 2x − 3e −2t + 6e −2t = 0.
¨x − (−2 − 1) ˙x + (2 − 15)x − 3e −2t + 6e −2t = 0
¨x − T ˙x + D x = −3e −2t
c1e λ1t + c2e λ2t
−2
qe −2t q ∈ R e −2t 4q − 6q − 13q = −3 q = 1/5
x(t) = c1e λ1t + c2e λ2t + 1
5 e−2t
˙x(t) = c1λ1e λ1t + c2λ2e λ2t − 2
5 e−2t .
y(t) = − 1
3
c1λ1e λ1t + c2λ2e λ2t − 2
= 1 1
e− 2(3+
6 √ 61)t
−
⎧
1
⎨
−
x(t) = c1e
c1, c2 ∈ R
⎩y(t)
= 1
6
1 + √ 61
5 e−2t
+ 2
c1e
√ 61t +
2(3− √ 1
61)t −
+ c2e 2(3+ √ 61)t e + −2t
5
1
e− 2(3+ √ 61)t
c1e λ1t
+ c2e λ2t 1
+
5 e−2t
√
61 − 1 c2 + 6e 1
2( √ 61−1)t
.
− 3e −2t
− 1 + √ 61 c1e √ √
61t + 61 − 1 c2 + 6e 1
2( √ 61−1)t
a ∈ R
x ′ (t) = t − x 2 (t),
x(0) = a,

a
(−∞, 0] a
[0, +∞)
˙x = 0
R := {(t, x(t)) : ˙x(t) > 0} = {(t, x) : t < x 2 },
˙t = 1
˙x = t − x 2
R (˙t, ˙x) (t,x)∈R = (1, 0) R
˙x
x → +∞
x ′′ = 1 − 2xx ′ = 1 − 2x(t − x 2 )
γ := {(t, x) : x ′′ = 0} = {(t, x) : 1 − 2xt + 2x 3 } = {(t, x) : t = 1/(2x) + x 2 }.
γ t = t(x) = 1/(2x) + x 2 t = x 2 x → +∞ x → 0 ±
±∞ x > 0 R x
− 3 1/2 x x < 0 R 2 \ γ
Q0 := {(t, x) : x ′′ < 0, x > 0, t > 0} = {(t, x) : 1 − 2xt + 2x 3 < 0 x > 0, t > 0}
Q1 := {(t, x) : x ′′ > 0} = {(t, x) : 1 − 2xt + 2x 3 > 0}
Q2 := {(t, x) : x ′′ < 0, x < 0} = {(t, x) : 1 − 2xt + 2x 3 < 0, x < 0}.
γ
F (t, x) = 0 F (t, x) = 1 − 2xt + 2x 3 (t, x) ∈ γ
ˆn(t, x) = ±∇F (t, x) = ∓(2x, 2t − 6x 2 )
Q0 Q1 Q2
Q0 Q0
∂Q0 x > 0 Q0 ˆn0(t, x) =
(2x, 2t − 6x 2 )
Q2 Q2
∂Q2 x < 0 Q2 ˆn2(t, x) =
(2x, 2t − 6x 2 ) = ˆn0(t, x)
Q1 ˆn1(t, x) = −n0(t, x) = −(2x, 2t − 6x 2 )
ˆni · (1, ˙x) ∂Qi
ˆn0 · (1, ˙x) t=1/(2x)+x 2 = [2x + (t − x 2 )(2t − 6x 2 )] t=1/(2x)+x 2 = 2x + 1
2x
= 2x + 1 1 3
+ −
2x2 x x
1
= > 0.
2x2 ˆn2 · (1, ˙x) t=1/(2x)+x2 = ˆn0 · (1, ˙x) t=1/(2x)+x2 = 1
> 0.
2x2 ˆn1 · (1, ˙x) t=1/(2x)+x2 = −ˆn0 · (1, ˙x) t=1/(2x)+x2 = − 1
< 0.
2x2
1
x + 2x2 − 6x 2

Q0 Q2 Q1
Q1 Q1
−(1, ˙x)
ˆn1 · (−1, − ˙x) t=1/(2x)+x2 = 1
> 0,
2x2 Q1 t < 0
y(t) = x(−t) x(t)
y(t) ˙y = − ˙x(−t) = −(−t − y 2 (t)) = t + y 2 (t)
¯ε = ¯ε(a) 0 < ε < ¯ε(a) y(t) |t| < ε 2
t > ε 2
˙y > ε 2 + y 2 ˙z = z 2 + ε 2 z(0) = a
1 z
1
arctan = t + C, C =
ε ε ε
z(t)
t ∗ =
π/2 − arctan a
ε
ε
a
arctan , z(t) = ε tan
ε
> 0, lim z(t) = +∞,
t→t∗−
εt + arctan a
.
ε
y(t) 0 < t1 < t ∗ x(t)
t = −t1 < 0
x(t) {(t, x) : t < 0}
+∞ −∞
t ≥ 0 a ≥ 0
t > 0 t = x 2 R
γ R +∞ Q0
R
t > 0 √ t
˙x
t
= 1 − x2
t .
t ˙x(t) x(t)
Q0 x 2 (t)/t 1 x(t)
√ t
Q2 a < − 3 1/2 t > 0
R −∞
Q2 −∞
Q2 t > 0
0 < t < 1 ˙x < 1 − x 2 ˙z = 1 − z 2 z(0) = a
t = t∗ < 1
1 1 1 1 1 1
= dz + dz =
1 − z2 2 1 − z 2 1 + z 2 log
1 − z
1
+ z
1
2 log
1 + z
1
− z = t + C.
z → ±∞ t + C → 0 t∗ = −C
C = 1
2 log
1 + a
1
− a
.
0 < − 1
2 log
1 + a
1
− a
< 1

0 < log
1 − a
1
+ a
< 2
|1 − a| > |1 + a| a < 0
|a| a < 0
1 a → −∞ a ∗ < 0 a < a ∗
a < a ∗ −∞ 0 < t ∗ a < 1 x(t)
t0 > 0 Q2
I −∞ t1 > t0 t1 ∈ I x(t0) = b < a ∗
v(t) = x(t − t0) x(t) t0 < t < t0 + 1
t > 0
t > 0 −∞
a1 < a2 < 0
x1(t) x2(t) x1(t) < x2(t) t ≥ 0
A := {a ∈ R : ta ≥ 0 ta = x 2 (a, ta)}
B :=
a ∈ R : ta ≥ 0 ta =
1
2x(a, ta) + x2 (a, ta), x(a, ta) < 0
x(a, t) a t A ⊃
[0, +∞[ B ⊃] − ∞, a ∗ ] A ∩ B = ∅ A
B a + = inf A ≤ 0 a − = sup B ≥ −1
a + /∈ A a + ∈ A ¯x(t)
¯t > 0 ¯t = ¯x 2 (¯t) ā ≤ 0 ¯x(¯t) = − √ ¯t
˙x = t − x 2 x(t0) = − √ ¯t R Q
[0, ¯t] ˙x < 0
¯t ¯x(¯t) = − √ ¯t 0 α +
α + a − /∈ B
[a − , a + ] A B
a − = a + a − < a1 < a2 < a +
d
dt (x(a1, t) − x(a2, t)) = x 2 (a1, t) − x 2 (a2, t) > 0
a1 a2
a1 − a2 > 0 t = x 2 t = 1/(2x) + x 2
a − = a + = α
R Q
α
− √ t t → +∞
Γ(z) =
+∞
0
α = − 31/3 Γ(2/3)
Γ(1/3)
−0, 729011,
t z−1 e −t dt
⎧
⎪⎨ −6∂tu(t, x) + 4∂xxu(t, x) + 3u = 0, (t, x) ∈]0, +∞[×[0, π],
∂xux(t, 0) = ∂xux(t, π) = 0
⎪⎩
u(0, x) = x(π − x).
,

˙x = t − x 2 x(0) = a a ∈ R
u(t, x) = U(t)X(x)
−6 ˙ U(t, x) + 4U(t) ¨ X(x) + 3U(t)X(x) = 0
U(t)X(x) λ ∈ R
− −6 ˙ U(t, x) + 3U(t)
U(t)
= 4 ¨ X(x)
X(x)
λ ∈ R
4 ¨ X(x) − λX(x) = 0
−6 ˙ U(t) + (3 + λ) U(t) = 0.
X 4µ 2 −λ = 0 ∆ = 16λ
λ > 0 µ1 = √ λ/2 µ2 = −µ1
Φ(c1, c2, x) = c1e µ1x + c2e µ2x d
dx Φ(c1, c2, x) = c1µ1e µ1x +
c2µ2e µ2x ˙ X(0) = d
dx Φ(c1, c2, 0) = 0 c1µ1 + c2µ2 = 0
d
dx Φ(c1, c2, x) = c1µ1(e µ1x − e µ2x ) µ1 = µ2
c1 = c2 = 0
λ = 0 X(x) X(x) = c0 + c1x
c1 = 0 X0(x) = c0 λ = 0
λ < 0 X(x) ω = |λ|2 X(x) = c1 cos ωx + c2 sin ωx
˙ X(x) = ω(−c1 sin ωx + c2 cos ωx) 0 c2 = 0 π
=: λ

ω = n ∈ Z ω λ < 0 λ
−4n 2
λ λ0 = 0 X0(x) = c0 λn = −4n 2
n ∈ N \ {0} Xn(x) = cn cos nx
U λn
Un(t) = dne 3−4n2
6
an = cndn Un(t)Xn(x)
un(t, x) = ane 3−4n2
6
t .
t cos nx.
∞
u(0, x) = x(π − x) = a0 + an cos nx
aj j > 1 x(π − x)
[0, π] [−π, π 2π R a0
0
a0 = 1
2
2 π 0π x(π − x) dx = π2
6 .
an = 2
π
π(x − x
π 0
2 ) cos nx dx = − 2(1 + (−1)n )
n2 .
a2k = −1/k2 k ∈ N \ {0}
u(t, x) = π2
6 et/2 ∞ e
−
3−16k2
t 6
k2 cos(2kx) = e t/2
⎛
⎝ π2
6 −
∞ e −8k2
t 3
k2 ⎞
cos(2kx) ⎠ .
k=1
1/k 2
n=1
k=1

2π
xt sin t
F (x) = e dt x ∈ R
0 t
x log(1 + xt)
G(x) =
0 1 + t2 dt x ∈ R
F (x, y) =
x
y
e −(x−t)2
dt
F (x) =
2π
e
0
x sin y dy
x = 0
y
F (y) =
0
log(1 + xy)
1 + x2 dx
F (x) =
y
x
0
e −xt2
dt − x
x = 0
x > 0
F (x) =
π
x
0
sin(xy)
y
f : R 2 → R
dy
f(x, y) = arcsin(xy).
f f
(x 2 + y 2 + 12ax + 9a 2 ) 2 = 4a(2x + 3a) 3 a > 0
y 4 − x 4 + ay 2 + bx 2 = 0 a, b ∈ R
f(x, y) = log arctan e (x2 +y4 ) 3
+ 1
(0, 0)
f(x, y) = 1/x + 1/y Ea =
{(x, y) ∈ R 2 : 1/x 2 + 1/y 2 = 1/a 2 , x = 0, y = 0} a > 0

f(x, y, z) = x 2 +y 2 +z 2 x+y+z+1 =
0
f(x, y, z) = (x + y + z) 2 x 2 + 2y 2 + 3z 2 = 1
f(x, y, z) = x2 + cos(y)
≤ 10}.
C = {(x, y, z) ∈ R3 : x2 + y2 + ez2 f(x, y, z) = y √ 1 + z2 C = {(x, y, z) ∈ R3 : x2 − 2x + y2 + z2 ≤ 3}
f(x, y, z) = (1 + x2 )ez2 C = {(x, y, z) ∈ R3 : x2 + y4 − 2y2 + z2 ≤ 0}
f(x, y) = cos2 (x)+y 3 −3y2
Q = {(x, y) ∈ R2 : −π < x < 2π, |y| < 3π}
a ∈ R
fa(x, y, z) = x + ax2 − cos(y) + z2ex C = {(x, y, z) ∈ R3 : 0 ≤ x ≤ 2, −π ≤
y ≤ π, −1 ≤ z ≤ 1}
a ∈ R
fa(x, y, z) = ea cos(z)+y2
+ sin 2 (x) C = {(x, y, z) ∈ R 3 : |y| ≤ 1}
D x2 + y2 dx dy D D
2 3
x D
2
x2 +y2 dx dy D
y = x y = −x x = 1
D = {(x, y) ∈ R2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}
y
1 + xy dxdy
D
D = B((0, 0), 3) \ B((0, 0), 2) ∩ {y ≥ 0}
x2 + y2 dxdy
D
D = B((1, 0), 1) \ B((0, 0), 1) ∩ {y ≥ 0}
x
D
2 + y 2 dxdy
x sin |y − 2x| dx dy
D (0, 0) (1, 0) (0, 1)
D 0 ≤ x ≤ 1 0 ≤ y ≤ 1
R 2
D
D
1
x + y dxdy
1
(1 + x 2 )(1 + y 2 ) dxdy
T
z 1 − y 2 dx dy dz,
T 1 z
1

xz 3 dx dy dz,
T
T y =
4x2 + 9z2 y = 1
e
T
(x2 +y2 +z2 ) 3/2
T 4x2 + y2 + z2 ≤ r2
dx dy dz,
(x
T
2 + y 2 ) dx dy dz,
T z
r z = h
D := {(x, z) ∈ R 2 : x 2 + z 2 ≤ 2, x 2 + (z − 1) 2 ≥ 1, z ≥ 0}
D
D + := {(x, z) ∈ D : x ≥ 0}
S D
z
I = [0, 1] α ∈ R fα : I × I → R
0 x = y
fα(x, y) =
1
|x−y| α
x = y
α fα I × I α
fα(x, y) dx dy
I×I
D := {(x, y) ∈ R 2 : (x + y) 2 ≤ e −(x−y)2
I =
}
e −(x−y)2
dxdy
D
z = arcsin x
(x, y) D y2 = x2 (1 −
x2 )
x2 + y2 − z2 = 0
x ≥ 0 y ≥ 0 z ≥ 0 x + y + z ≤ 1
γ ρ = Aθ θ ∈ [0, 4π]
I := θ 3 ds.
x(θ) = ρ(θ) cos θ = Aθ cos θ
y(θ) = ρ(θ) sin θ = Aθ sin θ
ϕ(θ) = (x(θ), y(θ))
˙x A cos(θ) − Aθ sin(θ)
Jac ϕ(θ) = =
˙y A sin(θ) + Aθ cos(θ)
γ

