The integrating factor of the differential equation, $x^2 \frac{dy}{dx} + xy = log_e x$ is equal to

The differential equation is $x^2 \frac{dy}{dx} + xy = \log_e x$.

The standard form of a linear differential equation is:

$\frac{dy}{dx} + P(x) \cdot y = Q(x)$

Dividing the entire equation by $x^2$:

$\frac{dy}{dx} + \frac{y}{x} = \frac{\log_e x}{x^2}$

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Comparing with the standard form $\frac{dy}{dx} + P(x) \cdot y = Q(x)$:

$P(x) = \frac{1}{x}$

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The integrating factor formula is:

$\text{I.F.} = e^{\int P(x)dx}$

Substituting $P(x) = \frac{1}{x}$:

$\text{I.F.} = e^{\int \frac{1}{x}dx}$

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Since $\int \frac{1}{x}dx = \log_e x$:

$\text{I.F.} = e^{\log_e x}$

Using the property $e^{\log_e x} = x$:

$\text{I.F.} = x$

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Therefore, the integrating factor is $x$.