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ODE about osscilating body

  1. Aug 24, 2009 #1
    1. The problem statement, all variables and given/known data

    the deflection y of a body oscillating about a fixed reference point satisfies the ordinary differential equation

    [math] d^2/dt^2 + 16y = 3sin(2t) [/math]

    Where t is time

    given the initial conditions y = 0 and dy/dt = 3/2 at time t= 0 , solve the differential equation to obtain an expression for the deflection y as a function of t.


    2. Relevant equations

    [math] d^2/dt^2 + 16y = 3sin(2t) [/math]

    3. The attempt at a solution

    I understand that i must let

    [math] y = e^alphax [/math]

    Therefor alpha^2 + 16 = 3sin(2t)

    however i then know that the general solution is (x + 4)(x - 4)

    however isnt this for a -16 not a positive 16?

    but then i goto x^2 + 4x - 4x -16

    Again -16?

    but right there im stuck

    ive never solved one of these and im at my limit with this one it really is beyond me.

    Any help would be appreciated so much

    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 24, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    so your equation is r2+16=0 to give you the homogeneous roots

    use the quadratic equation formula and the fact that i=√-1 (imaginary unit) to get your roots in the form of λ±μi.

    so yh=eλt(cos(μt)+sin(μt))

    Now find yp, then add that to yh and you can go on.
     
  4. Aug 24, 2009 #3

    Mark44

    Staff: Mentor

    Let's look at the related, homogeneous diff. eqn: y'' + 16y = 0.
    The characteristic equation is r2 + 16 = 0, which has roots +/- 4i. This means that two solutions to the homogeneous diff. eqn. are e^4t and e-4t). The solutions to the homogeneous problem consist of all linear combinations of these two functions; i.e., the sum of constant multiples of these two functions.

    For the nonhomogeneous diff. eqn. y'' + 16y = 3cos(2t), a good choice for a particular solution is yp = Acos(2t). Substitute this function into your differential equation to solve for the constant A.

    The general solution consists of the solutions to the homogeneous problem plus the particular solution.
     
  5. Aug 25, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, you mustn't. You must look at the associated homogeneous equation,
    [tex]\frac{d^1y}{dt^2}+ 16= 0[/tex] which gives characteristic equation [itex]\alpha^2+ 16= 0[/itex]

    No, you don't know that. In the first place, that is not a "solution" it is an attempt to factor. In the second place, its product is NOT "[itex]x^2+ 16[/itex]", it is "[itex]x^2- 16[/itex]" so your factoring is wrong. In the third place, it should be "[itex]\alpha[/itex]", not x. If you write the equation as [itex]\alpha^2= -16[/itex] it should be easy to solve.

    You will also want to use the fact that [itex]e^{i\alpha x}= cos(\alpha x)+ i sin(\alpha x)[/itex]

    And, on this forum, use [ tex] or [ itex ] for LaTex.
     
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