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How can I solve differential equation $y''=y$ by Fourier series when $y(0)=0$ and $y'(0)=-2$?

First, I consider the Fourier series $y=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n \cos nx+\sin nx)$, then I get $y''=\sum_{n=1}^{\infty}(-n^2 a_n \sin nx -n^2 b_n \cos nx)$. After that I have $$ \frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n \cos nx+\sin nx)=\sum_{n=1}^{\infty}(-n^2 a_n \sin nx -n^2 b_n \cos nx)$$.

But now I don't know what to do for obtaining $a_n$ and $b_n$.

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  • $\begingroup$ Are you sure the DE is $y''=y$. If it is $y''=-y$ then you can get a solution in terms of sine ans cosine functions. $\endgroup$ – Kavi Rama Murthy Oct 25 '19 at 5:30
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This DE does not have a periodic solution valid on $\mathbb R$, so this method fails. The actual solutions is $e^{-x}-e^{x}$

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  • $\begingroup$ Thank you for the answer. Could you tell me why we can't solve this DE by Fourier series? $\endgroup$ – M.Ramana Oct 26 '19 at 12:13
  • $\begingroup$ Fourier series is for periodic functions. Since there is no periodic solution to the DE we cannot use Fourier series. $\endgroup$ – Kavi Rama Murthy Oct 26 '19 at 12:15
  • $\begingroup$ How can I understand a solution is periodic when it is unknown? $\endgroup$ – M.Ramana Oct 27 '19 at 4:41

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