Answer
$$y=t- \frac{\pi}{3} \cos 3t -\frac{\pi}{3} $$
Work Step by Step
Given$$ \frac{d y}{d t}=1+\pi\sin (3t) , \quad y(\pi)=\pi $$
Since :\begin{align*}
y&= \int (1+\pi\sin (3t) ) dt\\
&=t- \frac{\pi}{3} \cos 3t+c
\end{align*}
at $t=\pi,\ \ \ y= \pi $, then $c= -\pi/3$, hence
$$y=t- \frac{\pi}{3} \cos 3t -\frac{\pi}{3} $$