1
Jac ϕ(θ) 1
ds = ˙x 2 + ˙y 2 = (A sin(θ) + Aθ cos(θ)) 2 + (A cos(θ) − Aθ sin(θ)) 2 = A 2 (1 + θ 2 ) = |A| 1 + θ 2 .
t = θ 2 dt = 2θ dθ z = 1 + t dz = dt
I =
4π
0
= |A|
2
= |A|
2
θ 3 |A| 1 + θ2 dθ = |A|
2
16π 2 +1
1
z 5/2
5/2
z=16π 2 +1
z=1
16π 2
0
16π2 +1
z 3/2 dz − |A|
2 1
− |A|
z
2
3/2
3/2
t √ 1 + t dt = |A|
2
z 1/2 dz
z=16π 2 +1
z=1
= |A|
5 ((16π2 + 1) 5/2 − 1) − |A|
3 ((16π2 + 1) 3/2 − 1).
16π 2 +1
F · n dσ, F := (xz, xy, yz)
S
1
(z − 1) √ z dz
S := ∂{(x, y, z) ∈ R3 : x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 1}
f dσ, f(x, y, z) := (z + 1) 1 + x2 + y2 + x 3 + y 2
C
C := {(x, y, z) ∈ R 3 : x 2 + y 2 = z 2 , 0 < z < 1}
α > 0
⎧
| sin(xy) − xy|
⎪⎨
f(x, y) =
⎪⎩
α
(x2 + y2 ) 3 (x, y) = (0, 0),
0 (x, y) = (0, 0).
α f (0, 0)
sin(xy)−xy
β > 0
sin(s) − s
lim
s→0 sβ
sin(s) − s
lim
s→0 sβ cos(s) − 1
= lim
s→0 βsβ−1 = lim
s→0
− sin(s)
,
β(β − 1)sβ−2 β − 2 = 1 β = 3
sin(s) − s
lim
s→0 s3 = − 1
6 .
α ≤ 0 α ≤ 0 (x, y) → 0
| sin(xy) − xy| α
(x 2 + y 2 ) 3
≥
1
(x 2 + y 2 ) 3

α > 0
lim
(x,y)→(0,0)
(x,y)=(0,0)
f(x, y) = lim
(x,y)→(0,0)
(x,y)=(0,0)
| sin(xy) − xy|
|xy| 3
α
|xy| 3α
(x2 + y2 1
=
) 3 6α |xy| ≤ 1
2 (x2 + y 2 )
0 ≤ |xy|α
x2 1
≤
+ y2 2α (x2 + y 2 ) α−1 .
α > 1
lim
(x,y)→(0,0)
(x,y)=(0,0)
|xy| α
x2 = 0, lim
+ y2 (x,y)→(0,0)
(x,y)=(0,0)
⎛
⎜
⎝ lim
(x,y)→(0,0)
(x,y)=(0,0)
f(x, y) = 0 = f(0, 0),
|xy| α
x2 + y2 α > 1 f α ≤ 1 y = mx
|xy| α
x2 |m|α
=
+ y2 m2 + 1
|x| 2α
,
x2 α < 1 x → 0 +∞ α = 1 |m|/(m 2 + 1) m
f f α > 1
lim
(x,y)→(0,0)
(x,y)=(0,0)
|xy| α
x2 =
+ y2 lim
ρ→0 +
x=ρ cos θ
y=ρ sin θ
|ρ2 sin θ cos θ| α
ρ 2 = lim
ρ→0 +
x=ρ cos θ
y=ρ sin θ
1
2 α ρ2α−2 | sin 2θ| α .
α > 1 θ α = 1 +∞ α < 1
f : R2 → R f(x, y) = (x2 + y2 )e−y2
C = {(x, y) ∈ R 2 : y ≥ 1 − x 2 |x| ≤ 1 y ≥ |x| − 1 |x| ≥ 1}.
lim f(x, y), lim
|(x,y)|→∞
|(x,y)|→∞
(x,y)∈C
f C
f
lim
|(x,y)|→∞
y=0
f(x, y) = lim
|x|→∞ x2 = +∞, lim
f(x, y).
f(x, y) = lim
|(x,y)|→∞
|y|→∞
x=0
y2e −y2
= 0.
lim f(x, y)
|(x,y)|→∞
0 ≤ f(x, y) ≤ (1 + y2 )e−y2 |x| ≤ 1 |x| > 1 (x, y) ∈ C y − 1 ≥ |x|
(y − 1) 2 ≥ x2 (x, y) ∈ C |x| > 1 0 ≤ f(x, y) ≤ ((y − 1) 4 + y2 )e−y2 (x, y) ∈ C
0 ≤ f(x, y) ≤ (1 + y 2 )e −y2
+ ((y − 1) 4 + y 2 )e −y2
.
|(x, y)| → ∞ (x, y) ∈ C y → +∞ 0
lim
|(x,y)|→∞
(x,y)∈C
f(x, y) = 0.
⎞
⎟
⎠
3

f f(x, y) > 0 f |(x, y)| → ∞ (x, y) ∈ C
∂xf(x, y) = 2xe −y2
∂yf(x, y) = 2ye −y2
(1 − x 2 − y 2 )
∂xf(x, y) = ∂yf(x, y) = 0 (x, y) = (0, 0) /∈ C (x, y) = (0, ±1)
C C
γ1(θ) = (cos θ, sin θ), θ ∈ [0, π]
γ2(t) = (t, t − 1), t > 1
γ3(t) = (−t, t − 1), t > 1
f(γ1(θ)) = e− sin2 θ θ = 0, π θ = π/2 f(±1, 0) = 1
f(0, 1) = 1/e f(0, y) = y2e−y2 y ≥ 1 (0, 1)
f C
f(γ2(t)) = f(γ3(t)) = (t2 + (t − 1) 2 )e−(t−1)2
d
dt f ◦ γ2(t) = d
dt f ◦ γ3(t) = −2te −(t−1)2
(2t 2 − 4t + 1)
t = 0 γ2(0), γ3(0) /∈ C t = t1 := (2 − √ 2)/2 < 1 γ2(t1), γ3(t1) /∈ C
t = t2 := (2 + √ 2)/2 f ◦ γ2(t) f ◦ γ3(t) 1
(0, ±1) f C γ2(t2) = ((2 + √ 2)/2, √ 2/2)
γ3(t2) = (−(2 + √ 2)/2, √ 2/2)
f M := f(±(2+ √ 2)/2, √ 2/2) = (2+ √ 2)/ √ e (x, y) ∈ C
|x| ≤ y − 1
f(x, y) ≤ ((y − 1) 4 + y 2 )e −y2
,
C
Iα =
Cα
α x
Iα =
1
= α4
4
Iα
lim ,
α→+∞ α4 x2 y2 dx dy, Cα
:= (x, y) ∈ R 2 : 1 ≤ x ≤ α, 1
≤ y ≤ x .
x
− α2
2
1/x
Iα 1
lim =
α→+∞ α4 4
x2 dy
y2
1 1
− +
4 2 .
dx =
α
1
x 2
x − 1
dx =
x
α
F : R 2 → R C 1
F (x, y) = y 3 − 2xy 2 + cos(xy) − 2.
1
,
(x 3
x4 − x) dx =
4
α x2
−
2 1
F (x, y) = 0 ϕ :] − δ, δ[→]1 − σ, 1 + σ[
(0, 1) ϕ ′ (0) ϕ R

F (0, 1) =
0
∂yF (x, y) = 3y 2 − 4xy − x sin(xy).
∂yF (0, 1) = 3 = 0 ϕ
∂xF (x, y) = 2y 2 − y sin(xy).
ϕ ′ (0) = − ∂xF (0, 1)
= −2
∂yF (0, 1) 3 .
x ∈ R\] − δ, δ[
y ↦→ y 3 − 2xy 2 + cos(xy) − 2 := gx(y).
x ∈ R\] − δ, δ[ gx(y) R R
lim
y→+∞ gx(y) = +∞, lim
y→−∞ gx(y) = −∞
yx
gx(yx) = 0 x ↦→ yx
ϕ
(y, z) z = 1
z = √ y − 2 3 ≤ y ≤ 6
D
D z
S
S
M :=
6
3
S = {(y, z) : 1 ≤ z ≤ y − 2, 3 ≤ y ≤ 6}.
( y − 2 − 1) dy =
4
G = (Gy, Gz)
Gz = 1
z dydz =
M S
3
6
5 3 1
Gy = 1
y dydz =
M
3
6
5
1
( √ t − 1) dt = 2
3 [t3/2 ] 4 1 − 3 = 16
3
√ y−2
√ y−2
z dzdy = 3
10
y dzdy = 3
5
S
3 1
(y − 2 = t 2 ) = 3
2
2(t
5 1
2 + 2)t 2 dt = 326
25 .
V = 2πGyM = 652
15 π.
6
3
6
S = {(x, y, z) : z = 1 − x 2 + y 2 , z ≥ 0, y ≥ √ 2/2}.
3
2 5
− − 3 =
3 3 .
((y − 2) − 1) dy = 27
20
y( y − 2 − 1) dy
z ≥ 0 x 2 + y 2 ≤ 1 y ≤ 1 √ 2/2 ≤ y ≤ 1
|x| ≤ 1 − y2 z = f(x, y)
ω2(x, y) = 1 + |∇f(x, y)| 2
2 2
=
x
y
1 + + =
x2 + y2 x2 + y2 √ 2.

1
A =
√ 1−y2 √
2/2 − √ 1−y2 π/2
π/2
√ √
1 √
2dxdy = 2 2 √ 1 − y2 dy = 2 2
2/2
π/2
= 2 √ 2 cos
π/4
2 θ dθ = 2 √ cos(2θ) + 1
2
dθ =
π/4 2
√ 2
√ √ √
π
2 2
2
= π + cos z dz = (π − 2).
4 2
4
π/2
π/4
π/2
π/4
1 − sin 2 θ cos θ dθ
(cos(2θ) + 1) dθ
S = {(x, y, z) : x = s cos t, y = t, z = s sin t, t ∈ (0, π), s ∈
(1, 2)} f : R3 → R f(x, y, z) = yz
f dσ.
S
ϕ(t, s) = (s cos t, t, s sin t)
⎛
−s sin t
Jac ϕ = ⎝ 1
⎞
cos t
0 ⎠ .
s cos t sin t
2
2
ω2(∂tϕ, ∂sϕ) = det 2
−s sin t cos t
+ det
1 0
2
−s sin t cos t
+ det
s cos t sin t
2
1
s cos t
0
sin t
= 1 + s 2
π 2
fdσ = f ◦ ϕ(t, s)ω2(∂tϕ, ∂sϕ) ds dt
S
0 1
π 2
= ts sin t
0 1
1 + s2
ds dt = 1
π
t sin t dt ·
2 0
= π
3 (5√5 − 2 √ 2).
4
1
√ 1 + w dw
D = {(u, v) ∈ R 2 : u 2 + v 2 < 1} S ⊂ R 3
ϕ : D → R 3 ϕ(u, v) = (u + v, u − v, u 2 + v 2 )
(x
S
2 + y 2 ) √ 1 + 2z dσ.
⎛ ⎞
1 1
Jac ϕ = ⎝ 1 −1 ⎠ .
2u 2v
2
2
ω2(∂uϕ, ∂vϕ) = det 2
1 1
+ det
1 −1
2
1 1
+ det
2u 2v
2
1 −1
2u 2v
= 4 + (2v − 2u) 2 + (2v + 2u) 2 = 2 1 + u 2 + v 2

(u + v)
D
2 + (u − v) 2
1 + 2(u2 + v2 ) 2 1 + u2 + v2 dudv
= 4 (u 2 + v 2 ) 1 + 2(u2 + v2 ) 1 + (u2 + v2 ) dudv
= 4
(w = ρ 2 ) = 4π
= 4π
D
2π 1
0
1
0
1
0
0
ρ 2 1 + 2ρ 2 1 + ρ 2 ρ dρdθ
w (1 + 2w)(1 + w) dw
w 2w 2 + 3w + 1 dw
= 4π(44 + 132 √ 6 − 9 √ 2 log(3 + 2 √ 2) + 9 √ 2 log(7 + 4 √ 3)),
√ s 2 + 1 1/2(y 1 + y 2 + log(z + √ 1 + z 2 ))
a ≥ 0 R 3 Sa
ϕ(θ, y) = ( y 2 + a 2 cos θ, y, y 2 + a 2 sin θ), θ ∈ [0, 2π], |y| < 1,
F : R 3 → R 3 F (x, y, z) = (x 2 , y/2, x)
F
F
√ 1 + a 2 y = 1
γ(θ) = ( a 2 + 1 cos θ, 1, a 2 + 1 sin θ), θ ∈ [0, 2π].
F
ϕ 2
ϕ Sa
a = 0
a > 0 (a, 0, 0)
F Sa
ϕ(θ, y) = (ϕ1, ϕ2, ϕ3) F = (F1, F2, F3)
div F (x, y, z) = ∂xF1 + ∂yF2 + ∂zF3 = 2x + 1/2,
⎛
rot F = det ⎝ e1 ∂x x2 e2 ∂y y/2 ⎠ = (0, −1, 0).
e3 ∂z x
D =
{(x, 1, z) : x2 +z 2 ≤ (1+a 2 )} (0, −1, 0) (0, −1, 0) D
γ
rot
F · ˆn dσ = dσ = Area(D) = π(1 + a 2 ).
D
D
2π
F dγ = F ( a2 + 1 cos θ, 1, a2 + 1 sin θ) · (− a2 + 1 sin θ, 0, a2 + 1 cos θ) dθ
γ
=
0
2π
0
− (a 2 + 1) 3/2 cos 2 θ sin θ + (a 2 + 1) cos 2
θ dθ = π(1 + a 2 )
F
⎞

Jac ϕ(θ, y) =
⎛
⎜
⎝
− y 2 + a 2 sin θ
y cos θ
√ y 2 +a 2
0 1
y2 + a2 cos θ
√y sin θ
y2 +a2
ω2 = det 2 B1 + det 2 B2 + det 2 B3
B1 =
− y 2 + a 2 sin θ
⎛
y cos θ
√ y 2 +a 2
0 1
B2 = ⎝ −y2 + a2 sin θ
y2 + a2 cos θ
B3 =
0 1
y2 + a2 cos θ
√y cos θ
y2 +a2 √y sin θ
y2 +a2 √y sin θ
y2 +a2
⎞
⎞
⎟
⎠ .
, det 2 B1 = (y 2 + a 2 ) sin 2 θ.
⎠ , det 2 B2 = y 2 ,
, det 2 B3 = (y 2 + a 2 ) cos 2 θ.
ω2 = 2y2 + a2 a = 0 ω2 = √ 2|y|
S0
2π 1
dσ =
1 √ √
1
ω2dθ dy = 2π 2|y| dy = 4π 2 y dy = 2π
0 −1
−1
0
√ 2.
Sa
ϕ
(a, 0, 0) = ϕ(0, 0) (0, a, 0) (0, 0, a)
(±1, 0, 0)
⎛
det ⎝
±1 0 0
0 0 a
0 a 0
⎞
= ⎠ = ∓a 2 .
(a, 0, 0) (−1, 0, 0)
Φ(Sa, ⎛
2π F1 ◦ ϕ −
1 ⎜
F ) = det ⎜
⎝
0 −1
y2 + a2 y cos θ
sin θ √
y2 +a2 F2 ◦ ϕ 0 1
F3 ◦ ϕ y2 + a2 ⎞
⎟
⎠ dy dθ
y sin θ
cos θ √
y2 +a2 ⎛
2π (y
1 ⎜
= det ⎜
⎝
0 −1
2 + a2 ) cos2 θ − y2 + a2 ⎞
y cos θ
sin θ √
y2 +a2 ⎟
y/2 0 1 ⎟
⎠ dy dθ
y2 + 1 cos θ y2 + a2 y sin θ
cos θ √
y2 +a2 ⎛
2π 1
= (−y/2)det ⎝
0 −1
−y2 + a2 ⎞
y cos θ
sin θ √
y2 +a2 ⎠
y2 + a2 y sin θ dy dθ+
cos θ √
y2 +a2 2π 1
(y2 + a2 ) cos2 θ − y2 + a2 + (−1)det
sin θ
dy dθ
0 −1
y2 + 1 cos θ y2 + a2 cos θ
2π 1
= y 2 1 2π
/2 dy dθ + (y 2 + a 2 ) 3/2 cos 3 θ + (y 2 + a 2
) sin θ cos θ dθ dy
0
−1
−1
0

