Question

Solve : $y^2+\left(x-\frac{1}{y}\right)\frac{dy}{dx}=0$

Answer 1

$y^2+(x-\frac{1}{y})\frac{dy}{dx}=0$

$y^2=-(x-\frac{1}{y})\frac{dy}{dx}$

$y^3=-\frac{-(xy-1)}{y}\frac{dy}{dx}$

$y^3=\left(1xy\right)\frac{dy}{dx}$

$\int \frac{1}{1-xy}dx=\int \frac{1}{y^3}dy$

Solving $LHS\to \int \frac{1}{1-xy}dx$

Substitute $u=1-yx$

$\Rightarrow \int -\frac{1}{yu}du=\frac{-1}{y}\int \frac{du}{u}$

$\Rightarrow \frac{-1}{y}du=\frac{-1}{y}\int -\frac{du}{u}$

as $\int 1/udu=\ln |u|$

$\Rightarrow \frac{-1}{y}\ln |u|=\frac{-1}{y}\ln |1-yx|$

$\Rightarrow \frac{-1}{y}\ln |1-yx|+C_1$

solving $RHS.\int \frac{1}{y^3}dy$

$\Rightarrow \int y^{-3}dy$

as $\int x^{2}dx=\frac{x^{n+1}}{n+1}+C$

$\therefore \int y^{-3}dy\Rightarrow \frac{-1}{2}{y}^{-2}+C_2$

Hence, $\frac{-1}{y}\ln |1-yx|+C_1=\frac{-1}{2}{y}^{-2}+C_2$

$\frac{-1}{y}\ln |1-yx|+\frac{1}{2}{y}^{-2}+C$