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Finding general solutions of forced equations

  1. Sep 18, 2011 #1
    I am having trouble after finding the complimentary solution to this problem, if possible I would like some guidance on how to proceed with the particular solution

    Find a general solution to the following equation: x'' + 2x' + x = 3


    x'' + 2x' + x = 3

    complimentary solution when = 0

    x'' + 2x' + x = 0

    m^2 + 2m + 1 = 0

    so this has only one root; m=-1

    complimentary solution must then be in the form

    x(t) = Ae^mt + Bte^mt

    particular solution when = 3

    x'' + 2x' + x = 3

    From this point I am unsure how to proceed and would value any help
     
    Last edited: Sep 19, 2011
  2. jcsd
  3. Sep 18, 2011 #2

    LCKurtz

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    Look for a solution of the form yp = C, a constant.
     
  4. Sep 19, 2011 #3
    Thank you LCKurtz, in this example because I have x''+2x'+x=3

    If I plug a value of x=3 into my original equation I end up with x=3 as a final answer so

    x(t)particular=3

    giving me x(t)general= x(t)complimentary + x(t)particular:

    x(t)=Ae^-1t + Bte^-1t + 3 ?
     
  5. Sep 19, 2011 #4

    Hootenanny

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    Spot on :approve:

    There are some general "rules" when finding particular solutions to ODEs, depending on the form of the forcing term. You will likely find them in your textbook.
     
  6. Sep 19, 2011 #5
    Thanks Hootenanny, I have a more complex forcing term to move onto now but with this basis I think I will get there.
     
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