α =
= 2
3 π.
2π
0
cos θ sin θ dθ = 1
2
2π
0
cos 3 θ dθ =
=
=
2π
0
3π/2
−π/2
π/2
−π/2
1
−1
sin(2θ) dθ = 1
4
cos 3 θ dθ =
4π
0
3π/2
−π/2
(1 − sin 2 θ) cos θ dθ +
(1 − w 2 ) dw +
−1
1
sin w dw = 0.
(1 − sin 2 θ) cos θ dθ
3/2π
π/2
(1 − w 2 ) dw = 0.
(1 − sin 2 θ) cos θ dθ
x = 0 γ(y) = y 2 (1 − y 2 ) + 1 t ∈ [0, α]
(1 + √ 5)/2 S γ z
(0, 0, 1)
F : R 3 → R 3
F (x, y, z) = (e 4y , x − x 2 , z).
F
γ
S 2
F S
F ˜γ
F
z = 0
F = (F1, F2, F3)
div F (x, y, z) = ∂xF1 + ∂yF2 + ∂zF3 = 1,
rot ⎛
F = det ⎝ e1 ∂x e4y e2 ∂y x − x2 ⎞
⎠ = (0, 0, 1 − 2x − 4e
e3 ∂z z
4y ).
γ(0) = 1 γ(α) = 0
˙γ(y) = 2y(1 − y 2 ) + y 2 (−2y) = 2y − 4y 3 = 2y(1 − 2y 2 ),
y ∈ [0, α] y = 0 y =
√
2
0 < y < 2 √ 2
2 < y ≤ α 0
1 √ 2/2 5/4 y = α
0 ¨γ(y) = 2 − 12y2 0 < y < 1/ √ 6
1/ √ 6 < y < α
S
ϕ(ρ, θ) = (ρ cos θ, ρ sin θ, ρ 2 (1 − ρ 2 ) + 1),
⎛
cos θ
Jac ϕ(ρ, θ) = ⎝ sin θ
ρ sin θ
ρ cos θ
2ρ − 4ρ3 ⎞
⎠ .
0
√ 2
2

ω2 = det 2 B1 + det 2 B2 + det 2 B3
cos θ ρ sin θ
B1 =
, det
sin θ ρ cos θ
2 B1 = ρ 2 .
cos θ ρ sin θ
B2 =
2ρ − 4ρ3
, det
0
2 B2 = ρ 2 (2ρ − 4ρ 3 ) 2 sin 2 θ,
sin θ ρ cos θ
B3 =
2ρ − 4ρ3
, det
0
2 B3 = ρ 2 (2ρ − 4ρ 3 ) 2 cos 2 θ.
ω2 = ρ2 + ρ2 (2ρ − 4ρ3 ) 2 = ρ 1 + 4ρ2 + 16ρ6 − 16ρ4 .
Σ = {(x, y, 0) : x2 +y2 ≤ α2 } ˆn = (0, 0, −1)
Σ ∪ S Ω S Σ Ω
div F dxdydz = Φ(S, F ) + Φ(Σ, F ).
α
0
Ω
Σ F (x, y, 0) · (0, 0, −1) = 0
Φ(Σ,
F ) = F · ˆn dσ = 0,
div
Ω
F dxdydz = dxdydx = Volume(Ω).
Ω
Ω
Σ
ψ(ρ, θ, z) = (ρ cos θ, ρ sin θ, z),
0 < ρ < α 0 < z < ρ 2 (1 − ρ 2 )
ρ Ω
2π α ρ2 (1−ρ2 )+1
α
ρ dzdρdθ = 2π (ρ
0 0 0
0
2 (1 − ρ 2
α2 ) + 1)ρ dρ = 2π
2
⎛
2π
det ⎝
0
F1 ◦ ϕ cos θ ρ sin θ
F2 ◦ ϕ sin θ ρ cos θ
F3 ◦ ϕ 2ρ − 4ρ3 ⎞
⎠ dθ dρ =
0
⎛
⎞
=
=
α 2π
e
det ⎝
0 0
4ρ sin θ ρ cos θ − ρ
cos θ ρ sin θ
2 cos2 θ sin θ ρ cos θ
ρ2 (1 − ρ2 ) + 1 2ρ − 4ρ3 ⎠ dθ dρ
α 2π
(ρ
0
0 0
2 (1 − ρ 2 ) + 1)ρ dρdθ+
α 2π
− (e
0 0
4ρ sin θ ρ cos θ − ρ cos θ − ρ 2 cos 2 θρ sin θ) dθ dρ
α2 α 2π
α4 α6
1 d
+ − −
2 4 6 0 0 4 dθ (e4ρ sin θ α 2π
) dθdρ +
0 0
= 2π
α2 α4 α6
= 2π + − ,
2 4 6
+ α4
4
ρ 3
3
α6
− .
6
d
dθ (cos3 θ) dθdρ

F D z = 0
(0, 0, ±1)
rot
F · ˆn dσ = (1 − 2x − 4e 4y ) dxdy,
F dγ1 =
D
˜γ
D
D
(1 − 2x − 4e 4y ) dxdy,
˜γ D
D := {(x, y, 0) :
|x| ≤ 1, |y| ≤ 1} ˆn = (0, 0, 1)
rot
F · ˆn dσ = (1 − 2x − 4e 4y ) dxdy
D
D
= Area(D) −
1
= 4 − 2
−1
e4y = 4 − 8
4
1 1
−1
2x dx − 2
y=1
y=−1
(2x + 4e 4y ) dxdy
−1
1
−1
4e 4y dy
= 4 − 2(e − e −1 ) = 0
D
x y z = 0 2
4 − 2(e − e −1 )
1
ω(x, y) = (3x 2 y − y 2 ) dx + (x 3 − 2xy + 1) dy
ω(x, y) = [sin(x + y) + x cos(x + y)] dx + x cos(x + y) dy
ω(x, y) =
y 2
x2x2 dx −
+ y2 y
x x2 dy
+ y2 x > 0 (1, 1)
0
ω(x, y) = y log y
− 1 dx + x log
x y
+ 1 dy
x
y 2 dx + 2xy dy
ω
A = (1, 1)
y = x x ∈ [0, 1]
y = x 2 x ∈ [0, 1]
y = √ x x ∈ [0, 1]
y = 2x 3 − x x ∈ [0, 1]

ω1(x, y) = yx y−1 dx + x y log x dy, ω2(x, y) =
x
y x2 dx −
+ y2
(x 2 + y 2 + 2x) dx + 2y dy = 0
3x 2 y 4 + 1
1 + x2
y(1) = 0
x 2
y2x2 dy
+ y2 dx + (4x 3 y 3 + cos y) dy = 0 y(x)
2x
1 3x2
dx + −
y3 y2 y4
dy = 0 (2, 1)
(y 2 − 1) dx + xy(1 − x 2 ) dy = 0
(x + 2y + 1) dx + (x + 2y + 2) dy = 0
(x + y − 1) 2 dx − 4x 2 dy = 0
3x + y x + 3y
√ dx − √ dy = 0
x + y x + y
(3y 2 − x) dx + 2y(y 2 − 3x) dy = 0 e f(x+y2 )
y ′′ − y = (2x + 1)e 3x
y ′′ − 4y ′ + 3y = e 2x
y ′′ − 6y ′ + 5y = e 5x
y ′′ − y = 2x 2 + 5 + 3e 2x
2y ′′ − y ′ − y = x 2 − 3e x
y ′′ + 4y ′ + 5y = cos x
y ′′ + y = sin x
y ′′ + y ′ = 5 sin 3x − 2 cos 3x
y ′ = sgn(y) · |y|, y(0) = 0
y ′ = cos x · y − 1
y ′ = y 2/3 (1, 0)
y ′ = (x + y) 2
y ′ + 1
x y = x3 y(x)
y(1) = −3
y ′ + y tan x = sin 2x
y ′ − 2y tan x = 2 √ y
y ′ = xy + x 3 y 2
dy
= 2xy − x.
dx
y(0) = 3/2
R

y ′ = 2xy − x y(0) = 3/2
y(0) = 3/2
ω(x, y) = p(x, y) dx + q(x, y) dy = (−2xy + x) dx + dy = 0.
∂yp − ∂xq = −2x = −2x q
h(x) = e −2x dx = e−x2 . hω R2
V
(x0, y0) ∈ R2 γ(t)
x0
V (x0, y0) = hω = e −x/2 y0
x dx +
γ
V
0
0
e −x/2 dy = − 1
2 e−x2
+ ye −x2
= (y − 1/2)e −x2
.
h(x)ω(x, y) = e −x2
(−2xy + x) dx + e −x2
dy = d(ye −x2
) + e −x2
x dx
= d(ye −x2
) + d − 1
2 e−x2
= d (y − 1/2)e −x2
.
(y − 1/2)e−x2 = c c ∈ R
y(x) = cex2 + 1/2 y(0) = 3/2 c = 1 y(x) = ex2 + 1/2
R +∞ x → ±∞
x > 0
x < 0 x = 0 3/2
dy
dx = − 1 − x2y x2 .
y − x3
lim |y(x)| = +∞ 0
x→0
ω(x, y) = (1 − x 2 y) dx + (x 2 y − x 3 ) dy = 0.
∂yp(x, y) − ∂xq(x, y) = 2x 2 − 2xy = − 2
q(x, y),
x
λ(x, y) = e − 2
x dx = 1
.
x2

λω = ( 1
x2 − y) dx + (y − x) dy
G(x,
1
y) = − y, y − x
x2
1
− y dx + (y − x) dy =
x2 1
dx + y dy − (x dy + y dx) = d −
x2 1
y2
+ − xy .
x 2
V (x, y) := − 1 y2
x + 2 − xy
− 1 y2
+ − xy = c, c ∈ R,
x 2
y(x) = x2 ± √ 2cx 2 + x 4 + 2x
x = 0 2x −2 + y 2 x − 2x 2 y = 2cx
|y| x → 0
−2 0 |y| → +∞ x → 0
x
.

Γ
f(x, y) = 0 f : R 2 → R f ∈ C 1 (R 2 )
(x0, y0) (x0, y0) ∈ R 2 Γ
f(x0, y0) = 0
x = ρ cos θ y = ρ sin θ g(ρ, θ) =
f(ρ cos θ, ρ sin θ)
{(ρ cos θ, ρ sin θ) : g(ρ, θ) = 0, ρ ≥ 0, θ ∈ [0, 2π]}.
g(ρ, θ) = 0
ρ ≥ 0 θ g(ρ, θ) = 0 ρ < 0
g(ρ, θ) = 0 ρ = h(θ)
A ⊆ [0, 2π] h(θ) ≥ 0 θ
θ ∗ ∈ A
cos(θ ∗ )y = sin(θ ∗ )x Γ α ∈ A h(α) = 0
cos(α)y = sin(α)x Γ
h ρ
f Γ f ρ Γ
Γ
g(ρ, θ) = ρ 3 − ρ 2 (cos 2 θ + 1)
g(ρ, θ) = 0 ρ = h(θ) h(θ) = cos 2 θ + 1 ρ ≥ 0
(0, 0) ρ = 0 g(0, θ) = 0
ρ = h(θ) g(ρ, θ) = ρ 3 − ρ 2 (cos 2 θ − 1)
g(ρ, θ) = 0 ρ = h(θ) h(θ) = cos 2 θ − 1 ρ ≥ 0
θ = 0, π
A = [0, π/2] A = [0, π/3]
π/3
y = tan(π/3)x π/2 /∈ A Γ
π/2 /∈ A 3π/2 /∈ A Γ

f
Γ
f(x, y) = f(y, x) Γ y = x
x = y ′ y = x ′ f(x, y) = f(y ′ , x ′ ) = f(x ′ , y ′ ) (x, y) f
(x ′ , y ′ ) f
f(x, y) = −f(y, x)
f(x, y) = f(x, −y) Γ
x = x ′ y = −y ′ f(x, y) = f(x ′ , −y ′ ) = f(x ′ , y ′ ) (x, y)
f (x ′ , y ′ ) f
f(x, y) = −f(x, −y)
f(x, y) = f(−x, y) Γ
x = −x ′ y = y ′ f(x, y) = f(−x ′ , y ′ ) = f(x ′ , y ′ ) (x, y)
f (x ′ , y ′ )
f f(x, y) = −f(−x, y)
f(x, y) = f(−x, −y) Γ
x = −x ′ y = −y ′ f(x, y) = f(−x ′ , −y ′ ) = f(x ′ , y ′ ) (x, y)
f (x ′ , y ′ ) f
f(x, y) = −f(−x, −y)
g(ρ, θ) = 0
ρ ≥ 0 g Γ
0 < α < 2π g(ρ, θ + α) = ±g(ρ, θ) Γ
nα n ∈ Z
y = mx
f(x, mx) = 0 x = k(m) m
x = k(m) y = mk(m)
m∗ ∈ R lim k(m) = ±∞
m→m∗ y = m∗x + q
q = lim
m→m∗ mk(m) − m∗k(m) ∈ R
(x0, y0) ∈ R2 Γ
f(x0, y0) = 0
df(x0, y0) = ∂xf(x0, y0) dx + ∂yf(x0, y0) dy.
Γ (x0, y0)
∂xf(x0, y0)x + ∂y(x0, y0)y = q q
(x0, y0) q = ∂xf(x0, y0)x0 +∂y(x0, y0)y0
∇f(x0, y0) = (0, 0)
∇f(x0, y0) · (x − x0, y − y0) = 0,
(x0, y0) y−y0 = m(x−x0) x−x0 = m(y−y0)
y = ϕ(x) = mk(m) x = k(m)

∇f(x0, y0) = (0, 0) (x0, y0)
f
Γ
(x0, y0)
∂yf(x0, y0) = 0
x = x0
x0 ϕ ϕ ∈ C 1 ϕ(x0) = y0 ϕ x0
ϕ ′ (x0) = − ∂xf(x0, y0)
∂yf(x0, y0) .
∂xf(x0, y0) = 0 y = y0
y0 ψ ψ ∈ C 1
ψ(y0) = x0 ψ y0
ϕ ′ (y0) = − ∂yf(x0, y0)
∂xf(x0, y0) .
∂yf(x0, y0) = 0 ∂xf(x0, y0) = 0 (x0, y0)
y = ϕ(x)
∂xf(x0, y0) = 0 ∂yf(x0, y0) = 0 (x0, y0)
x = ϕ(y)
f ∈ C 1
C 1
P Γ
q(x, y) = y p(x, y) = x Γ
y x Γ
f ∇f
Γ F : R 2 → R C 1 (R 2 )
F Γ
(¯x, ¯y)
λ ⎧
⎪⎨ ∂x(F (x, y) + λf(x, y)) = 0,
∂y(F (x, y) + λf(x, y)) = 0,
⎪⎩
f(x, y) = 0.
F
Γ
F Γ
Γ Γ
ρ(θ) = h(θ)
˜F (θ) = F (ρ(θ) cos θ, ρ(θ) sin θ),
θ ∈ A := {θ ∈ [0, 2π] : h(θ) ≥ 0} F Γ
˜ F A ˜ F ′ (θ) = 0
˜ F ′′ (θ)

