Replacing 'n' given in equation with 'a' and solving.
( 1 − x 2 ) y ′ ′ − 2 x y ′ + b y = 0 with b = a ( a + 1 ) \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+b y=0 \text { with } b=a(a+1) ( 1 − x 2 ) y ′′ − 2 x y ′ + b y = 0 with b = a ( a + 1 )
When x is small, this is y ′ ′ + b y = 0 y^{\prime \prime}+b y=0 y ′′ + b y = 0 which has solutions y = u sin ( x b ) + v cos ( x b ) y=u \sin (x \sqrt{b})+v \cos (x \sqrt{b}) y = u sin ( x b ) + v cos ( x b ) . So, this seems like what the solutions look like.
Proceeding mechanically, let y ( x ) = ∑ n = 0 ∞ a n x n = a 0 + a 1 x + ∑ n = 2 ∞ a n x n y(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=a_{0}+a_{1} x+\sum_{n=2}^{\infty} a_{n} x^{n} y ( x ) = ∑ n = 0 ∞ a n x n = a 0 + a 1 x + ∑ n = 2 ∞ a n x n . The reason for this will appear later.
x y ′ ( x ) = x ∑ n = 1 ∞ n a n x n − 1 = ∑ n = 1 ∞ n a n x n = a 1 x + ∑ n = 2 ∞ n a n x n = ( 1 − x 2 ) ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − ∑ n − 2 ∞ n ( n − 1 ) a n x n = ∑ n − 0 ∞ ( n + 2 ) ( n + 1 ) a n + 2 x n − ∑ n − 2 ∞ n ( n − 1 ) a n x n = 2 a 2 + 6 a 3 x + ∑ n = 2 ∞ ( n ( n − 1 ) a n − ( n + 2 ) ( n + 1 ) a n + 2 ) x n \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} x y ′ ( x ) = x n = 1 ∑ ∞ n a n x n − 1 = n = 1 ∑ ∞ n a n x n = a 1 x + n = 2 ∑ ∞ n a n x n = ( 1 − x 2 ) n = 2 ∑ ∞ n ( n − 1 ) a n x n − 2 = n = 2 ∑ ∞ n ( n − 1 ) a n x n − 2 − n − 2 ∑ ∞ n ( n − 1 ) a n x n = n − 0 ∑ ∞ ( n + 2 ) ( n + 1 ) a n + 2 x n − n − 2 ∑ ∞ n ( n − 1 ) a n x n = 2 a 2 + 6 a 3 x + n = 2 ∑ ∞ ( n ( n − 1 ) a n − ( n + 2 ) ( n + 1 ) a n + 2 ) x n
Adding these up, and separating the terms for n<2,
0 = ( 1 − x 2 ) y ′ ′ − 2 x y ′ + b y = 2 a 2 + 6 a 3 x − 2 a 1 x + b ( 2 a 0 + 6 a 1 x ) + ∑ n = 2 ∞ ( n ( n − 1 ) a n − ( n + 2 ) ( n + 1 ) a n + 2 − 2 n a n + b a n ) x n = 2 ( a 2 + b a 0 ) + 2 x ( 6 a 3 − 2 a 1 + 6 b a 1 ) + ∑ n = 2 ∞ ( ( n ( n − 1 ) − 2 n + b ) a n − ( n + 2 ) ( n + 1 ) a n + 2 ) x n = 2 ( a 2 + b a 0 ) + 2 x ( 6 a 3 − 2 a 1 ( 1 − 3 b ) ) + ∑ n = 2 ∞ ( ( n ( n − 3 ) + b ) a n − ( n + 2 ) ( n + 1 ) a n + 2 ) x n \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} 0 = ( 1 − x 2 ) y ′′ − 2 x y ′ + b y = 2 a 2 + 6 a 3 x − 2 a 1 x + b ( 2 a 0 + 6 a 1 x ) + n = 2 ∑ ∞ ( n ( n − 1 ) a n − ( n + 2 ) ( n + 1 ) a n + 2 − 2 n a n + b a n ) x n = 2 ( a 2 + b a 0 ) + 2 x ( 6 a 3 − 2 a 1 + 6 b a 1 ) + n = 2 ∑ ∞ ( ( n ( n − 1 ) − 2 n + b ) a n − ( n + 2 ) ( n + 1 ) a n + 2 ) x n = 2 ( a 2 + b a 0 ) + 2 x ( 6 a 3 − 2 a 1 ( 1 − 3 b ) ) + n = 2 ∑ ∞ ( ( n ( n − 3 ) + b ) a n − ( n + 2 ) ( n + 1 ) a n + 2 ) x n
Equating the coefficients to zero,
a 2 + b a 0 = 0 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 ) or a 3 = a 1 ( 1 3 − b ) ; and, for n ≥ 2 , 0 = ( n ( n − 3 ) + b ) a n − ( n + 2 ) ( n + 1 ) a n + 2 or a n + 2 = n ( n − 3 ) + b ( n + 2 ) ( n + 1 ) a n . From this, we see that there are two independent solutions: One with a 0 = 0 (the odd solution) and one with a 1 = 0 (the even solution). We also see that if b = 1 3 , the odd solution is just a 1 x / 3 , since all the higher odd coefficients are zero. Similarly, if b = − n ( n − 3 ) for some n , the even solution is a polynomial, since a n + 2 = 0 for the n such that b = n ( n − 3 ) and all greater even n . \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} a 2 + b a 0 = 0 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 ) or a 3 = a 1 ( 3 1 − b ) ; and, for n ≥ 2 , 0 = ( n ( n − 3 ) + b ) a n − ( n + 2 ) ( n + 1 ) a n + 2 or a n + 2 = ( n + 2 ) ( n + 1 ) n ( n − 3 ) + b a n . From this, we see that there are two independent solutions: One with a 0 = 0 (the odd solution) and one with a 1 = 0 (the even solution). We also see that if b = 3 1 , the odd solution is just a 1 x /3 , since all the higher odd coefficients are zero. Similarly, if b = − n ( n − 3 ) for some n , the even solution is a polynomial, since a n + 2 = 0 for the n such that b = n ( n − 3 ) and all greater even n .
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