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Initial Value Problem with Laplace Transforms

  1. Jul 21, 2010 #1
    Solve the following given y(0) = 0 & y'(0)=1:
    y′′+3y′+2y = u2(t), such that u2(t) is a heaviside step function

    Here's what I've got so far,
    =>s2Y(s)−sy(0)−y′(0) + 3sY(s)−3y(0) + 2Y(s)= exp(−2s)/s

    Y(s) = (exp(−2s) + s) / (s(s2+3s+2))
    Y(s) = exp(−2s)/(s(s2+3s+2))* + 1/(s2+3s+2)**

    The second part, **, I was able to solve with partial fractions => 1/(s+1) − 1/(s+2) which transforms to exp(−t) − exp(−2t).

    However I don't know how to solve the first part, *, since the step function isn't by itself,

    Any push in the right direction would be great,

    Thanks in advance
  2. jcsd
  3. Jul 21, 2010 #2
    Are you required to use laplace transforms to solve this problem? Are you allowed to use any other methods?
  4. Jul 21, 2010 #3


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    For the first term, just ignore the exponential for the moment and take the inverse Laplace of what's left. Then shift your result in time, replacing t with t-2, to take into account the exponential.
  5. Jul 21, 2010 #4
    In an attempt to learn these types of problems I've made an attempt at the solution.

    I'm not sure if I'm correct or not but hopefully somebody else on these forums (vela?), will let me know if I'm on the right track. Maybe you can better understand the problem after seeing my attempt.

    (see figures)

    Anyone see any problems with my attempt?

    Attached Files:

  6. Jul 21, 2010 #5


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    Looks good, though you didn't propagate the sign correction you made into the boxed answer. You can always check your answer by plugging it back into the differential equation as well as verifying that the initial conditions are met.
  7. Jul 21, 2010 #6


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    Looks good to me. Wikipedia has a good list of Laplace transform you might want to take a look at for future reference.
  8. Jul 21, 2010 #7
    :biggrin: Whoops!

    Glad to hear I came to a resonable solution, this is my first attempt at a problem like this. Hopefully the OP will benefit from my work.

    (Next time I'll try not to "hand out" the solution, I just had to know if I was right :wink: )
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