Cross posted from Physics Stack Exchange because this seems to mostly be a mathematical problem.
Consider a particle such that at $t=0$ (time), $x=0$ (x-coordinate) and $v=0$ (speed). Let $a(x)=x^2$ (acceleration).
Intuitively speaking, the particle should forever be at origin. It is at origin initially, and won't move till it acquires a speed. It won't acquire a speed since at origin, acceleration would be 0.
However, $a(x)=x^2$ can be solved to get $x(t)=\cfrac{6}{t^2}$.
What mistake am I making here?
While $x = \frac{6}{t^2}$ is a solution to the DE, it does not satisfy the given initial conditions. Notice that it's not even defined at $t = 0$, so it's impossible for $x(0) = \dot{x}(0) = 0$ to be true.
On the other hand, $x = 0$ is a solution to the DE, and it does satisfy the initial conditions. You can verify this directly through substitution.
In this system, $x = 0$ is an equilibrium solution, meaning that it's a valid solution to the DE that results in no movement. But it's an unstable equilibrium, meaning that a small perturbation to the system - e.g. starting the particle at $x = \epsilon$, or with a non-zero velocity - will cause it to travel on a completely different trajectory. This is different to stable equilibria where small perturbations give a solution that tries to return to rest at that point (it's the difference between balancing a boulder at the peak of a mountain versus the bottom of a valley).