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Laplace equations w/ discont. forcing functions

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]y'' + 3y' + 2y = u_2(t); y(0)=0, y'(0)=1[/tex]


    2. Relevant equations
    [tex]L\{u_c(t)h(t-c)\}=e^{-cs}H(s)[/tex]


    3. The attempt at a solution
    I've actually tried two laplace equation with discontinuous forcing function problems. I get answers somewhat close to the back of the book, but still off. Maybe I'm making a dumb algebra mistake? Could someone check my work over?

    [tex]
    \begin{aligned}
    & L\{f(t)\}(s^2+3s+2)-1=\frac{e^{-2s}}{s}\\
    & L\{f(t)\}=\frac{1+e^{-2s}}{(s)(s^2+3s+2)}
    \end{aligned}
    [/tex]

    [tex]
    \begin{aligned}
    f(t) & =L^{-1}\{\frac{1}{(s)(s^2+3s+2)}\} + L^{-1}\{\frac{e^{-2s}}{(s)(s^2+3s+2)}\} \\
    f(t) & =L^{-1}\{\frac{1}{(s)(s^2+3s+2)}\} + u_2(t)h(t-2)
    \end{aligned}
    [/tex]

    Where [tex]H(s) = \{\frac{1}{(s)(s^2+3s+2)}\}[/tex]

    Partial fraction decomposition:
    [tex]
    L^{-1}\{\frac{1}{(s)(s^2+3s+2)}\} = L^{-1}\{\frac{1/2}{s}\} - L^{-1}\{\frac{1}{s+1}\} + L^{-1}\{\frac{1/2}{s+2}\}[/tex]

    [tex]
    \Rightarrow f(t) = \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2} + u^2(t)h(t-2)[/tex]

    Where [tex]h(t)= \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2}[/tex]

    My book gives the answer as:

    [tex]f(t)=e^{-t}-e^{-2t}+u_2(t)h(t-2)[/tex] where

    [tex]h(t)= \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2}[/tex]

    Any ideas where I messed up? :(
     
    Last edited: Oct 5, 2011
  2. jcsd
  3. Oct 5, 2011 #2
    Sorry about editing the post so many times, I'm still figuring out MathJax.
     
  4. Oct 5, 2011 #3

    vela

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    You made an algebra mistake when you moved the 1 from the LHS to the RHS. The numerator should end up as s+e-2s.
     
  5. Oct 5, 2011 #4
    Thank you! I make the dumbest mistakes. :grumpy:
     
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