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An ODE involving delta function

  1. Nov 23, 2011 #1
    x''+2x'+x=t+delta(t) x(0)=0 x'(0)=1

    The textbook, "Elementary differential equations" by Edwards and Penney, gives the answer as -2+t+2exp(-t)+3t exp(-t)

    It is clearly wrong, as in this case x'(0)=2, not x'(0)=1.
     
  2. jcsd
  3. Nov 24, 2011 #2
    I got -2u(t)+t+2exp(-t)+3t exp(-t), with u(t) being the Heaviside function. I got this using a Laplace transform and then doing partial fractions. I think they just left off the u(t) for their solution on accident.
     
  4. Nov 24, 2011 #3
    What do mean by saying that x'(0)=1、 Do we mean x'(0+)=1 or x'(0-)=1?
     
  5. Nov 24, 2011 #4
    EDIT: Whoa, I was way off for my reasoning for x'(0)! Sorry about that. I'd suspect it would be x'(0-) since that is what you use when you take the transform.
     
    Last edited: Nov 24, 2011
  6. Nov 24, 2011 #5
    But there is a delta(t) in the input so x'(0-) and x'(0+) cannot be the same. x'(0+)=x'(0-)+1
     
  7. Nov 26, 2011 #6

    hunt_mat

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    Why not take Laplace transforms?
     
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