Solve $y e^{y} d x = (y^{3} + 2 x e^{y}) d y$

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Solution

$y e^{y} d x = (y^{3} + 2 x e^{y}) d y$
$\frac{d x}{d y} = \frac{y^{3} + 2 x e^{y}}{y e^{y}}$ by dividing the above equation by $y e^{y}$
$\Rightarrow \frac{d x}{d y} - \frac{2}{y} x = y^{2} e^{- y}$ by seperating the variables
$\frac{d x}{d y} - \frac{2}{y} x = y^{2} e^{- y}$ is of the form $\frac{d x}{d y} + P x = Q$
where $P = \frac{2}{y}$ and $Q = y^{2} e^{- y}$
Integrating Factor $=$ I.F $= x \times$ I.F $= \int Q \times I . F d y + c$
$\frac{x}{y^{2}} = \int y^{2} e^{- y} \times \frac{1}{y^{2}} d y + c$
$= - e^{- y} + c$ where $\int e^{- x} d x = - e^{- x} + c$
$\Rightarrow \frac{x}{y^{2}} = - e^{- y} + c$
$\Rightarrow x = - y^{2} e^{- y} + c y^{2}$

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