1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laplace Transform to Find Solution

  1. May 17, 2014 #1
    Use Laplace transfer to find the solution of the following initial value problem:

    y''+4y'+4y=f(t)
    where f(t) = cos(ωt) if 0<t<π and f(t)=0 if t>π ?

    Also, y(0) = 0, y'(0) = 1

    Currently, I have gotten to here, but not sure how to perform inverse Laplace:

    (s+2)² * F(s) − 1 = [s/(s²+w²)] + e^(−πs) * [s/ (s²+ω²)]
     
    Last edited: May 18, 2014
  2. jcsd
  3. May 18, 2014 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I suggest you put appropriate parentheses in that last line so we don't have to guess what that right hand side is supposed to really be.
     
  4. May 18, 2014 #3
    Sorry - Edited!
     
  5. May 18, 2014 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    After this point, you do some algebra to isolate F(s) by itself as the LHS of the equation.
    Then you examine the resulting RHS and refer to a table of Laplace transforms to see if the resulting expressions in 's' can be algebraically manipulated into one or more of the forms in the table.
     
  6. May 18, 2014 #5
    I isolated F(s):
    F(s) = [[s/(s²+w²)] + e^(−πs) * [s/ (s²+ω²)]] + 1 / ((s+2)²)

    But I have no idea how to manipulate it to look like the ones in the table...
     
  7. May 18, 2014 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    I think you have left out a factor of (1/s) when you formed the Laplace transform of the Unit Step Function, so check the F(s) equation.

    Also, be careful when writing complicated rational terms: you want to make sure you use parentheses to correctly group the terms in the numerator and the denominator.

    In order to analyze the F(s) expressions, split up a complicated sum into its component parts and analyze each term. After all, the Laplace transform of a sum = sum of the Laplace transforms. You will need to do even more algebra here.
     
  8. May 18, 2014 #7
    @ted s:

    For the Right Hand Side, I got
    L[coswt]+L[cos(ωt+π)], or s/(s^2+w^2) + L[cos(ωt+π)],
    where cos(ωt+π), using the identity for cos(x+y), =cos(−ωt)
    Therefore, the LHS is:
    s/(s^2+w^2)+e^(−πs)⋅s/(s^2+ω^2)
    Is that not correct?
     
  9. May 18, 2014 #8

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    I'm a little rusty on this, but bear with me:

    The original RHS for the ODE was:

    f(t) = cos(ωt) if 0<t<π and
    f(t) = 0 if t>π

    which can be re-written as:

    f(t) = cos(ωt) * [U(t) - U(t-π)],

    where U(t-c) is the delayed Unit Step Function at t = π

    I believe you have used the delayed unit impulse function δ(t-π) instead of the Unit Step Function in your re-writing of the RHS.

    The delayed unit impulse function δ(t-π) = 1 at t = π and zero everywhere else.

    The Unit Step Function U(t) = 1, when t > 0 and U(t) = 0 when t < 0, which also implies
    that the Delayed Unit Step Function U(t-π) = 1 for t > π and zero everywhere else.
     
  10. May 18, 2014 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Using Maple, I get
    [tex] {\cal L}(f)(s) = \int_0^{\pi} e^{-st} \cos(\omega t) \, dt
    = \frac{s}{s^2+\omega^2} -\frac{\cos(\pi \omega) \, s\, e^{-\pi s}}{s^2+\omega^2}
    +\frac{\omega \sin(\pi \omega) \, e^{-\pi s}}{s^2 + \omega^2} [/tex]
    Wolfram Alpha gives the same result.
     
  11. May 18, 2014 #10
    I too plugged it into Wolfram Alpha and got that result. I guess I can agree to that, though I still do not see the direct process used to derive the second and third terms.
    If somebody can assist, that would be great.

    Secondly, this is only half the problem - once we have this, we have to inverse Laplace to find y(t), which I am still having trouble with. There are no easy ways to group it, once dividing everything by (s + 2)^2...
     
  12. May 18, 2014 #11

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Do you have the formula ##\mathcal L(f(t)u(t-a)) = e^{-as}\mathcal L(f(t+a))##? Your function is
    $$\cos(\omega t)(1-u(t-\pi)) = \cos(\omega t) - \cos(\omega t)u(t-\pi)$$Use the above formula on that second term.
     
    Last edited: May 18, 2014
  13. May 19, 2014 #12
    Right, but that just means the second term is e^(−πs)*L[cos(ωt+π)],
    where cos(ωt+π), using the identity for cos(x+y), = cos(−ωt), no?
     
  14. May 19, 2014 #13

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Not quite. If ##f(t) = \cos(\omega t)## then ##f(t+\pi) = \cos(\omega(t+\pi))=\cos(\omega t + \omega \pi)##. Not quite what you wrote and this will get you to the correct answer.
     
  15. May 19, 2014 #14
    OK I almost got it. One last thing: Why is it cos(ω(t+π)) and not cos(ω(t-π))?
     
  16. May 19, 2014 #15

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Because of the above formula. If what you are really asking is where that formula comes from, try working out$$
    \mathcal L(f(t)u(t-a))=\int_0^\infty e^{-st}f(t)u(t-a)~dt$$
     
  17. May 19, 2014 #16
    Right, I understand the formula, but I thought that at the end, when I have completed the inverse Laplace and put the UnitStep function back into the formula for y(s), wherever there's a "t," I should put a "t-pi"? Which appears contradictory to placing "t+pi" here.
     
  18. May 19, 2014 #17

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That is a different question. For doing inverses you need the formula for ##\mathcal L^{-1}
    (e^{-as}F(s))##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Laplace Transform to Find Solution
Loading...