Mathematics - Differential Equation Question with Solution | TestHub

(i) put $e^x=X$ and $e^y=Y$

So that $e^{x}dx=dX$ and $e^{y}dy=dY$

$\Rightarrow \frac{e^y}{e^x}\cdot \frac{dy}{dx}=\frac{dY}{dX}$ or $\frac{Y}{X}p=P($ say $)$

or $p=\frac{X}{Y}\cdot P$

∴ The given equation becomes

$X^3\left(\frac{X}{Y}P-1\right)+\frac{X^3}{Y^3}\cdot Y^2{P}^3=0$

or $XP-Y+P^3=0$

or $Y=PX+P^3$

Which is of Clairaut's form.

∴ the solution is $Y=cX+c^3$

or $e^y=c{e}^x+c^3$