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Verify unsolvable ODE on Midterm

  1. Nov 7, 2008 #1
    Long story short,

    Professor was replaced with a replacement professor on the day of the midterm.
    The replacement prof. announces, after 3/4 of the time allowed has passed that the first question might contain a typo. He doesn't suggest how to fix the problem or anything. He just claims there might be a problem with the first question.

    Can someone take a look at it?
    This is for Ordinary Differential Equations I

    Here is the first question:

    Solve the initial value problem:
    y' ' ' + y' ' + 4y' + 4 = 0 y (0) = 5; y' (0)=7; y' ' (0)= -5;

    Thank you very much
     
  2. jcsd
  3. Nov 7, 2008 #2

    gabbagabbahey

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    I don't think there is any typo, it seems easily solvable to me; just use the method of undetermined coefficients, you have a fairly simple 3rd degree inhomogeneous ODE.
     
  4. Nov 7, 2008 #3
    y' ' ' + y' ' + 4y' + 4 = 0
    would be come
    y' ' ' + y' ' + 4y' = -4

    m^3 + m^2 + 4m = 0
    m(m^2 + m + 4) = 0

    m1 = 0 and m2 = a complex root

    is that right?
     
  5. Nov 7, 2008 #4

    gabbagabbahey

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    That looks like a good start for your complimentary solution.What form does that portion of your solution take?
     
  6. Nov 7, 2008 #5
    Hmm...

    e^a(c1cos(bx)+c2sin(bx))
     
  7. Nov 7, 2008 #6

    gabbagabbahey

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    That's part of it, what are the values of 'a' and 'b', and what happened to your m1=0 root?
     
  8. Nov 7, 2008 #7
    a would be -1/2 and b sqrt(15)/2

    We have never seen the case were m1 and m2 are real and complex.. Where would I go from here?
     
  9. Nov 7, 2008 #8

    gabbagabbahey

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    a real root m1 just adds an c3*e^(m1 x) term, in this case m1=0 and e^0=1, so it adds a constant term and your complimentary solution is:

    [tex]y_c(x)=e^{\frac{-x}{2}} \left( c_1 cos \left( \frac{\sqrt{15}}{2} x \right)+c_2 sin \left(\frac{\sqrt{15}}{2} x \right) \right) +c_3[/tex]

    Do you follow?
     
  10. Nov 7, 2008 #9
    Yes, I follow... Where would I go from here?
     
  11. Nov 7, 2008 #10

    gabbagabbahey

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    Now you need to find a particular solution....Your inhomogeneous term is just '-4'. Suppose you had only a second order ODE, what would you guess as a particular solution there? Can you guess the form of a particular solution for the 3rd order ODE?
     
  12. Nov 7, 2008 #11
    Ok I get it from this point on, however, weve never really had any practice with m1 = 0 and m2 = complex

    I wonder why the prof said it was a typo now.. lol..
     
  13. Nov 7, 2008 #12

    gabbagabbahey

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    maybe because he thought that the sqrt(15)/2 was a little too ugly for an exam question, but who knows :shrug:
     
  14. Nov 7, 2008 #13
    We weren't allowed calculators.. I don't think that changes much though
     
  15. Nov 7, 2008 #14

    gabbagabbahey

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    My guess is your replacement prof didn't have his morning coffee, and made an error in his attempt to solve the problem, and couldn't find it so he concluded something might b wrong with the question.

    But that's just speculation on my part :smile:
     
  16. Nov 7, 2008 #15

    HallsofIvy

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    He probably looked at y' ' ' + y' ' + 4y' + 4 = 0 and thought perhaps it should be
    y' ' ' + y' ' + 4y' + 4y = 0 which would be more "standard form" but distinctly harder than the original problem!
     
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