Question

The curve satisfying the differential equation, $(x^2-y^2)dx+2xydy=0$ and passing through the point $(1,1)$ is :

• A. a circle of radius two.
• B. a circle of radius one.
• C. a hyperbola.
• D. an ellipse.

Answer 1

The correct option is B a circle of radius one.

$(x^2-y^2)dx+2xydy=0$

$\Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}$

Put $y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{v^2{x}^2-x^2}{2v{x}^2}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{v^2-1}{2v}$

$\Rightarrow x\frac{dv}{dx}=\frac{-v^2-1}{2v}$

$\Rightarrow \frac{2vdv}{v^2+1}=-\frac{dx}{x}$

Integrating we get;

$ln|v^2+1|=-ln\left|x\right|+lnc$

$\frac{y^2}{x^2}+1=\frac{c}{x}$

Putting (1,1)

$c=2$

$x^2+y^2-2x=0$

hence its is a circle of radius 1