I'm doing some special relativity exercises. I have to find $x(t)$ and $v(t)$ of a charged particle left at rest in $t=0$ in an external constant uniform electric field $\vec{E}=E_{0} \hat{i}$, then with that velocity I should find the Liénard–Wiechert radiated power.
I will show you what I did but I feel that it is wrong.
We should solve the equation of motion given by
$$ \tag{1}\frac{dp^{\mu}}{d\tau} = \frac{q}{c} F^{\mu \nu}u_{\nu} $$
The four-velocity is given by
$$ u^{\mu} = (u^{0},u^{1},u^{2},u^{3}) = \gamma (c,v^{1},v^{2},v^{3}) $$
where $v^{\alpha}$ are the components of the three-velocity. The four-momentum is
$$ p^{\mu} = mu^{\mu} $$
This will give us four equtions where two of them will give a constant velocities and the other two are
$$ \tag{2}\frac{d\gamma}{d\tau} = -\frac{qE_{0}}{mc^{2}}\gamma v_{1} $$
$$ \tag{3}\frac{d\gamma}{d\tau} v_{1} + \gamma \frac{dv_{1}}{d\tau} = \frac{qE_{0}}{m} \gamma $$
Replacing $(2)$ in $(3)$ gives
$$ \tag{4}\frac{dv_{1}}{d\tau} = -\frac{qE_{0}}{mc^{2}} (v_{1})^{2} + \frac{qE_{0}}{m} $$
The solution of the ODE $(4)$ gives something like
$$ \tag{5}v_{1}(\tau) = A\tanh{(B\tau)} $$
This component of the three-velocity is in terms of the proper time $\tau$ and the problem ask me to find the velocity in terms of the time $t$. So my attempt was to solve
$$ \tag{6}\frac{dt}{d\tau} = \gamma (\tau) = \frac{1}{\sqrt{1 - \frac{(v_{1}(\tau))^{2}}{c^{2}}}} $$
and then replacing this solution for $\tau$ in $(5)$. But the solution of $(6)$ is this. Which doesn't make any sense to me.
I think that I'm misunderstanding something or missing something that will give me a easier solution to this problem. I thought it because in the Liénard–Wiechert radiated power I sould do $dv_{1}/dt$ which is almost impossible to do it without WolframAlpha.
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