Find a differential equation whose solutions are the circles $x^2+y^2=2ay$

I had to find differential equation of the family of circles which touch the $$x$$-axis at origin. Centering the variable circle at $$(0,a)$$, its equation will be $$x^2+y^2=2ay .$$ Solving conventionally by differentiating once and substituting the value of a in the equation of circle, the differential equation is $$(x^2-y^2)y_1=2xy$$ Now I might be missing some concept because I've just started with ODE, but what is wrong with the following process? Differentiating the circle equation $$x+yy_1=ay_1 ,$$ dividing both sides by $$y_1$$, and differentiating both sides gives the equation $$y_1-xy_2+{y_1}^3=0 .$$ Does this not represent the family of circles? In that case the differential equations are of different orders.

• If I understand your notation, you are writing $y_1$ for $y'=\frac{dy}{dx}$ and $y_2$ for $y'' = \frac{d^2y}{dx^2}$? Aug 21 '19 at 16:25
• Yes you are right Aug 21 '19 at 16:27

An ODE of order $$n$$ has a solution with $$n$$ independent parameters. So your second equation represents a double family of circles, not just the given one (you might differentiate once again, and describe a triple family…).

For a quick solution, write

$$\frac{x^2+y^2}{2x}=a$$ and differentiate, giving

$$(2x+2yy')2x-(x^2+y^2)2=0$$ (denominator omitted).

Differentiating an equation in general discards information.

A first-order differential equation $$\phantom{(\ast)} \qquad F(x, y, y_1) = 0 \qquad (\ast)$$ (with nice $$F$$) has a $$1$$-parameter family of solutions. And in the example in the question, this is exactly what we want to produce---a first-order differential equation whose solutions are precisely the given family of curves. This is exactly what we achieve when isolating $$a$$ as $$\frac{x^2}{2 y} + \frac{y}{2} = a$$ and differentiating w.r.t. $$x$$ to produce (after clearing denominators) a first-order o.d.e. $$(x^2 - y^2) y_1 = 2 x y$$ that does not depend on $$a$$, that is, for which every equation $$x^2 + y^2 = 2 a y$$ is a solution.

Regardless of whether $$F$$ in $$(\ast)$$ depends on a parameter like $$a$$, however, differentiating $$(\ast)$$ gives a second-order o.d.e., $$\frac{d}{dx}F(x, y, y_1) = 0,$$ that is, $$\phantom{(\ast\ast)} \qquad \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} y_1 + \frac{\partial F}{\partial y_1} y_2 =0 \qquad (\ast\ast)$$ which (again under nice conditions) has a $$2$$-parameter family of solutions. So, many---in a sense that can be made precise, most---solutions of $$(\ast\ast)$$ are not solutions of $$(\ast)$$. So, in general producing a second-order equation---if you like, differentiating twice---means we have produced a differential equation with too many solutions to satisfy our original prescription.

As a toy example if we want to find a differential equation whose solutions are (all) the horizontal lines $$y = c ,$$ the parameter $$c$$ is already isolated, and differentiating gives the desired equation $$(\ast)$$: $$y_1 = 0 .$$ But differentiating again gives $$y_2 = 0$$ (our $$(\ast\ast)$$), whose solutions are precisely the affine functions, $$y(x) = a x + b .$$ In particular, the solutions of $$(\ast\ast)$$ with $$a \neq 0$$ are not solutions of the original differential equation $$(\ast)$$.

It's not so much that the two equations are of different orders. We could differentiate both sides of the first equation and get a second-order equation. The difference is that you have treated $$a$$ differently in the two processes.

In the first, $$a$$ is a parameter which depends on $$x$$ and $$y$$ (Each point on the plane, other than the origin, is on exactly one circle tangent to the $$x$$-axis at the origin). In the second, you're treating $$a$$ like a constant.

If you solve the second equation (before you differentiate a second time) by separating variables, you get \begin{align*} \frac{x}{y'} + y &= a \\ \implies x &= (a-y) y' \\ \implies x \,dx &= (a-y)\,dy \\ \implies x^2 &= -(a-y)^2 + C \\ \implies x^2 + (a-y)^2 &= C \end{align*} So for each $$a$$, the family of solutions is the set of circles centered at $$(0,a)$$. That's quite different from the family of circles tangent to the $$x$$-axis at $$(0,0)$$.

• So the order two differential equation I got at the end is invalid ? Aug 21 '19 at 17:14
• Like Yves says, if you differentiate twice, the solution set is a two-parameter family, also not what you want. Aug 21 '19 at 17:19
• Sorry, this answer is simply wrong. Not every circle centered at $(0,a)$ for some value of $a$ touches the $x$ axis at the origin. Aug 21 '19 at 18:23
• That's my point. OP asked what was wrong with his proposed solution, and I wanted to point out that it doesn't solve the problem he was asked. I'll edit the answer to underscore. Aug 22 '19 at 14:18

$$x^2+y^2 = 2ay$$ Take the derivative wrt. $$x$$ $$2x + 2yy' = 2ay' \implies y' = \frac{x}{a-y}$$ This will have one constant too many; eliminate that by eliminating $$a$$ using $$x^2+y^2 = 2ay \implies a = \frac{x^2+y^2}{2y}$$ So $$y' = \frac{x}{a-y} = \frac{x}{\frac{x^2+y^2}{2y}-y} = \frac{2xy}{(x^2+y^2)-2y^2}$$

$$y' =\frac{2xy}{x^2-y^2}$$

As a check, note that when $$y = \pm x$$ the slope $$y'$$ is infinite, as woould be expected since a $$45^\circ$$ line from the origin intersects the circle at a place where the tangent is vertical.