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Q)

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

$\begin{array}{1 1}y=\log\frac{|2x+1|}{|x-1|} \\y=\log\frac{|2x-1|}{|x+1|} \\ y=\log\large\frac{|2x+1|}{|x+1|} \\y=\log\large\frac{|2x-1|}{|x-1|} \end{array}$

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A)
Toolbox:
• The linear differential equationof the form $\large\frac{dy}{dx}$$= F(x,y), where F(x,y) is of the form g(x).h(y), where g(x) is a function of x and h(y) is a function of y,is called variables seperable type. This can be solved by separating the variables. Step 1: Given : (x+1)\large\frac{dy}{dx}$$ = 2e^{-y }- 1$
Using the information in the tool box we understand that the equation is variables seperable type.
$\large\frac{dy}{(2e^{-y} - 1)} = \large\frac{dx}{(x+1)}$
Multiply and divide LHS by $e^y$ we get,
$\large\frac{e^ydy}{2-e^y}=\frac{dx}{(x+1)}$
Step 2:
Integrating on both sides,
$\int\large\frac{e^ydy}{2-e^y}=\int\frac{dx}{x+1}$
LHS can be integrated by substitution method
Let $2-e^y=t$
$-e^ydy=dt$
$\int \large\frac{-dt}{t}$$=-\log t Substituting for t we get, -\log (2-e^t) Step 3: integrating RHS we get \int \large\frac{dx}{(x+1)}$$=\log(x+1)$
$- \log(2-e^t) =\log(x+1) + \log C$
$\log|x+1| +\log|2-e^y| = \log C$
$\log(2-e^y).(x+1)=\log C$
$(2 - e^y)(x+1) = C$-------(1)
Step 4:
To evaluate C, let us substitute for $x = 0$ and $y = 0$
$(2 - e^0)((0+1) = C$
$(2-1)(1)=C$
$C=1$
Step 5:
Substitute $C$ in equ(1) we get,
$(2-e^y)(x+1)=1$
$2-e^y=\large\frac{1}{x+1}$
$e^y=2-\large\frac{1}{x+1}$
$e^y=\large\frac{(2x+1)}{(x+1)}$
$y=\log\large\frac{|2x+1|}{|x+1|}$
This is the required solution.