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.