Solution
= Solution
Set $y=tz$ and $u=(x,z)^T$. The system becomes
$$
tu'=\begin{pmatrix}4&-6\\3&-5\end{pmatrix}u+t\binom11+\binom21.
$$
The <matrix> has eigenpairs $1,(2,1)^T$ and $-2,(1,1)^T$. A particular solution is $(-t/3-2,-1)^T$. Therefore
$$
x=2C_1t+C_2t^{-2}-\frac t3-2,
$$
$$
y=C_1t^2+C_2t^{-1}-t.
$$