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Given the system:

$\begin{cases}x''=2y \\ y''=-2x\end{cases} $

I found the (I think) equivalent linear equation $x^{IV}+4x=0$

  • First question: is the equation actually equivalent to the system?

And then the eigenvalues of the equation, $\pm(1\pm i)$.

Which gives the general solution:

$$x=e^t(c_1\cos t + c_2\sin t)+e^{-t}(c_3\cos t + c_4\sin t)$$

  • Second question: Is this solution right?

Thanks in advance!

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    $\begingroup$ You are completely right. $\endgroup$ – Michael Galuza Aug 6 '15 at 8:59
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    $\begingroup$ If you prefer, you can rewrite this system for vector $X=\langle x, x', y, y'\rangle$, and $X' = AX$, where $$A = \begin{pmatrix}0 & 1 & 0 & 0\\0&0&2&0\\0&0&0&1\\-2&0&0&0\end{pmatrix}$$ For this it's obvious that depends on four arbitrary constants. $\endgroup$ – Michael Galuza Aug 6 '15 at 9:13

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