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Homework Help: Second order differential equation via substitution

  1. May 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Substitute [tex] p = \frac{dx}{dt} [/tex] to solve [tex]x\prime\prime + \omega^2x = 0 [/tex]


    2. Relevant equations

    [tex] \frac{dp}{dx} = v + x\frac{dv}{dx} [/tex]

    [tex] v = \frac{p}{x}[/tex]

    3. The attempt at a solution

    [tex] p = \frac{dx}{dt}, \frac{dp}{dt} = \frac{d^2x}{dt^2} [/tex]

    [tex] \frac{dp}{dt} = \frac{dp}{dx}\frac{dx}{dt} = \frac{dp}{dx}p [/tex]

    [tex] \frac{dp}{dx} + \frac{\omega^2x}{p} = 0 [/tex]

    [tex] v + x\frac{dv}{dx} = \frac{-\omega^2}{v} [/tex]

    [tex] \frac{-v}{\omega^2 + v^2}dv = \frac{1}{x} dx [/tex]

    [tex] \frac{-1}{2}ln(\omega^2 + v^2) = ln|x| + C [/tex]

    [tex] \frac{1}{\sqrt{\omega^2 + v^2}} = x + C [/tex]

    I get tripped up here and I'm not sure how to go forward, with regards to all the various substitutions I've made! I see the beginnings of an integral involving trigonometric substitution, so I hope I may be on the right track. A hint would be much appreciated.
     
  2. jcsd
  3. May 20, 2010 #2

    gabbagabbahey

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    I don't understand why you are making a second substitution. You have a separable 1st order ODE for [itex]p(x)[/itex]:

    [tex]\frac{dp}{dx} + \frac{\omega^2x}{p} = 0 [/tex]

    Just solve it.
     
  4. May 20, 2010 #3
    :rofl: :cry: Wow, so I do. That's what I get for doing this when I'm tired!
     
  5. May 20, 2010 #4
    To finish up:

    [tex] p dp = -\omega^2x dx [/tex]

    [tex] p^2 = -\omega^2 x^2 [/tex]

    [tex] p = i \omega x + C_1 [/tex]

    [tex] t = \int \frac{1}{i\omega x + C_1} dx [/tex]

    [tex] i \omega t = ln|i\omega x + C_1| + C_2 [/tex]

    [tex] e^{i\omega t} = C_2(i\omega x + C_1) = C_2x + C_1C_2 [/tex]

    [tex] \frac{1}{C_2}cos(\omega t) + \frac{1}{C_2}isin(\omega t) - C_1 [/tex]

    [tex] x = Acos(\omega t) + Bsin(\omega t) [/tex] if we set [tex] \frac{1}{C2} = A, [/tex] [tex] \frac{1}{C1} = B, [/tex] and let [tex] C_1 = \frac{1}{C_2}isin(\omega t) - \frac{1}{C_1}sin(\omega t). [/tex]
     
  6. May 21, 2010 #5

    gabbagabbahey

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    Why didn't the constant make an appearance in your second line?

    [tex] p dp = -\omega^2x dx \implies p^2=-\omega^2x^2+C_1 \implies p=\sqrt{C_1-\omega^2 x^2}[/tex]
     
  7. May 21, 2010 #6
    Carelessness. I'll try it again.
     
  8. May 21, 2010 #7
    I think I'm a little closer now. Continuing from the dropped constant and substituting into p = dx/dt we have:

    [tex] \frac{1}{\sqrt{C_1 - \omega^2 x^2}} dx = dt [/tex]

    Make the substitution [tex] x = \frac{\sqrt{C_1}}{\omega} sin\theta [/tex]

    [tex] \int \frac{\sqrt{C_1} cos\theta}{\omega C_1 cos\theta} d\theta = \int \frac{\sqrt{C_1}}{\omega C_1} d\theta = \frac{\sqrt{C_1}}{\omega C_1}\theta + C_2 [/tex]

    So [tex] t = \frac{\sqrt{C_1}}{\omega C_1} sin^-1(\frac{x \omega}{\sqrt{C_1}} + C_2) [/tex]

    And solving for X in terms of t I get:

    [tex] \frac{\sqrt{C_1}}{\omega} sin (\frac{\omega t C_1 - \omega C1 C2}{\sqrt{C_1}})
    [/tex]

    Not sure this is entirely correct, it's closer though! The first term we can call B because the constant swallows up the omega and square root, but as for the terms inside the sin() I'm not so sure. I know I can find the second solution by substituting y2 = y1*v into the original equation, where y1 = B sin (wt).
     
  9. May 21, 2010 #8

    gabbagabbahey

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    The [itex]C_1[/itex] in the denominator should be [itex]\sqrt{C_1}[/itex], so you get

    [tex]\int\frac{dx}{\sqrt{C_1 - \omega^2 x^2}}=\frac{\theta}{\omega}+C_2[/tex]


    Why is [itex]C_2[/itex] inside the arcsine?

    [tex]\frac{\theta}{\omega}+C_2=\frac{1}{\omega}\sin^{-1}\left(\frac{\omega x}{\sqrt{C_1}}\right)+C_2[/tex]
     
  10. May 21, 2010 #9
    I see it now. It's been months since I've worked on this stuff, and as you can tell I'm pretty out of practice. Ugly careless mistakes. :frown:
     
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