The differential equation for which sqrt{1+y^ | vedclass

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Question

(D) Given the equation: $\sqrt{1+y^2}=C x e^{	an ^{-1} x}$ Differentiating both sides with respect to $x$: $\frac{1}{2\sqrt{1+y^2}} \cdot 2y \frac{dy

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$x y\left(1+x^2\right) d y-\left(1+y^2\right)\left(1+x+x^2\right) d x=0$

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(D) Given the equation: $\sqrt{1+y^2}=C x e^{	an ^{-1} x}$ Differentiating both sides with respect to $x$: $\frac{1}{2\sqrt{1+y^2}} \cdot 2y \frac{dy}{dx} = C \left( e^{	an ^{-1} x} + x e^{	an ^{-1} x} \cdot \frac{1}{1+x^2} ight)$ $\frac{y}{\sqrt{1+y^2}} \frac{dy}{dx} = C e^{	an ^{-1} x} \left( 1 + \frac{x}{1+x^2} ight)$ Since $C e^{	an ^{-1} x} = \frac{\sqrt{1+y^2}}{x}$, substitute this into the equation: $\frac{y}{\sqrt{1+y^2}} \frac{dy}{dx} = \frac{\sqrt{1+y^2}}{x} \left( \frac{1+x^2+x}{1+x^2} ight)$ $\frac{xy}{\sqrt{1+y^2}} \frac{dy}{dx} = \frac{\sqrt{1+y^2}(1+x+x^2)}{1+x^2}$ $xy(1+x^2) dy = (1+y^2)(1+x+x^2) dx$ $xy(1+x^2) dy - (1+y^2)(1+x+x^2) dx = 0$

The differential equation for which $\sqrt{1+y^2}=C x e^{\tan ^{-1} x}$ is the general solution, is

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