I am having trouble solving a question.

The question is: Determine the general solution of the ODE system $$\dot{x_1} - 2x_1 + x_2 = e^{2t}$$ $$\dot{x_2} + x_1 - 2x_2 = 2e^{2t}$$

I am trying to solve this by using the matrix $\begin{bmatrix}-2 & 1 \\ 1 & -2 \end{bmatrix}$ and finding the associated eigenvalues and eigenvectors. I have found the eigenvalues to be $\lambda$ = -1 and $\lambda$ = -3. The eigenvectors are $\begin{bmatrix}1 \\ 1 \end{bmatrix}$ and $\begin{bmatrix}-1 \\ 1 \end{bmatrix}$ respectfully. I found the complementary solution to be $\dot{x_c}$ = A$\begin{bmatrix}1 \\ 1 \end{bmatrix}e^{-t}$ + B$\begin{bmatrix}-1 \\ 1 \end{bmatrix}e^{-3t}$.

We see that the forcing term is not proportional to either of the complementary solutions so we need to find a particular solution of the form $\dot{x_p}$ = C$e^{2t}$. I'm stuck at this point and don't know how to find C. My thinking is I need to find a matrix such that (Matrix M)C$e^{2t}$ = b$e^{2t}$ where b = $\begin{bmatrix}1 \\ 2 \end{bmatrix}$.

Hence (Matrix M) C = b and I can find C using row reduction on Matrix M. My problem is I can't find such a matrix to satisfy the equation. I don't know how to progress from this point.

Any help would be appreciated thanks.

  • $\begingroup$ I'm not sure I follow. I have copied the question which states it as $\dot{x_1}$ and $\dot{x_2}$, even then how would writing it as ${x_1}'$ and ${x_2}'$ change it? $\endgroup$ May 22, 2017 at 18:07
  • $\begingroup$ I think what you mean is to find the eigenvalues you do A + $\lambda$ I where A is a given matrix and I is the identity matrix. I am aware of this method but I have been taught instead to do A - $\lambda$ I. Hence why my eigenvalues being the negative of yours. $\endgroup$ May 22, 2017 at 18:16
  • $\begingroup$ I see what you mean. Even then only the eigenvalues would change, not the eigenvectors unless I am mistaken? In which case I have $e^{t}$ and $e^{3t}$ $\endgroup$ May 22, 2017 at 18:26

1 Answer 1


Try solutions proportional to $\exp{(2t)}$, $\textit{i.e}$: $$x_1=c_1\exp{(2t)}$$ $$x_2=c_2\exp{(2t)}$$ To arrive to the system $$\left\{2\begin{bmatrix}1&0\\0&1\end{bmatrix}+\begin{bmatrix}-2&1\\1&-2\end{bmatrix}\right\}\begin{bmatrix}c_1\\c_2\end{bmatrix}=\begin{bmatrix}1\\2\end{bmatrix}$$ Solve for $c_1,c_2$

  • $\begingroup$ Thanks for the quick answer, it seems simple enough now that I look at it. I was trying to do it by I + A and got to $\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}$ = $\begin{bmatrix}1 \\ 2 \end{bmatrix}$ and was stuck then. $\endgroup$ May 22, 2017 at 18:30
  • 1
    $\begingroup$ Forget that you have a system, treat each variable separately, and then form the system (this leads naturally to the eigenvalue system for the homogeneous case). $\endgroup$
    – HBR
    May 22, 2017 at 18:35

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