If $y = e^{ an^{-1} x}$, prove that $(1 + x^{2})\dfrac{d^{2}y}{dx^{2}} + (2x - 1)\dfrac{dy}{dx} = 0$.

Question

If $y = e^{	an^{-1} x}$, prove that $(1 + x^{2})\dfrac{d^{2}y}{dx^{2}} + (2x - 1)\dfrac{dy}{dx} = 0$.

Accepted Answer

>

> [!TLDR]
> Get $(1+x^2)y'=y$, then differentiate once more to reach $(1+x^2)y''+(2x-1)y'=0$.
>
> **Concept.** Differentiate $y=e^{	an^{-1}x}$, clear the denominator to get a first relation, then differentiate that relation again.
>
> $$y=e^{	an^{-1}x}\ \Rightarrow\ \frac{dy}{dx}=e^{	an^{-1}x}\cdot\frac{1}{1+x^2}=\frac{y}{1+x^2}.$$
> So
> $$(1+x^2)\,y'=y.\qquad(\ast)$$
> Differentiate $(\ast)$ with respect to $x$ (product rule on the left): …

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

Q.If $y = e^{\tan^{-1} x}$ , prove that $(1 + x^{2})\dfrac{d^{2}y}{dx^{2}} + (2x - 1)\dfrac{dy}{dx} = 0$ .

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Q.If y=etan⁡−1xy = e^{\tan^{-1} x}y=etan−1x, prove that (1+x2)d2ydx2+(2x−1)dydx=0(1 + x^{2})\dfrac{d^{2}y}{dx^{2}} + (2x - 1)\dfrac{dy}{dx} = 0(1+x2)dx2d2y​+(2x−1)dxdy​=0.

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Q.If $y = e^{\tan^{-1} x}$ , prove that $(1 + x^{2})\dfrac{d^{2}y}{dx^{2}} + (2x - 1)\dfrac{dy}{dx} = 0$ .

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