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Differential Equation problem

  1. May 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Compute the general solution of y'' + 6y' + 8y = 2t + et

    2. Relevant equations



    3. The attempt at a solution

    after using determinants, I found the general solution to be y(t) = k1e-2 + k2e-4 + yp

    To find yp, I would make it equal to yp = aet find its first and second derivates, and substitute them back into y'' + 6y' + 8y , but what do I do about the 2t?
     
  2. jcsd
  3. May 6, 2009 #2

    Cyosis

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    You can do some clever guess work. You can make sure that y'' is 0 by taking some linear function. This simplifies your problem to 6y'+8y=2t. Since we took a linear function y' will be a constant so this will reduce your problem to constant+8y=2t. A function of the form b+ct should suffice.
     
  4. May 6, 2009 #3
    would yp = t be a good guess?

    since y'p = 1 and y''p = 0
     
  5. May 6, 2009 #4

    Cyosis

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    But [itex]6+8t \neq 2t[/itex], so no. Try b+ct as suggested in my previous post.
     
  6. May 6, 2009 #5
    how about [tex]\frac{1}{4}t - \frac{3}{16}[/tex]

    so 6(1/4) + 8([tex]\frac{1}{4}t - \frac{3}{16}[/tex]) = 3/2 + 2t - 3/2 = 2t

    so when considering et, I would let yp =[tex]\frac{1}{4}t - \frac{3}{16} + e^t[/tex] ?
     
  7. May 6, 2009 #6

    Cyosis

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    The linear term is correct now, however if we plug in e^t we get, e^t+6e^t+8e^t=15e^t!=e^t. So you have to divide by 15.
     
  8. May 6, 2009 #7
    oops

    it should be [tex]\frac{1}{4}t - \frac{3}{16} + ae^t[/tex]

    and a would be 1/15
     
  9. May 6, 2009 #8

    Cyosis

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    Correct.
     
  10. May 6, 2009 #9
    Thank you for all your help.
     
  11. May 6, 2009 #10

    Cyosis

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    You're welcome.
     
  12. May 6, 2009 #11

    Mark44

    Staff: Mentor

    Pretty sure you meant to write y(t) = k1e-2t + k2e-4t + yp here.
     
  13. May 6, 2009 #12
    Oops (again).

    Yeah, I did. Thanks.
     
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