Verify that both y₁(t) = 1 - t and y₂(t) = -t²/4 are solutions of the initial value problem Where are these solutions valid? Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of Show that y = ct + c², where c is an arbitrary constant, satisfies the differential equation in part 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).