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I want to find the solution of

\begin{cases} x'= -5x-y+e^t \\ y'= 2x-3y \end{cases}

$$A = \begin{bmatrix} -5 & -1 \\ 2 & -3 \end{bmatrix}$$

I calculated the exponential matrix $e^{tA}$ and found the solution of the homogeneous system

$$\begin{bmatrix} x\\ y\end{bmatrix} = e^{tA} \begin{bmatrix} k_1\\ k_2\end{bmatrix}$$

where $k_1,k_2 \in \mathbb R$. How do I find a particular solution?

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  • $\begingroup$ You use the formula $$e^{tA}\int_0^t e^{-sA}\cdot\begin{pmatrix}e^s\\ 0\end{pmatrix}\,ds$$ $\endgroup$
    – user228113
    Jun 12, 2016 at 9:42

1 Answer 1

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Laplace-transforming the system of ODEs,

$$\begin{bmatrix}s+5 & 1\\ -2 & s+3\end{bmatrix} \begin{bmatrix} X (s) \\ Y (s)\end{bmatrix} = \begin{bmatrix}\frac{1}{s-1}\\ 0\end{bmatrix}$$

Hence,

$$\begin{array}{rl} \begin{bmatrix} X (s) \\ Y (s)\end{bmatrix} &= \begin{bmatrix}s+5 & 1\\ -2 & s+3\end{bmatrix}^{-1} \begin{bmatrix}\frac{1}{s-1}\\ 0\end{bmatrix}\\ &= \dfrac{1}{(s+5)(s+3)+2} \begin{bmatrix}s+3 & -1\\ 2 & s+5\end{bmatrix} \begin{bmatrix}\frac{1}{s-1}\\ 0\end{bmatrix}\\ &= \dfrac{1}{s^2 + 8 s + 17} \begin{bmatrix}\frac{s+3}{s-1}\\ \frac{2}{s-1}\end{bmatrix}\\ &= \dfrac{1}{(s+4)^2 + 1} \begin{bmatrix}\frac{s+3}{s-1}\\ \frac{2}{s-1}\end{bmatrix}\end{array}$$

Now do partial fraction expansion and then take the inverse Laplace transform.

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