˜ F
A
A A
(0, 0) ∈ Γ h(θ) = 0 θ ∈ A
Γ y = mx
x f(x, mx) = 0 x = k(m) y = mk(m)
¯F (m) = F (k(m), mk(m)),
k(m) K
K
K
y = mx Γ ∩ {(0, y) :
y ∈ R} Γ
x = my
Γ
x0 ∈ R
ϕi = ϕi(x) Γ x0
f(x0, y) = 0 y yλ
Γ x0
∂yf(x0, yλ) = 0 λ f(x0, y)
px0 (y)
y → ±∞
y px0 (y) = 0
px0
y
x0 ∂yf(x0, y) = 0 ∂xf(x0, y) = 0
Γ
px0 Γ
ρ(θ) = h(θ) θ ∈ A x(θ) = ρ(θ) cos θ
y(θ) = ρ(θ) sin θ θ ∈ A x y Γ
A
y = mx
x = k(m) y = mk(m)
Γ

ϕ : I ×J → R 3
I, J R F , G : R 3 × R 3 ϕ
u ∈ I v ∈ J ϕ ϕ = (ϕ1, ϕ2, ϕ3) F
F = (F1, F2, F3)
Σ f(x, y, z) =
0 ∇f = (0, 0, 0) Σ
Σ Σ
F div F = ∂F1
∂x
F
rot F = ∇ × ⎛
⎞
i j k
⎜
⎟
⎜
⎟
F = det ⎜ ∂x ∂y ∂z ⎟
⎝
⎠
F1 F2 F3
+ ∂F2
∂y
+ ∂F3
∂z
= i (∂yF3 − ∂zF2) + j (∂zF1 − ∂xF3) + k (∂xF2 − ∂yF1)
= (∂yF3 − ∂zF2, ∂zF1 − ∂xF3, ∂xF2 − ∂yF1) .
∂uϕ(u, v) ∂vϕ(u, v)
⎛
⎞
Jacϕ(u, v) =
⎝ ∂uϕ1 ∂vϕ1
∂uϕ2 ∂vϕ2
∂uϕ3 ∂vϕ3
2
2 × 2 Jac ϕ
∂uϕ2 ∂vϕ2
B1 =
,
∂uϕ3 ∂vϕ3
∂uϕ1
B2 =
∂uϕ3
∂vϕ1
∂vϕ3
,
∂uϕ1
B3 =
∂uϕ2
∂vϕ1
∂vϕ2
Σ R 3 x 2 + y 2 + z 2 − 1 = 0 f(x, y, z) =
x 2 + y 2 + z 2 − 1 ∇f(x, y, z) = 2(x, y, z) ∇f(x, y, z) = (0, 0, 0) (x, y, z) = (0, 0, 0) (0, 0, 0) /∈ Σ
f(0, 0, 0) = −1 = 0 Σ
Σ 2
z = ± 1 − x 2 − y 2 ϕ1(u, v) = (u, v, √ 1 − u 2 − v 2 )
ϕ2(u, v) = (u, v, √ 1 − u 2 − v 2 ) u 2 + v 2 ≤ 1
⎠

dσ = det 2 B1 + det 2 B2 + det 2 B3 du dv
Jac ϕ
dσ = |∂uϕ(u, v) ∧ ∂vϕ(u, v)| du dv
(x0, y0, z0)
(u0, v0) ∈ I×J ϕ(u0, v0) = (x0, y0, z0)
ˆn(x0, y0, z0) = ∂uϕ(u0, v0) ∧ ∂vϕ(u0, v0)
|∂uϕ(u0, v0) ∧ ∂vϕ(u0, v0)|
(u0, v0)
v v/|v|
Σ ϕ
ˆn(x, y, z)
(u, v) ∈ I × J ˆn ◦ ϕ(u, v) · ∂uϕ(u, v) ˆn ◦ ϕ(u, v) · ∂vϕ(u, v)
det
⎛
⎝ n1 ◦ ϕ(u, v) ∂uϕ1(u, v) ∂vϕ1(u, v)
n2 ◦ ϕ(u, v) ∂uϕ2(u, v) ∂vϕ2(u, v)
n3 ◦ ϕ(u, v) ∂uϕ3(u, v) ∂vϕ3(u, v)
(u, v)
Σ (ū, ¯v)
P (¯x, ¯y, ¯z) Σ (ū, ¯v)
ϕ(ū, ¯v) = P (¯x, ¯y, ¯x) (ū, ¯v)
Σ f(x, y, z) = 0 ∇f(x, y, z)
Σ Σ
Σ ∇f
Σ
ϕ f = 0
∇f
∂uϕ ∧ ∂vϕ
F
⎛
F · ˆn dσ = det ⎝ F1
⎞
◦ ϕ ∂uϕ1 ∂vϕ1
F2 ◦ ϕ ∂uϕ2 ∂vϕ2 ⎠ du dv
S
I
J
F3 ◦ ϕ ∂uϕ3 ∂vϕ3
γ G γ : [0, T ] → R 3 C 1
γ
G · ds =
T
0
G(γ(t)) · ˙γ(t) dt.
Σ ψ : [a, b]×[c, d] → R 3
γ1(t) = ψ(t, c) t ∈ [a, b]
γ2(t) = ψ(b, t) t ∈ [c, d] γ3(t) = ψ(b + a − t, d) t ∈ [a, b] γ4(t) = ψ(a, d + c − t)
t ∈ [a, b] ψ [a, b] × [c, d]
ψ(a, t) = ψ(b, t) t ∈]c, d[ ψ(t, c) = ψ(t, d) t ∈]a, b[
⎞
⎠ .

ψ(a, t) = ψ(b, t) t ∈]c, d[ ψ(t, c) = ψ(t, d) t ∈]a, b[
γ1 γ3
ψ(a, t) = ψ(b, t) t ∈]c, d[ ψ(t, c) = ψ(t, d) t ∈]a, b[
γ2 γ4
ψ(a, t) = ψ(b, t) t ∈]c, d[ ψ(t, c) = ψ(t, d) t ∈]a, b[
F Σ
F Σ
rot
F · ˆn dσ = F · ds.
Σ
F
R3
Ω C F
F · ˆn dσ = div F (x, y, z) dx dy dz.
C
Ω
Ω
S div F = 0 S
γ Π Π Σ
Σ ∩ S = γ Σ S
S ∪ Σ = C Ω
F · ˆn dσ + F · ˆn dσ = F · ˆn dσ = div F (x, y, z) dx dy dz = 0,
S
Σ
C
Σ Π
F Σ F
Σ Σ
Π
f : R → R T f(x) = f(x + T ) x ∈ R
a ∈ R
T
0
f(x) dx =
T/2
−T/2
f(x) dx =
γ
Ω
a+T
f : R → R f(x) = f(−x)
a
f(x) dx
a
g : R → R g(−x) = −g(x)
f(x) dx = 2
−a a
−a
a
0
g(x) dx = 0
f(x) dx

p, q ∈ N q
2π
0
cos p θ sin q θ dθ =
π
−π
cos p θ sin q θ dθ = 0
2π
p ∈ N
2π
cos 2 pθ dθ =
0
2π =
2π
0
2π
sin 2 pθ dθ =
2π
sin 2 (pθ + π/2) dθ =
0
cos 2 pθ dθ = π
5/2π
0
π/2
2π
(cos
0
2 pθ + sin 2 2π
pθ) dθ = 2
2π
0
m, n ∈ N m = n
0
sin 2 pσ dσ =
cos 2 pθ dθ
cos mθ sin nθ dθ = 0
2π
0
sin 2 (pσ) dσ,
z = tan(θ/2)
2π
0
cos 4 θ dθ =
= 1
2 4
e iθ + e −iθ
2π
4
dθ =
0 2
1
24 2π
(e
0
2iθ + e −2iθ + 2) 2 dθ
2π
(e
0
4iθ + e −4iθ + 4 + 2 + 4e 2iθ + 4e −2iθ ) dθ = 1
2π
8 0
sin θ = 2z
1 − z2
2 dz
, cos θ = , dθ = .
1 + z2 1 + z2 1 + z2 (cos(4θ) + 3 + 4 cos(2θ)) dθ = 3π
4 .

[
]
˙x(t) = f(t, x(t)),
x(t0) = x0.
K Y K = R C
Y = R n
I R I t0 ϕ : I → Y
ϕ C1 I ϕ(t0) = x0 I
ϕ I ϕ I
ψ J ⊂ R J ⊃ Ī Ī
I ψ = ϕ I
f t f = f(x)
Y K I R f : I × Y → Y
y ∈ Y t ∈ I
L > 0
f(t, y1) − f(t, y2)Y ≤ Ly1 − y2Y
t ∈ I y1, y2 ∈ Y t0 ∈ I y0 ∈ Y
ϕ ∈ C 1 (I, Y ) ϕ ′ (t) = f(t, ϕ(t)) I ϕ(t0) = y0
Y K I R f : I × Y → Y
K ⊆ I f y ∈ Y

t ∈ I LK > 0
f(t, y1) − f(t, y2)Y ≤ LKy1 − y2Y
t ∈ K K ⊆ I y1, y2 ∈ Y t0 ∈ I y0 ∈ Y
ϕ ∈ C 1 (I, Y ) ϕ ′ (t) = f(t, ϕ(t)) I ϕ(t0) = y0
Y K I R K I f : I × Y → Y
K × Y
∂Y f(t, y) L(Y ) ≤ LK < +∞
(t, y) ∈ K × Y Y K n
∂y1 f(t, y), ..., ∂ynf(t, y) K × Y
Y K Ω R × Y f : Ω → Y
(t0, y0) ∈ Ω L, δ0, r0 > 0 B(t0, δ0] × B(y0, r0] ⊆ Ω
f(t, y1) − f(t, y2)Y ≤ Ly1 − y2Y
(t, y) ∈ B(t0, δ0] × B(y0, r0] (t0, y0) ∈ Ω δ > 0 ϕ ∈
C 1 (B(t0, δ], Y ) y ′ = f(t, y) y(t0) = y0 ψ ∈
C 1 (B(t0, δ], Y ) t0 ϕ(t) = ψ(t)
t0
Y K Ω R×Y f : Ω → Y
f
∂Y f(t, y) Ω Y Kn ∂ykf(t, y)
k = 1, ..., n Ω
y ′ = f(t, y)
I R φ, ψ : I → Y y ′ = f(t, y)
I
y ′ = f(t, y)
y(t0) = y0
Φ(t, t0, y0) t y(t0) = y0
˙y = f(t, y) y(t0) = y0 (t0, y0) Φ
D ⊃ I × I × Ω I t0 Ω
y0 Φ : D → Y
f t t ↦→ y(t)
t ↦→ y(t + c) Φ(t, y0) =
φt(y0) t y(0) = y0
φ0(y0) = y0 φs ◦ φt(y0) = φs+t(y0)
Y = R n y = f(y)
Y n = 1, 2, 3

Y K Ω R × Ω
I R f : Ω → Y
ϕ : I → Y
β = sup I α = inf I c ∈ I ϕ ′ (t)
[c, β[ ]α, c] β = +∞ α = −∞
lim t→β − ϕ(t) = yβ lim t→α + ϕ(t) = yα Y
(β, yβ) /∈ Ω (β, yβ) /∈ Ω
K Ω U a = inf I
V b = sup I t ∈ U ∪ V (t, ϕ(t)) /∈ K
Ω
K Ω δ = δ(K) > 0 K f
(t0, y0) ∈ Ω [t0 − δ, t0 + δ]
A Y g : A → Y
ϕ : I → A y ′ = g(y) C A
b = sup I
V b ϕ(t) /∈ C t ∈ V ϕ
C
b = +∞
a = inf I
y ′ = g(y) g : A → Y
A Y
E ∈ C 1 (A, R) ϕ : I → A E ◦ ϕ
I R µ : I × [0, +∞[→
[0, +∞[ Y ϕ : I → Y u : I → [0, +∞[
ϕ(t0)Y ≤ u(t0)
ϕ ′ (t)Y < µ(t, ϕ(t)Y ) µ(t, u(t)) ≤ u ′ (t) t ≥ t0 t ∈ I t
ϕ(t)Y ≤ u(t)
ϕ ′ (t)Y < µ(t, ϕ(t)Y ) µ(t, u(t)) ≤ −u ′ (t) t ≤ t0 t ∈ I t
ϕ(t)Y ≤ u(t)
t = t0 ϕ(t)Y < u(t)
ϕ ′ (t)Y ≤ µ(t, ϕ(t)Y ) µ(t, u(t)) < u ′ (t) ϕ ′ (t)Y ≤ µ(t, ϕ(t)Y )
µ(t, u(t)) < −u ′ (t)
Ω R × R f : Ω → R I
R t ↦→ y(t) t ↦→ u(t) I t0 ∈ I y(t0) ≤ u(t0)
t > t0 t ∈ I y ′ (t) ≤ f(t, y(t)) f(t, u(t)) ≤ u ′ (t)
t ≥ t0 y(t) ≤ u(t) t ∈ I t ≥ t0
t0
I R f, g : I × R → R
x : I → R y : I → R
˙x ≤ f(t, x(t)) ˙y ≥ g(t, y(t)) t ∈ I
f(t, x(t)) ≤ g(t, y(t)) t ∈ I
x(t0) ≤ y(t0) x(t) ≤ y(t) t ∈ I t ≥ t0
x(t0) ≥ y(t0) x(t) ≥ y(t) t ∈ I t ≤ t0
a > 0 x : [a, +∞[→ R
limt→+∞ x(t) ∈ R limt→+∞ ˙x(t) = γ ∈ R ∪ {±∞} γ = 0
I R to ∈ I Y

ϕ ∈ C1 (I, Y ) ϕ ′ (t) ≤ a0 + a1ϕ(t) t ∈ I a1 > 0 a0 ≥ 0
a0
ϕ(t) ≤ + ϕ(t0) e
a1
a1|t−t0| a0
− .
a1
ψ ∈ C0 (I, R) L, M ≥ 0 t ∈ I
t
|ψ(t)| ≤ L
ψ(τ) dτ
+ M,
t ∈ I |ψ(t)| ≤ Me L|t−t0|
t0
Y = R n Ω R × Y f : Ω → Y
I t0 R ϕ ∈ C 1 (I, Y ) I
y ′ = f(t, x) y(t0) = y0

1 ω(x, y) = M(x, y) dx + N(x, y) dy
M N D R 2
ω(x, y) = 0
F (x, y) λ(x, y)
dF (x, y) = λ(x, y)ω(x, y) λ(x, y) = 0 D
F (x, y) = c c ∈ R S = {(x, y) ∈ D : M(x, y) = N(x, y) = 0}
D \ S
ω(x, y) = 0
ω(x, y) = M(x, y) dx + N(x, y) dy ,
ω F (x, y) dF = ω
∂F
(x, y) = M(x, y),
∂x
∂F
(x, y) = N(x, y).
∂y
F (x, y) ω F (x, y) = c c ∈ R
λ(x, y) ≡ 1
∂M
∂y
∂N
(x, y) = (x, y).
∂x
dF (x, y) = ω(x, y) = 0
ω(x, y) = 0
ω(x, y) = M(x) dx + N(y) dy
f(x) M g(y) N
f(x) + g(y) = c c ∈ R
ω(x, y) =
0
ω(x, y) = ϕ(x)ψ(y) dx + ϕ1(x)ψ1(y) dy
ψ(y) = 0 ϕ1(x) = 0 ψ(y)ϕ1(y)
ω(x, y) = M(x) dx+N(y) dy
M N D α ∈ R k > 0
M(kx, ky) = k α M(x, y), N(kx, ky) = k α N(x, y),

