# Solve second order differential equation with cosh using frobenios method

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.

• mathworld.wolfram.com/MathieuFunction.html – Nosrati Oct 10 '18 at 2:36
• $\cosh\neq\cos$ – Julián Aguirre Oct 10 '18 at 13:35
• If the writing of the cosh series is not a typo, then this might be the error that you are stumped upon. – 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.