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How to solve the following differential equation?

$$ \dot x = \begin{bmatrix} 4 & 5 \\ -2 & -2 \end{bmatrix} x + \begin{bmatrix} 4 e^t \cos(t)\\ 0\end{bmatrix}, \qquad x(0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

I started by finding eigenvalues for matrix A and I have that $\lambda_1 = 1+i$ and $\lambda_2 = 1-i$ but I don't know what I should do now :/

Lutz Lehmann
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El3na
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    Review the course material? Surely there’s an example of a system with complex eigenvalues somewhere in there. You could use the method described here, but I suspect that you’re supposed to go through the exercise of computing eigenvectors and generating real solutions from those. – amd Apr 19 '20 at 21:20

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Next you find left a left eigenvector for $λ_1=1+i$, that is, a left kernel vector for $$ \begin{bmatrix} 3-i&5\\-2&-3-i \end{bmatrix} $$ which is for instance $[2,\,3-i]$. According to that, consider the linear combination $u=2x_1+3x_2$ and express the system in $u,x_2$, \begin{align} \dot u&=2x_1+4x_2+8e^t\cos t&&=u+x_2+8e^t\cos t\\ \dot x_2&=-2x_1-2x_2&&=-u+x_2 \end{align} Now the system matrix is the representation of a complex number, allowing to treat $u,x_2$ as real and imaginary part of a complex valued function, $$ \dot u+i\dot x_2=(1-i)(u+ix_2)+4(e^{(1+i)t}+e^{(1-i)t}). $$ This can be solved like any other linear DE of first order.

Lutz Lehmann
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