P3 - Equações Diferenciais - 2023-2

EDA - 1ª prova individual (30 pts, sem consulta) 1. Resolva os problemas de valor inicial. \ \ a) (6 pts) \left\{ \begin{array}{l} y' = 4x \sqrt{y - 1} \\ y(0) = 2 \end{array} \right. \ \ b) (6 pts) \left\{ \begin{array}{l} y' + \frac{2}{x} y = x^3 \\ y(1) = 1 \end{array} \right. \ \ c) (6 pts) \left\{ \begin{array}{l}- y - x + (2y - x) y' = 0 \\ y(0) = 0 \end{array} \right. 2. (a) (9 pts) Resolva o problema de valor inicial. \ \ \left\{ \begin{array}{l} y'' + y' - 2y = 0 \\ y(0) = 2, y'(0) = -1 \end{array} \right. (b) (3 pts) Construa o gráfico de y(x) para x \geq 0. EDA - 1ª prova 1 (a) y' = 4x \sqrt{y - 1} , y(0) = 2 ∫ (y - 1)^{-1/2} \,dy = ∫ 4x \,dx \rightarrow 2(y - 1)^{1/2} = 2x^2 + c\quad(k = \frac{c}{2}) (y - 1)^{1/2} = x^2 + k \rightarrow y(x) = 1 + (x^2 + k)^2 y(0) = 1 + k^2 = 2 \rightarrow k = 1 \rightarrow\framebox{y(x) = 1 + (x^2 + 1)^2} (b) y' + \frac{2}{x^2} y = x^3 , y(1) = 1 m(x) = \int \frac{2}{x} \, dx = 2\ln x \rightarrow x^2 \Rightarrow x^2 y + 2xy = x^5 \frac{d}{dx}(x^2 y) = x^5 \rightarrow x^2 y = \frac{x^6}{6} + c \rightarrow\framebox{y(x) = \frac{x^4}{6} + \frac{c}{x^2}} y(1) = \frac{1}{6} + c = 1 \rightarrow c = \frac{5}{6} \rightarrow\y(x) = \frac{x^4}{6} + \frac{5}{6x^2} \rightarrow \framebox{y(x) = \frac{x^4}{6} + \frac{5}{6x^2}} (c) -y - x + (2y - x) y' = 0 , y(0) = 0 M = -y - x \rightarrow \frac{\partial M}{\partial y} = -1\quad ① \quad\text{eq. exacta} N = 2y - x \rightarrow \frac{\partial N}{\partial x} = -1\quad ②  \quad 1 \quad\text{ exacta }\quad \frac{\partial \psi}{\partial x} = -y - x \rightarrow \psi(x, y) = -xy - \frac{x^2}{2} + f(y) \frac{\partial \psi}{\partial y} = -x + f'(y) = 2y - x \rightarrow f'(y) = 2y \rightarrow\quad f(y) = y^2 + c \rightarrow \psi(x, y) = -xy - \frac{x^2}{2} + y^2\quad sq^2 + y^2 = k \quad\text{cond. initial}\{x=0 \}'', \begin{array}{l} y=0 \rightarrow k = 0\end{array} \} -xy - \frac{x^2}{2} + y^2 = 0 \rightarrow (y - \frac{x}{2)^2} - \frac{x^2}{4} - \frac{x^2}{2} = 0\rightarrow (y - \frac{x^2}{2))^2 = \frac{3x^2}{4} \rightarrow y = \frac{x}{2} \pm \frac{\sqrt{3}}{2} x \rightarrow\begin{array}{l y}=\frac{x}{2} \pm \frac{\sqrt{3}}{2} x \rightarrow\ y(x) = \frac{1 \pm \sqrt{3}}{2} x \text{ duas solu\c{c}\~{o}\'es}} y'' + y' - 2y = 0, y(0) = 2, y'(0) = -1 (a) y(t) = e^{rt} (r = ?) (r^2 + r - 2) e^{rt} = 0 \text{eq. caracteristica:} r^2 + r - 2 = 0, \Delta = 9, r = \frac{-1 \pm 3}{2} \rightarrow 1 \ -2 \text{sol. geral:} y(t) = c_1 e^{t} + c_2 e^{-2t} y(t) = c_1 e^{t} - 2c_2 e^{-2t} \text{cond. iniciais:} \{ y(0) = c_1 + c_2 = 2 \ y'(0) = c_1 - 2c_2 = -1 (-) -------------------\} 3c_2 = 3 \rightarrow c_2 = 1, c_1 = 1 \text{sol. PVI: } y(t) = e^{t} + e^{-2t} (b) \lim_{t \rightarrow \infty}(e^{t} + e^{-2t}) = +\infty , y(0) = 2, y'(0) = -1 (\text{decrescente}) (y(t) = e^{t}, ze^{-2t} = 0 \rightarrow e^{t} = ze^{-2t} \rightarrow e^{3t} = z^2 3t = \ln z \rightarrow t = \frac{1}{3} \ln z \quad e \acute{o \uacute}{\uacute}nico \text{ponto cr\uacute}tico\y(\frac{1}{3} \ln z) = e^{\frac{5}{3} \ln 2} + e^{-\frac{2}{3} \ln 2} = (2^{\frac{1}{3}} + 2^{- rac{2}{3}}) 2^{\frac{5}{3} + 2^{1}) 2^{2^{\frac{1}{3}} = 3.2^{2^{2/3} = \frac{3}{3^{\frac{\sqrt{4}}{4}}}}'