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  • Find a particular solution of the differential equation (x + 1) dy/dx = 2 e ^-y - 1, given that y = 0 when x = 0

Q14.    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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Given equation is
(x+1)\frac{dy}{dx} = 2e^{-y} -1
we can rewrite it as
\frac{e^ydy}{2-e^y}= \frac{dx}{x+1}\\
Integrate both the sides
\int \frac{e^ydy}{2-e^y}= \log |x+1|\\
\int \frac{e^ydy}{2-e^y}
Put
 2-e^y = t\\ -e^y dy = dt
\int \frac{-dt}{t}=- \log |t|
put t = 2- e^y  again
\int \frac{e^ydy}{2-e^y} =- \log |2-e^y|
Put this in our equation
\log |2-e^y| + C'= \log|1+x| + C''\\ \log (2-e^y)^{-1}= \log (1+x)+\log C\\ \frac{1}{2-e^y}= C(1+x)

Now, by using boundary conditions we will find the value of C
It is given that  y = 0 when x = 0
at   x = 0
\frac{1}{2-e^0}= C(1+0)\\ C = \frac{1}{2}
Now, put the value of C
\frac{1}{2-e^y} = \frac{1}{2}(1+x)\\ \\ \frac{2}{1+x}= 2-e^y\\ \frac{2}{1+x}-2= -e^y\\ -\frac{2x-1}{1+x} = -e^y\\ y = \log \frac{2x-1}{1+x}
Therefore, the particular solution is  y = \log \frac{2x-1}{1+x}, x\neq-1

Posted by

Gautam harsolia

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