Given: (1+ex)dy+yexdx=0 where y>0
This is a separable differential equation.
(1+ex)dy=−yexdx
ydy=1+ex−exdx
Integrating both sides:
∫ydy=∫1+ex−exdx
lny=−ln(1+ex)+lnC
lny+ln(1+ex)=lnC
ln[y(1+ex)]=lnC
y(1+ex)=C
Therefore, the general solution is C=y(1+ex) where C is an arbitrary constant.