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. 

You have
$(1-x^2)y' - xy - 1 = 0
$
and you want to show that
$(1-x^2)y^{(n+2)} - (2n+3)xy^{(n+1)} - (n+1)^2y^{(n)} = 0
$.
Rewrite these as
$(1-x^2)y' 
= xy + 1 
$
and
$(1-x^2)y^{(n+2)} 
= (2n+3)xy^{(n+1)} + (n+1)^2y^{(n)} 
$.
Differentiating the first one,
$(1-x^2)y''-2xy' 
= xy'+y 
$
or
$(1-x^2)y'' 
= 3xy'+y 
$,
which is what you want to show
for $n=0$.
Differentiating that,
$(1-x^2)y'''-2xy'' 
= 3(xy''+y')+y' 
$,
or
$(1-x^2)y''' 
= 5xy''+4y' 
$,
which is what you want to show
for $n=1$.
This establishes the basis
for the induction.
Suppose
$(1-x^2)y^{(n+2)} 
= (2n+3)xy^{(n+1)} + (n+1)^2y^{(n)} 
$.
Differentiating this,
$\begin{array}\\
(1-x^2)y^{(n+3)}-2xy^{(n+2)} 
&= (2n+3)(xy^{(n+2)}+y^{(n+1)}) + (n+1)^2y^{(n+1)}\\
\text{or}\\
(1-x^2)y^{(n+3)}
&= (2n+5)xy^{(n+2)}+(2n+3)y^{(n+1)} + (n+1)^2y^{(n+1)}\\
&= (2n+5)xy^{(n+2)} + (n^2+2n+1+2n+3)y^{(n+1)}\\
&= (2n+5)xy^{(n+2)} + (n^2+4n+4)y^{(n+1)}\\
&= (2(n+1)+3)xy^{(n+2)} + (n+2)^2y^{(n+1)}\\
\end{array}
$
This is the induction step,
and shows that the result is true.
Actually,
looking at this,
I didn't need to show that
the result was true
for $n=1$.
