Final answer:
Both y1(t)=1-t and y2(t)=-t^2/4 satisfy their initial value conditions when their derivatives are evaluated, thus verifying them as solutions to the initial value problem.
Step-by-step explanation:
To verify that both y1(t)=1-t and y2(t)=-t^2/4 are solutions of the initial value problem, we need to verify that each function satisfies the given differential equation and initial conditions.
Let's start with y1(t) = 1 - t. We differentiate y1 with respect to t to derive the first derivative. The first derivative of y1 is y1'(t) = -1. Now, we check the initial value condition. When t = 0, y1(0) = 1 - 0 = 1, this satisfies the initial condition of y(0) = 1.
Next, we consider y2(t) = -t^2/4. Its first derivative y2'(t) = -t/2. Evaluating at t = 0, y2(0) = -(0)^2/4 = 0, fulfills the initial condition of y(0) = 0. Therefore, both functions meet their respective initial value conditions, verifying them as solutions.