Using the Leibniz formula for a differential equation

Question

How would you use the Leibniz formula on the following DE:

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

for some constant $l\ge0$ to show that for $n\ge0$,

$$y^\left(n+2\right)(0)=[n(n+1)-l(l+1)]y^\left(n\right)(0)$$

Answer 1

Let $z=y'$ and $t(x)=1-x^2$, then notice how: $$ (1-x^2)y''(x)-2xy'(x) = (tz)'(x) $$ Thus, your equation may be written as: $$ (tz)'(x)+l(l+1)y(x)=0 $$ Then we derive $n$ times: $$ (tz)^{(n+1)}=-l(l+1)y^{(n)} $$ And by Leibniz formula (since $t^{(k)}=0$ for $k\geq 3$): $$ ty^{(n+2)}+(n+1)t'y^{(n+1)}+\frac{n(n+1)}{2}t''y^{(n)}=-l(l+1)y $$

Then, evaluating this expression in zero (given that $t'(x)=-2x$ and $t''(x)=-2$): $$ y^{(n+2)}(0)=[n(n+1)-l(l+1)]y^{(n)}(0) $$

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