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.