Question

请高手写出下列微分方程的通解！

Answer 1

解:举例子，请把具体题目发过来，解微分方程为dy/dx+(1+xy³)/(1+x³y)=0，(1+x³y)dy+(1+xy³)dx=0，dy+x³ydy+dx+xy³dx=0，dy+dx+x³ydy+y³xdx=0，d(x+y)+x³y³(dy/y²+dx/x²)=0，d(x+y)-x³y³(-dy/y²-dx/x²)=0，d(x+y)=x³y³d(1/y+1/x)，d(x+y)=x³y³d[(x+y)/xy]；设x+y=u，xy=v，方程化为du=v³d(u/v)，再设u=zv，方程化为d(zv)=v³dz，zdv+vdz=v³dz，zdv=(v³-v)dz，dv/(v³-v)=dz/z，vdv/(v²-1)-dv/v=dz/z，0.5ln|v²-1|-ln|v|=ln|z|+0.5ln|a|(a为任意非零常数)，ln|v²-1|=ln|av²z²|，v²-1=av²z²，有v²-1=au²，微分方程的解为x²y²-1=a(x+y)²请参考

Answer 2

微分方程的通解(xy^2+x)dx+(y-x^2y)dy=0。书上答案是(1+y^2)/(1-x^2)=C (xy^2+x)dx+(y-x^2y)dy=0

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

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

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

两边积分，得

ln(y^2+1)=ln(x^2-1)+lnc

y²+1=c【x²-1】

即

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