Question

The solution curve of $\frac{dy}{dx}=\frac{y^2-2xy-x^2}{y^2+2xy-x^2}$, $y\left(-1\right)=1$ is

• A. a straight line
• B. parabola
• C. circle
• D. ellipse

Answer 1

Answer: (A)

Substitute $y=vx$
$\Rightarrow \frac{dy}{dx}=\frac{xdv}{dx}+v$
Now, given equation becomes
$\frac{xdv}{dx}+v=\frac{v^2-2v-1}{v^2+2v-1}$
$\Rightarrow \frac{xdv}{dx}=\frac{v^2-2v-1}{v^2+2v-1}-v=-\frac{\left(v^2+1\right)\left(v+1\right)}{v^2+2v-1}$
$\Rightarrow \frac{dx}{x}=\frac{-\left(v^2+2v-1\right)dv}{\left(v^2+1\right)\left(v+1\right)}=-\left(\frac{2v}{v^2+1}-\frac{1}{v+1}\right)dv$
$\Rightarrow lnx+lnc=ln\left(\frac{v+1}{v^2+1}\right)$
$\Rightarrow xc=\frac{v+1}{v^2+1}$
$\Rightarrow xc=\frac{x\left(x+y\right)}{x^2+y^2}$
$\Rightarrow c\left(x^2+y^2\right)=x+y$
$x=1$,
$y=1$
$\Rightarrow c=0$
$\Rightarrow x+y=0$.