Question

Find the solution of $\frac{dy}{dx}=\frac{x+y+1}{x+y-1}$ when $y=\frac{1}{3}$ at $x=\frac{2}{3}.$

• A. $\therefore y-x+\frac{1}{3}=\log (x+y)$
• B. $\therefore y+x+\frac{1}{3}=\log (x+y)$
• C. $\therefore y-x+\frac{1}{3}=\log (x-y)$
• D. $\therefore y+x+\frac{1}{3}=\log (x-y)$

Answer 1

The correct option is A $\therefore y-x+\frac{1}{3}=\log (x+y)$

Given equation is
$\frac{dy}{dx}=\frac{x+y+1}{x+y-1}$ ....(1)

Put $x+y=v$
$\Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}$

So, eqn (1) becomes
$\frac{dv}{dx}-1=\frac{v+1}{v-1}$

$\Rightarrow \frac{dv}{dx}=\frac{2v}{v-1}$

$\Rightarrow \left(\frac{v-1}{v}\right)dv=2dx$

$(1-\frac{1}{v})dv=2dx$

Integrating, we get
$v-\log \left(v\right)=2x+c$

$\Rightarrow (y-x)=\log (x+y)+c$

Now, put $x=\frac{2}{3},y=\frac{1}{3}$

$(\frac{1}{3}-\frac{2}{3})=\log (\frac{2}{3}+\frac{1}{3})+c$

$\Rightarrow -\frac{1}{3}=\log 1+c$

$\Rightarrow c=-\frac{1}{3}$

$\therefore y-x+\frac{1}{3}=\log (x+y).$