Solution
= Solution
Differentiate $y'=-y+3x$ and use the other equation:
$$
y''=-y'+3x'=-y'+12x-6y.
$$
Since $3x=y'+y$, this becomes $y''-3y'+2y=0$. Its roots are one and two, so $y=Ae^t+Be^{2t}$. The given data imply $y'(0)=-2+3=1$, hence $A+B=2$ and $A+2B=1$. Thus $A=3$, $B=-1$, giving
$$
\boxed{y(t)=3e^t-e^{2t}}.
$$
Recover the other dependent variable from $x=(y'+y)/3$:
$$
\boxed{x(t)=2e^t-e^{2t}}.
$$
These expressions satisfy both original equations and the two initial values.