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 $x^{(2)}$.



1 Answer 1


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 $$x_2(t)=(v_1 t+v_2)e^{\lambda t}$$


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