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.

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.