C R 2
x = ξ, y = ξη
1
N(1, η)
dξ +
dη = 0.
ξ M(1, η) + ηN(1, η)
M, N α = −1
D
F (x, y) = 1
[x · M(x, y) + y · N(x, y)].
α + 1
A ω = 0
g : A → R C1 gω ω = 0
gω = 0 g > 0 g = ef f : A → R
ω = p(x, y) dx + q(x, y) dy
ef ω
∂yp − ∂xq = −p∂yf + q∂xf
∂yp − ∂xq = h(x)q e h(x) dx
∂yp(x, y) − ∂xq(x, y)
q(x, y)
x e h(x) dx
h(x) = ∂yp(x, y) − ∂xq(x, y)
;
q(x, y)
∂yp − ∂xq = k(y)p e − k(y) dy
∂yp(x, y) − ∂xq(x, y)
−p(x, y)
y e k(y) dy
k(y) = ∂yp(x, y) − ∂xq(x, y)
;
−p(x, y)
∂yp − ∂xq = f(x)q(x, y) − g(y)p(x, y)
f, q, p C1
x y
h(x, y) = exp f(t)dt + g(t)dt
ω
x0
x r y s (my dx + nx dy) + x ρ y σ (µy dx + νx dy) = 0
r, s, ρ, σ, m, n, µ, ν mν − nµ = 0 x α y β
α, β
M(x, y) dx + N(x, y) dy = yf(xy) dx + xg(xy) dy = 0
1
f = g
Mx − Ny
C ⊆ R 2 R 2 (x, y) ∈ C (kx, ky) ∈ C
k > 0
y0

M(x, y) dx + N(x, y) dy = 0
1
M, N Mx+Ny = 0
Mx + Ny
x dy − y dx
x dy − y dx
x dy − y dx
x dy − y dx
x dy + y dx
x dy + y dx
1
x 2
1
y 2
1
xy
1
x 2 + y 2
1
(xy) n
1
(x 2 + y 2 ) n
y
d
x
d − x
y
d ln y
x
d arctan y
x
⎧
⎪⎨
d(ln(xy)) n = 1
⎪⎩
1
d −
(n − 1)(xy) n−1
n = 1
⎧
1
⎪⎨
d
2 ln(x2 + y 2
)
⎪⎩ d −
1
2(n − 1)(x 2 + y 2 ) n−1
n = 1
n = 1
λω = 0
P (x0, y0) ∈ D N(x0, y0) = 0
dy y)
= −M(x,
dx N(x, y)
P λω F
F ∂yF (x0, y0) = λ(x0, y0)N(x0, y0) = 0 F
P (x0, y0) y = y(x) y0 = y(x0) M, N ∈ C 1
λ = 0 N(x, y) = 0 P (x0, y0)
y = y(x) C 1
dy y)
= −M(x,
dx N(x, y) .
M(x0, y0) = 0
dx
dy
= − N(x, y)
M(x, y)
P
y ′ = f(x, y)
f(x, y) dx − dy = 0.

F
Fc := {(x, y) ∈ A : F (x, y) = c}
N(x, y(x)) dy
dx
+ M(x, y(x)) = 0, M(x(y), y) + N(x(y), y) = 0.
dx dy
ω = 0 γ C1
P (x0, y0) (x, y) ∈ D ω 0 P
F (x, y) = ω
D
P (x0, y) (x, y) P (x, y0) (x, y)
F (x, y) =
F (x, y) =
x
x0
x
x0
γ
M(t, y) dt +
y
y0
y
M(t, y0) dt +
y0
N(x0, s) ds.
N(x, s) ds.
ω(x, y) C l ω G ω = 0
C l+1 D λ ∈ C l (A, R)
∂xG(x, y) = λ(x, y)p(x, y)
∂yG(x, y) = λ(x, y)q(x, y)
λ ∈ C l (D, R) λω λω
ω = 0
R 3
R 3
ω(x, y, z) = P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz
R 3 ω = 0
P (∂zQ − ∂yR) + Q(∂xR − ∂zP ) + R(∂yP − ∂xQ) = 0
ω(x, y, z) = dF (x, y, z) F (x, y, z) = C ∈ R
ω(x, y, z) λ(x, y, z)
λω = dF F (x, y, z) = C ∈ R λ(x, y, z) = 0
z
φ(z)
φ(z)
R 3
ω1(x, y, z) = 0 ω2(x, y, z) = 0 F1(x, y, z) = C1 ∈ R
F2(x, y, z) = C2 ∈ R
ω1 ω2 λ1 λ2
ω1 ω2 ω1 = 0 F1(x, y, z) = C
ω1 = 0 ω2 = 0

x, y
z dy dz
ω1 = P1 dx + Q1 dy + R1 dz = 0
⎧⎪ ⎨
⎪⎩
ω2 = P2 dx + Q2 dy + R2 dz = 0
P1 Q1
det
det
λ = 0
X = λ det
Q1 R1
Q2 R2
P2 Q2
R1 P1
R2 P2
Q1 R1
dx − det
dz = 0
Q2 R2
P1 Q1
dx − det
dy = 0
dx
X
= dy
Y
= dz
Z
R1 P1
, Y = λ det
R2 P2
P2 Q2
P1 Q1
, Z = λ det
P2 Q2
Y dx = X dy, Y dz = Z dy, X dz = Z dx.
dx dy dz
= =
X Y Z = l1 dx + m1 dy + n1 dz
l1X + m1Y + n1Z = l2 dx + m2 dy + n2 dz
l2X + m2Y + n2Z
l1, m1, n1, l2, m2, n2
lX + mY + nY = 0
l dx + m dy + n dz = 0
.

K
Y K = R C Y = R n
I R
t ∈ R A ∈ Matn×n(C)
e tA = (tA) j
A
j!
j∈N
λ1, ..., λn e tA
eλ1t , ..., eλnt P P AP −1 = D P etAP −1 = etD K[x]
K N, D ∈ R[x]
f(x) = N(x)/D(x) N D
f
D N
N D
Q, R ∈ R[x] f(x) = Q(x) + R(x)/D(x)
f Q
R(x)/D(x) R D
f(x) = N(x)/D(x) N, D
D N
x1, ..., xd ∈ R D α1 + iβ1, α1 − iβ1, ..., αh +
iβh, αh − iβh ∈ C \ R D D
Akjk , Bℓ sℓ , Cℓ sℓ ∈ R f
xk νk
Ak1
+
x − xk
Ak2
+ ... +
(x − xk) 2 Akνk
(x − xk) νk
αℓ + iβℓ αℓ − iβℓ
µℓ
Bℓ 1x + Cℓ 1
(x − αℓ) 2 + β 2 ℓ
+
Bℓ 2x + Cℓ 2
((x − αℓ) 2 + β2 ℓ )2 + ... + Bℓ µℓx + Cℓ µℓ
((x − αℓ) 2 + β2 .
)µℓ
ℓ

f(x) = N(x) Ak1
= +
D(x) x − xk
k
Ak2
+ ... +
(x − xk) 2 Akνk + νk (x − xk)
+ Bℓ 1x + Cℓ 1
+
ℓ
(x − αℓ) 2 + β 2 ℓ
Bℓ 2x + Cℓ 2
((x − αℓ) 2 + β2 ℓ )2 + ... + Bℓ µℓx + Cℓ µℓ
((x − αℓ) 2 + β2 .
)µℓ
ℓ
D(x)
Akjk , Bℓ sℓ , Cℓ sℓ ∈ R f
Akjk , Bℓ sℓ , Cℓ sℓ ∈ R
⎧
1
⎪⎨ −
+ C, n ∈ N, n > 1;
dx (n − 1)(x + a) n−1
=
(x + a) n
⎪⎩
log |x + a| + C, n = 1.
x
(x2 dx =
+ 1) n
⎧
1
⎪⎨
−
2(n − 1)(x2 + C, n ∈ N, n > 1;
+ 1) n−1
⎪⎩ 1
2 log |x2 + a| + C, n = 1.
In =
dx
(x2 I1 = arctan x + C n > 1
, n ∈ N
+ 1) n
x
In = −
2(n − 1)(1 + x2 2n − 3
+
) n−1 2n − 2 In−1,
I × Y
y ′ (t) = A(t)y(t) + b(t),
A ∈ C 0 (I, LK(Y )) b ∈ C 0 (I, Y ) LK(Y )
Y Y n LK(Y ) n × n
K A(t) n × n
I K b(t) = 0
t y ′ (t) = A(t)y(t) y ′ = A(t)y + b(t)
Y = R C
y ′ (t) = a(t)y
a ∈ C 0 (I, K) I R
c0 ∈ K A(t) ∈
y(t) = c0e A(t)
a(t) a(t)
y ′ (t) = a(t)y + b(t)

a, b ∈ C 0 (I, K) I R A(t) ∈ a(t) a(t) B(t) ∈ e −A(t) b(t)
e −A(t) b(t) c ∈ K
y(t) = ce A(t) + e A(t) B(t)
y(t0) = y0
t t t t
y(t) = y0 exp a(t) dt + exp a(t) dt exp − a(t) dt b(t) dt,
exp(x) = e x
t0
t0
Y t0 ϕ1, ϕ2 :
I → Y y ′ (t) = A(t)y(t) λ, µ ∈ K ϕ : I → Y
ϕ(t) = λϕ1(t) + µϕ2(t) y ′ (t) = A(t)y(t)
y ′ = A(t)y+b(t)
y ′ = A(t)y
w : I → Y
w + ϕ ϕ
y ′ = A(t)y + b(t) b(t) = b1(t) + ...bk(t) ϕi : I → Y
y ′ = A(t)y + bi(t) i = 1, ..., k ϕ : I → Y ϕ(t) =
ϕ1(t) + ... + ϕk(t) y ′ = A(t)y + b(t)
Y = R b(t) = f(t) cos(αt)
b(t) = 1
t0
2 f(t)eiαt + 1
2 f(t)e−iαt
b(t) = f(t) sin(αt)
1
2 f(t)e±iαt
ϕ1, ..., ϕr y ′ = A(t)y
ϕ1, ..., ϕr ∈ C 1 (I, Y )
C 1 (I, Y )
t0 ∈ I ϕ1(t0), ..., ϕr(t0) ∈ Y
t ∈ I ϕ1(t), ..., ϕr(t) ∈ Y
y ′ = A(t)y φ : I × I × Y → Y
φ(t, t0, y0) t t0 y0
t0 y0 t, t0 ∈ I y ↦→ φ(t, t0, y)
Y Y φ(t, t0, y0) = R(t, t0)y0 R(t, t0) ∈ LK(Y ) Y
n R(t, t0) n × n t t0
idY Y
t2, t1, t0 ∈ I
R(t0, t0) = idY
R(t2, t1) ◦ R(t1, t0) = R(t2, t0)
t0 ∈ I Rt0 (t) X′ = A(t)X X : I → LK(Y )
X n × n
Φ ∈ C1 (I, LK(Y )) Φ ′ = A(t)Φ(t) t y0 ϕy0 (t) =
Φ(t)y0 y ′ = A(t)y Φ(t) t ∈ I Φ(t)
t ∈ I y ′ = A(t)y Φ ∈ C1 (I, LK(Y )) Φ(t) ′ = A(t)Φ(t)
t ∈ I
t0

Φ y ′ = A(t)
y ′ = A(t)y + b(t) u(0) = y0
u(t) = R(t, t0)y0 +
t
t0
R(t, τ)b(τ) dτ
Φ R(t, τ) = Φ(t)Φ −1 (τ)
n
y (n) + ... + any = Q(x)
y(x) = c1y1(x) + ... + cnyn(x) ck ∈ R
ck ck = ck(x) n n c ′ k (x)
⎧
⎪⎨
⎪⎩
c ′ 1 (x)y1(x) + ... + c ′ n(x)yn(x) = 0,
c ′ 1 (x)y′ 1 (x) + ... + c′ n(x)y ′ n(x) = 0,
c ′ 1 (x)y(n−1) 1 (x) + ... + c ′ n(x)y (n−1)
n (x) = 0,
c ′ 1 (x)y(n) 1 (x) + ... + c′ n(x)y (n)
n (x) = Q(x).
⎛
y1(x)
⎜ y
⎜
⎝
... yn(x)
′ 1 (x) ... y′
y
n(x)
(n−1)
⎞ ⎛
⎟ ⎜
⎟ ⎜
⎟ ⎜
⎟ ⎜
⎟ ⎜
(x) ⎠ ⎝
1 (x) ... y (n−1)
n
y (n)
1
(x) ... y(n)
n (x)
c ′ 1 (x)
c ′ 2 (x)
c ′ n−1
c ′ n
⎞
⎛
⎟ ⎜
⎟ ⎜
⎟ ⎜
⎟ = ⎜
⎟ ⎜
⎠ ⎝
0
0
0
Q(x)
y1(x), ..., yn(x)
c ′ k (x) ck(x)
¯y(x) = c1(x)y1(x) + ... + cn(x)yn(x),
y(x) + ¯y(x) = (c1(x) + c1)y1(x) + ... + (cn(x) + cn)yn(x), c1, ..., cn ∈ R.
y ′′ + a(t)y ′ + b(t)y = f(t),
˜y = c1(t)y1(t) + c2(t)y2(t)
y1(t) y2(t)
y ′′ + a(t)y ′ + b(t)y = 0.
˜y ′ = c ′ 1y1 + c ′ 2y2 + c1y ′ 1 + c2y ′ 2.
c ′ 1y1 + c ′ 2y2 = 0 ˜y ′ = c1y ′ 1 + c2y ′ 2
˜y ′′ = c ′ 1y ′ 1 + c ′ 2y ′ 2 + c1y ′′
1 + c2y ′′
2.
(c ′ 1y ′ 1 + c ′ 2y ′ 2 + c1y ′′
1 + c2y ′′
2) + a(c1y ′ 1 + c2y ′ 2) + b(c1y1 + c2y2) = f
c1(y ′′
1 + ay ′ 1 + by1) + c2(y ′′
2 + ay ′ 2 + by2) + (c ′ 1y ′ 1 + c ′ 2y ′ 2) = f.
⎞
⎟
⎠

