separate the differential equation into variables
(x2-1)y,+2xy2=0
(x2−1)y′+2xy2=0(x2−1)dy+2xy2dx=0(x2−1)dy=−2xy2dxdyy2=−2xdxx2−1∫dyy2=∫−2xdxx2−1−1y=−ln∣x2−1∣−C1y=ln∣x2−1∣+C(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\\(x2−1)y′+2xy2=0(x2−1)dy+2xy2dx=0(x2−1)dy=−2xy2dxy2dy=x2−1−2xdx∫y2dy=∫x2−1−2xdx−y1=−ln∣x2−1∣−Cy1=ln∣x2−1∣+C
