The general solution of differential equation $y (x^{2} y + e^{x}) d x - e^{x} d y = 0$ is:

x^{3} y - 3 e^{x} = c y

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x^{3} y + 3 e^{x} = c y

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y^{3} x - 3 e^{y} = c x

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y^{3} x + 3 e^{y} = c x

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Solution

The correct option is B $x^{3} y + 3 e^{x} = c y$
We have $y (x^{2} y + e^{x}) d x - e^{x} d y = 0$

$\Rightarrow e^{x} \frac{d y}{d x} = x^{2} y^{2} + y e^{x}$

Dividing by $y^{2} e^{x}$ , we get $\frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} = x^{2} e^{- x}$

Substitute $\frac{1}{y} = V$ So that $\frac{- 1}{y^{2}} \frac{d y}{d x} = \frac{d V}{d x}$ .

We thus have $\frac{d V}{d x} + V = - x^{2} e^{- x}$ , which is linear.

$∴ I . F . = e^{\int 1 d x} = e^{x}$

Hence the solution is
$V . e^{x} = - \int x^{2} e^{- x} . e^{x} d x + \frac{c}{3}$
or $\frac{1}{y} e^{x} = - \frac{x^{3}}{3} + \frac{c}{3}$ or $x^{3} y + 3 e^{x} = c y$

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