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  • For the differential equation given below, find a particular solution satisfying the given condition <img alt="(x + 1) \frac{dy}{dx} = 2e^{-y} + 1 ; y = 0" src="https://learn.careers360.com/latex-image/?%28x%20&plus;%201%29%20%5Cfrac%7Bdy

For the differential equation given below, find a particular solution satisfying the given condition

(x + 1) \frac{dy}{dx} = 2e^{-y} + 1 ; y = 0 when x = 0.

 

 
 
 
 
 

Answers (1)

(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

Posted by

Safeer PP

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