The solution of the differential equation $\frac{y\, d x+ x\, dy}{ydx-x\, dy }=\frac{x^2e^{xy}}{y^4}$
Answer & explanation
Correct answer: option 2
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.