Question

The solution of the differential equation
$\frac{dy}{dx}=\frac{y^2-2xy-x^2}{y^2+2xy-x^2}$ given $y=1$ at $x=1$ is:

• A. $|x-y|=0$
• B. $x^2+y^2=|x+y|$
• C. $2x^2-y^2=|x-2y|$
• D. $x+2y^2=3$

Answer 1

The correct option is B $x^2+y^2=|x+y|$

put $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
The given equation becomes,
$\Rightarrow v+x\frac{dv}{dx}=\left(\frac{v^2-2v-1}{v^2+2v-1}\right)$
$\Rightarrow x\frac{dv}{dx}=-\frac{(v^3+v^2+v+1)}{v^2+2v-1}$
$\Rightarrow \int \frac{v^2+2v-1}{(v+1)(v^2+1)}dv=-\int \frac{dx}{x}\Rightarrow \int \frac{2v(v+1)-(v^2+1)}{(v+1)(v^2+1)}dv=-\int \frac{dx}{x}$
$\Rightarrow \ln (v^2+1)-\ln |v+1|=\ln |c|-\ln |x|\Rightarrow \frac{(v^2+1)|x|}{|v+1|}=|c|\Rightarrow \frac{x^2+y^2}{|y+x|}=|c|$

Given at $x=1,y=1\Rightarrow c=\pm 1.$
Hence the required equation is $x^2+y^2=|x+y|$