Another solution of the Legendre differential equation

Question

Legendre polynomials are solutions of the Legendre differential equation $${d \over dx} \left[ (1-x^2) {d \over dx} P_n(x) \right] + n(n+1)P_n(x) = 0$$ Since the ODE is of second order, it has a second, linearly independent solution from $P_n$. How to find a second solution? I could use change of variables to find it (lengthy method). But is there a way to use Rodrigues' formula to solve it?

Answer 1

Variation of parameters will do it. You can assume a form of solution $y=fP_{n}$ and substitute into the equation. The equation in expanded form is $$ (1-x^{2})P_{n}''-2xP_{n}'+n(n+1)P_{n}=0. $$ Now substitute $y=fP_{n}$ into the same equation $$ (1-x^{2})(fP_{n}''+2f'P_{n}'+f''P_{n})-2x(fP_{n}'+f'P_{n})+n(n+1)fP_{n}=0. $$ Subtracting $$ (1-x^{2})fP_{n}''-2xfP_{n}'+n(n+1)fP_{n}=0 $$ from the previous equation gives $$ (1-x^{2})(2f'P_{n}'+f''P_{n})-2xf'P_{n}=0. $$ This is a first order equation in $f'$ $$ (1-x^{2})P_{n}f''+\{2(1-x^{2})P_{n}'-2xP_{n}\}f'=0,\\ \frac{f''}{f'} = -2\frac{P_{n}'}{P_{n}}+\frac{2x}{1-x^{2}},\\ %% \ln|f'| = -2\ln|P_{n}|-\ln(1-x^{2})+C\\ %% f' = \frac{K}{(1-x^{2})P_{n}^{2}} \\ %% f = K\int\frac{1}{(1-x^{2})P_{n}^{2}}\,dx+L $$ A second solution has the following form for constant $K\ne 0$ and $L$: $$ fP_{n} = K\left(\int \frac{1}{(1-x^{2})P_{n}^{2}}\,dx\right)P_{n}+LP_{n}. $$