Relation between speed and potential I'm studying newtonian dynamical systems and they can be described by the differential equation
$$1)\space  m\ddot{x} = F(x)$$
supposing $F$ sufficiently regular we could define the potential $V$ as its primitive so
$$2)\space F(x) = -{dV\over dx}$$
We can also define the energy of the system $E$ as
$$3)\space E = \frac{1}{2}mv^2 + V(x)$$
as a results we have
$$4) \space v = \sqrt{\frac{2}{m}(E-V)}$$
What confuses me is that $(1)$ could also be described by the pair of equations
$$\dot{x} = v,\space \space \dot{v} = \frac{F(x)}{m}$$
for the sake of simplicity take $m=1$ and by the second equation above I think that
$$v(x) = -\int_{x_0}^{x}F(\zeta)d\zeta = V(x_0)-V(x)$$
So how $4)$ and the last equation could be the same thing? I was thinking about that and the conclusion I came is that I could not write the last relation, can't I? Every clarification is welcome, because in this moment I have a lot of confusion in my head. Thanks!
 A: Consider a simple example: the gravitational potential $V = -mgx$ where $x$ is the height.
Then $$F = -\dfrac{\text{d}V}{\text{d}x} = mg$$ which is indeed the weight force, as we expect.
In this way, an object subject to the gravitational potential, is free falling with a velocity $v = -gt$, indeed:
$$v(t) = \int_0^t \dfrac{F}{m}\ \text{d}t = - \int_0^t \dfrac{1}{m}\dfrac{\text{d}V}{\text{d}x}\ \text{d}t = -\int_0^t \dfrac{1}{m} mg\ \text{d}t = -gt$$
A: I would like to emphasize the fundamental relation
\begin{equation}
\text{d}x = v\,\text{d}t.
\end{equation}
The crucial point is
\begin{equation}
\dot{v} = \frac{\text{d}v}{\text{d}t} \ne \frac{\text{d}v}{\text{d}x}.
\end{equation}
And so you obtain for a conservative force $F(x)$
\begin{align}
\frac{\text{d}v}{\text{d}t} &= \frac{F(x)}{m} \\
\text{d}v &= \frac{F(x)}{m} \frac{\text{d}x}{v} \\
v\,\text{d}v &= \frac{F(x)}{m} \text{d}x \\
\int v\,\text{d}v &= \frac{1}{m} \int F(x)\, \text{d}x\\
\frac{m}{2}v^2 - \frac{m}{2}v_0^2 &= - V(x) + V_0
\end{align}
