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Solving x''(t)+Kx(t)=0 when x has the form sin(ωt+ø): the soln r LD!

  1. Feb 14, 2014 #1
    I have the ODE [itex]x''(t)+\frac{k}{m}x(t)=0[/itex].

    Given that the solution is of the form [itex]sin(ωt+ø)[/itex], I plug this form into the original ODE and obtain [itex]ω=+\sqrt\frac{k}{m},-\sqrt\frac{k}{m}[/itex].

    And hence, I obtain two solutions of the ODE as follows:

    [itex]x_{1}(t)=Asin(\sqrt\frac{k}{m}t+ø), x_{2}(t)=Bsin(-\sqrt\frac{k}{m}t+ø)[/itex]

    Next, I check for the linear independence of the two solutions using the 2x2 Wronskian matrix and obtain [itex]Wronskian=-2ωsin(ø)[/itex] which vanishes at ø=0 or ω=0, implying that the two solutions are not linearly independent.

    How can I find the second solution of the second order ODE?
     
  2. jcsd
  3. Feb 14, 2014 #2
    The two solutions are not necessarily with the same ø
    [itex]x_{1}(t)=Asin(\sqrt\frac{k}{m}t+ø_1), x_{2}(t)=Bsin(-\sqrt\frac{k}{m}t+ø_2)[/itex]
     
  4. Feb 14, 2014 #3

    pasmith

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    Homework Helper

    The general solution of [itex]x'' + \omega^2x =0[/itex] is
    [tex]x = A\cos (\omega t)+ B\sin (\omega t)[/tex]
    By use of the sum formulae for sine and cosine this can be written as either
    [tex]
    x = R \cos (\omega t + \phi)
    [/tex]
    where [itex]R^2 = A^2 + B^2[/itex], [itex]\cos \phi = A[/itex] and [itex]\sin\phi = -B[/itex], or as
    [tex]
    x = R \sin (\omega t + \theta)
    [/tex]
    where [itex]R^2 = A^2 + B^2[/itex], [itex]\cos \theta = B[/itex] and [itex]\sin\theta = A[/itex].

    The point is that the general solution has two parameters: either two amplitudes or an amplitude and a phase shift.
     
  5. Feb 14, 2014 #4

    PeroK

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    Science Advisor
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    Gold Member

    1) If two functions are linearly dependent, then the Wronskian will vanish everywhere.

    2) If the Wronskian is not zero at any point, then the functions must be linearly independent.

    So, your two functions are linearly independent unless w = 0 or ø is a multiple of pi.
     
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