Question

The solution of $\frac{dy}{dx}=\frac{x+y+1}{x+y-1}$ is

• A. $e^{y-x}=c(x+y)$
• B. $e^{y-x}=c(x-y)$
• C. $e^{y+x}=c(x+y)$
• D. $e^{y-x}=c(2x+y)$

Answer 1

Answer: (A)

$e^{y-x}=c(x+y)$

Let $x+y=t\Rightarrow 1+\frac{dy}{dx}=\frac{dt}{dx}$

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

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

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

Integrate on both sides

$\Rightarrow \frac{t}{2}-\frac{1}{2}lnt=x+c_1$

$\Rightarrow y-x=ln(x+y)+c_2$

$\Rightarrow e^{y-x}=c(x+y)$