Results 1 to 6 of 6

Math Help - DE's with discontinuous forcing functions

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    19

    DE's with discontinuous forcing functions

    I'm having trouble with a bunch of these kinds of problems:

    y''+4y=sin(t)-u_{2\pi}(t)sin(t-2\pi); \;\;\; y(0)=0, \; y'(0)=0

    So far I've got:

    s^2Y(s)-sy(0)-y'(0)+4Y(s)-y(0)=\frac{1}{s^2+1}-\frac{e^{-2\pi}}{s^2+1}

    \Rightarrow Y(s)=\frac{1}{(s^2+1)(s^2+4)}-\frac{e^{-2\pi}}{(s^2+1)(s^2+4)}

    And I'm laboring with the partial fraction decomposition:

    1=(As+B)(s^2+4)+(Cs+D)(s^2+1)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Observe that (s^2+4)-(s^2+1)=3.

    Pretty sure that someone else's gonna provide more help, 'cause now I'm going to sleep.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    19
    Still don't get it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by petition Edgecombe View Post
    I'm having trouble with a bunch of these kinds of problems:

    y''+4y=sin(t)-u_{2\pi}(t)sin(t-2\pi); \;\;\; y(0)=0, \; y'(0)=0

    So far I've got:

    s^2Y(s)-sy(0)-y'(0)+4Y(s)-y(0)=\frac{1}{s^2+1}-\frac{e^{-2\pi}}{s^2+1}

    \Rightarrow Y(s)=\frac{1}{(s^2+1)(s^2+4)}-\frac{e^{-2\pi}}{(s^2+1)(s^2+4)}

    And I'm laboring with the partial fraction decomposition:

    1=(As+B)(s^2+4)+(Cs+D)(s^2+1)
    (s^2+4) - (s^2+1) = 3

    So if you multiply top and bottom of both terms by 3:

    \Rightarrow Y(s)=\frac{3}{3(s^2+1)(s^2+4)}-\frac{3e^{-2\pi}}{3(s^2+1)(s^2+4)}


    \Rightarrow Y(s)=\frac{(s^2+4) - (s^2+1)}{3(s^2+1)(s^2+4)}-\frac{((s^2+4) - (s^2+1))e^{-2\pi}}{3(s^2+1)(s^2+4)}

    Now split up each fraction into 2 terms

    \Rightarrow \frac{s^2+4}{3(s^2+1)(s^2+4)} - \frac{s^2+1}{3((s^2+1)(s^2+4)} - \frac{(s^2+4)e^{-2 \pi}}{3(s^2+1)(s^2+4)} + \frac{(s^2+1)e^{-2 \pi}}{3(s^2+1)(s^2+4)}

    Easier now?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2008
    Posts
    19
    But what is the Inverse Laplace transform of \frac{1}{s^2+4} ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by petition Edgecombe View Post
    But what is the Inverse Laplace transform of \frac{1}{s^2+4} ?
     \mathcal L (\frac{k}{s^2+a^2}) = k. \frac{sin(at)}{a}

    In this case a = 2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 2nd order DE with periodic forcing
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: April 5th 2011, 06:30 AM
  2. Discontinuous Piecewise functions
    Posted in the Algebra Forum
    Replies: 4
    Last Post: August 30th 2010, 04:28 PM
  3. Discontinuous Functions
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: April 9th 2010, 02:54 PM
  4. Discontinuous functions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 12th 2008, 05:43 AM
  5. Continuous and discontinuous functions
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 24th 2007, 02:38 PM

Search Tags


/mathhelpforum @mathhelpforum