System of first order differential equation. Consider the linear system $y’=Ay+h$ where
$$
A=\begin{bmatrix}
1 & 1\\
4 & -2
\end{bmatrix}
$$and $$h=\begin{bmatrix}
3t+1\\
2t+5
\end{bmatrix}$$ Suppose $y(t)$ is a solution such that $$\lim_{t\to\infty}\frac{y(t)}{t}=k\in\mathbb R^2$$ What is the value of $k?$
$1$. $\begin{bmatrix}
\frac{-4}{3}\\
\frac{-5}{3}
\end{bmatrix}$.
$2$. $\begin{bmatrix}
\frac{4}{3}\\
\frac{-5}{3}
\end{bmatrix}$.
$3$. $\begin{bmatrix}
\frac{2}{3}\\
\frac{-5}{3}
\end{bmatrix}$.
$4$. $\begin{bmatrix}
\frac{-2}{3}\\
\frac{-5}{3}
\end{bmatrix}$.
I find eigen value of corresponding homogeneous system as $2$ and $-3$ and corresponding eigen vectors as $\begin{bmatrix}
1\\
1
\end{bmatrix}$ and $\begin{bmatrix}
1\\
-4
\end{bmatrix}$. Therefore general solution of corresponding homogeneous differential equation is given by $y_c=\Phi(x)c=\begin{bmatrix}
e^{2t} & e^{-3t}\\
e^{2t} & -4e^{-3t}
\end{bmatrix}c$, for arbitrary constant $c$.
Now as I know that by using variation of parameter general solution is given by $$y=\Phi(t)c+\Phi(t)\int {\Phi(x)}^{-1}h(x)dx= \begin{bmatrix}
e^{2t} & e^{-3t}\\
e^{2t} & -4e^{-3t}
\end{bmatrix}c+ \begin{bmatrix}
e^{2t} & e^{-3t}\\
e^{2t} & -4e^{-3t}
\end{bmatrix}\int {\begin{bmatrix}
e^{2x} & e^{-3x}\\
e^{2x} & -4e^{-3x}
\end{bmatrix}}^{-1} \begin{bmatrix}
3x+1\\
2x+5
\end{bmatrix}dx$$ Now I am unable to reach at final answer . Please help. Thank you.

We are given the linear system
$$y'(t) = Ay(t) + h(t) =\begin{bmatrix}1 & 1\\4 & -2\end{bmatrix}y(t) + \begin{bmatrix}3t+1\\2t+5\end{bmatrix}$$
I will use Example $2$ of these notes as a guide.
We find the eigenvalues and eigenvectors as
$$\lambda_1 = -3, v_1 = \begin{bmatrix} -1\\4 \end{bmatrix}\\ \lambda_2 = 2, v_2 = \begin{bmatrix} 1\\1 \end{bmatrix}$$
We can write the complimentary solution as
$$y_c(t) = c_1e^{-3t} \begin{bmatrix} -1\\4 \end{bmatrix} + c_2 e^{2t}\begin{bmatrix} 1\\1 \end{bmatrix}$$
From this we have $$X = \begin{bmatrix}
 -e^{-3 t} & e^{2 t} \\
 4 e^{-3 t} & e^{2 t} \\
\end{bmatrix}$$
The inverse is
$$X^{-1} = \begin{bmatrix}
 -\dfrac{e^{3 t}}{5} & \dfrac{e^{3 t}}{5} \\
 \dfrac{4 e^{-2 t}}{5} & \dfrac{e^{-2 t}}{5} \\
\end{bmatrix}$$
The multiplication inside the integral is
$$X^{-1} h =\begin{bmatrix}
 \dfrac{4 e^{3 t}}{5}-\dfrac{1}{5} e^{3 t} t \\
 \dfrac{14}{5} e^{-2 t} t+\dfrac{9 e^{-2 t}}{5} \\
\end{bmatrix}$$
Now do the integral
$$\int X^{-1}h dt = \int \begin{bmatrix}
 \dfrac{4 e^{3 t}}{5}-\dfrac{1}{5} e^{3 t} t \\
 \dfrac{14}{5} e^{-2 t} t+\dfrac{9 e^{-2 t}}{5} \\
\end{bmatrix}~dt = \begin{bmatrix}
 \dfrac{14 e^{3 t}}{135}-\dfrac{1}{45} e^{3 t} t \\
 \dfrac{7}{10} e^{-2 t} t+\dfrac{23 e^{-2 t}}{20} \\
\end{bmatrix}$$
Now we can get the particular solution
$$y_p(t) = X \int X^{-1}h~dt = \begin{bmatrix}
 -\dfrac{4 t}{3}-\dfrac{17}{9} \\
 -\dfrac{5 t}{3}-\dfrac{4}{9} \\
\end{bmatrix}$$
Next, use $y_p(t)$ (a solution of the DEQ) and find
$$\lim_{t\to\infty}\frac{y_p(t)}{t}=\lim_{t\to\infty}\begin{bmatrix}
 -\dfrac{4 }{3}-\dfrac{17}{9t} \\
 -\dfrac{5}{3}-\dfrac{4}{9t} \\
\end{bmatrix}$$
You arrive at choice $1.$
