The question is from Schaum's Theoretical Mechanics.
The electric field is given by $\underline E=E\hat k$
The magnetic field is given by $\underline B=-B\hat j$
Prove that the motion of a charged particle of charge $q$ and mass $m$ that starts from rest at the origin is given by
$x=b(\theta-\sin \theta), z=b(1-\cos \theta)$
$\theta=\frac{qBt}{m}, b=\frac{mE}{qB^2}$
My attempt:
The Lorentz force is given by
$\underline F=q(\underline E+\underline v\times\underline B)$
$\to\underline F=-qBv_3\hat i+q(E-Bv_1)\hat k$
Let $\underline v=v_1 \hat i+v_2\hat j+v_3\hat k$, where $\frac{dx}{dt}=v_1, \frac{dy}{dt}=v_2, \frac{dz}{dt}=v_3$
$\to\frac{dv_1}{dt}=-\frac{qBv_3}{m}$ equation 1
$\frac{dv_3}{dt}=\frac{q}{m}(E-Bv_1)$ equation 2
$\to \frac{dv_3}{dv_1}=\frac{Bv_1-E}{Bv_3}$
$\to Bv_3dv_3=(Bv_1-E)dv_1$
$\to \frac{1}{2}Bv_3^2=\frac{1}{2}Bv_1^2-Ev_1$
$\to v_3=\sqrt (v_1^2-\frac{2E}{B}v_1)$
Then, substituting the right hand side for $v_3$ into equation 2,
$\frac{Bv_1-E}{B\sqrt(v_1^2-\frac{2E}{B}v_1)}\frac{dv_1}{dt}=\frac{q}{m}(E-Bv_1)$
$\to \int \frac{1}{\sqrt(v_1^2-\frac{2E}{B}v_1)}dv_1=\int -\frac{qB}{m}dt$
$\to \int \frac{1}{\sqrt((v_1-\frac{E}{B})^2-\frac{E^2}{B^2})}dv_1=-\frac{qB}{m}t+c$
Let $v_1-\frac{E}{B}=\frac{E}{B}\sec \theta$
$\int \sec \theta d\theta= -\frac{qB}{m}t+c$
$\ln(\sec \theta+\tan \theta)=-\frac{qB}{m}t+c$
$\ln(\frac{Bv_1-E+\sqrt(B^2v_1^2-2BEv_1)}{E})=-\frac{qB}{m}t+c$
When $t=0, v_1=0$, so $c=\ln(-1)$. Then, the equation becomes
$\ln(\frac{Bv_1-E+\sqrt(B^2v_1^2-2BEv_1)}{E})=-\frac{qB}{m}t+\ln(-1)$
$\to \ln(\frac{E-Bv_1-\sqrt(B^2v_1^2-2BEv_1)}{E})=-\frac{qB}{m}t$
$\to E-Bv_1-\sqrt(B^2v_1^2-2BEv_1)=Ee^\frac{-qBt}{m}$
$B^2v_1^2-2BEv_1=(Bv_1-E(1-e^\frac{-qBt}{m}))^2$
$B^2v_1^2-2BEv_1=B^2v_1^2-2BEv_1+2BEv_1e^\frac{-qBt}{m}+E^2(1-e^\frac{-qBt}{m})^2$
$\to 2BEe^\frac{-qBt}{m}v_1=-E^2(1-e^\frac{-qBt}{m})^2$
$\to v_1=\frac{-E^2(1-e^\frac{-qBt}{m})^2}{2BEe^\frac{-qBt}{m}}$
But from this, I obtain that
$x=b\theta-b(\frac{e^\theta-e^{-\theta}}{2})$
I would be extremely grateful if someone could spot my mistake. I think I need some $i$'s somewhere to obtain $x=b\theta-b\sin \theta$, but I don't know where.
Thank you.