Question

ayuda urgente
Y^2dx+(1-x)dy=0

Answer 1

                        $y^2~dx+(1-x)dy=0\\ \\ y^2~dx=(x-1)dy\\ \\ \dfrac{dx}{x-1}=\dfrac{dy}{y^2}\\ \\ \\ \displaystyle \int\dfrac{dx}{x-1}=\int\dfrac{dy}{y^2}\\ \\ \\ \ln|x-1|=-\dfrac{1}{y}+C\\ \\ y=\dfrac{1}{C-\ln|x-1|}$