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Particular solution to linear nonhomogeneous equation

  1. Nov 3, 2015 #1
    1. The problem statement, all variables and given/known data
    I started learning about solving non homogeneous linear differential equations in class and I am a bit clueless on how to solve them since I've never had a prior experience with much of differential equation.

    I am trying to find the particular solutions to the equation L''+w^2L=cn^2sin(nt)+w^2b where w,c,n are constants.

    2. Relevant equations


    3. The attempt at a solution
    First, I split the equation into two:
    Lp1=> L''+w^2L=w^2b
    Lp2=>L''+w^2L=cn^2sin(nt)

    Not quite sure how to solve for the particular solution for the first one.

    Lp2:
    We guess that L=Acos(nt)+Bsin(nt).
    Then we have
    L'=-nAsin(nt)+nBcos(nt)
    L''=-n^2Acos(nt)-n^2Bsin(nt)=-n^2(Acos(nt)+Bcos(nt))

    Plugging back into the equation =>
    (-n^2(Acos(nt)+Bcos(nt)))+w^2(Acos(nt)+Bsin(nt))=cn^2sin(nt)

    What do we do from here?
    Any help will be appreciated.

    EDIT:
    For the second particular solution, do we just solve for Acos(nt)+Bsin(nt) because we supposed that is L?
    In that case, I get Lp2=(cn^2sin(nt))/(w^2-n^2).

    Also, for the first solution, do we guess that Lp1 is a constant, so when we take the second derivative, L'' becomes 0, so it just becomes w^2L=w^2b, hence L=b?
     
    Last edited: Nov 3, 2015
  2. jcsd
  3. Nov 3, 2015 #2

    epenguin

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    Last edited: Nov 3, 2015
  4. Nov 3, 2015 #3

    Ray Vickson

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    It may be easiest to eliminate the constant on the right, by writing the DE for ##M = L + \text{constant}## and figuring out what the appropriate constant would be in order to have the simpler right-hand-side ##cn^2 \sin(nt)##.
     
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