For the differential equation given below, find a particular solution satisfying the given condition when x = 0.

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

$\frac{d y}{2 e^{-y}+1}=\frac{d x}{x+1}$

$\frac{e^yd y}{2 + e^{y}}=\frac{d x}{x+1}$

Integrate both side

$\int \frac{e^{y}}{2+e^{y}} d y=\int \frac{d x}{x+1}$

$\log \left|2+e^{y}\right|=\log |x+1|+\log c$

$2+e^{y}=c(x+1)\\$

At $x=0$ , $y=0$

$2+e^{0}=c(0+1)\\$

$c=3$

$2+e^{y}=3(x+1)$

The particular solution:

$e^{y}=3 x+1$

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