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A Initial value ODE with shifting forcing function

  1. Jul 21, 2016 #1
    Use laplace Transform to solve this ode:
    upload_2016-7-21_21-49-59.png

    So I got:

    sV(s)-V(0)-12V(s)=U(s+5)
    V(s)(s-12)=U(s+5)+1
    V(s)=[U(s+5)+1]/(s-12)

    Now to go back to time domain with Inverse Laplace Transform...My question is, how to transform U(s+5)/(s-12)?

    Any help?

    Thanks guys
     
  2. jcsd
  3. Jul 22, 2016 #2

    Twigg

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    Gold Member

    I think it may help you to know that u(t) is the standard symbol for the step function, which has a known Laplace transform. Check your tables.
     
  4. Jul 24, 2016 #3
    Thanks
     
  5. Jul 24, 2016 #4
    @Twigg , can you shed some lights on Laplace Transform of e^(at)*u(t)?

    Thanks
     
  6. Jul 25, 2016 #5

    Twigg

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    Gold Member

    I'm pretty sure you just apply the shifting property to the Laplace transform of the Heaviside step function. The Laplace transform of ##u(t)## is ##\frac{1}{s}##, so the Laplace transform of ##e^{-5t} u(t)## is ##\frac{1}{s + 5}##. Just like you did in your first post.
     
  7. Jul 25, 2016 #6
    Thanks
     
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