# Give the general solution of the $2\times 2$ inhomogeneous system of differential equations

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.

• Have you tried a particular solution of the form $e^{2t}$ times a constant vector? – user147263 Nov 28 '15 at 22:24
• Thanks a lot, I'm new to this whole Stack Exhange thing! – Victor Pinto Nov 28 '15 at 22:25
• It can't be since $e^{2t}$is in the homogenous solution. – Victor Pinto Nov 28 '15 at 22:27

The eigenvalues of the matrix are $2$ and $4$, one of which is in $e^{2t}$. So try $$X(t)=\left(\begin{array}{c}ate^{2t}\\bte^{2t}\end{array}\right)$$
• shouldn't it be $\begin{pmatrix} (at+b)e^{2t} \\ (ct+d)e^{2t} \end{pmatrix}$? – Victor Pinto Nov 28 '15 at 22:27
• 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. – Empy2 Nov 28 '15 at 22:30
• 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 $2e^{2t}$ . – Victor Pinto Nov 28 '15 at 22:35