Question

The general solution of $\frac{dy}{dx}=\frac{1-3y-3x}{1+x+y}$ is
(where ${}^{\prime }{c}^{\prime }$ is constant of integration)

• A. $x+3y+2\ln |1+x+y|=c$
• B. $x+y+2\ln |1-x-y|=c$
• C. $3x+y+2\ln |1+x+y|=c$
• D. $3x+y+2\ln |1-x-y|=c$

Answer 1

The correct option is D $3x+y+2\ln |1-x-y|=c$

Assuming $x+y=t$
$1+\frac{dy}{dx}=\frac{dt}{dx}$
Now,
$\frac{dt}{dx}-1=\frac{1-3t}{1+t}\Rightarrow \frac{dt}{dx}=\frac{2(1-t)}{1+t}\Rightarrow \int 2dx=\int \frac{1+t}{1-t}dt\Rightarrow 2x+c_1=\int \frac{2-(1-t)}{1-t}dt\Rightarrow 2x+c_1=-2\ln |1-t|-t\therefore 3x+y+2\ln |1-x-y|=c$