1. (20 pts) Solve the IVP: x1' = -6x1 + 5x2, x2' = -5x1 + 4x2, x1(0) = 2, x2(0) = 4.

    Solution

    We'll be dealing with the matrix A:

    -6   5
    -5   4

    First of all let's find it's eigenvalues by considering the determinant of

    -6-r   5
    -5     4-r

    This leads to the quadratic eqution -(6 + r)(4 - r) + 25 = r2 + 2r + 1 = 0, which has two roots equal -1. So we got a case of a repeated eigenvalue. It's easy to find one eigenvector z: -5z1 + 5z2 = 0. The simplest solution is z = (1, 1)T.

    Clearly, it is impossible to find a second independent eigenvector. Instead, we'll try solving (A - I)t = z, which is represented by the augmented matrix:

    -5   5   1
    -5   5   1

    This is obviously solved, for example, by t = (0 1/5)T. Thus we have the following general solution:

    x(t) = C1e-t(1 1)T + C2e-t(t(1 1)T + (0 1/5)T).

    For t = 0, we get x1(0) = C1 = 2 and x2(0) = C1 + C2/5 = 4. Which implies

    Answer

    x(t) = 2e-t(1 1)T + 10e-t(t(1 1)T + (0 1/5)T).