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Homework Help: Find the general solution of a coupled differential equation:

  1. Sep 2, 2010 #1
    1. The problem statement, all variables and given/known data

    I want to find the general solution of these two equations,


    [tex]\ddot{z}=\omega\left(\frac{\mathbf{E}}{\mathbf{B}} - \dot{y}\right)[/tex]

    2. Relevant equations
    These two equations are the result of quantitatively solving to find the trajectory of a charged particle in a uniform electric field, perpendicular to a magnetic one,

    Suppose that the B field points in the x direction, and the E field in the z direction, and the particle is released from rest,

    the lorentz force law,
    [tex] \mathbf{F}_{mag} = Q(\mathbf{v} \times \mathbf{B}) [/tex]
    3. The attempt at a solution

    (i'm just going to show you how I got to the coupled differential equations

    there is no force in the x direction due to the magnetic field being in that direction,

    the position of the particle at t can be described by the vector, (0, y(t), z(t))

    so [tex] \mathbf{v}=(0,\dot{y},\dot{z}) [/tex]
    (the dots are the derivatives with respect to time)
    calculating the cross product from the Lorentz force law,

    [tex] \mathbf{v} \times \mathbf{B} = \left| \begin{array}{ccc} \mathbf{\hat{x}} & \mathbf{\hat{y}} & \mathbf{\hat{z}} \\ 0 & \dot{y}& \dot{z} \\ \mathbf{B}& 0& 0 \\ \end{array} \right| =B \dot{z}\mathbf{\hat{y}} - B\dot{y}\mathbf{\hat{z}} [/tex]

    and applying newtons second law, F=ma=QE=Q(v\timesB)=Q(E+(v\timesB))
    [tex] \mathbf{F} = Q(\mathbf{E}+(\mathbf{v}\times\mathbf{B})) = Q(E\mathbf{\hat{z}} + B \dot{z}\mathbf{\hat{y}} - B\dot{y}\mathbf{\hat{z}} = m\mathbf{a} = m(\ddot{y}\mathbf{\hat{y}}+\ddot{z}\mathbf{\hat{z}} [/tex]

    separating the [tex] \hat{y} & \hat{z} [/tex] components

    [tex] \mathbf{QB}\dot{z} = m\ddot{y} [/tex] & [tex] \mathbf{QE - QB}\dot{y} = m\ddot{z} [/tex]

    substituting in the cyclotron frequency formula, [tex] \omega = \frac{\mathbf{QB}}{m} [/tex]

    I get the two coupled differential equations that I showed earlier,

    1: [tex]\ddot{y}=\omega\dot{z}[/tex]

    2: [tex]\ddot{z}=\omega\left(\frac{\mathbf{E}}{\mathbf{B}} - \dot{y}\right)[/tex]

    I want to figure out the general solution, but I can never get it,
    I'm supposed to Differentiate the first function and then use the second to eliminate [tex] \ddot{z} [/tex]


    well if I differentiate the first equation,

    I get a triple dot y, [tex] \dot{\ddot{y}} = w\ddot{z} [/tex]
    Is that how I differentiate the first equation?
    Last edited: Sep 3, 2010
  2. jcsd
  3. Sep 3, 2010 #2

    Yes. And you substitute the value of [tex] \ddot{z} [/tex] from second equation there.
  4. Sep 3, 2010 #3

    [tex] \dot{\ddot{y}} = \omega \omega(\frac{\mathbf{E}}{\mathbf{B}}-\dot{y})=\omega^2 (\frac{\mathbf{E}}{\mathbf{B}}-\dot{y}) [/tex]

    [tex] \omega^2 = (\frac{QB}{m})^2 [/tex]

    [tex] \mathbf{QE - QB}\dot{y} = mw(\frac{\mathbf{E}}{\mathbf{B}}-\dot{y}) = QB*(\frac{\mathbf{E}}{\mathbf{B}}-\dot{y} [/tex]

    I don't understand what to do from here
  5. Sep 3, 2010 #4
    Differentiate 2nd equation. Plug in y[tex]^{..}[/tex] from first equation. It looks like a sinusoidal equation, with z[tex]^{.}[/tex] as your variable. Find z[tex]^{.}[/tex] from here, and z. Plug it back in the equations and similarly find y.

