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Struggling with a second-order differential equation

  1. Nov 14, 2005 #1

    For my Physics coursework I'm trying to show how the equation to describe simple harmonic motion, this is where I'm up to now:

    x displacement;
    t is time;
    k is spring constant;
    m is mass.

    A is an unknown constant;
    b is an unknown constant.

    Could somebody check through my working, see if I've gone wrong and hopefully point me in the right direction as to what to do (I've tried respresting the function of e in terms of trigonometric functions but this never made sense.)


    PS. I know it's possible, from the differential equation given, to solve it straight off using the method by assuming it's a quadratic (or something similar) but I want this proof to be rigourous and therefore everything (to some degree) must be explained by me.
  2. jcsd
  3. Nov 14, 2005 #2


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    Have you checked the solution? Does it satisfy the original DE?

    EDIT: Just realized that you do not have a complete statement of the DE. What are your initial conditions?
    Last edited: Nov 14, 2005
  4. Nov 14, 2005 #3


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    You have correctly determined that [itex]Ae^{\sqrt{\frac{k}{m}}it}[/itex]
    and [itex]Be^{-/sqrt{\frac{k}{m}}it}[/itex] are solutions to the differential equation. However, you have no reason to assert that
    [tex]Ae^{\sqrt\frac{k}{m}}it}-A\frac{k}{m}e^{-\sqrt\frac{k}{m}}it}= 0[/tex]. Use those two solutions to form the general solution to the differential equation. Do you know how to do that?

    Also do you know that [itex]e^{ix}= cos(x)+ i sin(x)[/itex]?
  5. Nov 18, 2005 #4
    Not sure what you mean by initial conditions and having a complete statement of the DE.

    I don't know what you mean by the general solution the differential equation, I haven't started University and my knowledge of differential equations is limited to first-order differential equations.

    I know that [itex]e^{ix} = cos(x) + i sin(x)[/itex] which I've tried to use but failed. I end up with an imaginary part which doesn't make sense.

  6. Nov 18, 2005 #5
    Initial conditions are usually given values of a function evaluated at a specific point.
    For example, given:
    y(0) = 3pi
    y'(0) = 2
    That would transform your problem into an initial-value problem, and you'll end up with a solution particular, i.e. satisfying specifically, to/for your differential equation.

    A general solution is pretty much a solution for your differential equation, but not limited to that differential equation; i.e. it could have constants, like the A and B, and thus satisfy many differential equations.

    I suggest you read up on http://tutorial.math.lamar.edu/AllBrowsers/3401/3401.asp if you feel that your knowledge is somewhat lacking.
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