# Legendre Differential Equation

• Sep 30th 2009, 08:55 PM
Fel
Legendre Differential Equation
I'm supposed to somehow solve the following differential equation using the power series method, but I''m not sure on how to approach it.

(1-x^2)y'' - 2xy' + 2y = 0, y(0) = 1, y'(0) = 0

I was thinking of dividing through with the term in front of the y'', and then doing the problem like any normal power series. Any help on how to do it is appreciated.
• Sep 30th 2009, 11:46 PM
chisigma
If we suppose that the general solution is analytic in $\displaystyle x=0$, it can be written as...

$\displaystyle y= \sum_{n=0}^{\infty} a_{n}\cdot x^{n}$ (1)

With this substitution the DE becomes...

$\displaystyle (1-x^{2}) y^{''} -2xy^{'} + 2y= (1-x^{2}) \sum_{n=2}^{\infty} n(n-1)a_{n} x^{n-2}-2 \sum_{n=1}^{\infty} n a_{n} x^{n} + 2 \sum_{n=0}^{\infty} a_{n} x^{n}=0$ (2)

The fact that the coefficient of every power of $\displaystyle x$ in (2) must vanish means that...

$\displaystyle (n+2)\cdot(n+1)\cdot a_{n+2} - (n-2)\cdot(n-1)\cdot a_{n} =0$ (3)

... so that is...

$\displaystyle a_{n+2} = \frac{(n-2)\cdot (n-1)}{(n+2)\cdot (n+1)}\cdot a_{n}$ (4)

The (4) is very useful because once You know $\displaystyle a_{0}$ and $\displaystyle a_{1}$ [that can be derived fron the 'initial conditions'...] with it You can derive all the $\displaystyle a_{n}$. The general solution is on the form...

$\displaystyle y(x)= y_{e} (x) + y_{o}(x)$ (5)

... where $\displaystyle y_{e} (x)$ is an 'even' function the coefficients of which are derived from $\displaystyle a_{0}$ and $\displaystyle y_{o} (x)$ is an 'odd' function the coefficients of which are derived from $\displaystyle a_{1}$, in both cases using (4)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$