Question

Solve the differential equation : $(x+1)\frac{dy}{dx}=2e^{-y}-1;y\left(0\right)=0.$

Answer 1

Given differential equation is
$(x+1)\frac{dy}{dx}=2e^{-y}-1$
$\Rightarrow \frac{dy}{2e^{-y}-1}=\frac{dx}{x+1}\Rightarrow \frac{e^y}{2-e^y}dy=\frac{1}{x+1}dx$

Integrating both sides, we get
$\int \frac{e^y}{2-e^y}dy=\int \frac{1}{x+1}dx\Rightarrow -\log |2-e^y|=\log |1+x|+\log |C|(\because \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\log f\left(x\right))\Rightarrow \log \frac{1}{|2-e^y|}=\log |C(1+x)|\Rightarrow \frac{1}{2-e^y}=C(1+x)\text{Given,}y\left(0\right)=0\Rightarrow C=1$
Therefore, the particular solution of the given differential equation is $\frac{1}{2-e^y}=1+x$
$\Rightarrow (1+x)(2-e^y)=1$