(a) Verify that both y1(t)=1−tandy2(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) fort≥−2c.Ifc=−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).