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Solving differential equations (circular motion)

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data

    I have a differential equation of the form

    [itex]\frac{dZ}{d\theta} + cZ = a cos \theta + b sin \theta [/itex]

    Where [itex]Z = \frac{1}{2}\dot{\theta}^{2}[/itex]

    I need to find the general solution of this equation. a, b and c are all constants.

    2. Relevant equations

    The questions suggests using this to help:

    [itex]\int e^{\lambda x} (a cos x + bsin x ) = \frac{1}{1+\lambda^2}e^{\lambda x}(\lambda (a cos x + b sin x) a sin x - b cos x) + C[/itex]

    3. The attempt at a solution

    I just dont know how that integral is supposed to help me solve the equation. How does e become relevant to this function?

    Im also a bit unsure about this.. If I integrate

    [itex]\frac{1}{2}C \dot{\theta}^2[/itex]

    with respect to θ, do I get

    [itex]\frac{1}{2}C \theta ^2[/itex] ?
     
  2. jcsd
  3. Apr 4, 2013 #2

    Curious3141

    User Avatar
    Homework Helper

    The first step involves solving for ##Z## in terms of ##\theta##. Are you familiar with the technique of using an integrating factor? You can read more about it here: http://en.wikipedia.org/wiki/Integrating_factor

    Multiply by the appropriate integrating factor, and the reason for the hint should become very clear.

    No. Remember that ##\dot \theta## signifies a derivative wrt time. There's no (simple) relationship between the two expressions.
     
  4. Apr 4, 2013 #3
    Thank you.

    Ah. Yes, we've done integrating factors earlier in the course.

    The integrating factor will be

    [itex]e^{\int {c}} = e^{c\theta}[/itex]

    multiplying by this gives

    [itex]e^{c\theta} \frac{dZ}{d\theta} + C e^{c\theta} Z = e^{c\theta}(a cos \theta + b sin \theta)[/itex]

    And then

    [itex]\frac{d}{d\theta}(e^{c\theta}Z) = e^{c\theta}(a cos \theta + b sin \theta)[/itex]

    And then I can integrate both sides and use the integral given as well as substituting back in for Z at the end and everything is finished.

    However, I largely just followed a 'process solution' for this from my textbook. Im unsure how the LHS goes from

    [itex]e^{c\theta} \frac{dZ}{d\theta} + C e^{c\theta} Z[/itex]

    to

    [itex]\frac{d}{d\theta}(e^{c\theta}Z)[/itex]

    Can someone explain that to me please?
     
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