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i need to show that the differential equation

$y^{''}+(\cosh(2x)-4)y = 0$

has the solution:

$ y(x) = x+\frac{1}{2}x^3-\frac{1}{40}x^5 -... $

using Frobenius method.

I started by writing cosh(2x) in the form $\sum{\frac{4^nx^{2n}}{{2n}!}}$ and assuming a solution of the form $y(x)= \sum{C_nx^{n+s}}$ but once i substitute everything back into the original equation i can't see a way of simplifying all the terms.

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    $\begingroup$ mathworld.wolfram.com/MathieuFunction.html $\endgroup$ – Nosrati Oct 10 '18 at 2:36
  • $\begingroup$ $\cosh\neq\cos$ $\endgroup$ – Julián Aguirre Oct 10 '18 at 13:35
  • $\begingroup$ If the writing of the cosh series is not a typo, then this might be the error that you are stumped upon. $\endgroup$ – LutzL Oct 10 '18 at 14:12
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It is just computation. \begin{align} \cosh(2\,x)-4&=-3+2\,x^2+\frac23\,x^4+\dots\\ y(x)&=x+a_2\,x^2+a_3\,x^3+a_4\,x^4+a_s\,x^5+\dots\\ y''(x)&=2\,a_2+6\,a_3\,x+12\,a_4\,x^2+20\,a_5\,x^3+\dots \end{align} Multiplying the first two series we get $$ (\cosh(2\,x)-4)\,y(x)=-3\,x - 3\,a_2\,x^2 + (2 - 3\,a_3)\,x^3 + \dots $$ Now sum the last series to the one of $y''$, set the result equal equal to $0$ and solve for the coefficients.

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  • $\begingroup$ Thank you so much! P.S on the second derivative it should be 20 and not 15 i think $\endgroup$ – xFugtree Oct 10 '18 at 14:56
  • $\begingroup$ Right. I have edited. $\endgroup$ – Julián Aguirre Oct 10 '18 at 16:54

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