= Solution
Write the <linear system of differential equations> as
$$
\mathbf z'=\frac1tA\mathbf z+\mathbf b,\qquad
A=\begin{pmatrix}4&-2\\-1&5\end{pmatrix},\quad\mathbf b=\binom{-9}{3}.
$$
A homogeneous power $t^\lambda\mathbf v$ solves this <Cauchy-Euler differential system> precisely when $A\mathbf v=\lambda\mathbf v$. The <eigenvalues> are $3$ and $6$, with <eigenvectors> $(2,1)^T$ and $(1,-1)^T$. The homogeneous solution is therefore $C t^3(2,1)^T+D t^6(1,-1)^T$.
For a particular solution, put $\mathbf z_p=t\mathbf c$. Then $(I-A)\mathbf c=\mathbf b$, which gives $\mathbf c=(3,0)^T$. At $t=1$, the two initial conditions become $2C+D+3=0$ and $C-D=0$, so $C=D=-1$. Thus
$$
\boxed{x(t)=3t-2t^3-t^6,\qquad y(t)=t^6-t^3,\qquad t\geq1}.
$$
The two powers correspond to the two <eigenvalues>; the lower-degree particular term accounts for the constant forcing. Substitution gives both original right sides and the zero initial data.
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