Question

Solve the following lengendre differential equation.

(1-x²)y"-2xy'+l(l+1)y=0

Answer 1

Solution

(1-x²)y"-2xy'+l(l+1)y=0

Can be rewritten as;

y"-x²y"-2xy'+l(l+1)y=0

Solve using power series;

By definition;

y=$\displaystyle\sum_{n=0}^ ∞$ a_nxⁿ

And;

y'=$\displaystyle\sum_{n=1}^ ∞$ na_nx^n-1

y"=$\displaystyle\sum_{n=2}^ ∞$ n(n-1)a_nx^n-2

Replace the values in the equation to obtain;

$\displaystyle\sum_{n=2}^∞$ n(n-1)a_nx^n-2-x²$\displaystyle\sum_{n=2}^∞$ n(n-1)a_nx^n-2-2x$\displaystyle\sum_{n=1}^∞$ na_nx^n-1+l(l+1)$\displaystyle\sum_{n=0}^∞$ a_nxⁿ

Make all powers of X equal

$\displaystyle\sum_{n=0}^∞$ (n+2)(n+1)a_n+2xⁿ+$\displaystyle\sum_{n=2}^∞$ -n(n-1)a_nxⁿ+$\displaystyle\sum_{n=1}^∞$ -2na_nxⁿ+$\displaystyle\sum_{n=0}^∞$ l(l+1)a_nxⁿ

n=0,1,2,3...

When;

n=0

((2.1)a₂+l(l+1))x⁰=0

n=1

3.2a₃-2a₁+l(l+1)x¹=0

n$=\geq2$

There is a general trend of

((n+2)(n+1)a_n+2-n(n-1)a_n-2na_n+l(l+1)a_n))xⁿ=0

All the coefficients of x must equal to zero, therefore;

Using

n=0;a₂=$\frac{-l(l+1)a_0}{2.1}$

n=1;a₃=$\frac{-(l^2+l-2)a_1}{3.2}$

n$\geq$ 2

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

This is the recurrence relation.

Using it ;

n=2,

a₄=$\frac{(l+3)(2-l)a_2}{4.3}$ =$\frac{(l+3)(l+1)(l)(l-2)a_0}{4!}$

n=3

a₅=$\frac{(l+4)(3-l)a_3}{5.4}$ =$\frac{(l+4)(l+2)(l-1)(l)(l-3)a_1}{5!}$

And so on

A pattern for even number is ;

$a_{2n}=\frac{(-1)^n(l+2n-1)(l+2n-3)...(l+1)l(l-2)...(l-(2n-2))}{(2n)!}$

A pattern for the odd numbers is;

$a_{2n+1}=\frac{(-1)^n(l+2n)(l+2n-2)...(l+2)(l-1)(l-3)...(l-(2n-1))a_1}{(2_{n+1})!}$

Since y=$\displaystyle\sum_{n=0}^∞$ a_nxⁿ=a₀$\displaystyle\sum_{n=0}^∞$ $\frac{(-1)^n(l+2n-1)(l+2n-3)...(l+1)l(l-2)..(l-(2n-2))}{(2n)!}x^{2n}+a_1\displaystyle\sum_{n=0}^∞\frac{((-1)^n(l+2n)(l+2n-2)..(l+2)(l-1)(l-3)...(l-(2n-1))}{(2n+1)!}x^{2n+1}$