The general solution of the differential equation

$y(x^2+y+e^x) dx-e^x dy = 0 , $ is

$x^3y+3e^x=3Cy$

The correct answer is option (2) : $x^3y+3e^x=3Cy$

We have,

$y(x^2+y+e^x) dx-e^x dy = 0  $

$⇒x^2y^2d\, x + y \, e^x dx-e^x dy =0$

$⇒x^2dx +\frac{ye^xdx-e^x\, dy}{y^2}= 0 $

$⇒x^2dx+d\left(\frac{e^x}{y}\right) = 0 $

On integrating, we get

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

$⇒x^3 y + 3e^x = 3 Cy$