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.
Best Answer
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.