The general solution of the differential equation $\frac{xdy}{dx}+ 4y = x^3, (x≠0)$ is:

$y=\frac{x^3}{7}+cx^{-4}$: c is arbitrary constant.

The correct answer is Option (1) → $y=\frac{x^3}{7}+cx^{-4}$: c is arbitrary constant.

Given differential equation:

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

Rewrite in standard linear form:

$\frac{dy}{dx} + \frac{4}{x} y = x^2$

Integrating factor (I.F.) = $e^{\int \frac{4}{x} dx} = e^{4\ln x} = x^4$

Multiplying both sides by $x^4$:

$x^4 \frac{dy}{dx} + 4x^3 y = x^6$

$\frac{d}{dx}(x^4 y) = x^6$

Integrating both sides:

$x^4 y = \int x^6 dx = \frac{x^7}{7} + C$

General solution:

$y = \frac{x^7}{7x^4} + \frac{C}{x^4} = \frac{x^3}{7} + \frac{C}{x^4}$