y'=1/(x+e^y)
x‘=x+e^y
x=Ce^y+ye^y哦y'=1/x+e^ye^(-y)y'=e^(-y)/x+1xe^(-y)y'=e^(-y)+xxe^(-y)dy-e^(-y)dx=xdx通xe^(-y)=-x^2/2+C