Question

verify that y=-tcost-t is a solution of the initial-value problem t(dy/dt)=y+t^2sint, y(pi)=0

Answer 1

\[ty'=y+t^2\sin{t}\quad,\quad y(\pi)=0\]\[ty'-y=t^2\sin{t}\]\[\frac{ty'-y}{t^2}=\sin{t}\]\[\left(\frac{y}{t}\right)'=\sin{t}\]\[\frac{y}{t}=-\cos{t}+C\]\[y=t(C-\cos{t})\]\[y(\pi)=\pi(C+1)=0\quad,\quad C=-1\]\[y=t(-1-\cos{t})=-t-t\cos{t}\]