Legendre Differential Equation(#3): A friendly introduction

This post is just a note on the notation that is used across internet sources and books while referring to the LDE.

$(1-x^2)y^{''} - 2xy^{'} + l(l+1) y = 0$

If one takes $p(x) = 1-x^2$ , then it follows that $p^{'}(x) = -2x$ . The differential equation can be rewritten as follows:

$p(x)y^{''} + p^{'}(x)y^{'} + l(l+1) y = 0$

Now the first two terms must seem familiar to you from the chain rule. ( $(py)^{'} = py^{'} + yp^{'}$ ). Ergo,

$(py^{'})^{'} + l(l+1)y = 0$

$\frac{d}{dx}(p \frac{dy}{dx}) + l(l+1)y = 0$

Now, putting back the value of p :

$\frac{d}{dx}\left((1-x^2) \frac{dy}{dx} \right) + l(l+1)y = 0$

And you will see this form of the LDE also in many places and I thought it was worth mentioning how one ended up in that form.

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