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Find particular solution

  1. Oct 21, 2009 #1
    I have to find the particular solution to the differential equation:

    (-21/4)y''+2y'+y=4xe^(3x)

    First, I chose my trial function to be yp=(Ax+B)*e^(3x). Is this correct???
    so yp'=3(Ax+B)*e^(3x)
    yp''=9(Ax+B)*e^(3x)

    So I plug these into the differential equation and I get:
    (-189/4)Axe^(3x)-(189/4)Be^(3x)+6Axe^(3x)+6Be^(3x)+Axe^(3x)+Be^(3x)=4xe^(3x)

    I group like terms and I get A=-16/161 and B=0
    So yp=(-16/161)Axe^(3x)

    This is not correct.
    Can someone please tell me where I'm going wrong? Thanks!
     
  2. jcsd
  3. Oct 21, 2009 #2

    Mark44

    Staff: Mentor

    The particular solution you choose is affected by the solutions to the homogeneous equation. For your problem, I'm going to guess that e3x is not a solution to the homogeneous problem, but I don't know that for a fact.

    In any case, and assuming that y = e3x is not a solution to the homogeneous problem, your choice for a particular solution is good, but you made a mistake in both of your derivatives. differentiation. =(Ax+B)*e3x is a product, a fact that you seem to have completely overlooked.
     
  4. Oct 21, 2009 #3
    Wow. Implied multiplication on my calculator. Next time I will just work that out by hand. Thanks for your help!
     
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