Prove using induction the following equation is true.

Question

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.

Answer 1

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.

Answer 2

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 $n=1$.

Answer 3

$(1−x^2)\frac{dy}{dx}−xy−1=0$

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

Answer 4

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.