Question 1 (6 pts (1.5 + 1.5 + 1.5 + 1.5 ) ). In the following questions, y is a function of t. 1. Consider the following differential equation: y" + y = 0, y(0) = 1, y(2) = 1 Find the solution(s) of this initial value problem. 2. Show that we do not have any solutions if the conditions is replaced by y(0) = 1, y(2) = 2. 3. Solve the following IVP: y (4) 5y (2) + 4y = 0, - 4. Solve the following IVP: y(4)+5y (2)+4y = 0, y(0) = 5, y'(0) = 9, y () (0) = 11, y (3) (0) = 33. y(0) = 3, y'(0) = 4, y() (0) = 6, y(3) (0) = 10.