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Second order DE involving repeat roots

  1. Apr 21, 2010 #1
    ok, for 2nd order differential equations, if we have repeat roots, like for example y" +4y = cos2x

    , we would have repeat roots +2i,-2i.

    so how do we try for the particular solution?

    i tried Acos2x + Bsin2x, it all cancelled out.
    i tried (Ax+B)cos2x + (Cx+D)sin2x it all cancelled out too

    then i tried just (Ax2+Bx)cos2x, but apparently that is wrong.

    apparently we have to substitute y = Re(Z), and let z" +4z = ei2x to solve

    so my question is,

    1)how do we know when we have to substitute, and when we can try AcosX +BsinX, also, for repeat roots, is the only way to solve is to let it be the real part of some substituted equation?

    2)when the RHS is only cos 2x, do we have to try Acos2x +Bsin2x, or is Acos2x suffice? why?

    help appreciated!
  2. jcsd
  3. Apr 23, 2010 #2


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    Try Axsin(2x) + Bxcos(2x). It will work.

    And your effort with (Ax+B)cos2x + (Cx+D)sin2x should have worked too.
  4. Apr 25, 2010 #3
    oh isee. i just have to assume my B and D take on the values of 0? for convenience?

  5. Apr 25, 2010 #4


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    It isn't a matter of convenience. If you equate like terms on both sides after you substitute into the equation, those values must be 0 to make it work. In other words, if you don't have an x cosx on the right side you can't have one on the left, etc.
  6. Apr 27, 2010 #5
    oh i see ha.. ok thanks a lot!
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