Question is to find the general solution of the given system of equations below.
$$
x'
=\left(\begin{array}{rr}\frac{-3}{2} & \frac{-1}{4} \\ 1 & \frac{-1}{2}\end{array}\right)x
$$
My solution is:
det($\begin{bmatrix}
\frac{-3}{2}-r & \frac{-1}{4}\\
1 & \frac{-1}{2}-r\
\end{bmatrix}) = (r+1)^2$
$r_1=-1$ and $r_2=-1$.
Eigenvector of $r=1$ is $\begin{bmatrix}
1\\
-2
\end{bmatrix}$
Therefore
$x^{(1)}(t) = \begin{bmatrix}
1\\
-2
\end{bmatrix}e^{-t}$
But what is $x^{(2)}$? I got stuck there, I know that this is an elementary question but even though I looked for solutions of similar problems, I dont understand how to find $x^{(2)}$.
Thanks.
Best Answer
Let $v_1$ be the eigenvector you found. We need to solve the equation $(A-\lambda I)v_2=v_1$ to get a second vector(which is not an eigenvector). Here $A$ is your matrix and $\lambda$ your eigenvalue. Then the second solution is $$x_2(t)=(v_1 t+v_2)e^{\lambda t}$$