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Solving a coupled nonhomegenoues ode

  1. Dec 6, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\dot{ω_{1}}[/itex] = λ[itex]ω_{2}[/itex] +μ
    [itex]\dot{ω_{2}}[/itex] = -λ[itex]ω_{1}[/itex]

    2. Relevant equations

    λ and μ are real, positive constants
    [itex]ω_{1}[/itex](0) ≠ 0
    [itex]ω_{2}[/itex](0) ≠ 0

    3. The attempt at a solution

    I know that the general solution will be in the form
    ω1(t) = A sin ωt + B cos ωt + C
    ω2(t) = D sin ωt + E cos ωt
    but I'm not sure how to solve it

    Eqn 2 becomes:
    [itex]\ddot{ω_{2}}[/itex] = -λ[itex]\dot{ω_{1}}[/itex]
    Substituting equation 1:
    [itex]\ddot{ω_{2}}[/itex] = -λ[itex]^{2}[/itex][itex]ω_{2}[/itex] +λμ

    I'm really just unsure of how to handle the constant when you solve it as a nonhomegenous equation.
     
    Last edited: Dec 6, 2012
  2. jcsd
  3. Dec 6, 2012 #2

    Dick

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    Take another derivative of the first equation and substitute the second one into it.
     
  4. Dec 6, 2012 #3
    Yeah I know how to do that part. I'm really just not sure what to do about the μ on the end. When I take the derivative it should disappear but I don't think it should.
     
  5. Dec 6, 2012 #4

    dextercioby

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    Well, if it's a constant, then it should...

    You can of course take the derivative of the second equation and use the first.
     
  6. Dec 6, 2012 #5
    So does μ affect this problem?

    the answers I came up with were

    [itex]ω_{1}[/itex](t) = [itex]ω_{2}[/itex](0) sin ωt + [itex]ω_{1}[/itex](0) cos ωt
    [itex]ω_{2}[/itex](t) = [itex]ω_{1}[/itex](0) sin ωt + [itex]ω_{2}[/itex](0) cos ωt

    but these were wrong
     
  7. Dec 6, 2012 #6

    dextercioby

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    Of course they're wrong, there's no evidence of mu and lambda.
     
  8. Dec 6, 2012 #7

    Dick

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    I think the μ will disappear when you take the derivative. That's ok. Just keep working. It will come back into the solution later. Show your work.
     
  9. Dec 6, 2012 #8
    Obviously I know they're wrong. Why else would I post this here? I posted it here looking for help with the solution not conformation that I was incorrect.
     
  10. Dec 6, 2012 #9

    Dick

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    Can't say why you got incorrect solutions until you show how you got them.
     
  11. Dec 6, 2012 #10
    Would it be a good idea to take the second derivative of :
    ω2(t) = D sin ωt + E cos ωt
    and plug them them into:
    ω2¨ = -λ^2ω2 +λμ
    and then use the initial conditions?
     
  12. Dec 6, 2012 #11

    Dick

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    It would be a good idea to say how you got ω2(t) = D sin ωt + E cos ωt. It looks like you are just plugging in a guess for the answer. And the guess is wrong. Why don't you start with the part we agreed on? Solve (ω1)''=(-λ^2ω1).
     
  13. Dec 6, 2012 #12
    I did just guess at that.

    okay so I set up the characteristic equation and got sqrt(λ) as the answer. would that make it:
    ω1(t) = A sin sqrt(λ)t + B cos sqrt(λ)t + C ?
     
  14. Dec 6, 2012 #13

    Dick

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    Good, you're getting started. Try to check that, plug it into (ω1)''=(-λ^2ω1). You should realize the the sqrt(λ) isn't quite right. I get a characteristic equation of x+λ^2=0. And does the C really belong there?
     
    Last edited: Dec 6, 2012
  15. Dec 6, 2012 #14
    Okay so ω=-λ
    working that out I came up with:

    ω1(t) = w1(0) cos -λt - [w2(0) + μ/λ] sin -λt
    ω2(t) = w1(0) sin -λt + [w2(0) + μ/λ] cos (-λt) - μ/λ
     
  16. Dec 6, 2012 #15

    Dick

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    That works. It would maybe look a little nicer if you would get rid of the signs on λ using trig, but it works. You've certainly gotten a lot better at this since your last post.
     
    Last edited: Dec 6, 2012
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