The Legendre differential equation is given by,
$$(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+(k)(k+1)y=0$$
whereby $k$ is a constant.
a) show that $x=0$ is an ordinary point of the differential equation. and find two linearly independent power series solution of x.
My work
$$y=\sum_{n=0}^\infty c_nx^n$$
$$\frac{dy}{dx}=n\sum_{n=1}^\infty c_nx^{n-1}$$
$$\frac{d^2y}{dx^2}=(n)(n-1)\sum_{n=2}^\infty c_nx^{n-2}$$
$$(1-x^2)[(n)(n-1)\sum_{n=2}^\infty c_nx^{n-2}]-2x[n\sum_{n=1}^\infty c_nx^{n-1}]+(k)(k+1)[\sum_{n=0}^\infty c_nx^n]=0$$
$$(n+1)(n+2)\sum_{n=0}^\infty c_{n+2}x^{n}-(n)(n-1)\sum_{n=2}^\infty c_nx^{n}-2n\sum_{n=1}^\infty c_nx^{n}+(k)(k+1)[\sum_{n=0}^\infty c_nx^n]=0 $$
$$2c_2+6c_3x-2c_1x+(k)(k+1)c_0+(k)(k+1)c_1x+[(n+1)(n+2)c_{n+2}-(n)(n-1)c_n-nc_n+(k)(k+1)c_n]\sum_{n=2}^\infty x^{n}=0$$
$$2c_2+(k)(k+1)c_0=0$$
$$6c_3-2c_1+(k)(k+1)c_1=0$$
$$c_{n+2}=\frac{(n^2-n-k^2-k)c_n}{(n+1)(n+2)}$$
b) show that if the value of $k$ is a non negative integer then one of the solution found in part (a) is a polynomial of degree k.
How to show that?
Best Answer
Letting the coefficients be $c_n$, let us find (mentally) the terms of degree $n$ obtained by plugging the series in the equation and equate them to zero:
$$(n+2)(n+1)c_{n+2}-n(n-1)c_n-2nc_n+k(k+1)c_n=0.$$
This yields the recurrence
$$c_{n+2}=\frac{n(n+1)-k(k+1)}{(n+2)(n+1)}c_n.$$
For the "base case", we want to obtain the constant and linear terms, which depend on the four first terms, let $a+bx+cx^2+dx^3$. Plugging in the equation,
$$(1-x^2)(2c+6dx)-2x(b+2cx+3dx^2)+k(k+1)(a+bx+cx^2+dx^3)=(ak(k+1)+2c)+(b(k(k+1)-2)+6d)x+\cdots$$
Thus in general $a$ and $b$ are free parameters from which the next coefficients follow, and you get independent even and and odd series.
If $k(k+1)=0$, the even series reduces to a constant. If $k(k+1)=2$, the odd series reduces to a constant.
More generally, when $k$ is an integer, there will be an $n$ such that $n(n+1)=k(k+1)$ and the recurrence stops at $c_n$.