TITLE: Does non-uniqueness of solution to 1st order ODE implies the existence of infinitely many solutions?
QUESTION [6 upvotes]: Consider the following initial value problem (IVP) to the first order ODE: 
$$\tag{1} \dot x = f(t, x), \ \ \ x(t_0) = x_0.$$
The Existence and Uniqueness Theorem describes when one has exactly one solutions. This is true (e.g.) when $f$ is continuously differentiable. 
There are examples where uniqueness fails and existence fails (here or here): 
In all of the examples where I am aware of, whenever one has more than one solutions, one actually has infinitely many. Hence my question: 
Can an IVP has more than one solutions, but only finitely many solutions?

REPLY [7 votes]: The answer is positive: in general (apart of some artificial examples) if an IVP for an ODE
$$
\dot x=f(t,x),\quad x(t)\in \mathbb R^n
$$
 have non unique solution it implies that there are uncountably many solutions. In these general setting this is a very nontrivial theorem which can be found in Hartman's book (Kneser's theorem). If, however, you are dealing with an ODE
$$
\dot x=f(t,x),\quad x(0)=x_0,
$$
where $x(t)$ is one-dimensional, it is a good (and simple) exercise to prove that if there are two solutions to this problem then there are infinitely (uncountably) many solutions.