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. 
 A: 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. 
A: 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 $n=1$.
A: $(1−x^2)\frac{dy}{dx}−xy−1=0$
differentiate
$-2x \frac{dy}{dx} + (1+x^2)\frac{d^2y}{dx^2}−y - x\frac{dy}{dx}=0\\
(1+x^2)\frac{d^2y}{dx^2} -3x \frac{dy}{dx} - y=0$
This covers the base case $n=0$
Suppose,
$(1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+3)x\frac{d^{n+1}y}{dx^{n+1}} - (n+1)^2\frac{d^ny}{dx^n} = 0$
(this is the inductive hypothesis)
$\frac {d}{dx} \big((1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+3)x\frac{d^{n+1}y}{dx^{n+1}} - (n+1)^2\frac{d^ny}{dx^n} = 0\big)\\
-2x\frac{d^{n+2}y}{dx^{n+2}} + (1-x^2)\frac{d^{n+3}y}{dx^{n+3}} - (2n+3)\frac{d^{n+1}y}{dx^{n+1}} - (2n+3)x\frac{d^{n+2}y}{dx^{n+2}} - (n+1)^2\frac{d^{n+1}y}{dx^{n+1}} = 0\\
(1-x^2)\frac{d^{n+3}y}{dx^{n+3}} - (2n+5)x\frac{d^{n+2}y}{dx^{n+2}} - (n^2 + 4n +4)\frac{d^{n+1}y}{dx^{n+1}} = 0\\
(1-x^2)\frac{d^{n+3}y}{dx^{n+3}} - (2(n+1)+3)x\frac{d^{n+2}y}{dx^{n+2}} - (n+2)^2\frac{d^{n+1}y}{dx^{n+1}} = 0\\$
QED
A: Writing it as $\,Y_n,\,$ the obvious thing to try is to differentiate $\,Y_n\,$  and compare it to $\,Y_{n+1}$  
$$\begin{align}  Y_n'\ =&\ \ \  \left(f_n\, y^{(n+2)}\! + g_n\, y^{(n+1)}\! + h_n\, y^{(n)}\right)'\\[0.5em]
=&\quad f_n y^{(n+3)}\! + (g_n\! + f_n') y^{(n+2)}\! + (h_n\! + g_n') y^{(n+1)}\ \, {\rm by}\ \ h_n' = 0\\[0.5em]
{\rm so}\quad Y_{n+1} =\, Y_n' \ \ \ {\rm if}\ \ \   f_{n+1} &\!= f_n,\ \  g_{n+1}\! = g_n + f_n',\ \   h_{n+1}\! = h_n + g_n'\end{align}$$
Our  $\ f_n = 1\!-\!x^2\  $ thus $\,\ f_{n+1} = f_n,\,$
and $\  g_n = -(2n\!+\!3)x \, \Rightarrow\, g_n + f_n' = g_n\! -2x = g_{n+1}$
and $\  h_{n} = -(n\!+\!1)^2 \ \Rightarrow\,\  h_n\! + g_n' = -(n\!+\!1)^2\!-(2n\!+\!3)= -(n\!+\!2)^2 = h_{n+1}$
This yields the induction step, so verifying the base case will complete the induction.
