Decide the smooth function $r : \mathbb R \rightarrow \mathbb R$ of the equation $r(t)^2 + r'(t)^2 = 1$. Suppose $r:\mathbb R \rightarrow \mathbb R$ is a smooth function and suppose $r(t)^2 + r'(t)^2 = 1$.
I want to determine the function $r(t)$.
I see that $r(t)^2 + r'(t)^2 = 1$, so I could take $r(t) = \pm \cos(t), \pm \sin(t)$.
However, are these all the solutions ? How do I see that the list above is exhaustive, if it is ?
 A: This is one of the rare occasions where it actually helps to differentiate the given ODE! In this way we obtain
$$2r'(t)\bigl(r(t)+r''(t)\bigr)=0\qquad\forall t\ .\tag{1}$$
This is satisfied when $r'(t)\equiv0$, or $r(t)=c$ for some $c\in{\mathbb R}$. From the original ODE it then follows that $c^2=1$, so that we obtain the two solutions $$r(t)\equiv\pm1\quad(-\infty<t<\infty)\ .$$ It will turn out that these are "special" solutions.
But there have to be more interesting solutions of $(1)$! If $r(\cdot)$ is a solution with $r'(t_0)\ne0$ for some $t_0$ then in a neighborhood $U$ of $t_0$ this solution has to satisfy the well known ODE
$$r(t)+r''(t)=0\qquad(t\in U)\ .\tag{2}$$
The general solution of $(2)$ is
$$r(t)=A\cos t+B\sin t\tag{3}$$
with $A$,  $B$ arbitrary real constants. From the original ODE it then  follows that necessarily $A^2+B^2=1$, and it is easily checked that this condition is also sufficient for $(3)$ to be a solution. In terms of graphs this means that the solution curves are arbitrary sinus waves of amplitude $1$. The "extra solutions" found earlier are the envelopes of these waves. Note that an IVP with initial point $(t_0,\pm1)$ has several solutions, four different solution germs in all.
