Aproxima la solución de los siguientes problemas con valores iniciales con los diferentes métodos provistos por las funciones del package pracma. Comenta la comparación entre las aproximaciones obtenidas y la solución exacta (en caso que se indique).
- \(y'=1+(t-y)^2\), \(2\leq t \leq 3\), \(y(2)=1\). Solución exacta \(y(t)=t+1/(1-t)\)
- \(y'=cos(2t)+sen(3t)\), \(0\leq t\leq 1\), \(y(0)=1\). Solución exacta \[\begin{equation} y(t)=\frac{1}{2}sen(2t)-\frac{1}{3}cos(3t)+\frac{4}{3}. \end{equation}\]
\[\begin{equation} y'=\frac{y^2+y}{t}, \quad 1\leq t \leq 3, \quad y(1)=-2 \end{equation}\]
\[\begin{equation} y'=\frac{y}{t}-(y/t)^2, \quad 1\leq t \leq 4, \quad y(1)=1 \end{equation}\]
Solución exacta.
\[\begin{equation} y(t)=\frac{t}{1+log\,t} \end{equation}\]
\[\begin{equation} y'=(t+2t^3)y^3-ty, \quad 0\leq t \leq 2, \quad y(0)=\frac{1}{3} \end{equation}\]
Solución exacta.
\[\begin{equation} y(t)=\frac{1}{\sqrt{3+2t^2+6e^{t^2}}} \end{equation}\]
\[\begin{equation} y'=\frac{2-2ty}{t^2+1}, \quad 0\leq t \leq 3, \quad y(0)=1 \end{equation}\]
Solución exacta.
\[\begin{equation} y(t)=\frac{2t+1}{t^2+1} \end{equation}\]
\[\begin{equation} \begin{aligned} y'_1&=y_2, &\qquad y_1(0)=1;\\ y'_2&=-y_1-2e^t+1, &\qquad y_2(0)=0;\\ y'_3&=-y_1-e^t+1, &\qquad y_3(0)=1; \end{aligned} \end{equation}\]
para \(0\leq t \leq 2\). Donde las soluciones exactas son:
\[\begin{equation} \begin{aligned} y_1(t)&=cos\,t+ sen\,t-e^t+1\\ y_2(t)&=-sen\, t+cos\, t-e^t\\ y_3(t)&=-sen\,t+cos\,t \end{aligned} \end{equation}\]
Solución real
\[\begin{equation} \begin{aligned} y'_1&=3y_1+2y_2-(2t^2+1)e^{2t}, &\qquad y_1(0)=1;\\ y'_2&=4y_1+y_2+(t^2+2t-4)e^{2t}, &\qquad y_2(0)=1; \end{aligned} \end{equation}\]
para \(0\leq t \leq 1\). Donde las soluciones exactas son:
\[\begin{equation} \begin{aligned} y_1(t)&=\frac{1}{3}e^{5t}-\frac{1}{3}e^{-t}+e^{2t}\\ y_2(t)&=\frac{1}{3}e^{5t}+\frac{2}{3}e^{-t}+t^2e^{2t} \end{aligned} \end{equation}\]
Solución real
\[\begin{equation} \begin{aligned} y'_1&=y_2-y_3+t, &\qquad y_1(0)&=1;\\ y'_2&=3t^2, &\qquad y_2(0)&=1;\\ y'_3&=y_2+e^{-t}, &\qquad y_3(0)&=-1; \end{aligned} \end{equation}\]
para \(0\leq t \leq 1\). Donde las soluciones exactas son:
\[\begin{equation} \begin{aligned} y_1(t)&=-0.05t^5+0.25t^4+t+2-e^{-t}\\ y_2(t)&=t^3+1\\ y_3(t)&=0.25t^4+t-e^{-t} \end{aligned} \end{equation}\]
Solución real
\[\begin{equation} \begin{aligned} y'_1&=3y_1+2y_2-y_3-1-3t-2\, sen\,t, &\qquad y_1(0)&=5;\\ y'_2&=y_1-2y_2+3y_3+6-t+2\,sen\,t+cos\, t, &\qquad y_2(0)&=-9;\\ y'_3&=2y_1+4y_3+8-2t, &\qquad y_3(0)&=-5; \end{aligned} \end{equation}\]
para \(0\leq t \leq 2\). Donde las soluciones exactas son:
\[\begin{equation} \begin{aligned} y_1(t)&=2e^{3t}+3e^{-2t}+t\\ y_2(t)&=-8e^{-2t}+e^{4t}-2e^{3t}+\,sen\,t\\ y_3(t)&=2e^{4t}-4e^{3t}-e^{-2t}-2 \end{aligned} \end{equation}\]
- \(y''-2y'+y=t\,e^t-t\), \(0\leq t\leq 1\), \(y(0)=y'(0)=0\).
Solución exacta:
\[\begin{equation} y(t)=\frac{1}{6}t^3e^t-te^t+2e^t-t-2. \end{equation}\]
- \(y'''+2y''-y'-2y=e^t\), \(0\leq t\leq 3\), \(y(0)=1\), \(y'(0)=2\), \(y''(0)=0\).
Solución exacta:
\[\begin{equation} y(t)=\frac{43}{36}e^t+\frac{1}{4}e^{-t}-\frac{4}{9}e^{-2t}+\frac{1}{6}te^t. \end{equation}\]
- \(t^3y'''+t^2y''-2ty'+2y=8t^3-2\), \(1\leq t\leq 2\), \(y(1)=2\), \(y'(1)=8\), \(y''(1)=6\). Solución exacta: \(y(t)=2t-t^{-1}+t^2+t^3-1\).
- Mayo 2022