Question

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

Answer 1

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

$\left[2y\right(x+1\left)\right]dy=(y^2-x)dx$

$(2xy+2y)dy=(y^2-x)dx$

$(x+1)y^2=y^{2}x-1$

$x{y}^2+y^2=x{y}^2-1$

$y^2+1=0$