Problem 11 | Separation of Variables

$(1 - x)y' = y^2$

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

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

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

$\displaystyle \int y^{-2} \, dy = -\int \dfrac{-dx}{1 - x}$

$-y^{-1} = -\ln (1 - x) - \ln c$

$\dfrac{1}{y} = \ln (1 - x) + \ln c$

$\dfrac{1}{y} = \ln c(1 - x)$

$1 = y \ln c(1 - x)$

$y \ln c(1 - x) = 1$           answer