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I need to solve this equation

$$ x'=x+y, y'=-2x+3y$$

I get the matrix rigor and the eigenvectors complex $2-i$ and $2+i$. When I try to apply the eigenvectors associated the solution for $x$ I simply can't find ( I do find the $y$ though).

The solution on the book for $x=e^{2t}(a\cos(t)+b\sin(t))$

Any help would be appreciated.

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What can't you find? Is it an eigenvector for one of the eigenvalues? – Git Gud Jun 9 '13 at 10:59
up vote 3 down vote accepted

For the system:

$$ x'=x+y,~y'=-2x+3y$$

We have: $A = \begin{bmatrix} 1 & 1\\ -2 & 3 \end{bmatrix}$.

As you found, setting up $|A- \lambda I| = 0$ and solving for the eigenvalue/eigenvector pairs yields:

  • $\lambda_1 = 2+i, ~v_1 = (\frac{1}{2}(1-i), 1)$
  • $\lambda_2 = 2-i, ~v_2 = (\frac{1}{2}(1+i), 1)$

So, for the first eigenvalue, we have:

$e^{\lambda_1 t}v_1 = e^{(2+i)t}\begin{bmatrix}\frac{1}{2}(1-i)\\1\end{bmatrix} = e^{2t}e^{it}\begin{bmatrix}\frac{1}{2}(1-i)\\1\end{bmatrix} = e^{2t}(\cos t + i \sin t)\begin{bmatrix}\frac{1}{2}(1-i)\\1\end{bmatrix} = \begin{bmatrix} \frac{1}{2}e^{2t}(\cos t + \sin t) + \frac{1}{2} i (\sin t-\cos t))\\ e^{2t}(\cos t + i \sin t) \end{bmatrix} $

So, our solution can be written as (because we know that the real and imaginary parts are both independent solutions):

$$W(t) = \begin{bmatrix}x(t)\\ y(t) \end{bmatrix} = a e^{2t}\begin{bmatrix}\frac{1}{2}(\sin t + \cos t)\\ \cos t \end{bmatrix} + b e^{2t}\begin{bmatrix}\frac{1}{2} (\sin t - \cos t)\\ \sin t \end{bmatrix}$$

If you want to write these out independently, we have:

  • $\displaystyle x(t) = \frac{1}{2}e^{2t}\left( a(\sin t + \cos t) + b(\sin t - \cos t)\right)$
  • $\displaystyle y(t) = e^{2 t}(a \cos t + b \sin t)$

Did you swap the your labels for $x(t)$ and $y(t)$?

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Thank you very much – Olinda Fernandes Jun 9 '13 at 15:34
@OlindaFernandes: You are very welcome. I would spend time to understand the approach that Babak S. outlined. Regards – Amzoti Jun 9 '13 at 15:38
Excellent work, like usual! +1 – amWhy Jun 10 '13 at 0:08
+1 for my friend. 23 hours ago we were walking on a same road here. Nice to meet you Amzoti. ;-) – Babak S. Jun 10 '13 at 13:14
@BabakS.: Ditto that my good friend! Hope all is well and +1 for your solution too! Regards – Amzoti Jun 10 '13 at 13:26

Let's rewrite your equations as follows,assuming that $Dx=x',~Dy=y'$: $$ \left\{ \begin{array}{ll} (D-1)x-y=0 \\ 2x+(D-3)y=0 \end{array} \right. $$ By solving this homogenous system of equations, we have:

$$ \left\{ \begin{array}{ll} (D-1)(D-3)x+2x=0 \\ (D-1)(D-3)y+2y=0 \end{array} \right. $$ which is equivalent to the following system: $$ \left\{ \begin{array}{ll} x''(t)-4x'(t)+5x(t)=0 \\ y''(t)-4y'(t)+5y(t)=0 \end{array} \right. $$ Now, I think you can do the rest.

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$+_+^+ 1 _+^++$ – amWhy Jun 10 '13 at 0:26

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