(21 pts) Solve the IVP:

Solve the IVP: y" + y' - 2y = 0, y(0) = 0, y'(0) = 3.

### Solution

This is a second order linear homogeneous equation. We are able to find its fundamental pair of solution and subsequently form a general solution by first solving its characteristic equation: r

^{2}+ r - 2 = 0.There are two roots to that quadratic equation:

r _{1, 2}= (-1 ±sqrt(1 + 2·4))/2 = (-1 ±3)/2.r

_{1}= 1, r_{2}= -2. A general solution to the equation could be written as y = C_{1}e^{t}+ C_{2}e^{-2t}. To use the initial conditions, we must also find the derivative: y' = C_{1}e^{t}- 2C_{2}e^{-2t}. The initial conditions now appear asC _{1}+ C_{2}= 0

C_{1}- 2C_{2}= 3.Let's, e.g., multply the first equation by 2 and add the result to the second equation: 3C

_{1}= 3. C_{1}= 1, C_{2}= -1.### Answer: y(t) = e

^{t}- e^{-2t}.