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$(Microsoft PowerPoint - 03_Gruppi_Alcani [modalit\340 compatibilit ...$

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)
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R R n
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2 × 2 SO(
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R a
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R F G R F G ¯ F ⊇ G
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R n x, y ∈ Rn x = (x1, ...
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(X, τ) D ⊆ X X
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f : D → R D ⊆ Rn γ : [
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A =]0, +∞[×]0, +∞[ f : A →
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lim (x,y)→(0,0) (x,y)=(0,0) |xy|
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E E E f −1 ([1, +∞[)
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s ↦→ 1 − e −s s ≥ 0 1
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Fn : [0, +∞[→ R Fn(ρ) = 2n ρ
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I =]a, b[ R {fn}n∈N fn : I
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K [0, +∞[ R > 0 B(0, R) ⊇
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an = 0 n = 2k an = 4/(πn2 ) n
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5 1 18 = + 2 4 π2 ∞ n=0 ∞ n=0
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g(x) := x(π−x) x ∈ [0, π]
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X Y T : X → Y T ℓ >
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X K E X a, b ∈ R f : E × [a
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|f(x, y)| ≤ log(1 + 3y
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(x, y) ∈ R 2 xy =
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X, Y D ⊆ X f : D → Y u ∈
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H A 2 × 2 λ 2
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(2kπ, π/2+2hπ) (π+2kπ, 3π/2+2
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∂xf(x, y, z) = −2 cos x sin x =
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(0, 0) f : R 2 → R f(x, y) =
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α ∈ R (0, 0) f(x, y) = 2
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Cα > 0 α /∈ [−1/2, −1/4]
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x = cos θ y = sin θ θ ∈ [−π
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γ(t) = (t, 1/t) t → ±∞ L(x
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⎧ ⎪⎨ 1 + λ(2x − 2) = 0, 1
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f(x, y) = x 2 + y 2 V = {(x, y)
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⎧⎪ 2x + λ(2x + y) + 2µx = 0
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R 2 D ⊆ R 2 f : D → R (x0,
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x = 1 y = −1 z = 2 y z
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⎛ 1 ⎜ 3 F (m1, m2, p1, p2) =
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k = 3√ 4 ξ ζ
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x = ρ cos θ y = ρ sin θ ρ 2 (
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f(x, y, z) = ez+x2 f + αx + y
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ϕ(1) = 0 ϕ ′ (1) = − ∂xf(1
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y 3 − xy 2 + x 2 y = x − x 3
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X = {(ρ, θ) ∈ R 2 : ρ > 0, 0 <
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arcsin sin θ = θ θ ∈] − π
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I R r : I → Rd γ µ : r
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γ lunghezza(γ) = |r ′ (θ)|
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p = 2 |(x, y)| −p dx dy = R 2 \B
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ρ 3 cos θ = aρ 2 (cos 2 θ − s
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Γ := {(x, y) ∈ R 2 : (x 2 + y 2
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x Γ Q2 Q3 x Γ \
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(x 2 + y 2 )(y 2 + x(x + 1)) = 4xy
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4 ±1, ±2, ±4 ±1, ±2 4
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−2 + 8t − 4t 2 t = 1/2(2
Page 121 and 122: a ∈ R R 2 (x 2 + y 2 ) 3 = 4
Page 123: (x 2 + y 2 ) 3 = 4x 2 y 2
