Answer to Question #212953 in Differential Equations for stacia

Question

find a series solution in powers of x of the equation

2x²d²y/dx²+xdy/dx+(x²-1)y=0

Answer 1

Solution

Let

$y\left(x\right)=\sum_{n=0}^{\infty}{a_nx^n}$

$\frac{dy}{dx}=\sum_{n=0}^{\infty}{\left(n+1\right)a_{n+1}x^n}$

$\frac{d^2y}{dx^2}=\sum_{n=0}^{\infty}{\left(n+1\right)\left(n+2\right)a_{n+2}x^n}$

Substitution into equation:

$2\sum_{n=0}^{\infty}{\left(n+1\right)\left(n+2\right)a_{n+2}x^{n+2}}+\sum_{n=0}^{\infty}{\left(n+1\right)a_{n+1}x^{n+1}}+\sum_{n=0}^{\infty}{a_nx^{n+2}}-\sum_{n=0}^{\infty}{a_nx^n}=0$

$a_1x-a_0-a_1x+2\sum_{n=2}^{\infty}{n\left(n-1\right)a_nx^n}+\sum_{n=2}^{\infty}{na_nx^n}+\sum_{n=2}^{\infty}{a_{n-2}x^n}-\sum_{n=2}^{\infty}{a_nx^n}=0$

Coefficient near xⁿ are equal to zero.

n=0: a₀=0

n=1: 0=0 for any a₁  

n>1: 2n(n-1)a_n+na_n+a_n-2-a_n=0 => (2n²-n-1)a_n+ a_n-2 = 0 =>

a_n =-a_n-2/ (2n²-n-1)= -a_n-2/ (2n²-n-1) = -a_n-2/ [(2n+1)(n-1)]

So a_2k=0 for any k≥0

and finally a_2k+1  = -a_2k-1/ [(4k+3)2k] for k>0

Solution is

$y(x)=\sum_{k=1}^{\infty}{a_{2k-1}x^{2k-1}}$