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

Question

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

Accepted Answer

With $y=vx$: integrating gives $\dfrac{y^2}{x^2}+1=\dfrac{c}{x}$; using $(1,1)$: $c=2$, giving $x^2+y^2-2x=0$ --- a circle of radius 1.

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