= Solution
Let $Y(s)$ be the <Laplace transform> of $y$. The initial values give $\mathcal L(y')=sY$ and $\mathcal L(y'')=s^2Y-1$, while $\mathcal L(te^{-3t})=(s+3)^{-2}$. Therefore, for $\operatorname{Re}s$ sufficiently large,
$$
(s^2-4s+3)Y(s)-1=\frac1{(s+3)^2},\qquad
Y(s)=\frac{1+(s+3)^{-2}}{(s-1)(s-3)}.
$$
The <partial fraction decomposition> is
$$
Y(s)=\frac{37}{72(s-3)}-\frac{17}{32(s-1)}+\frac5{288(s+3)}+\frac1{24(s+3)^2}.
$$
Inverting the elementary <Laplace transforms> gives
$$
\boxed{y(t)=\frac{37}{72}e^{3t}-\frac{17}{32}e^t+\left(\frac5{288}+\frac t{24}\right)e^{-3t},\qquad t\ge0.}
$$
These coefficients give $y(0)=0$ and $y'(0)=1$; the transformed equation, or direct differentiation, verifies the forcing. Uniqueness follows from uniqueness for the corresponding linear initial-value <ordinary differential equation>.
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