Question

For the differential equations , find the general solution:$e^x tany\,\, dx+(1-e^x)sec^2y\,\,\, dy=0$

Answer 1

The given differential equation is:
$e^x tany\,\, dx+(1-e^x)sec^2y\,\,\, dy=0$
$(1-e^x)sec^2y\,\, dy=-e^x tany\,\, dx$
Separating the variables,we get:
$\frac{sec^2y}{tany} dy=\frac{-e^x}{1-e^x} dx$
Integrating both sides,we get:
$∫\frac{sec^2y}{tany} dy=∫\frac{-e^x}{1-e^x} dx...(1)$
Let $tany=u$
$⇒\frac{d}{dy}(tany)=\frac{du}{dy}$
$⇒sec^2y=\frac{du}{dy}$
$⇒sec^2 y\,\,dy=du$
$∴∫\frac{sec^2y}{tany}\,\, dy=∫\frac{du}{u}=logu=log(tany)$
Now,let $1-e^x=t.$
$∴\frac{d}{dx}(1-e^x)=\frac{dt}{dx}$
$⇒-e^x=\frac{dt}{dx}$
$⇒-e^x dx=dt$
$⇒∫\frac{-e^x}{1-e^x} dx=∫\frac{dt}{t}=logt=log(1-e^x)$
Substituting the values of $∫\frac{sec^2y}{tan y} dy$ and $∫\frac{-e^x}{1-e^x} dx$ in equation(1),we get:
$⇒log(tany)=log(1-e^x)+logC$
$⇒log(tany)=log[C(1-e^x)]$
$⇒tany=C(1-e^x)$
This is the required general solution of the given differential equation.