y1 y2
c ′ 1y ′ 1 + c ′ 2y ′ 2 = f
c ′ 1 c ′ 2
c ′ 1y1 + c ′ 2y2 = 0
c ′ 1y′ 1 + c′ 2y′ 2 = f.
y1 y2
y ′ 1 y′
2
y1 y2
−y2f
c ′ 1 =
y ′ 2y1 − y ′ 1y2 y1f
c ′ 2 =
y ′ 2y1 − y ′ 1y2 c ′ 1 c ′ 2
˙z = Az + b(t) R 2
A =
5 3
2 3
, b(t) =
t
e −t
2
˙z = Az
A
A u ∈ R 2 Au = λu λ ∈ C
u det(λid R 2 − A) = 0
0 = det
λ − 5 −3
−2 λ − 3
.
= (λ − 5)(λ − 3) − 6 = λ 2 − 8λ + 9 = 0,
λ1 = 4 + √ 7 λ2 = 4 − √ 7
u1, u2
(λiid R 2 − A)ui =
λi − 5 −3
−2 λi − 3
ui = 0, i = 1, 2.
−2u x 1 + (1 + √ 7)u y
1 = 0, −2ux2 + (1 − √ 7)u y
2 = 0
ui = (ux i , uy
i ) u1 =
(1 + √ 7, 2) u2 = (1 − √ 7, 2) P u1
u2
√
1 + 7
P =
2
√
1 − 7
,
2
√
1/(2 7)
P =
−1/(2
√ √
(−1 + 7)/(4 7)
√ 7) (1 + √ 7)/(4 √
,
7)

det P = 4 √ 7 = 0 P −1 AP
w = P −1 z ˙w = P −1 ˙z = P −1 AP w
˙w1
˙w2
=
4 + √ 7 0
0 4 − √ 7
w1
˙wi = λiwi wi(t) = Cieλit i = 1, 2
z = P w
√ √
z1(t) 1 + 7 1 − 7 C1e
z(t) = =
z2(t) 2 2
λ1t
= C1e λ1t
u1 + C2e λ2t
u2.
C2e λ2t
z0 z0 = C1u1 + C2u2 C1 C2
z0
C1
= P −1
C1
z0, z0 = P .
T (t) =
C2
e λ1t 0
0 e λ2t
w2
,
C2
, T −1
e−λ1τ 0
(τ) =
0 e−λ2τ
= T (−τ).
T (t)T −1 (τ) = T (t − τ) z0 ∈ R 2 z(t) =
P T (t)P −1 z0 ˙z = Az z(0) = z0
Φ(t) = P T (t)P −1 Φ −1 (τ) = P T −1 (τ)P −1 R(t, τ) = Φ(t)Φ −1 (τ) =
P T (t)T −1 (τ)P −1 = P T (t − τ)P −1
t
u(t) = R(t, 0)z0 + R(t, τ)b(τ) dτ = P T (t)P −1 t
z0 + P T (t − τ)P −1
τ
e−τ
dτ
0
= P T (t)P −1 z0 +
t
0
= P T (t)P −1 z0 + P T (t)P −1
t
= P T (t)P −1
z0 + P
P T (t)P −1 P T (−τ)P −1
τ
t
0
0
T (−τ)P −1
P T (−τ)P −1
τ
τ
e −τ
0
e−τ
dτ
e−τ
dτ
dτ .
y ′ = Ay + b(t) y ′ = Ay A ∈ LK(Y )
t
A
y(0) = y0 ϕ(t) = etAy0
e tA = (tA) j
j!
j∈N
u ∈ Y u = 0 λ ∈ K ϕ(t) = e λt u y ′ = Ay λ
A u A n
λ1, ..., λn
e tA = [e λ1t u1....e λnt un][u1...un] −1

uj n
y(t) = e (t−t0)A y0 +
t
t0
e (t−τ)A b(τ) dτ
y(0) = y0
n n ≥ 1
y (n) + an−1(t)y (n−1) + ... + a1(t)y ′ + a0(t)y = b(t)
b, a0, ..., an−1 ∈ C 0 (I, K) I R
y (n) + an−1(t)y (n−1) + ... + a1(t)y ′ + a0(t)y = 0
z1 = y z2 = y ′ zn = y (n−1) z ′ = A(t)z + B(t)
z ′ = A(t)z A(t) ∈ Matn×n(K) A
−a0(t), ..., −an−1(t)
1 0 B(t) b(t)
r ϕ1, ..., ϕr : I → K C m−1 I
m m × r K
i 0 m−1 ϕr r
n r
t0 r t ∈ I
n n ϕ1, ..., ϕn n
w(t)
n 0
wj(t) n n
j
n ϕ1, ..., ϕn
y (n) + an−1(t)y (n−1) + ... + a1(t)y ′ + a0(t)y = 0
y (n) + an−1(t)y (n−1) + ... + a1(t)y ′ + a0(t)y = b(t)
t0 ∈ I n − 1
n
ϕ0(t) = γj(t)ϕj(t)
w(t) wj(t)
γj(t) =
t
t0
j=1
(−1)
n+j wj(τ)
w(τ)
b(τ) dτ
n
y (n) + an−1y (n−1) + ... + a1y ′ + a0y = b(t),
y (n) + an−1y (n−1) + ... + a1y ′ + a0y = 0
p(z) = z n + an−1z n−1 + ... + a1z + a0 λ p(z) ν
e λt , te λt , ..., t ν−1 e λt
λ = αj + iβj ¯ λ = αj − iβj t k e λjt t k e ¯ λjt t k e αjt cos(βjt)
t k e αjt sin(βjt)
n = 2

y ′′ (t) + py ′ (t) + qy(t) = 0
α, β ∈ K ζ 2 + pζ + q = 0
y(t) = c1e αt + c2e βt α = β
y(t) = c1e αt + c2te αt α = β
c1, c2 ∈ K
y ′ + ω 2 y = 0 ω ∈ R
y(t) = c1 cos(ωt) + c2 sin(ωt) = A cos(ωt + φ)
c1, c2, A, φ ∈ K
y(t0) = y0 y ′ (t0) = y ′ 0 t0 ∈ I α, β ∈ K
ζ2 +pζ +q = 0 C2 (I, K) y ′′ (t)+py ′ (t)+
qy(t) = b(t)
α = β
y(t) = y′ 0 − βy0
α = β
α − β eα(t−t0) + y ′ 0
− αy0
β − α eβ(t−t0) +
t
y(t) = y0e α(t−t0) ′
+ (y 0 − αy0)(t − t0)e α(t−t0)
t
+
b(t) = 0 y0 = y ′ 0
= 0 y = 0
t0
eα(t−s) − eβ(t−s) b(s) ds
α − β
t0
(t − s)e α(t−s) b(s) ds
n
a(t) t
α ∈ C
y (n) + an−1y (n−1) + ... + a1y ′ + a0y = a(t)e αt .
α
c(t)e αt c a
α ν
t ν c(t)e αt c a
t ↦→ c(t)
y (n) + an−1y (n−1) + ... + a1y ′ + a0y = b(t).
b(t) 0
xp(t) b(t)
0 ν
xp(t) = t ν c(t) c(t) b(t)

(t) = a(t)e αt α ∈ R a(t) α
xp(t) = c(t)e αt
c(t) a(t) α
ν xp(t) = t ν c(t)e αt c(t)
a(t) α = 0
b(t) = a(t) cos βx b(t) = a(t) sin βx a(t)
iβ
xp(t) = c(t)(A cos βt + B sin βx) c(t) a(t)
iβ ν
xp(t) = t ν c(t)(A cos βt+B sin βx) c(t) a(t)
b(t)
b(t) = a(t)e αx cos βx b(t) = a(t)e αx sin βx a(t)
α + iβ
xp(t) = c(t)e αx (A cos βt + B sin βx) c(t)
a(t) iβ ν
xp(t) = t ν c(t)e αx (A cos βt + B sin βx) c(t)
a(t) b(t)
c(t) xp(t)
xp(t)
h(t) = h1(t) + .... + hk(t) xj(t)
y (n) + an−1y (n−1) + ... + a1y ′ + a0y = hj(t)
j = 1...k x(t) = x1(t) + ... + xk(t) a¨x + b ˙x + cx(t) = h1(t) + ... +
hk(t) = h(t) h(t)
b(t)
y ′′ + 2y ′ + y = sin(2t) y(0) = 1 y ′ (0) = 2
y ′′ + 2y ′ + 1 = 0 λ 2 + 2λ + 1 = 0
λ = −1 ν = 2 sin 2t = (e i2t − e −i2t )/(2i)
y ′′ +2y ′ +1 = e i2t /(2i) y ′′ +2y ′ +1 = e i2t /(2i)
c(t)e αt α = 2i c(t) = 1/2i α = 2i
y ′′ + 2y ′ + 1 = e i2t /(2i)
c3e 2it c3 ∈ C
y ′′ + 2y ′ + 1 = e −i2t /(2i) c4e −2it c4 ∈ C
y ′′ + 2y ′ + y = 0 c1e −t + c2te −t c1, c2
y ′′ + 2y ′ + y = sin(2t)
y(t) = c1e −t + c2te −t + c3e 2it + c4e −2it = c1e −t + c2te −t + d1 cos(2t) + d2 sin(2t)
= (c1 + tc2)e −t + d1 cos(2t) + d2 sin(2t),
c1 c2 d1 d2

y(0) = c1 + d1 = 1
˙y(t) = (−c1 + c2 − tc2)e t + 2d2 cos(2t) − 2d1 sin(2t)
˙y(0) = −c1 + c2 + 2d2 = 2
¨y(t) = (c1 − 2c2 + tc2)e t − 4d1 cos(2t) − 4d2 sin(2t)
¨y(t) + 2 ˙y + y = sin(2t) = (−3d1 + 4d2) cos(2t) − (4d1 + 3d2) sin(2t).
4d1 + 3d2 = −1 4d2 − 3d1 = 0 c1 + d1 = 1
−c1 + c2 + 2d2 = 2 d1 = −4/25 d2 = −3/25 c1 = 29/25 c2 = 85/25
y(t) = 1 −t
(29 + 85t)e − 4 cos(2t) − 3 sin(2t) .
25

y ′ = f(x, y)
A
A
F (x, y, y ′ ) = 0
y ′ = f(x, y)
˙y = p(t)q(y)
p : I → R q : J → R I, J R q(y0) = 0
y(t) = y0 q(y)
q(y(t)) = 0 η = y(t)
y(t0) = y0
y dη
y0 q(η) =
t
p(τ)dτ.
t0
y ′ = f(ax + by) f a, b ∈ R \ {0} z = ax + by
z ′ = a + bf(z) z(x)
y(x) = z(x)−ax
b
y ′
y
= f
x
0 z =
y
x z′ = f(z)−z
x
y(x) = xz(x)
M(x, y) dx + N(x, y) dy = 0
M, N y = xz
x = 0 x = yz y = 0
y ′
= f
f a, b, c, a ′ , b ′ , c ′
ax+by+c
a ′ x+b ′ y+c ′
ab ′ − a ′ b = 0 c, c ′ ax + by + c = 0

a ′ x + b ′ y + c ′ = 0 (α, β) = (0, 0)
x = u + α y = v + β
dv
= f
du
a + b v
u
a ′ + b ′ v
u
v = v(u) y = β + v(x − α)
y ′ + p(x)y = q(x)
p, q
y = e − p(x) dx
p(x) dx
q(x)e dx + c
y ′ + p(x)y = q(x)y n
p, q n ∈ R n = 0, 1 y 1−n = z z ′ + (1 −
n)p(x)z = (1 − n)q(x) y(x) = [z(x)] 1
1−n
n > 0 y = 0
y
y ′′ (t) = f(t, y ′ (t)) f
z(t) = y ′ (t) z ′ (t) = f(t, z(t))
y(t) = z(t) + C
f ′ (y)y ′ + f(y)P (x) = Q(x).
v = f(y) v ′ +P (x)v =
Q(x)
F (x, y, y ′ ) = 0 F (x, y, y ′ ) =
0 F A A
y = y(x) y = y(x) y ′ = y ′ (x)
A
F (x, y, y ′ ) = 0 Fy ′(x, y, y′ ) = 0 y ′
ϕ(x, y) = 0 y = y(x)
y = y(x) F (x, y, y ′ ) = 0
Fy ′(x, y, y′ ) = 0
y ′′ (t) = f(y(t), y ′ (t)) f
p(y) = y ′ (t(y)) y ↦→ t(y)
t ↦→ y(t)
p(y) dp
= f(y, p(y))
dy
y
F (t, y(t), y ′ (t), y ′′ (t)) = 0
F (t, αx, αy, αz) = α k F (t, x, y, z)
α > 0 y ′ (t) = y(t)z(t)
t n y (n) + cn−1t n−1 y (n−1) + ... + c1ty ′ + c0y = 0

t > 0 t = es
y(es ) = u(s) t dy(t) du(s)
dt = ds
q2(x) = 0 v(x) = y(x)q2(x)
Q(x) = q2(x)q0(x) P (x) = q1(x) +
v ′ (x) = (y(x)q2(x)) ′
y ′ (x) = q0(x) + q1(x)y + q2(x)y 2 (x).
v ′ (x) = v 2 (x) + P (x)v(x) + Q(x),
q ′
2 (x)
q2(x)
= y ′ (x)q2(x) + y(x)q ′ 2(x) = (q0(x) + q1(x)y(x) + q2(x)y 2 (x))q2(x) + v(x) q′ 2 (x)
q2(x)
= q0(x)q2(x) + q1(x) + q′ 2 (x)
v(x) + v
q2(x)
2 (x).
v(x) = −u ′ (x)/u(x) u(x)
u ′′ (x) − P (x)u ′ (x) + Q(x)u(x) = 0,
v ′ = −(u ′ /u) ′ = −(u ′′ /u)+(u ′ /u) 2 = −(u ′′ /u)+v 2 u ′′ /u = v 2 −v ′ = −Q−P v =
−Q+P u ′ /u u ′′ −P u ′ +Qu = 0 u
y(x) = −u ′ (x)/(q2(x)u(x))
y1(x)
y(x) = y1(x) + 1/z(x) z(x) z ′ (x) = −(Q(x) +
2y1(x)R(x))z(x) − R(x)
˙y = g(y) t ↦→ y(t) t ↦→ y(t + c)
˙t = 1 y ′ = g(y) g : J → R
J R
g
g(y0) = 0 ˙y = g(y) y(t0) = y0
t0
g(y0) = 0 ]t1, t2[×]η1, η2[ (t0, y0) g(η) = 0
η ∈]η1, η2[ η2
y0
dη
g(η) =
t2
t0
dt,
y0
η1
dη
g(η) =
t0
t1
ti ηi
g(η) > 0 y0 < η < η2 g(η2) = 0
η2
y0
dη
= T < +∞
g(η)
η2 t2 < +∞ t2 =
T + t0
g(η) > 0 y0 < η < η2 t → ∞
η2
η2 dη
= +∞
g(η)
y0
dt

η2 g
g(η) > 0 y0 < η < η2
T t2 = t0 + T +∞
+∞
y0
dη
= T < +∞.
g(η)
⎧
dy
⎪⎨
= f(x(t), y(t)),
dt
⎪⎩ dx
= g(x(t), y(t)).
dt
¯t g(x(¯t), y(¯t)) = 0 ¯t x = x(t)
t = t(x) ¯x = x(¯t) ¯t = t(¯x) x(t(x)) = x y(t(x)) = ˜y(x)
¯x
d˜y dy dt
1
f(x, ˜y(x))
= · = f(x(t(x)), y(t(x))) ·
=
dx dt dx g(x(t(x)), y(t(x))) g(x, ˜y(x))
g = 0
d˜y f(x, ˜y(x))
=
dx g(x, ˜y(x)) .
d˜y
dt
dx
dt
= d˜y f(x, y)
=
dx g(x, y) ,
d˜y f(x, ˜y(x))
=
dx g(x, ˜y(x)) ,
t
dt
dx =
1
g(x, ˜y(x))
g(x, ˜y(x)) = 0
⎧
dy
⎪⎨
= f(x(t), y(t))
dt
⎪⎩ dx
= g(x(t), y(t)).
dt
φt(·) ˙x = f(x)
P ⊆ Rn
φt(z) ⊆ P z ∈ P
φt(z) ⊆ P z ∈ P
φt(z) ⊆ P z ∈ P
t∈R
t≥0
t≤0

