Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    Why not take Laplace transforms?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook