Question

这个微分方程怎么求解啊。。求解答

Answer 1

令x=uy，则u=x/y, dx/dy=y*du/dy+u,推出dx=udy+ydu,所以有(1+2e^u)(udy+ydu)+2e^u(1-u)dy=0,
[（2e^u+1)/(u+2e^u)]du=[-1/y]dy, 积分号d(u+2e^u)/(u+2e^u)=积分号-dy/y,ln(u+2e^u)=-lny+C,所以有 y(u+2e^u)=C,即 x+2ye^(x/y)=C

Answer 2

可以代换x/y=u x=uy x'=u+yu'
u+yu'=2e^u(u-1)/(1+2e^u)
yu'=2e^u(u-1)/(1+2e^u)-u=-(u+e^u)/(1+2e^u)
(1+2e^u)du/(u+2e^u)=-dy/y
通解为：ln(u+2e^u)=-lny+lnC
或：u+2e^u=C/y
x+2ye^(x/y)=C

Answer 3