3D E and B Fields (Corrected)

I'm very impressed by the solution provided by @Karan Chatrath, in which a linear system is solved to get the Laplace equation for each dimension, and then the inverse transformations are applied. For the sake of completeness, I will upload the numerical solution as well. This results in the same trajectories and final answer as the Laplace method.

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import math

m = 2.0
q = 1.0

dt = 10.0**(-5.0)

##########################

Ex = 1.0
Ey = -2.0
Ez = 3.0

Bx = 4.0
By = 8.0
Bz = -2.0

##########################

t = 0.0
count = 0

x = 0.0
y = 0.0
z = 0.0

xd = -1.0
yd = 1.0
zd = 2.0

xdd = 0.0
ydd = 0.0
zdd = 0.0

ymax = 0.0

##########################

#print "t x y z r"

while t <= 5.0:

    x = x + xd*dt
    y = y + yd*dt
    z = z + zd*dt

    xd = xd + xdd*dt
    yd = yd + ydd*dt
    zd = zd + zdd*dt

    r = math.sqrt(x**2.0 + y**2.0 + z**2.0)

    FEx = q*Ex
    FEy = q*Ey
    FEz = q*Ez

    crossx = yd*Bz - zd*By
    crossy = -(xd*Bz - zd*Bx)
    crossz = xd*By - yd*Bx

    FBx = q*crossx
    FBy = q*crossy
    FBz = q*crossz

    Fx = FEx + FBx
    Fy = FEy + FBy
    Fz = FEz + FBz

    xdd = Fx/m
    ydd = Fy/m
    zdd = Fz/m

    t = t + dt
    count = count + 1

    if y > ymax:
        ymax = y

    if count % 1000 == 0:
        print t,x,y,z,r

##########################

print ""
print ""

print dt
print t
print ""
print x
print y
print z
print ""
print r

#>>> 
#0.0001
#5.0

#-7.03230936686
#-9.80703251134
#2.95387626591

#12.4240350412
#>>> ================================ RESTART ================================
#>>> 
#1e-05
#5.00000999998

#-7.03189266663
#-9.80753140053
#2.95597656461

#12.4246925152
#>>> 

Applying Newton's second law gives:

$m\vec{a} = q\vec{E} + q\left( \vec{v} \times \vec{B}\right)$

$2\left[\begin{matrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \end{matrix}\right] = \left[\begin{matrix} 1 \\ -2 \\ 3 \end{matrix}\right] + \left( \left[\begin{matrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{matrix}\right] \times \left[\begin{matrix} 4 \\ 8 \\ -2 \end{matrix}\right]\right)$

Simplifying gives:

$\ddot{x} = \frac{1}{2} - 4 \dot{z} - \dot{y}$ $\ddot{y} = -1 + \dot{x} + 2\dot{z}$ $\ddot{z} = \frac{3}{2} + 4 \dot{x} - 2\dot{y}$

Now, knowing the equations of motion and the initial conditions, the Laplace transform can be taken on both sides, which converts the above differential equations to linear equations in the s-domain. To perform the Laplace transform, one may search the internet for a 'table of standard Laplace transforms'. The equations are:

$s^2 X + sY+4sZ = \frac{1}{2s}-1$ $-s X + s^2Y-2sZ = -\frac{1}{s}+1$ $-4sX + 2sY+s^2Z = \frac{3}{2s}+2$

Here, $X$ , $Y$ and $Z$ are the laplace transforms of the functions $x(t)$ , $y(t)$ and $z(t)$ . Solving for $X$ , $Y$ and $Z$ gives:

$X = -\frac{2\,s^3+17\,s^2+10\,s+18}{2\,s^3\,\left(s^2+21\right)}$ $Y = \frac{2\,s^3+4\,s^2+7\,s-36}{2\,s^3\,\left(s^2+21\right)}$ $Z = \frac{4\,s^3-9\,s^2+8\,s+9}{2\,s^3\,\left(s^2+21\right)}$

Now, all that needs to be done is the computation of the inverse Laplace transform of the above expressions. This can be done by using:

https://www.wolframalpha.com/widgets/view.jsp?id=7c762190486dfb47dca59a9a1f8cb1a8

The resulting solutions of the differential equations are:

$x(t) = \frac{113\,\cos\left(\sqrt{21}\,t\right)}{294}-\frac{5\,t}{21}-\frac{16\,\sqrt{21}\,\sin\left(\sqrt{21}\,t\right)}{441}-\frac{3\,t^2}{14}-\frac{113}{294}$ $y(t) = \frac{t}{6}-\frac{20\,\cos\left(\sqrt{21}\,t\right)}{147}+\frac{5\,\sqrt{21}\,\sin\left(\sqrt{21}\,t\right)}{126}-\frac{3\,t^2}{7}+\frac{20}{147}$ $z(t) = \frac{4\,t}{21}+\frac{11\,\cos\left(\sqrt{21}\,t\right)}{49}+\frac{38\,\sqrt{21}\,\sin\left(\sqrt{21}\,t\right)}{441}+\frac{3\,t^2}{28}-\frac{11}{49}$

Substituting $t=5$ and simplifying and computing the distance between the origin and the point of interest, gives the required answer. The trajectory of the particle can be studied as such: