Solve the differential equation:    (x +1) \frac{dy}{dx} = 2e^{-y}-1; \ y(0) =0

 

 

 

 
 
 
 
 

Answers (1)

(x +1) \frac{dy}{dx} = 2e^{-y}-1; \ y(0) =0    [Given]

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

\Rightarrow \frac{dy}{2e^{-y}-1} = \frac{dx}{x +1}

\Rightarrow\int \frac{e^ydy}{2-e^{y}} = \int\frac{dx}{x +1}

\\\Rightarrow \text{put } 2 - e^y = t \\ -e^ydy = dt\\ e^ydy = -dt

\Rightarrow -\int \frac{dt}{t} = \int\frac{dx}{x +1}

\Rightarrow -\log t = \log |x+1| + \log c

\Rightarrow -\log t = \log \left [c|x+1| \right ]

\Rightarrow \frac{1}{t} = c\left [x+1 \right ]

\Rightarrow \frac{1}{2-e^y} = c\left [x+1 \right ] \qquad -(i)

Put x =0 & y - 0 in (i), we get

\Rightarrow \frac{1}{2} = c

\therefore \frac{1}{2-e^y} = \frac{1}{2}\left [x+1 \right ]

 

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