Exercícios - Equações Diferenciais 2022 1

1. (40 pontos) Resolva o problema de valor inicial abaixo. y'' + 9y = δ(t − 2) + 4e^t, y(0) = 1, y'(0) = 0 2. (20 pontos) Obtenha y(t), solução do problema abaixo. y'(t) + \frac{5}{4}\int_{0}^{t}e^{-(t−τ)}y(τ)dτ = 0, y(0) = 2 3. (40 pontos) Resolva o problema de valor inicial abaixo. { x' = 3x − y, x(0) = 0 { y' = 5x − 3y, y(0) = −4 \begin{array}{|c|c|} \hline f(t) = \mathcal{L}^{−1}\{F(s)\} & F(s) = \mathcal{L}\{f(t)\} \\ \hline 1 & \frac{1}{s}, s > 0 \\ e^{at} & \frac{1}{s−a}, s > a \\ t^n \ (n \ inteiro \ positivo) & \frac{n!}{s^{n+1}}, s > 0 \\ sen(at) & \frac{a}{s^2+a^2}, s > 0 \\ cos(at) & \frac{s}{s^2+a^2}, s > 0 \\ senh(at) & \frac{a}{s^2−a^2}, s > |a| \\ cosh(at) & \frac{s}{s^2−a^2}, s > |a| \\ e^{at}sen(bt) & \frac{b}{(s−a)^2+b^2}, s > a \\ e^{at}cos(bt) & \frac{s−a}{(s−a)^2+b^2}, s > a \\ u_c(t), & \frac{e^{-cs}}{s}, s > 0 \\ u_c(t)f(t−c), & e^{-cs}F(s) \\ e^{ct}f(t) & F(s−c) \\ f(ct) & \frac{1}{s}F\left(\frac{s}{c}\right), c > 0 \\ (f * g)(t) = \int_{0}^{t}f(t−τ)g(τ)dτ & F(s)G(s) \\ δ(t−c) & e^{-cs} \\ f^(n)(t) & s^nF(s) − s^{n−1}f(0) − ⋯ − f^{(n−1)}(0) \\ (−t)^n f(t) & F^{(n)}(s) \\ \hline \end{array} Questão 1 Temos a equação: 𝑦′′ + 9𝑦 = 𝛿(𝑡 − 2) + 4𝑒𝑡 Aplicando Laplace, temos: 𝐿(𝑦′′) + 9𝐿(𝑦) = 𝐿(𝛿(𝑡 − 2)) + 4𝐿(𝑒𝑡) 𝑠𝐿(𝑦′) − 𝑦′(0) + 9𝐿(𝑦) = 𝑒−2𝑠 + 4 𝑠 − 1 𝑠𝐿(𝑦′) + 9𝐿(𝑦) = 𝑒−2𝑠 + 4 𝑠 − 1 𝑠(𝑠𝐿(𝑦) − 𝑦(0)) + 9𝐿(𝑦) = 𝑒−2𝑠 + 4 𝑠 − 1 𝑠(𝑠𝐿(𝑦) − 1) + 9𝐿(𝑦) = 𝑒−2𝑠 + 4 𝑠 − 1 𝑠2𝐿(𝑦) − 𝑠 + 9𝐿(𝑦) = 𝑒−2𝑠 + 4 𝑠 − 1 (𝑠2 + 9)𝐿(𝑦) = 𝑠 + 𝑒−2𝑠 + 4 𝑠 − 1 𝐿(𝑦) = 𝑠 𝑠2 + 9 + 𝑒−2𝑠 1 𝑠2 + 9 + 4 𝑠 − 1 1 𝑠2 + 9 𝐿(𝑦) = 𝑠 𝑠2 + 9 + 𝑒−2𝑠 1 𝑠2 + 9 + 4 [ 𝐴 𝑠 − 1 + 𝐵𝑠 + 𝐶 𝑠2 + 9] 𝐿(𝑦) = 𝑠 𝑠2 + 32 + 1 3 𝑒−2𝑠 3 𝑠2 + 32 + 4 [ 𝐴 𝑠 − 1 + −𝐴𝑠 + 𝐶 𝑠2 + 9 ] 𝐿(𝑦) = 𝑠 𝑠2 + 32 + 1 3 𝑒−2𝑠 3 𝑠2 + 32 + 4 [ 𝐴 𝑠 − 1 + −𝐴𝑠 − 𝐴 𝑠2 + 9 ] 𝐿(𝑦) = 𝑠 𝑠2 + 32 + 1 3 𝑒−2𝑠 3 𝑠2 + 32 + 4𝐴 [ 1 𝑠 − 1 + −𝑠 − 1 𝑠2 + 9] 𝐿(𝑦) = 𝑠 𝑠2 + 32 + 1 3 𝑒−2𝑠 3 𝑠2 + 32 + 4 10 [ 1 𝑠 − 1 − 𝑠 + 1 𝑠2 + 9] 𝐿(𝑦) = 𝑠 𝑠2 + 32 + 1 3 𝑒−2𝑠 3 𝑠2 + 32 + 2 5 [ 1 𝑠 − 1 − 𝑠 𝑠2 + 9 − 1 𝑠2 + 9] 𝐿(𝑦) = 3 5 𝑠 𝑠2 + 32 + 1 3 𝑒−2𝑠 3 𝑠2 + 32 + 2 5 [ 1 𝑠 − 1 − 1 3 3 𝑠2 + 32] Aplicando a inversa, temos: 𝑦 = 3 5 cos 3𝑡 + 1 3 𝐿−1 [𝑒−2𝑠 3 𝑠2 + 32] + 2 5 [𝑒𝑡 − 1 3 sin 3𝑡] 𝑦 = 3 5 cos 3𝑡 + 1 3 𝑢(𝑡 − 2)𝐿−1 [ 3 𝑠2 + 32] 𝑡−2 + 2 5 [𝑒𝑡 − 1 3 sin3𝑡] 𝑦 = 3 5 cos 3𝑡 + 1 3 𝑢(𝑡 − 2)[sin3𝑡]𝑡−2 + 2 5 [𝑒𝑡 − 1 3 sin 3𝑡] 𝒚 = 𝟑 𝟓 𝐜𝐨𝐬 𝟑𝒕 + 𝟏 𝟑 𝒖(𝒕 − 𝟐) 𝐬𝐢𝐧(𝟑𝒕 − 𝟔) + 𝟐 𝟓 [𝒆𝒕 − 𝟏 𝟑 𝐬𝐢𝐧 𝟑𝒕] Questão 2 Temos a equação: 𝑦′ + 5 4 ∫ 𝑒−(𝑡−𝜏)𝑦(𝜏)𝑑𝜏 𝑡 0 = 0 Aplicando Laplace, temos: 𝐿(𝑦′) + 5 4 𝐿 [∫ 𝑒−(𝑡−𝜏)𝑦(𝜏)𝑑𝜏 𝑡 0 ] = 𝐿[0] = 0 𝑠𝐿[𝑦] − 𝑦(0) + 5 4 𝐿[𝑒−𝑡]𝐿[𝑦] = 0 𝑠𝐿[𝑦] − 2 + 5 4 1 𝑠 + 1 