Finding the Second Solution for a System with a Repeated Eigenvalue

[Buri]

Homework Statement

I'm told to solve the following:

X' = (1 1; -1 3)X

Where (1 1) is the first row and (-1 3) is the second row.

The Attempt at a Solution

Okay so I calculated the eigenvalues and I got repeated eigenvalue of 2. I calculated the eigenvector and got (1,1). So a solution would be X(t) = ce^(2t)(1,1).

My text considers the matrix (a 1; 0 a) before the problems and goes on to find a eigenvalue with eigenvector, but then adds another solution to it so getting:

X(t) = ce^(at)(1,0) + de^(at)(t,1)

I understand how they got it in that one, but I'm not exactly sure how to go about finding one for mine. Any help?

 

[quantumdude]

There's a procedure for this. If a system $\vec{X}^{\prime}=A\vec{X}$ has a repeated eigenvalue $\lambda$ and there is only one eigenvector $\vec{K}$ associated with it (as you have here), then the second solution $\vec{X}_2$ that you seek is found as follows.

$$\vec{X}_2=\vec{K}te^{\lambda t}+\vec{P}e^{\lambda t}$$,

where $\vec{P}$ satisfies:

$$\left(A-\lambda I\right)\vec{P}=\vec{K}$$.