P3 - Equações Diferenciais a - 2018-1

EDA - 3a prova individual (30 pts) - 28/06/2018 1. Considere f(t) = 4δ(t−1)+u2(t) t, definida para t ≥ 0. (a) (2 pontos) Calcule a transformada de Laplace de f(t). (b) (6 pontos) Resolva: y′′ = f(t), y(0) = 1, y′(0) = 1. (c) (2 pontos) Calcule y(1), y(2) e y(3). 2. (10 pontos) Resolva: y′′ − y = g(t), y(0) = 1, y′(0) = 0, g(t) = { 0 para 0 ≤ t < π sen t para t ≥ π 3. (10 pontos) Resolva o seguinte problema de valor inicial. { x′ = 2x + 4y, x(0) = 4 y′ = x + 2y, y(0) = −4 f(t) = L−1{F(s)} F(s) = L{f(t)} 1 1 s, s > 0 eat 1 s−a, s > a tn, n inteiro positivo n! sn+1 sen(at) a s2+a2, s > 0 cos(at) s s2+a2, s > 0 senh(at) a s2−a2, s > |a| cosh(at) s s2−a2, s > |a| uc(t), e−cs s , s > 0 uc(t)f(t − c), e−csF(s) ect f(t) F(s − c) f(ct) 1 sF(s c), c > 0 (f ∗ g)(t) = ∫ t o f(t − τ)g(τ) dτ F(s)G(s) δ(t − c) e−cs f (n)(t) snF(s) − sn−1f(0) − . . . − f (n−1)(0) (−t)n f(t) F (n)(s) EDA - 3ª PROVA INDIVIDUAL (1) \( f(t) = 4\delta(t-1) + u_2(t) \cdot t \quad t>0 \) (a) Reescrevemos \( f(t) = 4\delta(t-1) + u_2(t)[(t-2) + 2] \) onde identificamos \( g(t-2) = (t-2) + 2 \rightarrow g(t) = t+2 \rightarrow G(s) = \frac{1}{s^2} + \frac{2}{s} \) Assim \( F(s) = 4e^{-s} + \left( \frac{1}{s^2} + \frac{2}{s} \right)e^{-2s} \) (b) \( y'' = f(t), \ y(0) = 1, \ y'(0) = 1 \) \( s^2Y(s) - s - 1 = 4e^{-s} + \left( \frac{1}{s^2} + \frac{2}{s} \right)e^{-2s} \) \( Y(s) = \frac{1}{s^2} + \frac{1}{s} + 4e^{-s} + \left( \frac{1}{s^4} + \frac{2}{s^3} \right)e^{-2s} \) \( \frac{1}{s} + \frac{1}{s^2} + \frac{4}{s} e^{-s} + \frac{1}{s^4} e^{-2s} + \frac{2}{s^3} e^{-2s} \) Portanto \( y(t) = 1 + t + 4u_1(t)\cdot(t-1) + u_2(t) \left[ \frac{1}{6}(t-2)^3 + (t-2) \right] \) (c) \( y(1) = 1+1=2 , \ y(2) = 3+2+4=7, \ y(3) = 1+3+8 + \frac{7}{6} = \frac{79}{6} \) (2) \( y'' - y = g(t), \ y(0) = 1, \ y'(0) = 0, \ g(t) = \begin{cases} 0 & 0<t<\pi \\ \sin t & t>\pi \end{cases} \) \( g(t) = u_{\pi}(t)\cdot \sin t = u_{\pi}(t)\cdot\sin[(t-\pi) + \pi] = -u_{\pi}(t)\cdot\sin(t-\pi) \) Assim \( s^2Y(s) - s - Y(s) = -\frac{e^{-\pi s}}{s^2+1} \rightarrow Y(s) = \frac{s}{s^2-1} + \frac{1}{(s^2-1)(s^2+1)} e^{-\pi s} \) \( \frac{1}{(s^2-1)(s^2+1)} = \frac{As+B}{s^2-1} + \frac{Cs+D}{s^2+1} = \frac{(As+B)(s^2+1) + (Cs+D)(s^2-1)}{(s^2-1)(s^2+1)} \) \( s = 0: A + C = 0 \rightarrow A=C=0 \) \( s^2: B+D=0 \rightarrow B=\frac{1}{2}, D=-\frac{1}{2} \) \( s: A-C=0 \) \( 1: B-D=1 \) \( Y(s) = \frac{s}{s^2-1} + \frac{1}{2} \left( \frac{1}{s^2+1} - \frac{1}{s^2-1} \right)e^{-\pi s} \) Assim \( y(t)=\cosh t + \frac{1}{2}u_{\pi}(t) \left[ \sin(t-\pi)-\sinh(t-\pi) \right] = -\sin t \) (3) \{ \dot{x} = 2x + 4y \\ \dot{y} = x + 2y \} \( x(0) = 4 \) \( y(0) = -4 \) \( \frac{d}{dt} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \rightarrow \frac{d}{dt} x = Ax \) \( x = Ce^{\lambda t} \rightarrow (A-\lambda I)c=0 , \ c\neq 0 \rightarrow \det(A-\lambda I)=0 \) \( \det \begin{bmatrix} 2-\lambda & 4 \\ 1 & 2-\lambda \end{bmatrix} = (2-\lambda)^2 - 4 = \lambda(\lambda-4) = 0 \rightarrow \lambda_1 = 4, \lambda_2 = 0 \) \( \lambda_1 = 4 \) \( \begin{bmatrix} -2 & 4 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix} \rightarrow a - 2b = 0 \rightarrow c_1 = \alpha \begin{bmatrix} 2 \\ 1 \end{bmatrix} \) \( \lambda_2 = 0 \) \( \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} 0 \end{bmatrix} \rightarrow a + 2b = 0 \rightarrow c_2 = \beta \begin{bmatrix} -2 \\ 1 \end{bmatrix} \) Solução geral: \( \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \alpha_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} e^{4t} + \alpha_2 \begin{bmatrix} -2 \\ 1 \end{bmatrix} \) \( x(0) = 2\alpha_1 - 2\alpha_2 = 4 \) \( y(0) = \alpha_1 + \alpha_2 = -4 \) \( \alpha_1 - \alpha_2 = 2 \) \( 2\alpha_1 - 2 = -2 \rightarrow \alpha_1 = -1, \alpha_2 = -3 \) \( x(t) = 6 - 2e^{4t} \) \( y(t) = -3 - e^{4t} \)