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

Answer 1

Answer: (A)

a circle of radius one.

Answer 2

Given $\left(x^{2}-y^{2}\right) d x+2 x y d y =0$
$\Rightarrow \frac{d y}{d x} =-\frac{\left(x^{2}-y^{2}\right)}{2 x y}$
Let $y =v x, \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\int \frac{2 v}{1+v^{2}} d v =-\int \frac{d x}{x}+\ln c$
$\ln \left(1+y^{2}\right) =-\ln x+\ln c=\ln (c / x)$
$1+v^{2} =c / x$
$x^{2}+y^{2} =e x$
It is passing through $(1,1)$ and $c=2$. So, $x^{2}+y^{2}-2 x=0$.
Therefore, the curve is circle.