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

$(x+1)\frac{dy}{dx}=2e^{-y}-1\frac{dy}{2e^{-y}-1}=\frac{dx}{x+1}\frac{dy}{2e^{-y}-1}=\ln (x+1)+c$
Let $t=2e^{-y}-1$
So, $dt=-2e^{-y}dy$
Putting these in obtained equation
$\frac{-dt}{t(t+1)}=\ln (x+1)+c\ln \left(t\right)-\ln (t+1)=\ln (x+1)+c\ln (2e^{-y}-1)-\ln \left(2e^{-y}\right)=\ln (x+1)+c$
Putting $x=0,y=0$
$\ln \left(1\right)-\ln 2=\ln 1+cc=-\ln 2$
Particular solution is
$\ln (2e^{-y}-1)-\ln \left(2e^{-y}\right)=\ln (x+1)-\ln 2$