Reduction of Order - Legendre Eqn

[MidnightR]

Legendre's eq of order n>=0 is

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

You are given the soln y = P_n(x) for n=0,1,2,3 to be P_0(x)=1 ; P_1(x)=x ; P_2(x)=(3x^2-1)/2 ; P_3(x)=(5x^3 -3x)/2. Use reduction of order to find the second independent soln's Q_n(x)

OK I've found Q_1(x) = ln(1-x)(1+x)

I'm struggling with Q_2(x), the integrals get really horrible

Is there a faster way of doing this? Am I suppose to solve all four in this way or is there a way to do it for the general case (any n)?

Cheers

 

[fzero]

You'll always have to do an integral for any n. The most general result you can obtain is the following. Substitute

$$Q_n(x) = v_n(x) P_n(x)$$

into the Legendre equation. After some algebra, you find that it reduces to

$$( (1-x^2) P_n v_n' )' =0,$$

leading to an expression for the functions $$v_n(x)$$ as

$$v_n = c_n \int \frac{dx}{(1-x^2) P_n},$$

where $$c_n$$ is an integration constant. All of the integrals seem like they can be done by either partial fractions or trig substitutions, but I can see how they get tedious at higher order.

 

[MidnightR]

Cheers, I guess it's good "practice". Sigh.