Relativistic particle moving in a potential

[PhysicsRock]

Homework Statement

Consider a relativistic particle of mass $m_0$ in $(1+1)$-spacetime dimensions. In an inertial frame of reference with spacetime coordinates $(t,x)$, we define the potential as $V(x) = \alpha x$ with $\alpha > 0$. The particle is at rest at time $t=0$ and it's position is $x(0) = 0$. Determine the trajectory $x(t)$ of the particle.
Hint: Use the conservation of total energy.

Relevant Equations

$E_\text{tot} = \sqrt{ c^2 p^2 + m_0^2 c^4 } + \alpha x$

Since energy is conserved and the particle is initially at rest, we can determine that $E(0) = m_0 c^2$, so

$$
m_0 c^2 = \sqrt{ c^2 p^2 + m_0^2 c^4 } + \alpha x.
$$

Squaring this eqation gives

$$
m_0^2 c^4 = \alpha^2 x^2 + c^2 p^2 + m_0^2 c^4 + 2 \alpha x \sqrt{ c^2 p^2 + m_0^2 c^4 }
\Rightarrow 0 = \alpha^2 x^2 + c^2 p^2 + 2 \alpha x ( E - \alpha x ).
$$

Using $p = \gamma m_0 \dot{x}$, I was able to simplify this equation to

$$
0 = -\alpha^2 x^2 + \frac{E^2 \dot{x}^2}{c^2 - \dot{x}^2} + 2 \alpha E x
$$

This is the point where I'm stuck. I have doubled checked and I'm pretty sure that this final expression is correct, however, I cannot guarantee that it actually is. If it is, I have no clue how to solve this equation.

 

[renormalize]

$$0 = -\alpha^2 x^2 + \frac{E^2 \dot{x}^2}{c^2 - \dot{x}^2} + 2 \alpha E x$$

This expression is correct, and can be solved via integration using the following steps:

1. Solve for $\dot{x}$ to get the form $\dot{x}\equiv\frac{dx}{dt}=f\left(x\right)$.
2. Rewrite this as $\frac{dx}{f\left(x\right)}=dt$.
3. Integrate both sides.

 

[PhysicsRock]

This expression is correct, and can be solved via integration using the following steps:

1. Solve for $\dot{x}$ to get the form $\dot{x}\equiv\frac{dx}{dt}=f\left(x\right)$.
2. Rewrite this as $\frac{dx}{f\left(x\right)}=dt$.
3. Integrate both sides.

Thank you for the hint. I'll try to solve it.