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I am solving system of differential equations, $$ \begin{cases} \dot x = 2x + y +e^t; \\ \dot y = -2x + 2t. \end{cases} $$ It's matrix, eigenvalues and eigenvectors are as follows:

$\qquad A = \begin{pmatrix} 2 & 1 \\ -2 & 0 \end{pmatrix}, \qquad \lambda_{12} = 1 \pm i, \qquad V_1 = \begin{pmatrix} 1 \\ i - 1\end{pmatrix}, \qquad V_2 = \begin{pmatrix} 1 \\ -i -1\end{pmatrix}$.

Thus, the solution is $$ X_0 = e^{(1+i)t}\begin{pmatrix} 1 \\ i-1 \end{pmatrix} = e^t (\cos t+ i \sin t)\begin{pmatrix} 1 \\ i-1 \end{pmatrix} = \\ e^t \begin{pmatrix} \cos t \\ -\cos t - \sin t \end{pmatrix}+i e^t \begin{pmatrix} \sin t \\ \cos t - \sin t \end{pmatrix}. $$ And for each variable it yields \begin{cases} x_0 = C_1 e^t \cos t + C_2 e^t \sin t; \\ y_0 = C_1 e^t ( -\cos t - \sin t ) + C_2 e^t (\cos t - \sin t). \end{cases} When I diffepentiate it and plug into original system, I get wrong answer. But I can already see it is not right. The textbook says, the answer for homogenous system must be $$ \begin{cases} x_0 = C_2 e^t \sin t+ C_1 e^t (\sin t+\cos t); \\ y_0 = C_2 e^t (\cos t-\sin t)-2 C_1 e^t \sin t \end{cases} $$ or $$ X_0 = C_1 e^t \begin{pmatrix} \sin t+\cos t \\ -2 \sin t \end{pmatrix}+ C_2 e^t \begin{pmatrix} \sin t \\ \cos t-\sin t \end{pmatrix} $$ Where is my mistake? Thank you in advance.

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  • $\begingroup$ $A = \begin{pmatrix} 2 & 1 \\ -2 & 0 \end{pmatrix};$ $|A -\lambda E| = \begin{pmatrix} 1-i & 1 \\ -2 & -1 - i \end{pmatrix}$ then from the first line (lines are dependent) $ x_1 = \alpha, x_2 = (i-1) \alpha \Rightarrow V_1 = (1, i-1)$ $\endgroup$ – Dankevich Dec 22 '14 at 1:01
  • $\begingroup$ When I tried that one, it worked, but how can the answer depend on the vector I chose? $\endgroup$ – Dankevich Dec 22 '14 at 1:07
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probally it 'swrong after

'And for each variable it yields '

$X_0$ is a solution of $$ \begin{cases} \dot x = 2x + y; \\ \dot y = -2x . \end{cases} $$ So $C_1 X_0 $ is also a solution but not $C_1 Re(X_0) + C_2 Im(X_0) $

You have to find the second solution $X_1$ with the second eigen value and so $C_1 X_0 + C_2 X_1 $ will be a solution

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  • $\begingroup$ I suspect the $X_0$ itself, but do not know how to fix it, if the mistake is there $\endgroup$ – Dankevich Dec 22 '14 at 1:39

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