Question

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

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

Answer 1

Answer: (B)

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

First we take $x+y=t$

Now,differentiating the equation with respect to $x$ on both sides gives,

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

Replacing this in the main equation gives,

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

$⟹\frac{dt}{dx}=\frac{2t}{t+1}$

$⟹\frac{1}{2}(1+\frac{1}{t})dt=dx$

$⟹\frac{t}{2}+\ln \sqrt{t}=x+c$

Where $c$ is the integration constant

Substituting $t=x+y$ back gives,

$2x=x+y+\ln \left(c(x+y)\right)$

$⟹x+y=c{e}^{x-y}$