The solution of the differential equation $e^{x} (x + 1) d x + (y e^{y} - x e^{x}) d y = 0$ with initial condition $y (0) = 0,$ is

x e^{x} + 2 y^{2} e^{y} = 0

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2 x e^{x} + y^{2} e^{y} = 0

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x e^{x} - 2 y^{2} e^{y} = 0

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2 x e^{x} - y^{2} e^{y} = 0

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Solution

The correct option is B $2 x e^{x} + y^{2} e^{y} = 0$
$e^{x} (x + 1) d x + (y e^{y} - x e^{x}) d y = 0$
$\Rightarrow e^{x} (x + 1) \frac{d x}{d y} + y e^{y} - x e^{x} = 0$
Put $x e^{x} = t$
$\Rightarrow e^{x} (x + 1) \frac{d x}{d y} = \frac{d t}{d y}$
$\Rightarrow \frac{d t}{d y} - t = - y e^{y},$ which is a linear differential equation in $y .$
I.F. $= e^{- y}$
General solution is
$t \cdot e^{- y} = - \int y d y + C$
$\Rightarrow x e^{x - y} = - \frac{y^{2}}{2} + C$
Since $y (0) = 0,$
$∴ C = 0$
$∴ 2 x e^{x} + y^{2} e^{y} = 0$

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