Page 127 and 128: a > 0 x = at at2 , y = . 1
Page 129: cos(π − α) = − cos(α) π/4
Page 132 and 133: I := F · ˆn dσ ˆn Σ =
Page 134 and 135: S ψ(x, y) = (ψ1, ψ2, ψ3) = (
Page 136 and 137: C = {(x, y, z) : (x, y) ∈ D, λ <
Page 138 and 139: ˆn = ∇G = ((∇G)1, (∇G)2, (
Page 140 and 141: 2π 0 cos θ sin θ dθ = 1 2 2π
Page 142 and 143: L − F · ˆn dσ = − L− √
Page 145 and 146: K ⊆ R n χK : R n → {0, 1}
Page 147 and 148: f [1, +∞[ [0, 1] [0
Page 149 and 150: f α k = 1 x α + k α , sα n = n
Page 151 and 152: D R X n 1 1 X C ℓ
Page 153 and 154: γ2(1/2) = (1/8, 3/4) γ2(1) = (0,
Page 155 and 156: C = −B ∂yu(x, y) = C/x 1 + y2 y
Page 157 and 158: dy N(x, y) = − dx M(x, y) N M
Page 159 and 160: 2xy dx + (x 2 + 1) dy = 0 (x 2 + y
Page 161 and 162: x − c = arcsin(2y − 1) 0 < y
Page 163: h(x, y) = 1/(|x| √ x2 − 1)
Page 166 and 167: ˙x = e t−x /x x(α) = 1 α ∈ R
Page 168 and 169: ˙x = x 2 /(1 − tx) x(0) = α α
Page 170 and 171: y ′′′ − 6y ′′ + 11y ′
Page 175 and 176: y ′ + y 1 x = tan x sin x y ′
Page 177 and 178: λ 2 − 4λ − 5 = 0, A
Page 179 and 180: F (x, y, ˙y) = 0, F : A → R A
Page 181 and 182: F (x, y, p) = 0 c = 0 dy = p dx
Page 183: y = xy ′ + f(y ′ ) F (x, y, p
Page 186 and 187: y → 0 −∞
Page 188 and 189: x > 0 x y = ± 1 + x x < 0
Page 191 and 192: [0, π] utt + 2ut − uxx = 0,
Page 193 and 194: e |uk(x, t)| = C −t sin( 4k(1
Page 195: ∞ n=1 bne −x sin nx = x(
Page 198 and 199: T ˙ Tn(t) = −5n 2 Tn(t) Tn(
Page 200 and 201: ⎧ −utt + 3uxx = 0 ]0, π[×]0,
Page 202 and 203: ∆ = 0 λ = −1 µ1 = µ2 = −
Page 204 and 205: x = cos θ y = sin θ F F (co
Page 206 and 207: I := arctan y dx − xy dy γ γ
Page 208 and 209: f(ρ) f ′ (ρ) = 4ρ3 + 36ρ 8
Page 210 and 211: Jac ϕ(θ, y) = ⎛ ⎜ ⎝ − y
Page 212 and 213: z = (x, y) ˙z = Az + B(t)
Page 214 and 215: Q0 Q2 Q1 Q1
Page 216 and 217: ˙x = t − x 2 x(0) = a a ∈ R
Page 219 and 220: 2π xt sin t F (x) = e dt x ∈ R
Page 221 and 222: xz 3 dx dy dz, T T y
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α > 0 lim (x,y)→(0,0) (x,y)=(0,0
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F (0, 1) = 0 ∂yF (x, y) = 3y 2
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(u + v) D 2 + (u − v) 2 1 + 2(u2
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α = = 2 3 π. 2π 0
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F D z = 0 (0,
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y ′ = 2xy − x y(0) = 3/2
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Γ f(x, y) = 0 f : R 2 → R
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∇f(x0, y0) = (0, 0) (x0, y0)
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ϕ : I ×J → R 3 I, J R
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ψ(a, t) = ψ(b, t) t ∈]c, d[
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[
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Y K Ω R × Ω I R f :
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1 ω(x, y) = M(x, y) dx + N(x, y)
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M(x, y) dx + N(x, y) dy = 0 1 M, N
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x, y z dy dz
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f(x) = N(x) Ak1 = + D(x) x − xk
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Φ y ′ = A(t) y ′ = A(t)
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det P = 4 √ 7 = 0 P −1 AP
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y ′′ (t) + py ′ (t) + qy(t) =
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y(0) = c1 + d1 = 1 ˙y(t) = (−c1
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a ′ x + b ′ y + c ′ = 0
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η2 g g(η) > 0 y0 < η <
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2 × 2 F ∈ R[x] n
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2 × 2 t ↦→ x(
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∆ < 0 α = −d 2c ω = √ |
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a = 0 b = 0 c = 0 ⎧ ⎪⎨ b
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SO(3) U(3) := {O ∈
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SO(3) Aξ = ω × ξ ξ
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x 2 + y 2 = 1 x 2 − y 2 = 1
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1 arcsin(x) + x − x2 dx = √ 1
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z 1 + cos z cos = ± 2 2 z 1
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