I have a differential equation system

$x_1'(t) = -x_2(t)$

$x_2'(t) = -x_1(t)$

I see that I can write $\dot{x} = Ax$ where

$A = \begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}$

The complete real solution will be

$x(t) = c_1 e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix} + c_2 e^t \begin{pmatrix}-1 \\ 1\end{pmatrix}$

where $c_1,c_2 \in \mathbb{R}$.

It this correct?

Now I need to find the solution such that $x_1(0) = -1$ and $x_2(0) = 1$.

How can I do this?

I think I could split my complete solution in

$x_1(t) = c_1 e^t - c_2 e^{-t}$


$x_2(t) = c_1 e^t + c_2 e^{-t}$

and then find $c_1$ and $c_2$ by solving $x_1(0) = -1$ and $x_2(0) = 1$ with respect to $c_1$ and $c_2$ and finally substitute these values in the complete solution $x(t)$.

Is this correct?

  • $\begingroup$ I have edited the signs in my matrix. Thanks. In my exercise I am told to find all the real solutions to the system and not the complete solution. What is all the solutions? Are they just $x(t) = c_1 e^{\lambda_1 t} v_1$ and $x(t) = c_2 e^{\lambda_2 t} v_2$? $\endgroup$ – Jamgreen Sep 14 '14 at 13:21

The solution here is even simpler than that: $x_1''=x_1$. Solutions are linear combinations of $e^t$ and $e^{-t}$, or even better, linear combinations of $\cosh(t)$ and $\sinh(t)$.

In the general case, you are right to use a matrix etc. But no need to use that here.


In general form general solution should be:


According to my calculations, I find $\lambda_1=1$ and $\lambda_2=-1$. Thus, resulting eigenvectors are:

$$x(t) = c_1 e^{t} \begin{pmatrix}1 \\ -1\end{pmatrix} + c_2 e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix}$$

You should check your results. And yes you find $c_1$ and $c_2$ from initial values.

Edit: I didn't realize your edit. Your results are now true.


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