Legendre's equation of order n in differential equations The equation
$$(1-x^2)y'' - 2xy'+n(n+1)y = 0$$
is called Legendre's equation of order $n$. 
I need to show that this equation of order $1$ has $y=x$ as one solution and then I need to use this to find the general solution for $x$ greater than $-1$ but less than $1$ (not inclusive)
 A: Here is a solution using power series.
\begin{equation}
 (1-x^2)y'' - 2xy'+n(n+1)y = 0
 \end{equation}
    Let $y=\sum_{k=0}^{\infty}c_kx^k$, $y^\prime=\sum_{k=1}^{\infty}kc_kx^{k-1}$, $y^{\prime\prime}=\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}$
Then
\begin{equation}
 (1-x^2)\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}-2x\sum_{k=1}^{\infty}kc_kx^{k-1}+n(n+1)\sum_{k=0}^{\infty}c_kx^k=0
 \end{equation}
\begin{equation}
\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}-\sum_{k=2}^{\infty}k(k-1)c_kx^{k}-\sum_{k=1}^{\infty}2kc_kx^{k}+\sum_{k=0}^{\infty}n(n+1)c_kx^k=0
 \end{equation}
\begin{equation}
  \sum_{k=0}^{\infty}(k+2)(k+1)c_{k+2}x^{k}-\sum_{k=0}^{\infty}k(k-1)c_kx^{k}-\sum_{k=0}^{\infty}2kc_kx^{k}+\sum_{k=0}^{\infty}n(n+1)c_kx^k=0
  \end{equation}
\begin{equation}
\sum_{k=0}^{\infty}[(k+2)(k+1)c_{k+2}-k(k-1)c_k-2kc_k+n(n+1)c_k]x^k=0
\end{equation}
Therefore
\begin{equation}
(k+2)(k+1)c_{k+2}+(n(n+1)-k(k+1))c_k=0 \text{ for }k\ge0
\end{equation}
So
\begin{equation}
c_{k+2}=\frac{k(k+1)-n(n+1)}{(k+2)(k+1)}c_k
\end{equation}
Solving for the coefficients $c_k$ of the solution yields $c_2=-c_0, c_3=0$. Thus all odd powers of the series greater than $1$ will have coefficient zero and the general solution will be of the form
\begin{equation}
y=c_1x+c_0(1-x^2+\cdots)
\end{equation}
Therefore one linearly independent solution is $y_1=x$.
