Question

(xy^2-x)dx-(y+yx^2)dy=0,求这个微分方程的通解.

Answer 1

y(1+x^2)dy=x(y^2-1)dx

ydy/(y^2-1)=xdx/(x^2+1)

ln|y^2-1|/2=ln|x^2+1|/2+C

ln|y^2-1|=ln(x^2+1)+C

y^2-1=C(x^2+1)

y^2=C(x^2+1)+1

y=±√(C(x^2+1)+1)