Considering the differential equation

$$y'(x) = A y(x) + g(x) e^{ax}$$

where $A in Bbb R^{2 times 2}$ and $a$ is no eigenvalue of $A$, and $g(x)$ a polynomial vector, then there exists a particular solution

$$y_p(x) = h(x) e^{ax}$$

where $h(x)$ is a polynomial vector.

I tried some approaches, but I couldn’t solve it.