Answer
$$ y=\frac{\sin t}{t^3}$$
Work Step by Step
Given $$t^{3} \frac{d y}{d t}+3 t^{2} y=\cos t, \quad y(\pi)=0$$
Rewriting the equation as
$$ \frac{d y}{d t}+\frac{3}{t}y =\frac{\cos t}{t^3} $$
Since the equation is linear and $p(t)= \frac{3}{t}$ and $q(t)=\frac{\cos t}{t^3}$, then
\begin{align*}
\mu &=e^{\int p(x)}dx\\
&= e^{\int \frac{3}{t}dx}\\
&=e^{3\ln t}\\
&=t^3
\end{align*}
Hence the general solution given by
\begin{align*}
\mu(t)y&=\int \mu(t)q(t)dt\\
t^3 y&= \int t^3 \frac{\cos t}{t^3}dt\ \ \\
&= \int \cos t dt\ \ \\
&=\sin t +C
\end{align*}
Then
$$ y=\frac{\sin t}{t^3}+\frac{C}{t^3}$$
Since $y(\pi )=0 $, then $C=0$, hence
$$ y=\frac{\sin t}{t^3}$$
