Mathematics - Differential Equation Question with Solution | TestHub

(i) Put $x=\frac{1}{X}$ and $y=\frac{1}{Y}$

$dx=-\frac{1}{X^2}dX$ and $dy=-\frac{1}{Y^2}dY$

$\Rightarrow p=\frac{dy}{dx}=\frac{X^2}{Y^2}\frac{dY}{dX}=\frac{X^2}{Y^2}P$

the given equation becomes

$\frac{1}{Y^2}\left(\frac{1}{Y}-\frac{1}{X}\cdot \frac{X^2}{Y^2}P\right)=\frac{1}{X^4}\cdot \frac{X^4}{Y^4}{P}^2$

$\Rightarrow Y-XP=P^2$ or $Y=PX+P^2$

Which is the Claraut's form

∴ The solution is $Y=cX+c^2$ or $\frac{1}{y}=\frac{c}{x}+c^2$