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!

Finding an inverse Laplace Transform for a function - solving IVPs with Laplace

  1. Jul 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Use Laplace Transforms to solve the following initial value problems

    a. [tex]t\,y''\, - \,t\,y'\, + \,y\; = \;2\;\;\;y(0)\;=\;2\;\;\;y'(0)\;=\;-1[/tex]

    b. [tex]y''\,+\,2\,y'\,-3\,y\;=\;\delta(t\,-\,1)\,-\,\delta(t\,-\,2)\;\;\;y(0)\;=\;2\;\;\;y'(0)\;=\;-2[/tex]



    2. Relevant equations

    Laplace Transforms



    3. The attempt at a solution

    PART A

    [tex] - \frac{d}{{ds}}\left[ {s^2 \,Y\left( s \right)\, - \,s\,y\,\left( 0 \right)\, - \,y'\left( 0 \right)} \right]\, + \,\frac{d}{{ds}}\left[ {s\,Y\left( s \right)\, - \,y\,\left( 0 \right)} \right]\, + \,Y\left( s \right)\; = \;\frac{2}{s}[/tex]

    [tex] - \frac{d}{{ds}}\left[ {s^2 \,Y\left( s \right)\, - \,2\,s\, + \,1} \right]\, + \,\frac{d}{{ds}}\left[ {s\,Y\left( s \right)\, - \,2} \right]\, + \,Y\left( s \right)\; = \;\frac{2}{s}[/tex]

    [tex]\begin{array}{l}
    - \left[ {2\;s\;Y\left( s \right)\; + \;s^2 \;Y'\left( s \right)\; - 2} \right]\; + \;\left[ {Y\left( s \right)\; + \;s\;Y'\left( s \right)} \right]\; + \;Y\left( s \right)\; = \;\frac{2}{s} \\
    Y'\left( s \right)\;\left( { - s^2 \; + \;s} \right)\; + \;Y\left( s \right)\;\left( { - 2\;s\; + \;2} \right)\; = \;\frac{2}{s}\;\;\;\; \to \;\;\;\;Y'\; + \;Y\;\left( {\frac{{2\; - \;2\;s}}{{ - s^2 \; + \;s}}} \right)\; = \; - \frac{2}{{s^3 \; - \;s^2 }} \\
    \end{array}[/tex]

    [tex]\begin{array}{l}
    \mu \left( s \right)\; = \;e^{\int {\frac{2}{s}\;ds} } \; = \;s^2 \;\;\;\; \to \;\;\;\;\frac{d}{{ds}}\left\{ {\mu \left( s \right)\;Y\left( s \right)} \right\}\; = \;\mu \left( s \right)\;Q\left( s \right) \\
    \mu \left( s \right)\;Y\left( s \right)\; = \;\int { - \frac{2}{{s\; - \;1}}} \;ds\;\;\;\; \to \;\;\;\;Y\left( s \right)\; = \;\frac{{ - 2\;\ln \left( {s\; - \;1} \right)}}{{s^2 }}\; + \;\frac{C}{{s^2 }} \\
    \end{array}[/tex]

    How do I do an inverse transform for

    [tex]\frac{{ - 2\;\ln \left( {s\; - \;1} \right)}}{{s^2 }}\; + \;\frac{C}{{s^2 }}[/tex]


    PART B

    [tex]\begin{array}{l}
    \left[ {s^2 \,Y\left( s \right)\, - \,s\,y\,\left( 0 \right)\, - \,y'\left( 0 \right)} \right]\, + \,2\left[ {s\,Y\left( s \right)\, - \,y\,\left( 0 \right)} \right]\, - \,3\,Y\left( s \right)\; = \;e^{ - s} \; - \;e^{ - 2\,s} \\
    Y\,\left( s \right)\left( {s^2 \; + \;2\,s\; - \;3} \right)\; - \;2\,s\; - \;2\; = \;e^{ - s} \; - \;e^{ - 2\,s} \;\;\;\; \to \;\;\;\;Y\,\left( s \right)\; = \;\frac{{e^{ - s} \; - \;e^{ - 2\,s} \; + \;2\,s\; + \;2}}{{s^2 \; + \;2\,s\; - \;3}}\end{array}[/tex]

    [tex]Y\,\left( s \right)\; = \;\frac{{e^{ - s} \; - \;e^{ - 2\,s} \; + \;2\,s\; + \;2}}{{\left( {s\; - \;1} \right)\,\left( {s\; + \;3} \right)}}[/tex]

    How would I go about the partial fraction expansion of the last expression?
     
    Last edited: Jul 26, 2007
  2. jcsd
  3. Jul 26, 2007 #2
    [tex]f(t) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}F(s)e^{st} dt[/tex]
     
  4. Jul 26, 2007 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    These are meaningless. The Laplace transform of a derivative does not involve a derivative. You seem to be writing "d/dx" of the Laplace transform of the derivative. If that is true you do not want the "d/dx" in the expression.

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Finding an inverse Laplace Transform for a function - solving IVPs with Laplace
Loading...