# Lineare Differentialgleichung/(t 3 1 t)/(0,1)/Picard-Lindelöf/Bis dritte Iteration/Aufgabe/Lösung

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Die nullte Iteration ist die konstante Funktion

${\displaystyle {}\varphi _{0}(t)={\begin{pmatrix}0\\1\end{pmatrix}}\,.}$

Die erste Iteration ist

{\displaystyle {}{\begin{aligned}\varphi _{1}(t)&={\begin{pmatrix}x_{1}(t)\\y_{1}(t)\end{pmatrix}}\\&={\begin{pmatrix}0\\1\end{pmatrix}}+\int _{0}^{t}{\begin{pmatrix}s&3\\1&s\end{pmatrix}}{\begin{pmatrix}0\\1\end{pmatrix}}ds\\&={\begin{pmatrix}0\\1\end{pmatrix}}+\int _{0}^{t}{\begin{pmatrix}3\\s\end{pmatrix}}ds\\&={\begin{pmatrix}3t\\{\frac {1}{2}}t^{2}+1\end{pmatrix}}.\end{aligned}}}

Die zweite Iteration ist

{\displaystyle {}{\begin{aligned}\varphi _{2}(t)&={\begin{pmatrix}x_{2}(t)\\y_{2}(t)\end{pmatrix}}\\&={\begin{pmatrix}0\\1\end{pmatrix}}+\int _{0}^{t}{\begin{pmatrix}s&3\\1&s\end{pmatrix}}{\begin{pmatrix}3s\\{\frac {1}{2}}s^{2}+1\end{pmatrix}}ds\\&={\begin{pmatrix}0\\1\end{pmatrix}}+\int _{0}^{t}{\begin{pmatrix}{\frac {9}{2}}s^{2}+3\\{\frac {1}{2}}s^{3}+4s\end{pmatrix}}ds\\&={\begin{pmatrix}{\frac {3}{2}}t^{3}+3t\\{\frac {1}{8}}t^{4}+2t^{2}+1\end{pmatrix}}.\end{aligned}}}

Die dritte Iteration ist

{\displaystyle {}{\begin{aligned}\varphi _{3}(t)&={\begin{pmatrix}x_{3}(t)\\y_{3}(t)\end{pmatrix}}\\&={\begin{pmatrix}0\\1\end{pmatrix}}+\int _{0}^{t}{\begin{pmatrix}s&3\\1&s\end{pmatrix}}{\begin{pmatrix}{\frac {3}{2}}s^{3}+3s\\{\frac {1}{8}}s^{4}+2s^{2}+1\end{pmatrix}}ds\\&={\begin{pmatrix}0\\1\end{pmatrix}}+\int _{0}^{t}{\begin{pmatrix}{\frac {15}{8}}s^{4}+9s^{2}+3\\{\frac {1}{8}}s^{5}+{\frac {17}{6}}s^{3}+4s\end{pmatrix}}ds\\&={\begin{pmatrix}{\frac {3}{8}}t^{5}+3t^{3}+3t\\{\frac {1}{48}}t^{6}+{\frac {17}{24}}t^{4}+2t^{2}+1\end{pmatrix}}.\end{aligned}}}