$$
t^{3} y^{\prime}+4 t^{2} y=e^{-t}, \quad y(-1)=0, \quad t<0
$$
First divide by $t^3$ on both side of the equation, we get
$$
y^{\prime}+\frac{4}{t} y=\frac{e^{-t}}{t^{3}}
$$
Using the method of integrating factor we have equation for $u(t)$
$$
u(t)=e^{\int \frac{4}{t} d t}=e^{4 \ln (t)} =t^{4}
$$
Multiply both sides by $t^{4}$
$$
t^{4} y^{\prime}+4 t^{3} y=t e^{-t}
$$
$$
\left(t^{4} y\right)^{\prime}=t e^{-t}
$$
$$
t^{4} y=\int t e^{-t}
$$
$$
t^{4} y=-t e^{-t}-e^{-t}+C
$$
$$
y=-\frac{e^{-t}}{t^{3}}-\frac{e^{-t}}{t^{4}}+\frac{C}{t^{4}}
$$
to check $C,$ plug in condition $y(-1)=0$
$$
y(-1)=e-e+C=C=0
$$
Plug in $C=0$ gets
$$
y=-\frac{e^{-t}}{t^{3}}-\frac{e^{-t}}{t^{4}}
$$
