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Cauchy Problem with two solutions

Suppose $f:\mathbb{R}^2 \to \Bbb{R}$ is continuous and $t_0, x_0 \in \Bbb{R}$. Prove that if the Cauchy Problem $\begin{cases} \dot{x}=f(t,x) \\ x(t_0)=x_0 \end{cases}$ has two distinct solutions then it has infinitely many solutions.

Proof: Without loss of generality we may assume that $x_0=t_0=0$. Then there exist two solution $x_1,x_2$ of the Cauchy problem such that they are different in a point $a$ which we may assume is greater than $0$. Therfore assume $x_1(a) < x_2(a)$ and denote $P(a,h)$ a point on the segment $a \times (x_1(a),x_2(a))$ we can extend $x$ towards $0$. From Cauchy’s existence theorem, we can see that there exists a solution $x$ around $P$ of the differential equation $\begin{cases} \dot{x}=f(t,x) \\ x(a)=h \end{cases}$. Denote by $K$ the compact determined by $x_1,x_2$ and the line $x=a$. Since $x$ is in that compact in a left neighborhood of $a$, by the compact extension theorem, it can be extended until it reaches the boundary of $K$. From the intersection point of the graph of $x$ with the boundary of $K$ we can go on the graph of $x_1$ or $x_2$ until we reach $(0,0)$ and by the corollary of Lagrange’s theorem the graph we choose is the graph of a solution to the initial differential equation.

Therefore, we have found a solution $x_h$ for all $h \in (x_1(a),x_2(a))$, and therefore the intial equation has uncountably infinitely many solutions.