Nonhomogeneous system of ordinary differential equations

Question

I have been working on this problem for hours now and I don't know how to finish it $$ X'= \begin{bmatrix} -2 & -2 \\ 4 & -2 \\ \end{bmatrix}X\ +\ \begin{bmatrix} 3t \\ 1 \\ \end{bmatrix}\ $$ I have managed to eigenvalues and eigenvectors, which I got $\lambda=-2+(\sqrt2)i\ and \lambda=-2-(\sqrt2)i$ and then I managed to solve up to here $$ C_1e^{-2t}\begin{bmatrix} -\sqrt2sin(2\sqrt2t) \\ 2cos(2\sqrt2t) \\ \end{bmatrix}\ +\ C_2e^{-2t}\begin{bmatrix} \sqrt2cos(2\sqrt2t) \\ 2sin(2\sqrt2t) \\ \end{bmatrix}\ $$ Some help on to what to do next would be greatly appreciated. Thank you

Answer 1

I will give a quick solution.

Let us make a guess \begin{align} X(t) = w + tv \end{align} then it follows \begin{align} v = A(w+tv) = Aw + tAv +f \ \ \Rightarrow \ \ v-Aw - tAv = f \end{align} which mean \begin{align} \begin{pmatrix} v_1\\ v_2 \end{pmatrix} - \begin{pmatrix} -2 & -2\\ 4 & -2 \end{pmatrix} \begin{pmatrix} w_1\\ w_2 \end{pmatrix} = \begin{pmatrix} 0\\ 1 \end{pmatrix} \end{align} and \begin{align} t\begin{pmatrix} -2 & -2\\ 4 & -2 \end{pmatrix} \begin{pmatrix} v_1\\ v_2 \end{pmatrix} =t \begin{pmatrix} -3\\ 0 \end{pmatrix}. \end{align} Solving for $v, w$ yields \begin{align} v= \begin{pmatrix} \frac{1}{2}\\ 1 \end{pmatrix} \ \ \ \text{ and } \ \ \ w = \begin{pmatrix} -\frac{1}{12}\\ -\frac{1}{6} \end{pmatrix} \end{align} which means \begin{align} X(t) = \begin{pmatrix} -\frac{1}{12}+\frac{t}{2}\\ -\frac{1}{6}+t \end{pmatrix} \end{align} is a particular solution.