0
$\begingroup$

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

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

and

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

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

In general form general solution should be:

$$x(t)=c_{1}e^{\lambda_{1}}\overrightarrow{v_{1}}+c_{2}e^{\lambda_{2}}\overrightarrow{v_{2}}$$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.