Answer
$y = \frac{{\sin t}}{{{t^3}}}$
Work Step by Step
$$\eqalign{
& {t^3}\frac{{dy}}{{dt}} + 3{t^2}y = \cos t,{\text{ }}y\left( \pi \right) = 0 \cr
& {\text{Divide both sides by }}{t^3} \cr
& \frac{{dy}}{{dt}} + \frac{{3{t^2}y}}{{{t^3}}} = \frac{{\cos t}}{{{t^3}}} \cr
& \frac{{dy}}{{dt}} + \frac{3}{t}y = \frac{{\cos t}}{{{t^3}}} \cr
& {\text{The differential equation is in the form }}\frac{{dy}}{{dt}} + P\left( t \right)y = Q\left( t \right) \cr
& {\text{With }}P\left( t \right) = \frac{3}{t}{\text{ and }}Q\left( t \right) = \frac{{\cos t}}{{{t^3}}} \cr
& {\text{Find the integrating factor }}I\left( t \right) = {e^{\int {P\left( t \right)} dt}} \cr
& I\left( t \right) = {e^{\int {\frac{3}{t}dt} }} = {e^{3\left| {\ln t} \right|}} = {t^3} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& \left( {\frac{{dy}}{{dt}} + \frac{3}{t}y} \right){t^3} = \left( {\frac{{\cos t}}{{{t^3}}}} \right){t^3} \cr
& {t^3}\frac{{dy}}{{dt}} + 3{t^2}y = \cos t \cr
& {\text{Write the left side in the form }}\frac{d}{{dt}}\left[ {I\left( t \right)r} \right] \cr
& \frac{d}{{dt}}\left[ {{t^3}y} \right] = \cos t \cr
& d\left[ {{t^3}y} \right] = \cos tdt \cr
& {\text{Integrate both sides}} \cr
& {t^3}y = \int {\cos t} dt \cr
& {t^3}y = \sin t + C \cr
& {\text{Solve for }}y \cr
& y = \frac{{\sin t}}{{{t^3}}} + \frac{C}{{{t^3}}},{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( \pi \right) = 0 \cr
& 0 = \frac{{\sin \pi }}{{{\pi ^3}}} + \frac{C}{{{\pi ^3}}} \cr
& C = 0 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{{\sin t}}{{{t^3}}} \cr} $$