Question

Solution of $\left(\frac{x+y-a}{x+y-b}\right)\left(\frac{dy}{dx}\right)=\left(\frac{x+y+a}{x+y+b}\right)$ is:

• A. $\log \left[\right(x+y{)}^2-ab]=\frac{2}{b-a}[x-y]+k$
• B. $\log \left[\right(x+y{)}^2+ab]=\frac{1}{b-a}[x+y]+k$
• C. $(\frac{b-a}{2}\left)\right[\log \left(\right(x+y{)}^2-ab\left)\right]=x+c$
• D. $2\log (x+y)=\frac{x+y}{b-a}+k$

Answer 1

The correct option is A $\log \left[\right(x+y{)}^2-ab]=\frac{2}{b-a}[x-y]+k$

$\frac{dy}{dx}=\frac{(x+y+a)(x+y-b)}{(x+y-a)(x+y+b)}$

Let $x+y=v$

$\frac{dy}{dx}=\frac{dv}{dx}-1$

$\Rightarrow \frac{dv}{dx}-1=\frac{(v+a)(v-b)}{(v-a)(v+b)}$

$\frac{dv}{dx}=\frac{(v+a)(v-b)+(v-a)(v+b)}{(v-a)(v+b)}=\frac{v^2-ba+av-vb+bv-av+v^2-ab}{(v-a)(v+b)}$ $=\frac{2(v^2-ab)}{(v-a)(v+b)}$

$\Rightarrow \frac{(v-a)(v+b)}{2(v^2-ab)}dv=dx$

$\Rightarrow \frac{1}{2}[1-\frac{(a-b)v}{v^2-ab}]dv=dx$

$\Rightarrow \frac{1}{2}\int [1-\frac{(a-b)v}{v^2-ab}]dv=\int dx$

$\Rightarrow \frac{1}{2}\int dv-\int \frac{(a-b)v}{v^2-ab}dv=\int dx$

Let $v^2-ab=t\Rightarrow 2vdv=dt$

$\Rightarrow \frac{1}{2}\int dv-\frac{(a-b)}{2}\int \frac{1}{t}dt=\int dx$

$\Rightarrow \frac{1}{2}[v+\frac{(b-a)}{2}log(t\left)\right]=x$

$\Rightarrow \frac{1}{2}[v+\frac{(b-a)}{2}log(v^2-ab\left)\right]=x$

$\Rightarrow 2v+(b-a)log(v^2-ab)=4x$

$\Rightarrow (b-a)log(v^2-ab)=4x-2v$

$\Rightarrow (b-a)log\left(\right(x+y{)}^2-ab)=2(x-y)+c$ where $v=x+y$

$\Rightarrow \log \left[\right(x+y{)}^2-ab]$ $=\frac{2(x-y)}{(b-a)}+c$