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