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The given differential equation is $(x+1)\frac{dy}{dx} = 2e^{-y} - 1$. Rearranging the terms to separate $x$ and $y$, we get: $$ \frac{dy}{2e^{-y} - 1} = \frac{dx}{x+1} $$ Multiplying the numerator and denominator of the left side by $e^y$: $$ \frac{e^y dy}{2 - e^y} = \frac{dx}{x+1} $$
Integrate both sides: $$ \int \frac{e^y}{2 - e^y} dy = \int \frac{1}{x+1} dx $$ For the left integral, let $u = 2 - e^y$, then $du = -e^y dy$. The integral becomes: $$ -\int \frac{du}{u} = \ln|x+1| + C $$ $$ -\ln|2 - e^y| = \ln|x+1| + C $$
Given $y=0$ when $x=0$: $$ -\ln|2 - e^0| = \ln|0+1| + C $$ $$ -\ln|2 - 1| = \ln(1) + C $$ $$ -\ln(1) = 0 + C \implies C = 0 $$
Substituting $C=0$ back into the equation: $$ -\ln|2 - e^y| = \ln|x+1| $$ $$ \ln|2 - e^y|^{-1} = \ln|x+1| $$ $$ \frac{1}{2 - e^y} = x + 1 $$ $$ 2 - e^y = \frac{1}{x+1} \implies e^y = 2 - \frac{1}{x+1} = \frac{2x+1}{x+1} $$ $$ y = \ln\left(\frac{2x+1}{x+1}\right) $$
Final Answer: y = \ln\left(\frac{2x+1}{x+1}\right)
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