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Give the general solution of the system: $$X'(t) = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} X(t)+\begin{pmatrix} 2e^{2t} \\ 0 \end{pmatrix} $$

I manage to come to the general solution to the homogenous but when I get finding a particular solution I'm messing up.

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  • $\begingroup$ Have you tried a particular solution of the form $e^{2t}$ times a constant vector? $\endgroup$ – user147263 Nov 28 '15 at 22:24
  • $\begingroup$ Thanks a lot, I'm new to this whole Stack Exhange thing! $\endgroup$ – Victor Pinto Nov 28 '15 at 22:25
  • $\begingroup$ It can't be since $e^{2t}$is in the homogenous solution. $\endgroup$ – Victor Pinto Nov 28 '15 at 22:27
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The eigenvalues of the matrix are $2$ and $4$, one of which is in $e^{2t}$. So try $$X(t)=\left(\begin{array}{c}ate^{2t}\\bte^{2t}\end{array}\right)$$

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  • $\begingroup$ shouldn't it be $\begin{pmatrix} (at+b)e^{2t} \\ (ct+d)e^{2t} \end{pmatrix} $? $\endgroup$ – Victor Pinto Nov 28 '15 at 22:27
  • $\begingroup$ Yes, but I was thinking you already have $b$ and $d$ from the homogeneous solution. I might be wrong; I was thinking of the scalar diff eqn. $\endgroup$ – Empy2 Nov 28 '15 at 22:30
  • $\begingroup$ I'm not sure, because if it was indeed $\begin{pmatrix} ate^{2t} \\ bte^{2t} \end{pmatrix} $ a and b would be 0 since they never equal $ 2e^{2t}$ . $\endgroup$ – Victor Pinto Nov 28 '15 at 22:35
  • $\begingroup$ Remember the derivative on the left-hand side. $\endgroup$ – Empy2 Nov 28 '15 at 22:35
  • $\begingroup$ Ohh maybe that was it, I was able to get values for a and c and b in function of d. $\endgroup$ – Victor Pinto Nov 28 '15 at 22:48

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