Prove using induction the following equation is true. If $$(1-x^2)\frac{dy}{dx} - xy - 1 = 0$$
Using induction prove the following for any positive integer n$$(1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+3)x\frac{d^{n+1}y}{dx^{n+1}} - (n+1)^2\frac{d^ny}{dx^n} = 0$$
I know Leibtniz can be used to solve it easier but I need the proof to use induction. 

It obviously holds for $n=0$; now assume it holds for a general $n$. That is, we have $$(1-x^2)\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} - (2n+3)x\frac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}} - (n+1)^2\frac{\mathrm{d}^ny}{\mathrm{d}x^n} = 0$$
Let us differentiate (all differentiation in this answer is with respect to $x$) each term separately for clarity. The first term differentiates to (via use of the product rule) $$(1-x^2)\frac{\mathrm{d}^{n+3}y}{\mathrm{d}x^{n+3}}  - 2x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}}$$
The second term, again (via use of the product rule) differentiates to $$-(2n+3)\left(x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} + \frac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}}\right)$$
and the third term differentiates to $$-(n+1)^2 \frac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}}$$
The RHS of the equality differentiates trivially to $0$. So, putting all this together gives us
$$(1-x^2)\frac{\mathrm{d}^{n+3}y}{\mathrm{d}x^{n+3}}  - 2x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} - (2n+3)x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} - 2n\frac{\mathrm{d}^{n+1}y}{\mathrm{d} x^{n+1}} -3\frac{\mathrm{d}^{n+1}y}{\mathrm{d} x^{n+1}} - (n+1)^2\frac{\mathrm{d}^{n+1}y}{\mathrm{d} x^{n+1}} $$
(all of that equal to $0$) Cleaning up and grouping terms gives
$$(1-x^2)\frac{\mathrm{d}^{n+3}y}{\mathrm{d}x^{n+3}} - (2(n+1) +3)x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} - (n^2 + 4n + 4)\frac{\mathrm{d}y^{n+1}}{\mathrm{d}x^{n+1}} = 0$$
And hence, since $n^2 + 4n+4 = (n+2)^2$ we have $$(1-x^2)\frac{\mathrm{d}^{n+3}y}{\mathrm{d}x^{n+3}} - (2(n+1) +3)x\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} - (n +2)^2\frac{\mathrm{d}y^{n+1}}{\mathrm{d}x^{n+1}} = 0$$
as required, rather uneventfully. 
