Verify that y=-tcost-t is a solution of the initial-value problem t(dy/dt)=y+t^2sint, y(pi)=1

Ty' = y + t^2 sin t

ty' - y = t^ sin t



= sin t

y/t = - cos t + C

y = T (C - cos t)

y (π) = π (C + 1) = 0 , C =  -1

y = t (-1 - cos t) =  -t - t cos t

hope this helps