1. Limited time only! Sign up for a free 30min personal 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 Transforms - Convolution Theorem

  1. Apr 11, 2009 #1

    JJBladester

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    Use the Laplace transform to solve the given initial-value problem.

    [tex]y'' + 9y = cos(3t), y(0)=2, y'(0)=5[/tex]

    2. Relevant equations

    [tex]L\left\{y''\right\} = s^{2}Y(s)-sy(0)-y'(0)[/tex]

    [tex]L\left\{y\right\} = Y(s)[/tex]

    [tex]L\left\{cos(kt)\right\} = \frac{s^{2}}{s^{2} + k^{2}}, s>0[/tex]

    [tex]L\left\{sin(kt)\right\} = \frac{k}{s^{2} + k^{2}}, s>0[/tex]

    [tex]L^{-1}\left\{F(s-a)\right\} = e^{at}f(t)[/tex] (shifting property)

    [tex]L\left\{F(s)G(s)\right\} = f \ast g[/tex] (convolution of f and g)

    3. The attempt at a solution

    [tex]y'' + 9y = cos(3t), y(0)=2, y'(0)=5[/tex]

    [tex]L\left\{y''\right\} + 9L\left\{y\right\} = L\left\{cos(3t)\right\}[/tex]

    [tex]\left[s^{2}Y(s)-sy(0)-y'(0)\right] + 9Y(s) = \frac{s}{s^{2}+9}[/tex]

    [tex]Y(s)\left[s^{2} - 2s + 9\right] = 5 + \frac{s}{s^{2} + 9}[/tex]

    [tex]Y(s) = \frac{5}{s^{2} - 2s + 9} + \frac{s}{(s^{2} - 2s + 9)(s^{2} + 9)}[/tex]

    [tex]\frac{5}{(s-1)^{2} + 8} + \frac{s}{\left[(s-1)^{2} + 8)\right](s^{2} + 9)}[/tex]

    [tex]y(t) = \frac{5}{2\sqrt{2}}L^{-1}\left\{\frac{2\sqrt{2}}{s^{2} + (2\sqrt{2})^{2}}_{s\rightarrow(s-1)}\right\} + L\left\{F(s)G(s)\right\}[/tex] where [tex]F(s) =

    \frac{s}{(s-1)^{2} + 8}[/tex] and [tex]G(s) = \frac{1}{s^{2} + 9}[/tex]

    [tex]f(t) = L^{-1}\left\{\frac{(s-1) + 1}{(s-1)^{2} + (2\sqrt{2})^{2}}\right\} = L^{-1}\left\{\frac{s}{s^{2} + (2\sqrt{2})^{2}}_{s\rightarrow(s-1)}\right\} + \frac{1}{2\sqrt{2}}L^{-1}\left\{\frac{2\sqrt{2}}{s^{2} + (2\sqrt{2})^{2}}_{s\rightarrow(s-1)}\right\} = e^{t}cos(2\sqrt{2}t) + \frac{1}{2\sqrt{2}}\left[e^{t}sin(2\sqrt{2}t)\right][/tex]

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

    [tex]L^{-1}\left\{F(s)G(s)\right\} = f \ast g[/tex]

    And I get stuck here. The book's answer is:

    [tex]y = 2cos(3t) + \frac{5}{3}sin(3t) + \frac{1}{6}tsin(3t)[/tex]
     
  2. jcsd
  3. Apr 11, 2009 #2

    JJBladester

    User Avatar
    Gold Member

    Nevermind... Got it.
     
  4. Apr 11, 2009 #3

    JJBladester

    User Avatar
    Gold Member

    It was a good exercise in Latex, though. Count your positives! Happy Easter everyone.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Laplace Transforms - Convolution Theorem
  1. Laplace Transforms (Replies: 4)

  2. Laplace transformation (Replies: 2)

  3. Laplace transform (Replies: 3)

Loading...