Aufgabe: Es handelt sich um Differentialgleichungssysteme
Habe ich die Aufgabe richtig gelöst? Könnt ihr mir bitte eine Rückmeldung geben?
Problem/Ansatz:
I) \( y_{1}^{\prime}=y_{1}+2 y_{2}+e^{-2 x} \)
\( y_{2}^{\prime}=4 y_{1}-y_{2} \)
\( \left(\begin{array}{l}y_{1} \\ y_{2}\end{array}\right)^{\prime}=\left(\begin{array}{cc}1 & 2 \\ 4 & -1\end{array}\right)\left(\begin{array}{l}y_{1} \\ y_{2}\end{array}\right)+\left(\begin{array}{c}e^{-2 x} \\ 0\end{array}\right) \)
\( \operatorname{det}\left(\begin{array}{cc}1-\lambda & 2 \\ 4 & -1-\lambda\end{array}\right)=(1-\lambda) \cdot(-1-\lambda)-8 \)
\( =-1-x+\not x+\lambda^{2}-8 \)
\( =\lambda^{2}-9=0 \)
\( \Rightarrow \lambda=\pm 3 \)
\( \lambda=3\left(\begin{array}{cc|c}-2 & 2 & 0 \\ 4 & -4 & 0\end{array}\right) \sim\left(\begin{array}{cc|c}-2 & 2 & 0 \\ 0 & 0 & 0\end{array}\right) \)
\( \sim\left(\begin{array}{cc|c}1 & -\frac{1}{2} & 0 \\ 0 & 0 & 0\end{array}\right) \Longrightarrow E V=\left(\begin{array}{c}-\frac{1}{2} \\ -1\end{array}\right) \triangleq\left(\begin{array}{l}1 \\ 2\end{array}\right) \)
\( \lambda=-3 \)
\( \leadsto\left(\begin{array}{cc|c}1 & \frac{1}{2} & 0 \\ 0 & 0 & 0\end{array}\right) \Rightarrow E V=\left(\begin{array}{c}1 \\ -1 \\ -1\end{array}\right) \leqq\left(\begin{array}{c}1 \\ -2\end{array}\right) \)
\( y_{2}=c_{1} \cdot e^{3 x}\left(\begin{array}{l}1 \\ 2\end{array}\right)+c_{2} \cdot e^{-3 x}\left(\begin{array}{c}1 \\ -2\end{array}\right) \)
\( \psi=\left(\begin{array}{cc}e^{3 x} & e^{-3 x} \\ 2 e^{3 x} & -2 e^{-3 x}\end{array}\right)\left(\begin{array}{l}c_{1} \\ c_{2}\end{array}\right)= \)
\( \left(\begin{array}{cc}e^{3 x} & e^{-3 x} \\ 2 e^{3 x} & -2 e^{-3 x}\end{array}\right)\left(\begin{array}{l}c_{1}^{\prime} \\ c_{2}^{\prime}\end{array}\right)=\left(\begin{array}{c}e^{-2 x} \\ 0\end{array}\right) \)
\( c_{1}^{\prime}=\frac{\operatorname{det}\left(\begin{array}{cc}e^{-2 x} & e^{-3 x} \\ 0 & -2 e^{-3 x}\end{array}\right)}{\operatorname{det}\left(\begin{array}{cc}e^{3 x} & e^{-3 x} \\ 2 e^{3 x} & -2 e^{-3 x}\end{array}\right)}= \)
\( =\frac{-2 e^{-5 x}}{-2-2}=\frac{-2 e^{-5 x}}{-\not 42}=\frac{1}{2 e^{5 x}} \)
\( a_{1}=\int\left(\frac{1}{2} \cdot e^{-5 x}\right) d x=-\frac{1}{10} e^{-5 x} \)
\( q^{\prime}=\frac{\operatorname{det}\left(\begin{array}{ll}e^{3 x} & e^{-2 x} \\ 2 e^{3 x} & 0\end{array}\right)}{-4}=\frac{-2 e^{x}}{-4} \)
\( =\frac{1}{2} e^{x} \)
\( C_{2}=\int\left(\frac{1}{2} e^{x}\right) d x=\frac{1}{2} e^{x} \)
\( y_{p}=\left(\begin{array}{cc}e^{3 x} & e^{-3 x} \\ 2 e^{3 x} & -2 e^{-3 x}\end{array}\right)\left(\begin{array}{c}-\frac{1}{10} e^{-5 x} \\ \frac{1}{2} e^{x}\end{array}\right) \)
\( =\left(\begin{array}{c}-\frac{1}{10} e^{-2 x}+\frac{1}{2} e^{-2 x} \\ -\frac{2}{10} e^{-2 x}-e^{-2 x}\end{array}\right)=\left(\begin{array}{c}\frac{2}{5} e^{-2 x} \\ -\frac{6}{5} e^{-2 x}\end{array}\right) \)
\( -\frac{1}{10}+\frac{5}{10}=\frac{\not^{2}}{10_{5}}=\frac{2}{5} \)
\( -\frac{2}{10}-\frac{10}{10}=-\frac{42}{10}=-\frac{6}{5} \)
