The solution of the differential equation $\frac{y\, d x+ x\, dy}{ydx-x\, dy }=\frac{x^2e^{xy}}{y^4}$

$x^3=3y^3(1-e^{-xy})$

The correct answer is option (2) : $x^3=3y^3(1-e^{-xy})$

The given differential equation is

$(ydx+xdy)y^4 =x^2e^{x/y}(ydx-xdy)$

$⇒e^{-xy}d(xy) = \left(\frac{x}{y}\right)^2 d\left(\frac{x}{y}\right)$

Integrating, we obtain

$-e^{-xy}=\frac{1}{3}\left(\frac{x}{y}\right)^3 +C$ ................(i)

It is given that $y = 1 $ when x=0.

Putting $x=0$ and $y = 1 $ in (i), we get $C=-1$.

Putting $C=-1$ in (i) , we obtain

$-3y^3e^{-xy}= (x^3 -3y^3) $ or, $x^3 = 3y^3 (10e^{-xy})$ as the solution.