Showing that a function is not a solution to $y'=-\frac{x}{y}$ Consider the equation $x^2 + y^2 = 25$ of a circle with radius 5. The differential equation of this circle is 
$$y' = -\frac{x}{y}$$
Consider the piecewise defined function
$$\phi(x) = \begin{cases} 
      \sqrt{25-x^2},& -5 < x < 0 \\
      -\sqrt{25-x^2}, & 0 \leq x < 5 \\
   \end{cases}$$
I want to show that $\phi$ is not a solution to the differential equation. 
I know that a function $\psi$ defined on an interval $I$ and possessing at least $n$ derivatives that are continuous on $I$ is a solution to an $n$th order differential equation $F(x,y,y',\dotsc, y^{(n)})=0$, where $F$ is a real valued function of $n+2$ variables, if $F(x,\psi(x), \psi'(x),\dotsc , \psi^{(n)}(x)) = 0$.
Because $y'=-\frac{x}{y}$, we require $\phi$ to be differentiable on the interval of existence $(-5,5)$. But $\phi$ is discontinuous at $x=0$ and, therefore, not differentiable at this point, which means that $\phi'$ is not a derivative on the interval of existence $(-5,5)$ at the point $x=0$. Therefore, $\phi$ cannot be a solution to the differential equation $y' = -\frac{x}{y}$.
Is my reasoning correct?
 A: Suppose $\phi$ is a solution. Then:
$$\begin{align}
y'&=-\frac{x}{y}\\
\implies
\phi'(x)&=-\frac{x}{\phi(x)}\\
\implies
\phi'(0)&=-\frac{0}{\phi(0)}\\
\implies
\phi'(0)&=-\frac{0}{-5}\\
\implies
\phi'(0)&=0
\end{align}$$
But looking directly at your formula for $\phi$, $\phi$ is discontinuous at $0$. It jumps from $5$ to $-5$ instantaneously. Since is discontinuous, it is not differentiable there. So $\phi'(0)$ does not exist.
This contradiction implies $\phi$ is not a solution after all.
A: In $x^2+y^2=25$, it can be true that $y=0$. But when you write
$$
y'=-\frac{x}{y}\tag{1}.
$$
you are assuming implicitly that $y\neq 0$. Hence "The differential equation of this circle" would be incorrect in the first place. 

You are right that $\phi$ is not continuous on the whole interval $(-5,5)$ and thus is not a solution to (1) on $(-5,5)$. 
A: The sentence $\phi$ is a solution of the differential equation on the interval $(-5, 5)$ is false because this sentence implies the false statement that $\phi$ has at least one derivative on $(-5, 5)$ which in turn implies $\phi$ is continuous on $(-5, 5)$ which implies $\phi$ is continuous at $x=0$. Your reasoning is correct.
A: The differential equation for a circle with the center at the origin has the form 
$$ 
0 = F(x, y, y') = x / y + y' \quad (x \in I) \quad (*)
$$
if $I$ does not contain $\pm R$, where $R$ is the radius of the circle.
The function $u$ is a solution of $(*)$ on some interval $I\subset \mathbb{R}$ if $u$
is differentiable on $I$ and $$ 0 = F(x, u(x), u'(x)) \quad (x \in I)
$$
Here $R=5$, $I = (-5,5)$ and $\phi(x)$ is not differentiable on $I$, as it is not differentiable at $x = 0 \in I$. So it is no solution in the above sense.