𝐿[𝑦] = 0 (𝑠 + 5 4 1 𝑠 + 1) 𝐿[𝑦] = 2 (4𝑠2 + 4𝑠 4(𝑠 + 1) + 5 4(𝑠 + 1)) 𝐿[𝑦] = 2 (4𝑠2 + 4𝑠 + 5 4(𝑠 + 1) )𝐿[𝑦] = 2 𝐿[𝑦] = 8(𝑠 + 1) 4𝑠2 + 4𝑠 + 5 𝐿[𝑦] = 2 (𝑠 + 1) 𝑠2 + 𝑠 + 5 4 𝐿[𝑦] = 2 (𝑠 + 1 2 + 1 2) (𝑠 + 1 2) 2 + 1 𝐿[𝑦] = 2 (𝑠 + 1 2) (𝑠 + 1 2) 2 + 1 + 2 (1 2) (𝑠 + 1 2) 2 + 1 𝐿[𝑦] = 2 (𝑠 + 1 2) (𝑠 + 1 2) 2 + 1 + 1 (𝑠 + 1 2) 2 + 1 Tomando a inversa, temos: 𝑦 = 2𝐿−1 [ (𝑠 + 1 2) (𝑠 + 1 2) 2 + 1 ] + 𝐿−1 [ 1 (𝑠 + 1 2) 2 + 1 ] 𝑦 = 2𝑒−𝑡 2𝐿−1 [ 𝑠 𝑠2 + 1] + 𝑒−𝑡 2𝐿−1 [ 1 𝑠2 + 1] 𝑦 = 2𝑒−𝑡 2 cos 𝑡 + 𝑒−𝑡 2 sin 𝑡 𝒚 = 𝒆−𝒕 𝟐(𝟐𝐜𝐨𝐬 𝒕 + 𝐬𝐢𝐧𝒕) Questão 3 Temos o seguinte sistema: { 𝑥′ = 3𝑥 − 𝑦 𝑦′ = 5𝑥 − 3𝑦 Aplicando Laplace, temos: { 𝑠𝐿[𝑥] − 𝑥(0) = 3𝐿[𝑥] − 𝐿[𝑦] 𝑠𝐿[𝑦] − 𝑦(0) = 5𝐿[𝑥] − 3𝐿[𝑦] { 𝑠𝐿[𝑥] = 3𝐿[𝑥] − 𝐿[𝑦] 𝑠𝐿[𝑦] + 4 = 5𝐿[𝑥] − 3𝐿[𝑦] { (𝑠 − 3)𝐿[𝑥] = −𝐿[𝑦] (𝑠 + 3)𝐿[𝑦] + 4 = 5𝐿[𝑥] Juntando as equações, temos: (𝑠 + 3)𝐿[𝑦] + 4 = 5𝐿[𝑥] (𝑠 + 3)(−(𝑠 − 3)𝐿[𝑥]) + 4 = 5𝐿[𝑥] −(𝑠 + 3)(𝑠 − 3)𝐿[𝑥] + 4 = 5𝐿[𝑥] −(𝑠2 − 9)𝐿[𝑥] + 4 = 5𝐿[𝑥] −(𝑠2 − 9 + 5)𝐿[𝑥] + 4 = 0 −(𝑠2 − 4)𝐿[𝑥] = −4 (𝑠2 − 4)𝐿[𝑥] = 4 𝐿[𝑥] = 4 (𝑠2 − 4) 𝐿[𝑥] = 2 2 (𝑠2 − 22) 𝑥 = 2 sinh2𝑡 𝒙 = 𝒆𝟐𝒕 − 𝒆−𝟐𝒕 Assim, temos: 𝐿[𝑦] = −(𝑠 − 3)𝐿[𝑥] 𝐿[𝑦] = −(𝑠 − 2 − 1) 4 (𝑠2 − 4) 𝐿[𝑦] = −4 (𝑠 − 2 − 1) (𝑠2 − 4) 𝐿[𝑦] = −4 (𝑠 − 2) (𝑠2 − 4) + 4 1 (𝑠2 − 4) 𝐿[𝑦] = −4 1 𝑠 + 2 + 2 2 (𝑠2 − 22) 𝑦 = −4𝑒−2𝑡 + 2 sinh2𝑡 𝑦 = −4𝑒−2𝑡 + 𝑒2𝑡 − 𝑒−2𝑡 𝒚 = 𝒆𝟐𝒕 − 𝟓𝒆−𝟐𝒕