= Solution
The one-sided <Laplace transform> is $\bar u(p)=\int_0^\infty e^{-pt}u(t)dt$ for sufficiently large $\operatorname{Re}p$. <Integration by parts>, with the exponential suppressing the upper boundary, gives
$$
\overline{u'}=p\bar u-u(0),\qquad\overline{u''}=p^2\bar u-pu(0)-u'(0),\qquad\overline{u'''}=p^3\bar u-p^2u(0)-pu'(0)-u''(0).
$$
Differentiation under the integral gives $d\bar u/dp=-\int_0^\infty te^{-pt}u(t)dt$, proving $\overline{tu}=-d\bar u/dp$. Start with $\overline{e^{-t}}=1/(p+1)$ and apply this rule repeatedly:
$$
\boxed{\bar v_0=\frac1{p+1},\quad\bar v_1=\frac1{(p+1)^2},\quad\bar v_2=\frac2{(p+1)^3},\quad\bar v_3=\frac6{(p+1)^4}.}
$$
With all three initial data zero, transforming the forced equation gives $(p+1)^3\bar y=1/(p+1)$. Thus $\bar y=(p+1)^{-4}$ and the listed inverse transform gives
$$
\boxed{y(t)=\frac{t^3e^{-t}}6,\qquad t\geq0.}
$$
The cubic factor is also the <resonant exponential particular solution> associated with the triple characteristic root $-1$.
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