Ω ⊆ R n F : Ω → R n C 1
D ⊆ R n C 1 ˙x = F (x) F (x) · ˆn(x) ≤ 0
x ∈ ∂D ˆn D D
D t0 t > t0
D
x(t)
Ω
d
dt dist(x(t), ∂Ω) = ∇d(x(t), ∂Ω) · ˙x(t) = ∇d(x(t), ∂Ω) · F ( ˙x(t)).
Ω C 2 x(t) ¯x ∈ ∂Ω
d
dt dist(x(t), ∂Ω) −ˆn(¯x) · F (¯x) ≥ 0.
∂Ω C 2
x ′ (t) = f(t, x(t))
˙t 1
˙y = =
=:
˙x f(t, x)
F (y),
˙y = F (y) x ′ (t) = f(t, x(t))
˙t −1
˙y = =
=:
˙x −f(t, x)
G(y),
˙y = G(y) x ′ (t) =
f(t, x(t))
x ′ (t) = f(t, x(t))
f C 1
Γ = {(t, x) ∈ R 2 : f(t, x) = 0}.
Γ f(¯t, ¯x) = 0 Ω F (¯t, ¯x) G(¯t, ¯x)
(1, 0) (−1, 0) Γ
R 2 \ Γ
ˆn (¯t, ¯x)
F G
R n
R n
C 1

2 × 2
F ∈ R[x] n
F (x) = a0x n + ... + an, a0, ..., an ∈ R, a0 = 0.
y = y(t) Cn
d
(F (D)y)(t) = a0
ny dy
(t) + ... + an−1 (t) + any(t).
dtn dt
t F (D)y
F1(D)x + G1(D)y = f(t),
F2(D)x + G2(D)y = g(t).
F1, G1, F2, G2 f, g D
F1(D) G1(D)
det
F2(D) G2(D)
x y
2 × 2
˙x − ax − by = f(t)
˙y − cx − dy = g(t)
,
f, g : R → R C 1 (R) a, b, c, d ∈ R
x0 y0
x0 y0

2 × 2
b = 0 ˙x y
˙x = ax + f(t) φ(c1, t)
c1 ˙y = dy+cφ(c1, t)+g(t)
y
c1 c2
c = 0 x y
A B(t) H(t)
y B(t)
a b
f(t)
b f(t)
A = , B(t) = , H(t) =
.
c d
g(t)
d g(t)
T = tr(A) = a + d A D = det(A) = ad − bc
A h(t) = det(H(t)) = bg(t) − df(t)
b = 0 c = 0
x
by = ˙x − ax − f(t)
˙y = cx + dy + g(t)
b ˙y = ¨x − a ˙x − f ′ (t)
˙y
b(cx + dy + g(t)) = ¨x − a ˙x − f ′ (t).
¨x − a ˙x − bcx − bdy − f ′ (t) − bg(t) = 0
by
¨x − a ˙x − bcx − d( ˙x − ax − f(t)) − f ′ (t) − bg(t) = 0.
x
¨x − a ˙x − bcx − d ˙x + adx + df(t) − f ′ (t) − bg(t) = 0.
¨x − (a + d) ˙x + (ad − bc)x + df(t) − f ′ (t) − bg(t) = 0
¨x − T ˙x + D x = ψ(t) ψ(t) = f ′ (t) + h(t)
λ 2 − T λ + D = 0,
A det(A − λId) = 0
T 2 − 4D > 0 λ1 = (T −
√ T 2 − 4D)/2 λ2 = (T + √ T 2 − D)/2
Φ(c1, c2, t) = c1e λ1t + c2e λ2t c1, c2 ∈ R
T 2 − 4D < 0 λ1 =
T +iβ λ2 = T −iβ β = |T 2 − 4D|
Φ(c1, c2, t) = c1e T t cos(βt) + c2e T t sin(βt) c1, c2 ∈ R
T 2 − 4D = 0 λ = T
Φ(c1, c2, t) = c1e T t + c2te T t
c1, c2 ∈ R

2 × 2
t ↦→ x(t)
Φ(c1, c2, t) xp(t) ¨x − T ˙x + D x = ψ(t)
⎧
⎪⎨
x(t) = Φ(c1, c2, t) + xp(t)
⎪⎩ y(t) = 1
1 d
( ˙x − ax − f(t)) =
b b dt Φ(c1, c2, t) + x ′
p(t) − a(Φ(c1, c2, t) + xp(t)) − f(t) .
x
A = 0.
y
det A = 0 x = y = 0
det A = 0 (x, y)
ax + by = 0 det A = 0
λ1 λ2
λ1 = λ2
λ1 = λ2
λ1 = λ2
λ1 = λ2 = λ
λ1 = λ2 = λ
λ1 = α + iβ λ2 = α − iβ α > 0
λ1 = α + iβ λ2 = α − iβ α < 0
λ1 = iβ λ2 = −iβ
a = b = c = d = 0

a∂ttu(t, x) + b∂tu(t, x) + c∂xxu(t, x) + d∂xu(t, x) + eu = 0, (t, x) ∈]0, +∞[×[0, π],
a, b, c, d, e ∈ R c = 0
u(t, x) = U(t)X(x)
a Ü(t)X(x) + b ˙ U(t, x) + cU(t) ¨ X(x) + dU(t) ˙ X(x) + eU(t)X(x) = 0
U(t)X(x) λ ∈ R
− aÜ(t) + b ˙ U(t, x) + eU(t)
U(t)
= c ¨ X(x) + d ˙ X(x)
X(x)
λ ∈ R
c ¨ X(x) + d ˙ X(x) − λX(x) = 0
a Ü(t) + b ˙ U(t) + (e + λ) U(t) = 0.
=: λ
X
c ¨ X(x) + d ˙ X(x) − λX(x) = 0
cµ 2 + dµ − λ = 0 ∆ = d 2 + 4cλ
∆ > 0 µ1 = −d−√∆ 2c µ2 = −d+√∆ 2c
Φ(c1, c2, x) = c1e µ1x + c2e µ2x
∆ = 0 µ1 = µ2 = −d
2c Φ(c1, c2, x) =
c1e µ1x + c2xe µ1x

∆ < 0 α = −d
2c ω =
√
|∆|
2c
Φ(c1, c2, x) = e αx (c1 cos ωx + c2 sin ωx)
X
X(x)
u(t, 0) = u(t, π) = 0
X(0) = X(π) = 0 Φ(c1, c2, 0) = 0
Φ(c1, c2, π) = 0
c1, c2
∆
∆ > 0
∆ = 0
∆ < 0
c1 + c2 = 0
e µ1πc1 + e µ2πc2 = 0
c1 = 0
c1e µ1π + c2πe µ1π = 0
c1 = 0
eαπ (c1 cos ωπ + c2 sin ωπ) = 0
e µ2π − e µ1π = 0 c1 = c2 = 0
µ1 = µ2
πe µ1π = 0
c1 = c2 = 0
e απ sin ωπ c1 = 0, c2 ∈ R
0 = ω ∈ Z 0 = ω ∈ Z
ux(t, 0) = ux(t, π) = 0
˙ X(0) = ˙ X(π) = 0 Φ(c1, ˙ c2, 0) = 0
˙ Φ(c1, c2, π) = 0
c1, c2
λ = 0 µ1, µ2 = 0 λ = 0
∆
∆ > 0
µ1c1 + µ2c2 = 0
µ1e µ1πc1 + µ2e µ2π
µ1µ2(e
c2 = 0
µ2π − e µ1π ) = 0 c1 = c2 = 0
λ = 0 µ1 = µ2
µ1, µ2 = 0
∆ = 0
µ1c1 + c2 = 0
µ1e µ1πc1 + e µ1π
µ1πe
(1 + µ1π)c2 = 0
µ1π = 0 c1 = 0, c2 = 0
λ = 0 µ1 = 0
∆ < 0
αc1 + ωc2 = 0
−c1ω sin(πω) + c2α sin(πω) = 0
(α2 + ω2 λ = 0
) sin ωπ
0 = ω ∈ Z
d c1
c1 ∈ Rc2 = 2cω
0 = ω ∈ Z
λ = 0
d = 0
µ1 = 0 µ2 = 0 ∆ = 0 c1 ∈ R c2 = 0
λ = 0
d = 0
µ1 = µ2 = 0 ∆ = 0 c1 ∈ R c2 = 0
ω ∈ Z \ {0} ∆ < 0 2cn = |∆| ∆ < 0 n ∈ Z \ {0} λ
n ∈ N \ {0}
− d2 +4c 2 n 2
4c

λ λ
λn = − d2 + 4c 2 n 2
4c
d
− n ∈ N \ {0} Xn(x) = cne 2c x sin nx
λn = − d2 + 4c2n2 d
− n ∈ N \ {0} Xn(x) = cne 2c
4c
x cos nx + d
2nc sin nx .
λ = 0 X0(x) = c0
U U
a Ü(t) + b ˙ U(t) + (e + λ) U(t) = 0.
e+λ
− t a = 0 b = 0 U(t) = U(0)e b
a = 0 aµ 2 + bµ +
(e + λ) = 0 ˜ ∆ = b2 − 4a(e + λ)
˜ ∆ > 0 ν1 = −b−
√
˜∆
2a ν2 = −b+
√
˜∆
2a
Un(d1, d2, t) = d1eν1t + d2eν2x ˜ ∆ = 0 ν1 = ν2 = −b
2a Un(d1, d2, t) =
d1eν1t + c2teν1t √
| ∆| ˜
2a θ = 2a
˜ ∆ < 0 β = −b
Un(d1, d2, t) = eβt (d1 cos θt + d2 sin θt)
λ
λn Un(t) Xn(x)
λn un(t, x) = Un(t)Xn(x)
a = 0 un(t, x)
a = 0
dn d 1 n d 2 n
u(0, x) = f(x) ut(0, x) = g(x)
∞
f(x) = bn sin nx =
g(x) =
f(x) =
g(x) =
n=1
∞
dn sin nx =
n=1
∞
an cos nx =
n=0
∞
dn cos nx =
∞
Un(0)Xn(x)
n=1
∞
Un(0) ˙ Xn(x)
n=1
∞
Un(0)Xn(x)
n=1
∞
Un(0) ˙ Xn(x)
n=1
n=1
d1 n d2 n n
a = 0 u(0, x) = f(x)
dn
un(t, x) n

a = 0 b = 0 c = 0
⎧
⎪⎨ b ∂tu + c ∂xxu + d ∂xu + e u = 0 ]0, π[×]0, +∞[,
u(t, 0) = u(t, π) = 0,
⎪⎩
u(0, x) = f(x).
a = 0 b = 0 λ
λn = − d2 +4c 2 n 2
4c n ∈ N n > 0
X U
d2 + 4c2n2 − 4ce
un(t, x) = bn exp
4bc
d
−
t e 2c x sin nx.
u(0, x) = f(x)
∞
∞
∞
u(0, x) = f(x) = un(0, x) =
bn sin nx
n=1
n=1
d
−
bne 2c x d
−
sin nx = e 2c x
bn f(x)e d
2c x
[−π, π] 2π
bn = 2
π
u(t, x) = 2 d
e− 2c
π x
∞
n=1
π
0
π
0
f(s)e d
2c s sin ns ds
f(x)e d
2c x sin nx dx,
a = 0 b = 0 c = 0 d = 0
⎧⎪⎨
b ∂tu + c ∂xxu + e u = 0 ]0, π[×]0, +∞[,
ux(t, 0) = ux(t, π) = 0,
⎪⎩
u(0, x) = f(x).
n=1
d2 + 4c2n2 − 4ce
exp
t sin nx.
4bc
a = 0 b = 0 d = 0 λ
λn = −cn 2 n ∈ N n > 0
X U
cn2 − e
un(t, x) = an exp
b
t cos nx.
u(0, x) = f(x)
∞
∞
u(0, x) = f(x) = un(0, x) = a0 + an cos nx
n=1
an f(x) [−π, π]
2π
a0 = 1
π
π
f(x) dx,
n=1
0
an = 2
π
f(x) cos nx dx,
π 0

u(t, x) = 1
π
f(s) ds
π
0
e
−
e b t + 2
π
∞
n=1
e cn2−e b
π
t
0
f(s) cos ns ds cos nx.

SO(3)
U(3) := {O ∈ Mat3×3(C) : O T O = OO T = Id3},
O(3) := {O ∈ Mat3×3(R) : O T O = OO T = Id3},
SO(3) := {O ∈ O(3) : det O = +1}.
O(3) U(3)
O ∈ O(3) O R 3
x, y ∈ R 3 〈x, y〉 = 〈Ox, Oy〉
〈Ox, Oy〉 = 〈x, O T Oy〉 = 〈x, y〉,
O T O
U(3) C 3
O U(3) O(3)
v 2 = 〈v, v〉 = 〈Ov, Ov〉 = Ov 2 .
O ∈ O(3) det O = ±1
1 = det Id3 = det(O T O) = det O T · det O = det 2 O.
λ O ∈ O(3) det(O − λId3) = 0
O
det(O − λId3) 3 λ O ∈ O(3)
O ∈ O(3)
λ1 v
Ov = λ1v Ov = v v =
Ov = |λ1| · v |λ1| = 1
O ∈ SO(3)
1 1 +1
(λ1, λ2, λ3) ∈ {(1, 1, 1), (1, −1, −1), (−1, 1, −1), (−1, −1, 1)}
O
λ2 = e α+iβ λ3 = e α−iβ λ1λ2λ3
1 1 = λ1λ2λ3 = λ1e 2α e 2α > 0 λ1 = ±1

SO(3)
λ1 = 1 e 2α = 1 α = 0 1, e iβ , e −iβ U(3)
O ∈ SO(3)
⎛
D := ⎝
1 0 0
0 e iβ 0
0 0 e −iβ
vi λi i = 1, 2, 3 v1 λ1 = 1
O Ov1 = λ1v1 = v1
1 R 3
v1 v ⊥ 1
O
C 3 O v ⊥ 1 2 × 2
D
w ∈ v ⊥ 1 w = 1 cos α = 〈w, Ow〉 β
e ±iβ
O ∈ SO(3)
R : R → SO(3) R(t + s) = R(t)R(s)
(R, +) SO(3) t ↦→ R(t) SO(3)
R(0) = R(0 + 0) = R(0)R(0) R(0) −1 R(0) = Id3
˙R(s)
R(t + s) − R(s)
= lim
t→0 t
R(t) − R(0)
⎞
⎠ .
R(t)R(s) − R(s) R(t) − Id3
= lim
= lim
R(s)
t→0 t
t→0 t
= lim
R(s) =
t→0 t
˙ R(0)R(s) =: AR(s).
R(0) = Id3 R(s) = exp(sA)
A SO(3) A
A + AT = 0 R(s) T R(s) = Id3 R(s) ∈ S0(3)
˙R(s) T R(s) + R(s) T R(s) ˙ = 0 s = 0 R(s) = R(s) T = 0 AT + A = 0
ξ0 ∈ R 3 ξ(t) = R(t)ξ(0) ξ(t) ξ(0) = ξ0
˙ξ(t) = Aξ(t) ξ(t) = exp(tA)ξ0
R(t) R(t)R(s) = R(s)R(t) = R(t + s)
ω A
⎛
A = ⎝ 0 −ω3
ω3 0
ω2
−ω1
⎞
⎠ ,
−ω2 ω1 0
(ω1, ω2, ω3) = (0, 0, 0) ω = (ω1, ω2, ω3)
A A = −AT
det A = det(−AT ) = (−1) 3 det AT = − det A det A = 0
ω1, ω2, ω3 2×2 A ω1 = 0
ω3 = 0 ω2 = 0
A 2
1 Aω = 0 ω
R(t) ω ξ0 = ω
R(s)ω = exp(sA)ω = ω +
∞
n=1
(sA) n
n!
ω = ω +
∞
n=1
(sA) n−1
n!
s Aω = ω.
=0

