Question

Find the solution of
$\left(\frac{x+y-a}{x+y-b}\right)\frac{dy}{dx}=\frac{x+y+a}{x+y+b}.$

• A. $(b-a)\log \{{(x+y)}^2-ab\}=2(x-y)+k.$
• B. $(b-a)\log \{{(x-y)}^2-ab\}=2(x+y)+k.$
• C. $(b-a)\log \{{(x-2y)}^2-ab\}=2(x-y)+k.$
• D. $(b-a)\log \{{(x+y)}^2-ab\}=2(x+2y)+k.$

Answer 1

The correct option is A $(b-a)\log \{{(x+y)}^2-ab\}=2(x-y)+k.$

Given $\left(\frac{x+y-a}{x+y-b}\right)\frac{dy}{dx}=\frac{x+y+a}{x+y+b}.$ ....(1)
Put $x+y=v$
$\Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}$
So, eqn (1) becomes
$\therefore \frac{dv}{dx}-1=\frac{(v+a)(v-b)}{(v-a)(v+b)}$
$\frac{dv}{dx}=\frac{(v+a)(v-b)}{(v-a)(v+b)}+1$
$\frac{dv}{dx}=2\frac{(v^2-ab)}{v^2+(b-a)v-ab}.$
$\therefore \frac{v^2-ab+(b-a)v}{v^2-ab}dv=2dx$
$\Rightarrow (1+\frac{b-a}{2}.\frac{2v}{v^2-ab})dv=2dx.$
Integrating, we get
$v+\frac{b-a}{2}\log (v^2-ab)=2x+c$
$(b-a)\log \{{(x+y)}^2-ab\}=2(x-y)+k.$