Question

separate the differential equation into variables

(x²-1)y^,+2xy²=0

Answer 1

$(x^2 -1)y'+2xy^2=0\\ (x^2 -1)dy+2xy^2dx=0\\ (x^2 -1)dy=-2xy^2dx\\ \frac{dy}{y^2}=\frac{-2xdx}{x^2-1}\\ \int\frac{dy}{y^2}=\int\frac{-2xdx}{x^2-1}\\ -\frac{1}{y}=-\ln|x^2-1|-C\\ \frac{1}{y}=\ln|x^2-1|+C\\$