The general solution of the differential equation $(y^{2} + e^{2 x}) d y - y^{3} d x = 0$ (C being the constant of integration), is

y^{2} e^{- 2 x} + 2 l n y = c

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y^{2} e^{- 2 x} - 2 l n y = c

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y^{2} e^{- 2 x} - \frac{1}{2} y l n y = c

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y^{2} e^{- 2 x} - \frac{1}{2} l n y = c

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Solution

The correct option is A $y^{2} e^{- 2 x} + 2 l n y = c$
$\frac{d x}{d y} = \frac{y^{2} + e^{2 x}}{y^{3}} \frac{d x}{d y} = \frac{1}{y} + \frac{e^{2 x}}{y} \Rightarrow e^{- 2 x} \frac{d x}{d y} - e^{- 2 x} \frac{1}{y} = \frac{1}{y^{3}} L e t \frac{- e^{- 2 x}}{2} = u$
sothat $e^{- 2 x} \frac{d x}{d y} = \frac{d u}{d y} \frac{d u}{d y} + \frac{2}{y} u = \frac{1}{y^{3}}$
I.F. $= e^{\int \frac{2}{y} d y} = e^{2 l n y} = y^{2}$
Solution is $u . y^{2} = \int \frac{1}{y} d y + k$
$\Rightarrow u y^{2} = l n y + k \Rightarrow - \frac{e^{- 2 x}}{2} y^{2} = l n y + k \Rightarrow - e^{- 2 x} y^{2} = 2 l n y + 2 k ∴ e^{- 2 x} y^{2} + 2 l n y = c o n s t a n t$

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