Legendre Differential Equation

**Fel** · September 30th 2009, 08:55 PM

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.

**chisigma** · September 30th 2009, 11:46 PM

If we suppose that the general solution is analytic in $x=0$ , it can be written as...

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

With this substitution the DE becomes...

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

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

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

... so that is...

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

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

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

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

Kind regards

$\chi$ $\sigma$

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