The integrating factor of $x \frac{dy}{dx} - y = x^4 - 3x$ is

$\frac{1}{x}$

The correct answer is Option (3) → $\frac{1}{x}$ ##

Given that,

$x \frac{dy}{dx} - y = x^4 - 3x$

$\Rightarrow \frac{dy}{dx} - \frac{y}{x} = x^3 - 3$

which is a linear differential equation.

On comparing it with $\frac{dy}{dx} + P \cdot y = Q$

Here, $P = -\frac{1}{x}, Q = x^3 - 3$

$∴\text{I.F} = e^{\int P \, dx} = e^{-\int \frac{1}{x} \, dx} = e^{-\log x} = x^{-1}$

$= \frac{1}{x}$