(20 pts) Solve the IVP: x

_{1}' = -6x_{1}+ 5x_{2}, x_{2}' = -5x_{1}+ 4x_{2}, x_{1}(0) = 2, x_{2}(0) = 4.### Solution

We'll be dealing with the matrix A:

-6 5

-5 4First of all let's find it's eigenvalues by considering the determinant of

-6-r 5

-5 4-rThis leads to the quadratic eqution -(6 + r)(4 - r) + 25 = r

^{2}+ 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: -5z_{1}+ 5z_{2}= 0. The simplest solution isz = (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 1This is obviously solved, for example, by t = (0 1/5)

^{T}. Thus we have the following general solution:x(t) = C _{1}e^{-t}(1 1)^{T}+ C_{2}e^{-t}(t(1 1)^{T}+ (0 1/5)^{T}).For t = 0, we get x

_{1}(0) = C_{1}= 2 and x_{2}(0) = C_{1}+ C_{2}/5 = 4. Which implies### Answer

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