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General solution to a 6th order DE

  1. Oct 28, 2014 #1

    Problem: Given the DE: y^(6) + y = 0, find the general solution.

    Solution. I found the roots to be: (sqrt3 - i)/2, (sqrt3 + i)/2, i, -i, (-sqrt3 + i)/2, (-sqrt3 - i)/2

    Thus, my general solution led to:

    y = c1cost + c2sint + c3exp(t/2*(sqrt3))*cos(t/2) + c4exp(t/2*(sqrt3))*sin(t/2) + c5exp(t/2*-(sqrt3))*cos(t/2) + c6exp(t/2*(-sqrt3))*sin(t/2)

    I don't see any errors in this, but please let me know if there are.

    When I looked at the solution, the answer is:

    y = exp(t/2*(sqrt3))*(c1cos(t/2) + c2sin(t/2)) + c3cost + c4sint*exp(t/2*-(sqrt3))*(c5cos(t/2) + c6sin(t/2)

    How exactly did they derive the solution? Is my original answer wrong? The solutions in my original answer seem linearly dependent, and I don't quite see how they derived the solution they did...any help is great!
  2. jcsd
  3. Oct 28, 2014 #2


    Staff: Mentor

    I think there's an error in their solution. Notice that they have a cos(t) term but no sin(t) term. You don't get one without the other. You could check your solution, although that's going to be a lot of work.
  4. Oct 28, 2014 #3
    Yes, exactly! I just wasn't sure if they used a trig identity of sorts to find linear dependency or not b/w the terms.
    It also seems odd (I may be mistaken) that they included c4 and c5 and c6 in their solution despite those terms being multiplied to yield only 2 terms, making the extra constant factor redundant, no?
    Regardless, thank you for your comment.
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