Calculus: Early Transcendentals, Chapter 9, 9.1, Section 9.1, Problem 2

First we need to find derivative
dy/dt=-cos t+tsin t-1
Here we've used product rule to differentiate -tcos t.
Now we insert dy/dt and y into the initial value problem.
t(-cos t+tsin t-1)=-tcos t-t+t^2sin t
-tcos t+t^2sin t-t=-tcos t-t+t^2sin t
Obviously, the left and the right sides are equal to each other meaning that the solution satisfies the differential equation. Now we only need to check the initial condition.
y(pi)=-pi cos pi-pi=-pi(-1)+pi=pi-pi=0
Therefore, y=-tcos t-t satisfies both the differential equation and the initial condition meaning it is the solution to our initial value problem.