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

Given, differential equation is $(x+1)\frac{dy}{dx}=2e^{-y}-1$ ...(i)
On separating the variables
$\Rightarrow \frac{dy}{2e^{-y}-1}=\frac{dx}{x+1}\Rightarrow \frac{e^{y}dy}{2-e^y}=\frac{dx}{x+1}$
On integrating both sides, we get
$\int \frac{e^{y}dy}{2-e^y}=\int \frac{dx}{x+1}$ ...(ii)
Put $2-e^y=t\Rightarrow -e^y=\frac{dt}{dy}\Rightarrow e^{y}dy=-dt$
Then Eq. (ii) becomes
$\int \frac{-dt}{t}=\int \frac{dx}{x+1}\Rightarrow -log|t|=log|x+1|+logC\Rightarrow -log|2-e^y|=log|C(x+1)|\Rightarrow \frac{1}{2-e^y}=C(x+1)$ $[\because -logx=log{x}^{-1}=\frac{1}{x}]$
$\Rightarrow 2-e^{-y}=\frac{1}{C(x+1)}$
Now, at x=0 and y=0, $2-1=\frac{1}{C}\Rightarrow C=1$
On putting the value of C in Eq. (iii), we get
$2-e^y=\frac{1}{(x+1)}\Rightarrow e^y=2-\frac{1}{x+1}\Rightarrow e^y=\frac{2x+2-1}{x+1}\Rightarrow e^y=\frac{2x+1}{x+1}\Rightarrow y=log\left|\frac{2x+1}{x+1}\right|,(x\ne -1)$ $[\because Iflo{g}_{e}x=m\Rightarrow e^m=x]$