Find all solutions of the nonlinear differential equation

on the interval satisfying the initial conditions

From the previous exercise (Section 8.5, Exercise #13) we know that a function which is never zero on an interval is a solution of the initial value problem

if and only if on where is the unique solution of the initial-value problem

In the present problem we have the equation

Therefore, to apply the previous exercise we have

Hence, is a solution to the given equation if and only if where is the unique solution to

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

giving us

Therefore,

No 2 inside that integral there with Q.