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Homework Help: Solve Diff EQ using Laplace

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Given:

    dx/dt + 3*x = exp(-3*t) and all initial conditions are zero.



    2. Relevant equations
    Laplace



    3. The attempt at a solution

    L[dx/dt + 3*x = exp(-3*t)]

    s*X(s) + 3*X(s) = 1 / (s + 3)

    X(s) = 1 / (s + 3)^2

    So here is where I get mixed up. For some reason, I thought I was supposed to use a partial fraction expansion here. But those of you who know better will probably get a good chuckle out of hearing about how I did do the partial fraction expansion only to re-discover that

    X(s) = 1 / (s + 3)^2

    So....2 questions:

    1) Do you know why I thought I needed a PFE?

    2) I have a list of Laplaces and their inverses. 1 / (s + a)^n is NOT one of them.

    1 / (s + a) IS one of them.

    I presume I am supposed to use this rule in conjunction with some other rule to find the inverse Laplace of 1 / (s + a)^n .
    Can I get a hint on how to do this?
     
  2. jcsd
  3. Feb 13, 2010 #2

    rock.freak667

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    Homework Helper

    If you had

    X(s)=1/s2, you can easily get x(t)

    So you'd need to apply the shift property.

    You probably thought 1/(s+3)(s+3) would have given you something simpler, but it was already in its simplest form.:tongue: Don't worry, I've done that exacts same thing.
     
  4. Feb 13, 2010 #3
    I will have to Google the "shift theory." This is for a Controls class, so the treatment of the mathematics is really brief and incomplete. So...I'll be back in a moment.

    EDIT:So the shift theory is just

    L-1[F(s − a)] = exp(at)*f(t)

    I am not too sure how to use this. This is going to take a while :redface:


    Okay. I am looking at one of my old texts. I have that:

    [tex]L[t*u(t)] = \frac{1}{s^2}[/tex]

    where u(t) is the unit step function. What does that really mean? Is that just a fancy name for '1' ?

    I also have the Frequency Shift Theorem :

    [tex]L[e^{-at}f(t)] = F(s + a)[/tex]

    so.....
     
    Last edited: Feb 14, 2010
  5. Feb 14, 2010 #4

    rock.freak667

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    Homework Helper

    Say you need to get

    [tex]L^{-1} \left( \frac{3}{(s-1)^2+9} \right)[/tex]

    You easily know that


    [tex]L^{-1}\left( \frac{3}{s^2+9} \right)=sin3t[/tex]

    So to get the require inverse transform, you need to shift the 's' to 's-1', and to do this you multiply the usual transform by e1t.
     
  6. Feb 14, 2010 #5

    Hmmm...so it's simply

    [tex]L^{-1} \left( \frac{3}{(s-1)^2+9} \right) = e^t*\sin(3t)[/tex]

    ?
     
  7. Feb 14, 2010 #6
    So for my particular case:

    F(s) = 1 / s^2 --> f(t) = t

    and

    F(s+a) = 1 / (s+3)^2 --> f(t) = exp(-3*t)*t

    ?
     
  8. Feb 14, 2010 #7

    rock.freak667

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    Yep, the F(s-a) is your 3/((s-1)2+9), your f(t)=sin(3t) and your shift is -1, so it is et
     
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