# Solution of differential equation $\frac{dy}{dx} = \frac{y + 2x^2 + 2y^2}{x}$ is Option 1) $\tan^{-1}\left(\frac{y}{x} \right )- 2 x = c$ Option 2) $\tan^{-1}\left(\frac{x}{y} \right ) + x = c$ Option 3) $\tan^{-1}\left(\frac{x}{y} \right )- x = c$ Option 4) $\tan^{-1}\left(\frac{y}{x} \right ) + 2 x = c$

H Himanshu

As we have learnt,

General form of Variable Separation -

$d\left ( \tan ^{-1} \frac{y}{x}\right )= \frac{xdy-ydx}{x^{2}+y^{2}}$

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Equation can be written as -

$x\frac{dy}{dx} -y = 2(x^2 + y^2) \\*\Rightarrow \frac{xdy-ydx}{x^2 +y^2} =2dx \\*\Rightarrow d\left(\tan^{-1}\left(\frac{y}{x} \right ) \right ) = 2dx$

Integrating both sides, we get

$\tan^{-1}\left(\frac{y}{x} \right ) - 2x =c$

Option 1)

$\tan^{-1}\left(\frac{y}{x} \right )- 2 x = c$

Option 2)

$\tan^{-1}\left(\frac{x}{y} \right ) + x = c$

Option 3)

$\tan^{-1}\left(\frac{x}{y} \right )- x = c$

Option 4)

$\tan^{-1}\left(\frac{y}{x} \right ) + 2 x = c$

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