Question

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

Answer 1

$(x+1)\frac{dy}{dx}=2e^{-y}-1⟶eqn.1\frac{dy}{dx}+\frac{1}{1+x}=\frac{2e^{-y}}{1+x}$

Eqn.1 can be written as:

$\frac{1}{2e^{-y}-1}dy=\frac{1}{1+x}dx\Rightarrow \frac{1}{\frac{2}{e^y}-1}dy=\frac{1}{1+x}dx\Rightarrow \frac{-\left({-e}^y\right)}{2-e^y}dy=\frac{1}{1+x}dx\Rightarrow (-\log |2-e^y|)=\log |x+1|+Cy=0,x=0\Rightarrow -\log 2=\log 1+C\Rightarrow C=-\log 2$

Answer: $\log (x+1)-\log 2=-\log (2-e^{-y})\Rightarrow \log \left(\frac{(x+1)(2-e^{-y})}{2}\right)=0$