The curve amongst (x^2-y^2),dx+2xy,dy=0 passing through (1,1) is:

Question

The curve amongst $(x^2-y^2)\,dx+2xy\,dy=0$ passing through $(1,1)$ is:

Accepted Answer

With $y=vx$: solving gives $\dfrac{y^2}{x^2}+1 = \dfrac{c}{x}$, and using $(1,1)$: $c=2$, giving $x^2+y^2-2x=0$ --- a circle centered at $(1,0)$, on the $x$-axis.

Solutions · 1