1. (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: r2 + r - 2 = 0.

    There are two roots to that quadratic equation:

    r1, 2 = (-1 ±sqrt(1 + 2·4))/2 = (-1 ±3)/2.

    r1 = 1, r2 = -2. A general solution to the equation could be written as y = C1et + C2e-2t. To use the initial conditions, we must also find the derivative: y' = C1et - 2C2e-2t. The initial conditions now appear as

    C1 + C2 = 0
    C1 - 2C2 = 3.

    Let's, e.g., multply the first equation by 2 and add the result to the second equation: 3C1 = 3. C1 = 1, C2 = -1.

    Answer: y(t) = et - e-2t.