Question

x^2ydx=(1-y^2+x^2-x^2y^2)dy的通解,求详细过程.

Answer 1

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

所以

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

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

所以x-arctanx=lny-y^2/2+c