a) Verify that both y1(t)=1−tan dy 2(t)=−t2/4 are solutions of the initial value problem y=−t+t2+4y2,y(2)=−1.Where are these solutions valid?
b) Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of Theorem 2.4.2.
c) Show that y = ct+c2, where c is an arbitrary constant, satisfies the differential equation in part (a) for t≥−2c. If c=−1, the initial condition is also satisfied, and the solution y=y1(t) is obtained. Show that there is no choice of c that gives the second solution y=y2(t).