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Find a first-order differential equation having all circles through (1,0) and (-1,0) as integral curves

by RoRi

Find a first-order differential equation with all circles through the points (1,0) and (-1,0) as integral curves.

Circles that go through both the points (1,0) and (-1,0 must have center on the y-axis, at (0,C) say. Then the radius is given by

    \[ r = \sqrt{1+C^2}. \]

Therefore, they all satisfy the equation

    \[ x^2 + (y-C)^2 = 1+C^2. \]

Differentiating both sides of this equation with respect to x we have,

    \[ x^2 + (y-C)^2 = 1+C^2 \quad \implies \quad 2x + 2(y-C)y' = 0. \]

From the original equation we can also solve for the constant,

    \[ x^2 + (y-C)^2 = 1+C^2 \quad \implies \quad C = \frac{x^2 + y^2 - 1}{2y}. \]

Therefore we have,

    \begin{align*} && 2x + 2(y-c)y' &= 0 \\ \implies && x + yy' - \left( \frac{x^2+y^2-1}{2y} \right)y' &= 0\\ \implies && 2xy + 2y^2 y' - (x^2+y^2-1)y' &= 0\\ \implies && (y^2 - x^2 + 1)y' + 2xy &= 0 \\ \implies && (x^2-y^2-1)y' - 2xy &= 0. \end{align*}

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