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.
 A: 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 $(\ast)$.
A: 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 $(0,0)$.  
A: 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).
A: $$
x^2+y^2 = 2ay
$$
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.
