Question

Find a particular solution of the differential equation\((x+1)\frac{dy}{dx}=2e^{-y}-1\),given that \(y=0\), when \(x=0\)

Answer 1

The correct answer is: \(y=log|\frac{2x+1}{x+1}|,(x≠-1)\)
\((x+1)\frac{dy}{dx}=2e^{-y}-1\)
\(⇒\frac{dy}{2e^{-y}-1}=\frac{dx}{x+1}\)
\(⇒\frac{e^ydy}{2-e^y}=\frac{dx}{x+1}\)
Integrating both sides,we get:
\(∫\frac{e^ydy}{2-e^y}=log|x+1|+logC...(1)\)
Let \(2-e^y=t.\)
\(∴\frac{d}{dy}(2-e^y)=\frac{dt}{dy}\)
\(⇒-e^y=\frac{dt}{dy}\)
\(⇒e^ydt=-dt\)
Substituting this value in equation(1),we get:
\(∫\frac{-dt}{t}=log|x+1|+logC\)
\(⇒-log|t|=log|C(x+1)|\)
\(\implies -log|2-e^y|=log|C(x+1)|\)
\(⇒\frac{1}{2-e^y}=C(x+1)\)
\(⇒2-e^y=\frac{1}{C(x+1)}...(2)\)
Now,at \(x=0\) and \(y=0\),equation (2) becomes:
\(⇒2-1=\frac{1}{C}\)
\(⇒C=1\)
Substituting \(C=1\) in equation(2),we get:
\(2-e^y=\frac{1}{x+1}\)
\(⇒e^y=2-\frac{1}{x+1}\)
\(⇒e^y=\frac{2x+2-1}{x+1}\)
\(⇒e^y=\frac{2x+1}{x+1}\)
\(⇒y=log|\frac{2x+1}{x+1}|,(x≠-1)\)
This is the required particular solution of the given differential equation.