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

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
Page 3 and 4:
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
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a ∈ R R 2 (x 2 + y 2 ) 3 = 4
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(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 172 and 173: c ′ 1 (x) cos x + c′ 2 (x) sin
Page 175: y ′ + y 1 x = tan x sin x y ′
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
Page 223 and 224: α > 0 lim (x,y)→(0,0) (x,y)=(0,0
Page 225 and 226: 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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