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Laplace transform for differential equation

  1. Jan 18, 2015 #1
    1. The problem statement, all variables and given/known data
    use laplace transforms to solve the differential equation
    y"+2y'+17y = 1

    2. Relevant equations
    Initial conditions are
    y(0) = 0
    y'(0) = 0
    3. The attempt at a solution
    so it converts to Y(s) (s^2+2s+17) = 1/s
    which then ends up as;
    Y(s) = 1/s*1/(s^2+2s+17)

    i know i need to invert the equation from laplace again but not sure how.

    Im not sure if im on the right track by re-arranging to 1/s((s+1)^2+4^2)?

    Still getting used to spotting laplace eqns.
    and help would be much appreciated.
  2. jcsd
  3. Jan 18, 2015 #2
    Perhaps it would be easier to do two simple inverse Laplace transforms instead of one complicated one. Can you break that expression into a sum of two fractions using partial fraction decomposition?
  4. Jan 18, 2015 #3


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    ^complete the square
    what are a and b?
  5. Jan 18, 2015 #4


    Staff: Mentor

    The OP already has that figured out - a = 1 and b = 4

    @welcomereef, yes you are on the right track. axmls's advice is what I would follow if I were doing this problem.
  6. Jan 18, 2015 #5


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    The equation in latex:

    $$Y(s) = \frac{1}{s} \frac{1}{s^2+2s+17}$$

    You could always go the good old convolution theorem route:

    $$f(t) \star g(t) = \Lagr^{-1} \{F(s)\} \star \Lagr^{-1} \{G(s)\} = \Lagr^{-1}\{F(s) G(s)\}$$

    Then simply using the fact that:

    $$f(t) \star g(t) = \int_0^t f(t-x) g(x) \space dx$$

    You should get the same answer as any other method.
  7. Jan 18, 2015 #6


    Staff: Mentor

    Do you mean ##\mathcal{L}^{-1}## in your work above?

    The OP might not be aware of convolutions at this point. In any case, I think that decomposing the fraction above is quite a bit simpler.
  8. Jan 19, 2015 #7


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    Yes indeed I did. For some reason \Lagr was displaying properly yesterday, but not today by the looks of it. Guess I have to go with \mathcal{L}.

    I was also demonstrating there's more than one way to skin this problem, albeit a bit more tedious. I just thought it to be interesting.
  9. Jan 19, 2015 #8

    Ray Vickson

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    Do a partial fraction expansion of Y(s), just the way you would do it if you were integrating instead of inverting.
    He can also use the standard result (found in many Laplace transform tables) that
    [tex] f(t)\: \leftrightarrow\: F(s) \; \Longrightarrow \; \int_0^t f(\tau) \, d \tau \: \leftrightarrow \frac{F(s)}{s} [/tex]

    Of course, this follows right away from the "convolution" result, but does not require it, and maybe is even known before one has ever heard of convolution. It also follows more-or-less directly from
    [tex] g(t)\: \leftrightarrow \: G(s) \:\Longrightarrow \: g'(t) \: \leftrightarrow \: s G(s) - g(0+),[/tex]
    just by setting ##g(t) = \int_0^t f(\tau) \, d \tau##.
    Last edited: Jan 19, 2015
  10. Jan 21, 2015 #9

    rude man

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    Yes you are.
    Use two theorems:
    1. If g(t) ↔ G(s), then G(s+a) ↔ e-atg(t).
    That gives you a very simple G(s+a) = 1/((s+1)^2+4^2) to invert.
    So now Y(s) = G(s+a)/s.
    Then, to take care of the s, consider
    If f(t) ↔ F(s), then
    (1/s)F(s) ↔ ∫f(t')dt' with limits 0 and t.
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