## Calculus (3rd Edition)

$$t=\pm \sqrt{ \pi+4}$$
Since $$\sin y+y=t^{2}+C$$ is a solution and $y(2)=0$, then $$\sin (0) +0 =4+C\ \ \Rightarrow C=-4$$ Hence the particular solution is $$\sin y+y=t^{2}+4$$ We find the $t$ that satisfies $y( t)=\pi$, \begin{align*} \sin \pi+\pi&=t^{2}-4\\ \pi+4&=t^2 \end{align*} Then $$t=\pm \sqrt{ \pi+4}$$