An integrating factor of the differential equation $x d y - y d x + x^{2} e^{x} d x = 0$ is

\frac{1}{x}

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log \sqrt{1 + x^{2}}

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\sqrt{1 + x^{2}}

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x

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\frac{1}{1 + x^{2}}

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Solution

The correct option is A $\frac{1}{x}$
Given differential equation is
$x d y - y d x + x^{2} e^{x} d x = 0$
$\Rightarrow x d y + d x (x^{2} e^{x} - y) = 0$
$\Rightarrow \frac{d y}{d x} - \frac{y}{x} = - x e^{x}$
Therefore, $I.F. = e^{\int p d x} = e^{\int - \frac{1}{x} d x}$
$\Rightarrow e^{- log x} = \frac{1}{x}$

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