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Question

(A) Given differential equation: $(x+1) \frac{dy}{dx} = 2e^{-y} - 1$Separate the variables:$\frac{dy}{2e^{-y} - 1} = \frac{dx}{x+1}$Multiply numerator

Vedclass · Accepted Answer

$y = \log \left| \frac{2x+1}{x+1} ight|, (x 
eq -1)$

Vedclass · Answer

(A) Given differential equation: $(x+1) \frac{dy}{dx} = 2e^{-y} - 1$Separate the variables:$\frac{dy}{2e^{-y} - 1} = \frac{dx}{x+1}$Multiply numerator and denominator by $e^y$:$\frac{e^y dy}{2 - e^y} = \frac{dx}{x+1}$Integrating both sides:$\int \frac{e^y dy}{2 - e^y} = \int \frac{dx}{x+1}$Let $t = 2 - e^y$,then $dt = -e^y dy$,so $e^y dy = -dt$:$-\int \frac{dt}{t} = \log |x+1| + C$$-\log |2 - e^y| = \log |x+1| + C$Using the condition $y = 0$ when $x = 0$:$-\log |2 - e^0| = \log |0+1| + C$$-\log |1| = \log |1| + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$Thus,$-\log |2 - e^y| = \log |x+1|$$\log |2 - e^y|^{-1} = \log |x+1|$$\frac{1}{2 - e^y} = x+1$$2 - e^y = \frac{1}{x+1}$$e^y = 2 - \frac{1}{x+1} = \frac{2x+2-1}{x+1} = \frac{2x+1}{x+1}$$y = \log \left| \frac{2x+1}{x+1} ight|, (x 
eq -1)$

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

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