(x-y^2)dx+y(1+x)dy=0变形得dy/dx = (y^2-x)/(y+xy)设t = y^2-x ,则dy = dt/(2sqrt(t+x)) [注：sqrt表示开平方]带入原式：dt/(2t) = dx/(1+x)易得该微分方程的解为：2Ct = 1+x即2C(y^2-x) = 1+x [C为常数]
