Question

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

• 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. $\left(\frac{b-a}{2}\right)\left[log\left(\right(x+y{)}^2-ab)\right]=x+c$
• D. 0

Answer 1

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

Consider the problem

Let

$x+y=z$

then,

$1+\frac{dy}{dx}=\frac{dz}{dx}$

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

therefore,

$\left(\frac{z-a}{z-b}\right)(\frac{dz}{dx}-1)=\frac{z+a}{z+b}$

$\frac{dz}{dx}=1+\frac{(z+a)(z-b)}{(z-a)(z+b)}$

$=\frac{(z-a)(z+b)+(z+a)(z-b)}{(z-a)(z+b)}$

$=\frac{z^2+bz-az-ab+z^2-bz+az-ab}{(z-a)(z+b)}$

$\frac{dz}{dx}=\frac{2(z^2-ab)}{(z-a)(z+b)}$

$\begin{array}{c}\int \frac{\left(z-a\right)\left(z+b\right)}{\left(z^2-ab\right)}dz=\int 2dx+C=2x+C\\ \\ \int \frac{\left(z^2-ab\right)+z\left(b-a\right)}{\left(z^2-ab\right)}dz=2x+C\\ \\ \int \left\{1+\frac{z\left(b-a\right)}{\left(z^2-ab\right)}\right\}dz=2x+C\\ \\ z+\frac{\left(b-a\right)}{2}\ln \left(z^2-ab\right)=2x+C\\ \\ \frac{\left(b-a\right)}{2}\ln \left(z^2-ab\right)=2x+C-\left(x+y\right)\\ \\ =x-y+C\\ \\ 2\left(x-y+C\right)=\left(b-a\right)\ln \left[{\left(x+y\right)}^2-ab\right]\end{array}$