I want to find a differential equation, which has as solutions circles of radius $r$. And the equation needs to be of the form $F(y'',y',y)=0$.
The equation $$(x-x_0)^2 +(y-y_0)^2 = r$$
describes a circle of radius $r$ centered at the point $(x_0,y_0)$. If we differentiate this relation twice with respect to $x$, we have the equation $$2+2y''(y-y_0)+2(y')^2=0$$
which is in the correct form. I think this is too simple to be correct. Any suggestions?
Best Answer
Your differential equation is satisfied by any circle of any radius whose centre has $y$ coordinate $y_0$. I would interpret the question as saying that you want $x_0$ and $y_0$ to be arbitrary but $r$ to be fixed. So you want to take your original equation and its first and second derivatives and eliminate $x_0$ and $y_0$. I get
$$ (y')^6 + 3 (y')^4 - r^2 (y'')^2 + 3 (y')^2 + 1 = 0$$