Show the solution of this ODE has to be the trivial one I'm trying to solve this question:

Let $f$ be continuous in $\Omega=\{(t,x);|t|\leq a,|x|\leq b\}$. If
  $f(t,x)\lt 0$ when $tx\gt0$ and $f(t,x)\gt 0$ when $tx\lt0$. Show that $x'=f(t,x), x(0)=0$ has $\varphi=0$ as the unique solution.

I'm completely confused with this question, $x$ is a function or a variable?
I need a hint to begin to solve this question.
Thanks a lot!
 A: Here $x$ is a function of $t$, so e.g. the ODE, written more explicitly, is
$$x'(t)=f\big(t,x(t)\big).$$
To be clear, both sides are functions of $t.$
Here's a hint: suppose instead that $\varphi=0$ is not the unique solution, so that there's a nonzero function $\varphi$ with $\varphi'(t)=f(t,\varphi(t))$ and $\varphi(0)=0.$ Since we're supposing that $\varphi$ is not identically zero, it has to be nonzero somewhere, say at $t^\ast>0$ we have $\varphi(t^\ast)>0$ (but also consider separately the case $\varphi(t^\ast)<0$). From what we know about $f$, we have $\varphi'(t^\ast)<0.$ So in summary, we have a function $\varphi$ which is positive somewhere and is decreasing whenever it is positive. Try to show then that in such a scenario we can't also have $\varphi(0)=0.$
A: Expanding on Youler's answer: 
Assume we have a nonzero solution and $t^*>0$ is a point in which $\phi(t^*)>0$. By assumption this implies $\phi^\prime=f<0$ at $t^*$.
Observe $\phi$ is continuous, since $\phi(t)=\phi(t)-\phi(0)=\int_0^t f(\phi(s),s)\, ds.$
Select $m=\sup \{x<t^* : \phi(x)=0\}$. This exists and is less than $t^*$, otherwise we may select an $x_n\in (t-\frac{1}{n},t)$ such that $\phi(x_n)=0$. Due to continuity, $$0=\lim_{n\to \infty}\phi(x_n)=\phi(t^*)>0$$
which is a contradiction.
Without loss of generality allow $m=0$.
So, by the mean value theorem, there exists a $c\in(0,t^*)$ such that 
$$\phi(t^*)-\phi(0)=\phi^{\prime}(c)(t^*-0),$$
which is a contradiction because the left side is positive and the right is negative. Do the other cases similarly.
