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I Can you check the solution for this second order ODE?

  1. Nov 8, 2016 #1
    The second order ODE is,

    \begin{equation*}
    \frac{d^2 x}{dt^2} = -\omega^2_g \frac{dx}{dt}
    \end{equation*}

    I tried solving this by substitution of the second order derivative into a variable and transforming the equation into a second order polynomial, and I get the solution involving an exponential. But, the solution includes a sinusoidal function, which I don't know how it got a sinusoidal function.
     
  2. jcsd
  3. Nov 8, 2016 #2
    Hi ecastro:

    Consider a substitution of y(t) = dx/dt to get a first order ODE, dy/dt = -ωg2 y

    Hope this helps.

    Regards,
    Buzz
     
  4. Nov 8, 2016 #3
    Hi @ecastro:

    I was hoping my post would help you recognize that a sin or cos is NOT a solution. Did you get that? My guess is that the given sin or cos solution is a typo.

    You are correct that an exponential is a solution, but there is also a second solution: x = t + const.

    Regards,
    Buzz
     
  5. Nov 10, 2016 #4
    Mod note: Post referred to below is now deleted, as it contained a complete solution .
    Thank you for your answer. It turns out I misunderstood the equation; I am sorry. The original equation is this,

    \begin{equation*}
    \ddot{v}_x = -\omega_g^2 v_x.
    \end{equation*}

    The ##v_x## is the velocity of the particle along the ##x## direction, and a dot above the function means a derivative with respect to time. So this equation should have a sinusoidal function as a solution. If I am not mistaken, the solution is

    \begin{eqnarray*}
    v\left(t\right) &=& C_1 e^{\pm i \omega_g t} \\
    &=& C_1 \left(\cos \omega_g t \pm i \sin \omega_g t\right).
    \end{eqnarray*}

    From here, I can integrate the velocity component with respect to time to calculate the position of the particle; but it would seem that the position of the particle is real, how can I eliminate the imaginary component?
     
    Last edited by a moderator: Nov 14, 2016
  6. Nov 12, 2016 #5
    The general solution is
    ##v(t)=C_1 e^{i\omega_g t}+C_2 e^{-i\omega_g t}=C_1(\cos \omega_g t+i\sin\omega_g t)+C_2(\cos \omega_g t-i\sin\omega_g t)##.
    Alternatively, by defining the new constants ##A_1=C_1+C_2## and ##A_2=i(C_1-C_2)##:
    ##v(t)=A_1\cos \omega_g t+A_2\sin\omega_g t##.
    If ##A_1## and ##A_2## are chosen real, ##v(t)## will also be real.
     
  7. Nov 14, 2016 #6
    Is this acceptable? The first constants, ##C_1## and ##C_2## can be calculated if initial conditions are given, right?

    Here is the solution of the equation with respect to position,

    \begin{equation*}
    x - x_0 = r_g \sin \left(\omega_g t\right),
    \end{equation*}

    where ##\omega_g = q B / m##, ##r_g = m v_{\perp} / \left(|q| B\right)##, and ##x_0## is the initial position.
     
  8. Nov 14, 2016 #7
    The sets of constants, i.e. ##(C_1,C_2)## and ##(A_1,A_2)## are connected by a linear transformations. For given initial conditions you can either determine ##C_1## and ##C_2##, or ##A_1## and ##A_2##. I only introduced the constants ##A_1## and ##A_2## for convenience. As I said in my previous post ##A_1## and ##A_2## will be real if ##v(t)## is real. However, ##C_2## will not be. But, if you prefer you can keep using ##C_1## and ##C_2##.

    From your solution I suppose your initial conditions are
    $$x(0)=x_0$$,
    and
    $$v(0)=r_g\omega_g$$.

    By integrating the expression for ##v(t')## from 0 to to ##t## on both sides we have
    $$x(t)-x_0=(A_1/\omega_g)\sin\omega_g t -(A_2/\omega_g )(\cos\omega_g t -1) $$

    So, the solution you posted is satisfied if
    $$A_1=r_g \omega_g$$,
    and
    $$A_2=0$$.

    You can also use the set ##(C_1,C_2)## but the solution becomes slight more cumbersome, I guess.
     
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