Find $y(t)$ solution of the initial value problem

$t{y}^{\text{'}}-2y2t^{3}sin(3t)=0,y(\frac{}{3})=0,t>0$

$y(t)=\frac{3}{4}tsin(3t)$

$y(t)=sin(3t)$

$y(t)=\frac{2}{3}{t}^2(cos(3t)1)$