If we have two solutions to an ODE, we will have an infinity ones? Let $\dot x=f(t,x)$, $x(t_0)=x_0$ be an initial value problem with $f$ continuous. If this problem has two solutions, we can say it has an infinity ones?
It's true to linear case, because every linear combination of the solutions is a solution of the problem, but I would like to know if this is valid in the general case.
Thanks in advance
 A: Yes. Let $x$ and $y$ be two distinct solutions: that is, $x(t_0)=y(t_0)=x_0$ and $x(t_1)\ne y(t_1)$ for some $t_1>t_0$. Let $z$ be a solution of the initial value problem $z(t_1)=\alpha$ where $\alpha$ is strictly between $x(t_1)$ and $y(t_1)$. Moving backward in time from $t_1$, we observe that the solution $z$ exists as long as it stays between $x$ and $y$. At some moment $T\in [t_0,t_1)$ we must have $z(T)\in \{x(T),y(T)\}$. For definiteness assume $z(T)=x(T)$. Then the function 
$$\tilde z(t)=\begin{cases} x(t)\quad &t_0\le t\le T \\ z(t) & T\le t\le t_1 \end{cases}$$
solves the ODE and satisfies $\tilde z(t_0)=x_0$, $\tilde z(t_1)=\alpha$. Since there are uncountably many choices of $\alpha$, we have uncountably many solutions of the initial value problem.
That said, it is possible to have finitely many germs of solutions at $t_0$. For example, $\dot x=\frac32x^{1/3}$ with $x(0)=0$ has two germs:


*

*solutions that are $0$ in some neighborhood of $t=0$

*the solution $x(t)=t^{3/2}$ (with $x(t)=0$ for $t<0$)

