Question

General solution of the differential equation $\frac {dy} {dx} = \frac {x+y+1} {x+y-1}$ is given by

• A. $x + y = \log |x + y| + C$
• B. $x -y = \log |x + y| + C$
• C. $y = x + \log |x|y| + C$
• D. $y = x \log |x + y | + C.$

Answer 1

Answer: (C)

$\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.
put $x+y = z$
$\therefore \frac{dz}{dz}-1 =\frac{z+1}{z-1}$
$\therefore 1+\frac{dy}{dx} = \frac{dz}{dx}$
$\Rightarrow \frac{dz}{dx} = 1+\frac{z+1}{z-1}$
$= \frac{z-1+z+1}{z-1} = \frac{2z}{z-1}$
$\Rightarrow \frac{z-1}{2z}dz =dx= \frac{1}{2}\left[1-\frac{1}{z}\right]dz = dx$
$\Rightarrow z-log \,|z| = 2x + c$
$\Rightarrow x + y-log\, |x + y| =2x + c$
$\Rightarrow y - log\, |x + y| = x + c$