I was given a "classical" homework question where I have to prove that the Legendre polynomials solve the differential equation:

$\frac{d}{dx}[(1-x^2)\frac{d}{dx}P_n(x)] + n(n+1)P_n(x) = 0$

However, I was asked to show this using the mathematical induction. I tried to do this using Rodrigues' formula or with the recursion relation:

$(n+1)P_{n+1}(x) = (2n+1)xP_n(x) - nP_{n-1}(x)$ It's probably a quite simple problem, but I can't see the proper way to show that. I'll appreciate any hint!

Thanks in advance!


There's three recurrence relations that help here:

$$P_{n+1}^{'} -P_{n-1}^{'} = (2n+1)P_n$$

$$(n+1)P_{n+1} = (2n+1)xP_n -nP_{n-1}$$

$$P_{n+1}-P_{n-1} = (x^2-1)\cdot \frac{2n+1}{n(n+1)}\cdot P_n^{'}$$

The induction step then looks as follows:

$\newcommand{\partial}[1]{\left[#1\right]}$ $\newcommand{\bracket}[1]{\left(#1\right)}$ \begin{equation} \begin{split} [(1-x^2)P_{n+1}^{'}]^{'} &=[(1-x^2)(P_{n-1}^{'}+(2n+1)P_n)]^{'} \\ &=[(1-x^2)P_{n-1}^{'}]^{'}+(2n+1)\bracket{-2xP_n + (1-x^2)P_n^{'}} \\ &=-(n-1)nP_{n-1}-2\{(n+1)P_{n+1}+nP_{n-1}\} + n(n+1)(P_{n-1}-P_{n+1}) \\ &=\{-(n-1)n-2n+ n(n+1)\}P_{n-1}+\{-2(n+1)-n(n+1)\}P_{n+1} \\ &=n\{-(n-1)-2+(n+1)\}P_{n-1}-(n+1)(2+n)P_{n+1} \\ &=-(n+1)(n+2)P_{n+1} \end{split} \end{equation}



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