Questoes Lista Calc2 Resolvida-2022 1

3) Y'' + 2Y' + 2Y = x e^{-x} + e^{-x} \cos x r² + 2r + 2 = 0 Δ = 4 - 8 = -4 r = \frac{-2 ± 2i}{2} = -1 ± i Y_h(x) = e^{-x} (C_1 \cos x + C_2 \sin x) Y_{p1}(x) = (Ax + B) e^{-x} Y_{p2}(x) = C x e^{-x} \sin x + D x e^{-x} \cos x, pois e^{-x} (\sin x \cos x) foge parte de Y_h Y_p(x) = (Ax + B) e^{-x} + C x e^{-x} \sin x + D x e^{-x} \cos x E 1) Y[(3+2x)y - 2x - 2x² - 2y²] dx + [2(x+1)y - x² - 3y²] dy = 0 M_y = 3y + 2xy - 2x - 2x² - 2y² + 3y + 2xy - 4y² = -2x + 6y - 2x² - 6y² + 4xy N_x = 2y - 2x \frac{M_y - N_x}{N} = \frac{4y - 2x² - 6y² + 4xy}{2y + 2xy - x² - 3y²} = 2 \frac{du}{dx} = 2u => \frac{1}{u} du = 2 dx ln u = 2x u = e^{2x} = e^{2x} \left(-x²y² - y³ + y²(1+x)\right) = C => C = 0 Y(1) = 1 => e^{2}(-1 - 1 + 2) = C => C = 0 -x² - y² + y(1 + 2xy) = 0 -x² - y² + y + 2xy = 0 A x² + y² - 2xy - y = 0 2) Y'' + βy' + γy = e^{-β/2} x ln x r² + βr + γ = 0 Δ = β² - 4γ = 0 r = \frac{-β ± 0}{2} = -β/2 Y_h(x) = C_1 e^{-β/2 x} + C_2 x e^{-β/2 x} Y(x) = C_3 e^{-β/2 x} + C_2 x e^{-β/2 x} + 6e^{-β/2 x} x³ ln x - 5e^{-β/2 x} x³ = \frac{e^{-β/2 x}}{36} \left(K_1 + K_2 x + 6 x³ ln x - 5 x³\right), x70 F β^2 - 4γ < 0 e ω = √4γ - β^2 / 2 γ" + βγ' + γ = e^-β/2 t sen(ωt), com γ(0) = 0 e γ'(0) = 1 r^2 - βr + γ = 0 Δ = β^2 - 4γ < 0 r = -β ± √4γ - β^2 * i / 2 γₕ(x) = e^-β/2 x (C₁ cos ωx + C₂ sen ωxₕ) γₚ(x) = E^-β/2 x (A sen (ω₀x₀) + B cos (ω₀x₀) γ′ₚ(x) = -β/2 e^-β/2 x (A sen(ωx) + B cos (ωx)) + e^-β/2 x (2ωA cos(ωx) - 2ωB sen(ωx)) γ″ₚ(x) = e^-β/2 x [-β/2 B + 2ωA) cos(ωx)) + (-β/2 A - 2ωB) sen(ωx))] γ″ = -β/2 × e^-β/2 x [...] + e^-β/2 x [...](-2 ω) sen(ωx) + (-2 ω) cos(ωx) = e^-β/2 x [-β/2 (-β/2B + 2ωA) + 2ω (-β/2A - 2ωB) cos(ωx) + [-β/2 (-β/2A - 2ωB) + 2ω (-β/2B + 2ωA)] β = 0 -> β^2 - 4γ = -4 -β/2 = -1 C₁ γ = 2 γ = 2 ω = 1 β^2 - 4γ = 9 - 25 = -16 ω = √16/2 = 4/2 = 2 Com → 0 2ω = C fᵧ(0,0) = lim f(0,h) - f(0,0) / h = lim h^2/h^2 - 1 / h = lim 1/h - lim 0 = 0 ∂/∂y (0,0) = 0 Se f(x,y) ≠ (0,0), f(x,y) = y^2 x / (x^2 + y^2) ⇒ ∂/∂y f(x,y) = 2y(x^2 + y^2) - 2y(y^2 - x) / (x^2 + y^2)^2 = 2yx^2 + 2y^3 - 2y^3 + 2yx / (x^2 + y^2)^2 = 2xy(1 + x^2)/(x^2 + y^2)^2 C