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Very easy question

  1. Aug 6, 2010 #1
    I use the calculator

    What are the constants c1 and c2 appearing in the second order dif. eq. solutions?
    A guess of mine is that c1=f(0) and c2=f ' (0), but I am not sure.
  2. jcsd
  3. Aug 6, 2010 #2
    The constants can be every number you like, and in any case you get a solution of the aquation. That is to say, the equation has infinite solutions. To specify one of them, you have to give additional constraints. These are usually the initial conditions, the values of the functions and its derivatives at a certain initial time t0. The number of initial conditions needed equals the order od the equation, that is, the higher derivative of the function that appears in the equations, and it's also the number of constants you get in the solution. In your case f(0) = c1 + c2 and f'(0) = c1 - c2.
  4. Aug 6, 2010 #3
    But then e.g. for case that the acceleration is analogous to the distance travelled, e.g.
    f’’(t)=5f(t), where f(t)=(distance travelled), f''(t)=(acceleration),
    at the wolframalfa.com solution of this dif. eq.
    when replacing c1=c2=0 (because f(0)=0 and f'(0)=0)
    I get the nonsense that f(t)=0.
    What’s going on?
    Last edited: Aug 6, 2010
  5. Aug 6, 2010 #4


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    Homework Helper

    What's going on is that f(t) = 0 is the unique solution to any linear homogeneous differential equation will all initial conditions being zero. Your initial conditions do not describe what you think they describe. The acceleration depends on the displacement, so if at time t = 0 the object isn't displaced there's no acceleration. The only way the object could have a non-zero displacement and hence non-zero acceleration at a later time is if at t = 0 the object was moving, but it's not by your initial condition on the velocity. What you're describing is an object that's sitting still, and you got the correct solution for that initial condition.
  6. Aug 7, 2010 #5
    No. At the spring its motion dx is 0 at t=0, and the problem can be solved using dif.eq. just by setting that at t=0 its position is not zero, but A, i.e. just change the frame of reference. At t=0 its velocity x'(t=0) is 0.
    Last edited: Aug 7, 2010
  7. Aug 7, 2010 #6


    Staff: Mentor

    A differential equation + initial conditions is an initial value problem. For your differential equation, y'' - y = 0, the solution is y = c1et + c2e-2, just as given in WolframAlpha.

    If the initial conditions are y(0) = 0 and y'(0) = 0, the unique solution to the DE is y [itex]\equiv[/itex] 0. The initial conditions mean that at time t = 0, the object is at the origin and its velocity is 0. If the velocity is 0, the object isn't moving, so its acceleration is 0, so the object will remain at the origin. For these initial conditions, the solution y(t) = 0 is not nonsensical.
  8. Aug 8, 2010 #7


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    No, the differential equation in question breaks translation invariance and so is not invariant under a change of reference frame. If you have

    [tex]\ddot{x}(t) - x(t) = 0,[/tex]
    with initial conditions [itex]x(0) = x_0[/itex], [itex]\dot{x}(0) = \dot{x}_0[/itex], then changing reference frame amounts to setting [itex]x(t) = X(t) + x_0[/itex], so you get the initial conditions [itex]X(0) = 0[/itex] and [itex]dX/dt(0) = \dot{x}_0[/itex], but the differential equation is now

    [tex]\ddot{X}(t) - X(t) = x_0.[/tex]

    This is equation is no longer homogeneous, so the unique solution is not zero unless the initial conditions are both zero.
    Last edited: Aug 8, 2010
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