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Homework Help: Diff EQ - Imaginary identities

  1. Jun 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine the general solution of:
    y(6) + y''' = t

    3. The attempt at a solution

    r = 0, 0, 0, 1/2 +- 3i/√2, -1/2 + 3i/√2

    What do I do with that last r value? It turns into ce-t somehow, but I don't see it.

    edit: typed a number in wrong, fixed now~
  2. jcsd
  3. Jun 28, 2013 #2
    Well you have your values for the homogeneous solution, so now you need to find the values for your particular solution. There are a number of methods that apply to higher order differential equations that are the same for second order differential equations. Try using one of those methods. Then of course, add together the homogeneous solutions and the particular solutions. Your particular solution will have ce^-t in it. Just a guess but undetermined coefficients could work well
    Last edited: Jun 28, 2013
  4. Jun 28, 2013 #3
    Yes! The only method I know of that can deal with imaginary numbers is:
    r = λ +- μi → eλ(cos μt + sin μt),
    but my r value doesn't have the +-, only +.

    edit: yes, I don't understand how the last r value becomes ce^-t
  5. Jun 28, 2013 #4
    So your last value should fit the form C[itex]e^{rt}[/itex] for this particular problem. Since your value is r = -1/2 + 3i/√2, we get [itex]e^{(-1/2 + 3i/√2)t}[/itex].

    This still presents a problem though hmmm
  6. Jun 28, 2013 #5
    Are you sure your r values are correct? Now that I am looking back over it I would think that r = (-1) should be one of your values since in your characteristic equation you get

    r^3 = -1

    If r = -1 then were in business!
    Last edited: Jun 28, 2013
  7. Jun 28, 2013 #6


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    You are confused because you haven't solved the characteristic equation correctly and have come up with 3 complex solutions. In solving polynomials, complex solutions always occur in pairs: a complex number and its conjugate are solutions to the same polynomial equation.

    The value r = -(1/2) + 3i/Sqrt(2) is not a solution to this characteristic equation, but you have omitted a real solution (which should be easy to find).
  8. Jun 28, 2013 #7
    Yeah, r = 3√(-1), r = 3√(0) Which breaks down into r = 0, 0, 0, 1/2 +- 3i/√2, -1/2 + 3i/√2

  9. Jun 28, 2013 #8
    Well look at it this way. If you cube your solutions of r, you should get -1.

    What is (-1)^3? Well, (-1). That tells me that (-1) MUST be a solution
  10. Jun 28, 2013 #9
    Ohh, so when I get something along the lines of:
    rx = -1, if x is even, then I will get imaginary numbers in conjugate pairs that form solutions.

    If x is odd, then r = -1 is a solution, along with an imaginary pair, and an extra imaginary number that must be discarded.

    Thank you very much for the help!

    If you guys are still willing, could you look at this question I posted?
  11. Jun 28, 2013 #10


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    There is no extra imaginary number solution to r^n = -1, n odd, which must be discarded.

    Read the article, http://www.jimloy.com/algebra/roots.htm

    For r^n = -1, there must be n solutions. If n is odd, then r = -1 is one solution, and the other two solutions are a complex number and its conjugate. If n is even, then r will be two complex numbers and their conjugates as solutions.

    This should have been covered in a basic algebra course.
  12. Jun 28, 2013 #11
    Hm, I don't mean to say that a solution is discarded, but I'm getting an extra imaginary number. An extraneous value? Is that a better way to word it?

    Anyway, thank you very much for the help!
  13. Jun 28, 2013 #12


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    An incorrect solution method or a mistake in your arithmetic is more likely.

    Specifically, how did you arrive at this extra solution?
  14. Jun 28, 2013 #13

    Ray Vickson

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    It is much easier to work with the DE z''' + z = t, obtained by setting z = y''' .
  15. Jun 28, 2013 #14
    Oh? Sure, maybe I am making a mistake somewhere. Here are the steps:

    y(6) + y''' = t
    Characteristic Equation:
    r3(r3 + 1) = 0
    r3 = 0 and r3 + 1 = 0
    r = 0, multiplicity of 3 and r = 3√(-1)
    r = 0, 0, 0, -1, and two more
    Solving r = 3√(-1) for the remaining two solutions:
    First, rewrite -1 as an exponent,
    -1 = cos(π + 2mπ) + i sin(π + 2mπ)
    -1 = e(π + 2mπ)
    Substitute in:
    r = e(π + 2mπ)(1/3)
    r = cos(π/3 + 2mπ/3) + i sin(π/3 + 2mπ/3)
    with m = 0
    r = cos(π/3) + i sin(π/3) = 1/2 + 3i/√2
    with m = 1
    r = cos(π/3 + 2π/3) + i sin(π/3 + 2π/3) = -1


    Thank you SteamKing. I worked this problem three times on paper and for some reason kept getting:
    ∏/3 + 2∏/3 = 2∏/3

    Next time I'll type out all my steps in the original post instead of assuming that there's something complicated going wrong.
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