Solving the following Initial Value Problem

Question

Solve the initial value problem: $$y_1'=2y_1+2e^{2t}$$ $$y_2'=3y_1+2y_2+3e^{2t}$$ and $y_1(0)=2, y_2(0)=3$

I'm trying to start by using $y'=Ay+g$

Does $g=\begin{bmatrix}2e^{2t}\\3e^{2t}\end{bmatrix}$

Following, I get $det(A-\lambda I)$ as $\lambda^{2}-4\lambda+4=0$

How can I obtain the homogeneous solution by this equation?

Answer 1

Hint

I denote $$A=\begin{pmatrix}2&0\\ 3&2\end{pmatrix},$$ and $g$ as you did. Let $y=\begin{pmatrix}y_1\\y_2\end{pmatrix}.$ Unfortunately, $A$ is not diagonalizable. Nevertheless : \begin{align*} y'=Ay+g&\iff e^{-At}y'-e^{-At}Ay=e^{-At}g\\ &\iff \frac{\mathrm d }{\mathrm d t}\left(e^{-At}y\right)=e^{-At}g, \end{align*} where $e^{A}$ is the exponential matrix of $A$. I let you conclude.