    The approach you are using, certainly works. Its just slightly messier.
  6. Sep 3, 2010 #5
    ok if equation2= [tex] \ddot{z} = \omega (\frac{E}{B} - \dot{y}) [/tex]
    then expanding

    [tex] \ddot{z} = \omega (\frac{E}{B}) - \omega(\dot{y}) [/tex]

    and differentiating

    [tex] \dot{\ddot{z}} =\frac{E}{B} - \omega(\ddot{y}) [/tex]

    then plugging in [tex] \ddot{y} [/tex] from the first equation

    [tex] \dot{\ddot{z}} = \frac{E}{B} - \omega(\omega\dot{z}) = \frac{E}{B}-\omega^2\dot{z} [/tex]

    I don't see how this looks like a sinusoidal equation?

    so to get rid of those dots, I integrate?
  7. Sep 3, 2010 #6


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    Science Advisor

    Well, you can't just integrate. Do you know anything at all about solving "linear differential equations with constant coefficients"?
  8. Sep 3, 2010 #7
    Well I know how to integrate, and i've dealed with hard differential/coupled equations in physics before, but haven't started proper, differential equations in mathematics yet, I don't think?
    is it just the same as integrating normal linear equations?
    Last edited: Sep 3, 2010
  9. Sep 3, 2010 #8
    So it's essentially an ODE with constant coeff
    [tex]y'''+\omega ^2y'=K[/tex]
    Perhaps the easiest way to solve it is using LapLace transform. To make life further easier, assuming you have the system completely at rest for t=0, i.e. y(0)=y'(0)=y''(0). Take LT of both sides and rearrange one would get:

    So the solution is a ramp minus a sinusoid.
  10. Sep 4, 2010 #9


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    Science Advisor

    I really dislike the use of the Laplace transform for problems like this!

    If you were to try a solution of the form [itex]y= e^{rx}[/itex], then you would have [itex]y'= re^{rx}[/itex], y"=[itex]r^2e^{rx}[/itex], [itex]y'''= r^3e^{rx}[/itex] and the "associated homogeneous equation" (with the "K" dropped) becomes [itex]r^3e^{rx}+ \omega^2 r e^{rx}= (r^3+ \omega r)e^{rx}= r(r^2+ \omega)e^{rx}= 0[/itex]. Since [itex]e^{rx}[/itex] can't be 0, we must have [itex]r(r^2+ \omega)= 0[/itex] which gives r= 0 and [itex]r= \pm i\sqrt{\omega}[/itex]. Of course, [itex]e^{ix}= cos(x)+ i sin(x)[/itex] so we can write [itex]e^{it\sqrt{\omega}}[/itex] and [itex]it\sqrt{\omega}}[/itex] in terms of [itex]cos(t\sqrt{\omega})[/itex] and [itex]sin(t\sqrt{\omega})[/itex]. Of course, [itex]e^{0t}= 1[/itex]: the general solution to the associated homogeneous equation is
    [tex]C+ Dcos(t\sqrt{\omega})+ Esin(t\sqrt{\omega})[/tex].

    To find the solution to the entire equation, try a constant solution, y= A, and add to that previous solution.
  11. Sep 4, 2010 #10
    What should I do here Ivy? I think I'll just go down to the physics department on tuesday and ask one of the tutors to help me, I don't know how they expect me to solve this, sad thing is, I think everyone else in my class can do this really quick, but I can't, bugger it.
  12. Sep 4, 2010 #11
    Ivy's way is more "mathematician". The main idea is assuming the solution in form of complex exponential. Since God makes [tex](e^{st})'=se^{st}[/tex], the DE will be converted to a algebra equations. I think this hinted Oliver Heaviside to invent using LT to solve this kind of DE, by letting s=a+jb.

    As engineer, we were trained to use LT for linear ODE with constant coeffs which is the guts of LTI systems. We did learn Ivy's way before LT was taught. So I think you should know both.
    Last edited: Sep 4, 2010
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