Consider the initial value problem dy dt = −t + (t 2 + 4y) 1/2 2 , y(2) = −1. (a) Verify that both y 1 (t) = 1 − t and y 2 (t) = −t 2 /4 are both solutions to the initial value problem. (b) Explain why the existence of two solutions doesn’t violate the uniqueness part of the existence and uniqueness theorems discussed in class. (c) Show that for any c ∈ R, the function y(t) = ct + c 2 is a solution to the differential equation for t ≥ −2c; that if c = −1, then the solution y(t) satisfies the initial condition, and that in this case y(t) = y 1 (t); and that there is no choice of c that makes y(t) equal to y 2 (t)