Mathematics - Differential Equation Question with Solution | TestHub

(i) Let $e^x=X$ 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 $p=\frac{X}{Y}\cdot P$.

Therefore, 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. Therefore, the solution is $Y=cX+c^3$ or $e^y=c{e}^x+c^3$.