Question

Solve the following ODE using the power series method:
(x2 −1) y′′ + 3xy′ + xy = 0

Answer 1

Let $y=\sum\limits_{n=0}^{\infty}a_nx^n$

Then

$y'=\sum\limits_{n=1}^{\infty}na_{n}x^{n-1}$

$y''=\sum\limits_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}$

After reidexing the series, we will have

$y'=\sum\limits_{n=0}^{\infty}(n+1)a_{n+1}x^{n}$

$y''=\sum\limits_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}$

Plugging $y'$ and $y''$ into the equation will lead us to

$(x^2-1)\sum\limits_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}+3x\sum\limits_{n=0}^{\infty}(n+1)a_{n+1}x^{n}+x\sum\limits_{n=0}^{\infty}a_nx^n=0$

$\sum\limits_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n+2}-\sum\limits_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}+\sum\limits_{n=0}^{\infty}3(n+1)a_{n+1}x^{n+1}+\sum\limits_{n=0}^{\infty}a_nx^{n+1}=0$

Let's transform the series in the right hand side

$2a_2+(6a_3+3a_1+a_0)x+\sum\limits_{n=2}^{\infty}n(n-1)a_{n}x^{n}-\sum\limits_{n=2}^{\infty}(n+2)(n+1)a_{n+2}x^{n}+$

$+\sum\limits_{n=2}^{\infty}\left(3na_{n}+a_{n-1}\right)x^{n}=2a_2+(6a_3+3a_1+a_0)x+\sum\limits_{n=2}^{\infty}\left(n(n-1)a_{n}+(n+2)(n+1)a_{n+2}+\right.$

$\left.+3na_{n}+a_{n-1}\right)x^n$

As this series is supposed to be equal to 0, we know that the coefficients of the resulting series must be equal to 0. Therefore,

$2a_2=0$

$6a_3+3a_1+a_0=0$

$n(n-1)a_{n}+(n+2)(n+1)a_{n+2}+3na_{n}+a_{n-1}=0.\quad n\ge2$

$(n^2+2n)a_n+(n+2)(n+1)a_{n+2}+a_{n-1}=0$

$a_{n+2}=-\frac{(n^2+2n)a_n+a_{n-1})}{(n+2)(n+1)}$

Plug in $a_0=0;\quad a_1=1$ we find the first solution

$6a_3+3=0\quad a_3=-\frac{1}{2}$

$a_{4}=-\frac{(2^2+4)a_2+a_{1})}{(2+2)(2+1)}=-\frac{1}{12}$

$a_5=-\frac{(9+6)a_3+a_2}{(3+2)(3+1)}=-\frac{-\frac{15}{2}}{20}=\frac{3}{8}$

$a_6=-\frac{(14+8)a_4+a_3}{(4+2)(4+1)}=-\frac{-\frac{22}{12}-\frac{1}{2}}{30}=\frac{7}{90}$

Therefore

$y_1=x-\frac{1}{2}x^3-\frac{1}{12}x^4+\frac{3}{8}x^5+\frac{7}{90}x^6+...$

Plug in $a_0=1,\quad a_1=0$ we find the second solution

$6a_3+1=0\quad a_3=-\frac{1}{6}$

$a_{4}=-\frac{(2^2+4)a_2+a_{1})}{(2+2)(2+1)}=0$

$a_5=-\frac{(9+6)a_3+a_2}{(3+2)(3+1)}=-\frac{-\frac{15}{6}}{20}=\frac{1}{8}$

$a_6=-\frac{(14+8)a_4+a_3}{(4+2)(4+1)}=-\frac{-\frac{1}{6}}{30}=\frac{1}{180}$

Therefore

$y_2=1-\frac{1}{6}x^3+\frac{1}{8}x^5+\frac{1}{180}x^6+...$