SO(3)
Aξ = ω × ξ ξ ˙ ξ = Aξ
˙ξ = ω × ξ × ω
π AOB
π ′ A ′ OB ′ π ∩ π ′

R 2 γ 1 x 2 +y 2 = 1
θ
γ P cos θ P sin θ
P cos 2 θ + sin 2 θ = 1 P ∈ γ (x, y) = (cos θ, sin θ)
(x, y) (0, 0) (1, 0) θ/2
cos : R → [−1, 1] sin : R → [−1, 1]
2π cos(θ + 2π) = cos(θ) sin(θ + 2π) = sin(θ)
tan : R\{π/2+kπ : k ∈ Z} → R tan(x) = sin(x)
cos(x)
cot : R\{kπ : k ∈ Z} → R cot(x) = cos(x)
sin(x)
tan cot
π tan(x + π) = tan x cot(x + π) = cot x
sec : R \ {kπ : k ∈ Z} → R csc : R \ {π/2 + kπ : k ∈ Z} → R
sec(x) = 1
cos(x)
csc(x) = 1
sin(x) sec2 x + csc 2 x = csc 2 x · sec 2 x
x = kπ/2 k ∈ Z
y = arcsin x x = sin y x ∈ [−1, 1] y ∈ [−π/2, π/2]
y = arccos x x = cos y x ∈ [−1, 1] y ∈ [0, π]
y = arctan x x = tan y x ∈ R y ∈] − π/2, π/2[
y = arc sec x x = sec y y = arccos(1/x) |x| ≥ 1 y ∈]0, π[\{π/2}
y = arc csc x x = csc y y = arcsin(1/x) |x| ≥ 1 y ∈] − π/2, π/2[\{0}
y = arc cot x x = cot y y = arctan(1/x) x ∈ R y ∈]0, π[
d
sin x = cos x,
dx
d
cos x = − sin x,
dx
d
dx tan x = sec2 x = 1 + tan 2 x,
d
dx cot x = − csc2 x,
d
sec x = tan x sec x,
dx
d
csc x = − csc x cot x,
dx
d
arcsin x =
dx
1
√ 1 − x 2
d
−1
arccos x = √
dx 1 − x2 d
1
arctan x =
dx 1 + x2 d
−1
arc cot x =
dx 1 + x2 d
arc sec x =
dx
d
arc csc x =
dx
1
|x| √ x 2 − 1
−1
|x| √ x 2 − 1

x 2 + y 2 = 1
x 2 − y 2 = 1
R 2 Γ (±1, 0)
x 2 − y 2 = 1 γ
P = (x, y)
x (x, y) = (cosh θ, sinh θ)
θ/2 θ ≥ 0
θ/2 θ < 0
|θ|/2 cosh : R → [1, +∞[
sinh : R → R tanh : R → ] − 1, 1[
coth : R \ {0} → R\] − 1, 1[
tanh(x) = sinh(x)
cosh(x)
coth(x) = cosh(x)
sinh(x)
sech : R → [0, 1] csc : R \ {0} → R
sec(x) = 1
1
cosh(x) csc(x) = sinh(x)
arcsinh(x) = sinh −1
(x) = log x + x2
+ 1
arccosh(x) = cosh −1
(x) = log x + x2
− 1 , x ≥ 1
arctanh(x) = tanh −1 √
1 − x2 (x) = log
=
1 − x
1
2 ln
1 + x
, |x| < 1
1 − x
arcoth(x) = coth −1 √
x2 − 1
(x) = log
=
x − 1
1
2 ln
x + 1
, |x| > 1
x − 1
arcsech(x) = sech −1
1 ±
(x) = log
√ 1 − x2
, 0 < x ≤ 1
x
arccsch(x) = csch −1
1 ±
(x) = log
√ 1 + x2
.
x
d
sinh(x) = cosh(x)
dx
d
cosh(x) = sinh(x)
dx
d
dx tanh(x) = 1 − tanh2 (x) = sech 2 (x) = 1/ cosh 2 (x)
d
dx coth(x) = 1 − coth2 (x) = −csch 2 (x) = −1/ sinh 2 (x)
d
csch(x) = − coth(x)csch(x)
dx
d
sech(x) = − tanh(x)sech(x)
dx

d −1 1
sinh x = √
dx
x2 + 1
d −1 1
cosh x = √
dx
x2 − 1
d −1 1
tanh x =
dx
1 − x2 d −1 1
csch x = −
dx
|x| √ 1 + x2 d −1 1
sech x = −
dx
x √ 1 − x2 d −1 1
coth x =
dx
1 − x2
sinh ax dx = 1
cosh ax + C
a
cosh ax dx = 1
sinh ax + C
a
tanh ax dx = 1
ln(cosh ax) + C
a
coth ax dx = 1
ln(sinh ax) + C
a
du
√
a2 + u2 = sinh−1 u
+ C
a
du
√
u2 − a2 = cosh−1 u
+ C
a
du
a2 1
=
− u2 a tanh−1 u
+ C; u
a
2 < a 2
du
a2 1
=
− u2 a coth−1 u
+ C; u
a
2 > a 2
du
u √ a2 = −1
− u2 a sech−1 u
+ C
a
du
a csch−1
u
+ C
a
u √ a2 = −1
+ u2 dx
√ 1 − x 2
π
= arcsin(x) + c = −arccos(x) + + c
2
dx
√ x 2 + 1 = arcsinh(x) + c = ln(x + x 2 + 1) + c
dx
√ x 2 − 1 = arccosh(x) + c = ln(x + x 2 − 1) + c

1 arcsin(x) + x
− x2 dx = √ 1 − x2 + c
2
x arcsinh(x) + x 2 + 1 dx = √ x2 + 1
2
x −arccosh(x) + x 2 − 1 dx = √ x2 − 1
2
+ c = ln(x + √ x 2 + 1) + x √ x 2 + 1
2
+ c
+ c = − ln(x + √ x 2 − 1) + x √ x 2 − 1
2
dx
= arctan(x) + c
1 + x2 dx
1
= arctanh(x) + c =
1 − x2 2 ln
1 + x
+ c
1 − x
+ c

z ∈ C
e z ∞ z
= Re(z)(cos(Im(z)) + i sin(Im(z))) =
n
n!
cos z = eiz + e −iz
2
sin z = eiz − e −iz
2i
tan z =
cot z =
n=0
= cosh iz cosh z = ez + e −z
= cos iz
2
= −i sinh iz sinh z = ez − e −z
= −i sin iz
2
sin z
cos z = eiz − e −iz
eiz + e−iz = e2iz − 1
e2iz sinh z
= −i tanh(iz) tanh z =
+ 1 cosh z = ez − e −z
ez + e−z = e2z − 1
e2z = −i tan(iz)
+ 1
cos z
sin z = eiz + e −iz
eiz − e−iz = e2iz + 1
e2iz cosh z
= i coth(iz) coth z =
− 1 sinh z = ez + e −z
ez − e−z = e2z + 1
e2z = i cot(iz)
− 1
cos 2 z + sin 2 z = 1 cosh 2 z − sinh 2 z = 1
cos(z1 ± z2) = cos z1 cos z2 ∓ sin z1 sin z2
sin(z1 ± z2) = sin z1 cos z2 ± cos z1 sin z2
tan(z1 ± z2) =
cot(z1 ± z2) =
tan z1 ± tan z2
1 ∓ tan z1 tan z2
cot z1 cot z2 ∓ 1
cot z2 ± cot z1
cosh(z1 ± z2) = cosh z1 cosh z2 ± sinh z1 sinh z2
sinh(z1 ± z2) = sinh z1 cosh z2 ± cosh z1 sinh z2
tanh(z1 ± z2) =
coth(z1 ± z2) =
tanh z1 ± tanh z2
1 ± tanh z1 tanh z2
coth z1 coth z2 ± 1
coth z2 ± coth z1
cos(2z) = cos 2 z − sin 2 z = 2 cos 2 z − 1 = 1 − 2 sin 2 z cosh(2z) = cosh 2 z + sinh 2 z = 2 cosh 2 z − 1 = 1 + 2 sinh 2 z
sin(2z) = 2 sin z cos z sinh(2z) = 2 sinh z cosh z
tan(2z) =
2 tan z
1 − tan2 (z)
cot(2z) = cot2 (z) − 1
2 cot z
tanh(2z) =
2 tanh z
1 + tanh 2 (z)
coth(2z) = coth2 (z) + 1
2 coth z

z
1 + cos z
cos = ±
2
2
z
1 − cos z
sin = ±
2
2
z
1 − cos z sin z 1 − cos z
tan = ±
= =
2 1 + cos z 1 + cos z sin z
cot
z
1 + cos z 1 + cos z sin z
= ±
= =
2 1 − cos z sin z 1 − cos z
z1 +
z2 z1 −
z2
cos z1 + cos z2 = 2 cos cos
2
2
z1 +
z2 z1 −
z2
cos z1 − cos z2 = −2 sin sin
2
2
z1 +
z2 z1 −
z2
sin z1 + sin z2 = 2 sin cos
2
2
z1 +
z2 z1 −
z2
sin z1 − sin z2 = 2 cos sin
2
2
tan z1 ± tan z2 =
cot z1 ± cot z2 =
sin(z1 ± z2)
cos z1 cos z2
sin(z2 ± z1)
sin z1 sin z2
z
1 + cosh z
cosh = ±
2
2
z
1 − cosh z
sinh = ± −
2
2
z
1 − cosh z
tanh = ± −
2 1 + cosh z
z
1 + cosh z
coth = ± −
2 1 − cosh z
sinh z 1 − cosh z
= = −
1 + cosh z sinh z
= 1 + cosh z
sinh z
= − sinh z
1 − cosh z
z1 +
z2 z1 −
z2
cosh z1 + cosh z2 = 2 cosh cosh
2
2
z1 +
z2 z1 −
z2
cosh z1 − cosh z2 = 2 sinh sinh
2
2
z1 +
z2 z1 −
z2
sinh z1 + sinh z2 = 2 sinh cosh
2
2
z1 +
z2 z1 −
z2
sinh z1 − sinh z2 = 2 cosh sinh
2
2
tanh z1 ± tanh z2 =
coth z1 ± coth z2 =
sinh(z1 ± z2)
cosh z1 cosh z2
sinh(z2 ± z1)
sinh z1 sinh z2
2 cos z1 sin z2 = sin(z1 + z2) − sin(z1 − z2) 2 cosh z1 sinh z2 = sinh(z1 + z2) − sinh(z1 − z2)
2 sin z1 sin z2 = cos(z1 − z2) − cos(z1 + z2) 2 sinh z1 sinh z2 = cosh(z1 + z2) − cosh(z1 − z2)
2 cos z1 cos z2 = cos(z1 − z2) + cos(z1 + z2) 2 cosh z1 cosh z2 = cosh(z1 − z2) + cosh(z1 + z2)
cos z =
1 −
t2
z
, t = tan
1 + t2 2
sin z = 2t
z
, t = tan
1 + t2 2
tan z = 2t
z
, t = tan
1 − t2 2
cot z =
1 −
t2
z
, t = tan
2t
2
cosh z =
1 +
t2
z
, t = tanh
1 − t2 2
sinh z = 2t
z
, t = tanh
1 − t2 2
tanh z = 2t
z
, t = tanh
1 + t2 2
coth z =
1 +
t2
z
, t = tanh
2t
2

z
1 + cos z
cos = ±
2
2
z
1 − cos z
sin = ±
2
2
z
1 − cos z sin z 1 − cos z
tan = ±
= =
2 1 + cos z 1 + cos z sin z
cot
z
1 + cos z 1 + cos z sin z
= ±
= =
2 1 − cos z sin z 1 − cos z
z1 +
z2 z1 −
z2
cos z1 + cos z2 = 2 cos cos
2
2
z1 +
z2 z1 −
z2
cos z1 − cos z2 = −2 sin sin
2
2
z1 +
z2 z1 −
z2
sin z1 + sin z2 = 2 sin cos
2
2
z1 +
z2 z1 −
z2
sin z1 − sin z2 = 2 cos sin
2
2
tan z1 ± tan z2 =
cot z1 ± cot z2 =
sin(z1 ± z2)
cos z1 cos z2
sin(z2 ± z1)
sin z1 sin z2
z
1 + cosh z
cosh = ±
2
2
z
1 − cosh z
sinh = ± −
2
2
z
1 − cosh z
tanh = ± −
2 1 + cosh z
z
1 + cosh z
coth = ± −
2 1 − cosh z
sinh z 1 − cosh z
= = −
1 + cosh z sinh z
= 1 + cosh z
sinh z
= − sinh z
1 − cosh z
z1 +
z2 z1 −
z2
cosh z1 + cosh z2 = 2 cosh cosh
2
2
z1 +
z2 z1 −
z2
cosh z1 − cosh z2 = 2 sinh sinh
2
2
z1 +
z2 z1 −
z2
sinh z1 + sinh z2 = 2 sinh cosh
2
2
z1 +
z2 z1 −
z2
sinh z1 − sinh z2 = 2 cosh sinh
2
2
tanh z1 ± tanh z2 =
coth z1 ± coth z2 =
sinh(z1 ± z2)
cosh z1 cosh z2
sinh(z2 ± z1)
sinh z1 sinh z2
2 cos z1 sin z2 = sin(z1 + z2) − sin(z1 − z2) 2 cosh z1 sinh z2 = sinh(z1 + z2) − sinh(z1 − z2)
2 sin z1 sin z2 = cos(z1 − z2) − cos(z1 + z2) 2 sinh z1 sinh z2 = cosh(z1 + z2) − cosh(z1 − z2)
2 cos z1 cos z2 = cos(z1 − z2) + cos(z1 + z2) 2 cosh z1 cosh z2 = cosh(z1 − z2) + cosh(z1 + z2)
cos z =
1 −
t2
z
, t = tan
1 + t2 2
sin z = 2t
z
, t = tan
1 + t2 2
tan z = 2t
z
, t = tan
1 − t2 2
cot z =
1 −
t2
z
, t = tan
2t
2
cosh z =
1 +
t2
z
, t = tanh
1 − t2 2
sinh z = 2t
z
, t = tanh
1 − t2 2
tanh z = 2t
z
, t = tanh
1 + t2 2
coth z =
1 +
t2
z
, t = tanh
2t
2