Problem 21 | Separation of Variables

$x^2 \, dx + y(x - 1) \, dy = 0$

$\dfrac{x^2 \, dx}{x - 1} + y \, dy = 0$
 

By long division
$\dfrac{x^2 \, dx}{x - 1} = x + 1 + \dfrac{1}{x - 1}$
 

Thus,
$\left[ (x + 1) + \dfrac{1}{x - 1} \right] \, dx + y \, dy = 0$

$\displaystyle \int (x + 1) \, dx + \int \dfrac{dx}{x - 1} + \int y \, dy = 0$

$\frac{1}{2}(x + 1)^2 + \ln |x - 1| + \frac{1}{2}y^2 + \ln c = 0$

$(x + 1)^2 + 2\ln |x - 1| + y^2 + 2\ln c = 0$

$(x + 1)^2 + y^2 + (2\ln c + 2\ln |x - 1|) = 0$

$(x + 1)^2 + y^2 + 2(\ln c + \ln |x - 1|) = 0$

$(x + 1)^2 + y^2 + 2 \ln |c(x - 1)| = 0$           answer