Answer to Question #280153 in Differential Equations for Dawood

Question

Solve this Diffrential equation (1-x^2)y"-2xy'+n(n+1)y=0 using Power Series Method

Answer 1

Replacing 'n' given in equation with 'a' and solving.

$\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+b y=0 \text { with } b=a(a+1)$

When x is small, this is $y^{\prime \prime}+b y=0$ which has solutions $y=u \sin (x \sqrt{b})+v \cos (x \sqrt{b})$ . So, this seems like what the solutions look like.

Proceeding mechanically, let $y(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=a_{0}+a_{1} x+\sum_{n=2}^{\infty} a_{n} x^{n}$ . The reason for this will appear later.

$\begin{aligned} x y^{\prime}(x) &=x \sum_{n=1}^{\infty} n a_{n} x^{n-1} \\ &=\sum_{n=1}^{\infty} n a_{n} x^{n} \\ &=a_{1} x+\sum_{n=2}^{\infty} n a_{n} x^{n} \\ &=\left(1-x^{2}\right) \sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2} \\ &=\sum_{n=2}^{\infty} n(n-1) a_{n} x^{n-2}-\sum_{n-2}^{\infty} n(n-1) a_{n} x^{n} \\ &=\sum_{n-0}^{\infty}(n+2)(n+1) a_{n+2} x^{n}-\sum_{n-2}^{\infty} n(n-1) a_{n} x^{n} \\ &=2 a_{2}+6 a_{3} x+\sum_{n=2}^{\infty}\left(n(n-1) a_{n}-(n+2)(n+1) a_{n+2}\right) x^{n} \end{aligned}$

Adding these up, and separating the terms for n<2,

$\begin{gathered} 0=\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+b y \\ =2 a_{2}+6 a_{3} x-2 a_{1} x+b\left(2 a_{0}+6 a_{1} x\right)+\sum_{n=2}^{\infty}\left(n(n-1) a_{n}-(n+2)(n+1) a_{n+2}-2 n a_{n}\right. \\ \left.+b a_{n}\right) x^{n} \\ =2\left(a_{2}+b a_{0}\right)+2 x\left(6 a_{3}-2 a_{1}+6 b a_{1}\right)+ \\ \sum_{n=2}^{\infty}\left((n(n-1)-2 n+b) a_{n}-(n+2)(n+1) a_{n+2}\right) x^{n} \\ =2\left(a_{2}+b a_{0}\right)+2 x\left(6 a_{3}-2 a_{1}(1-3 b)\right)+\sum_{n=2}^{\infty}\left((n(n-3)+b) a_{n}-(n+2)(n+1) a_{n+2}\right) x^{n} \end{gathered}$

Equating the coefficients to zero,

$\begin{aligned} &a_{2}+b a_{0}=0 \text { or } a_{2}=-b a_{0} ; \\ &0=6 a_{3}-2 a_{1}+6 b a_{1}=6 a_{3}-2 a_{1}(1-3 b) \text { or } a_{3}=a_{1}\left(\frac{1}{3}-b\right) ; \\ &\text { and, for } n \geq 2,0=(n(n-3)+b) a_{n}-(n+2)(n+1) a_{n+2} \text { or } a_{n+2}=\frac{n(n-3)+b}{(n+2)(n+1)} a_{n} . \\ &\text { From this, we see that there are two independent solutions: One with } a_{0}=0 \text { (the odd solution) } \\ &\text { and one with } a_{1}=0 \text { (the even solution). } \\ &\text { We also see that if } b=\frac{1}{3}, \text { the odd solution is just } a_{1} x / 3, \text { since all the higher odd coefficients are } \\ &\text { zero. } \\ &\text { Similarly, if } b=-n(n-3) \text { for some } n, \text { the even solution is a polynomial, since } a_{n+2}=0 \text { for the } n \\ &\text { such that } b=n(n-3) \text { and all greater even } n . \end{aligned}$