# Q14.    Find a particular solution of the differential equation $(x+1)\frac{dy}{dx} = 2e^{-y} -1$, given that $y = 0$ when $x = 0$

G Gautam harsolia

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  $y = \log \frac{2x-1}{1+x}, x\neq-1$

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