Question

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

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

Answer 1

Answer: (C)

$y=x+\log \mid x+y\mid +c$

$\frac{dy}{dx}=\frac{x+y+1}{x+y-1}$

Put $x+y=t$

$\Rightarrow 1+\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{dt}{dx}-1$

$\therefore \frac{dt}{dx}-1=\frac{t+1}{t-1}$

$\Rightarrow \frac{dt}{dx}=\frac{t+1+t-1}{t-1}$

$\Rightarrow \frac{dt}{dx}=\frac{2t}{t-1}$

$\Rightarrow \left(\frac{t-1}{2t}\right)dt=dx$

$\Rightarrow \left(\frac{1}{2}-\frac{1}{2t}\right)dt=dx$
On integrating, we get
$\frac{1}{2}t-\frac{1}{2}\log t=x+c_1$
$\Rightarrow t-\log t=2x+2c_1$
$\Rightarrow x+y-\log (x+y)=2x+2c_1$
$\Rightarrow y=x+\log (x+y)+c$