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Ordinary Differential Equation - Comparing 2 Solutions

  1. Apr 27, 2014 #1
    1. The problem statement, all variables and given/known data

    [tex]{g}'\left ( s \right )+\mu g\left ( s \right )={f}'\left ( -s \right )+\mu f\left ( -s \right )[/tex]
    Integrate up to get
    [tex]g\left ( s \right )=-f\left ( -s \right )+2\mu e^{-\mu s}\int_{-s}^{\infty }e^{-\mu {s}'}f\left ( {s}' \right )d{s}'[/tex]

    2. Relevant equations

    As above

    3. The attempt at a solution

    I've seen a few attempts at this, one that arrives at the correct solution, which is the method i used and had no trouble with. However, another attempt i've seen, which doesn't yield the correct solution is causing me much headache as i can't work out where it's going wrong. Some pointing out here would be really helpful. Thanks

    Using the following integrating factor on the equation above
    [tex]e^{\int \mu ds}=e^{\mu s}[/tex]
    [tex]e^{\mu s}{g}'\left ( s \right )+e^{\mu s}\mu g\left ( s \right )=e^{\mu s}\left [ {f}'\left ( -s \right )+\mu f\left ( -s \right ) \right ][/tex]
    At this point i wrote
    [tex]\frac{d}{ds}\left [ e^{\mu s}g\left ( s \right ) \right ]=e^{\mu s}\left [ {f}'\left ( -s \right )+\mu f\left ( -s \right ) \right ][/tex]
    [tex]\left. e^{\mu {s}'}g\left ( {s}' \right ) \right|_{-\infty }^{s}=\int_{-\infty }^{s}e^{\mu {s}'}\left [ {f}'\left ( {-s}' \right )+\mu f\left ( {-s}' \right ) \right ]d{s}'[/tex]
    Then to skip a few steps i did by parts integration to get the following (can include missing steps if needed)
    [tex]e^{\mu s}g\left ( s \right )=-e^{\mu s}f\left ( -s \right )+ 2\mu \int_{-\infty }^{s}f\left ( {-s}' \right )e^{\mu {s}'}d{s}'[/tex]
    [tex]g\left ( s \right )=-f\left ( -s \right )+ 2\mu e^{-\mu s}\int_{-\infty }^{s}f\left ( {-s}' \right )e^{\mu {s}'}d{s}'[/tex]
    Swapping limits and changing 's' for '-s' gives
    [tex]g\left ( s \right )=-f\left ( -s \right )+ 2\mu e^{-\mu s}\int_{-s}^{\infty }f\left ( {s}' \right )e^{-\mu {s}'}d{s}'[/tex]
    As required


    The other route wrote took the following path
    [tex]\frac{d}{ds}\left [ e^{\mu s}g\left ( s \right ) \right ]e^{-\mu s}=\frac{d}{ds}\left [ f\left ( -s \right )e^{-\mu s} \right ]e^{\mu s}[/tex]
    [tex]\frac{d}{ds}\left [ e^{\mu s}g\left ( s \right ) \right ]=-\frac{d}{ds}\left [ f\left ( -s \right )e^{-\mu s} \right ]e^{2\mu s}[/tex]
    [tex]\left. e^{\mu {s}'}g\left ( {s}' \right ) \right|_{-\infty }^{s}=\int_{-\infty }^{s}-e^{2\mu {s}'}\frac{d}{ds}\left [ f\left ( {-s}' \right )e^{-\mu {s}'} \right ]d{s}'[/tex]
    [tex]e^{\mu s}g\left ( s \right )=\left. -e^{2\mu {s}'}f\left ( {-s}' \right )e^{-\mu {s}'}\right|_{-\infty }^{s}+ \int_{-\infty }^{s}2\mu e^{2\mu s}f\left ( {-s}' \right )e^{-\mu {s}'}d{s}'[/tex]
    [tex]e^{\mu s}g\left ( s \right )=-e^{\mu s}f\left ( -s \right )+ 2\mu e^{2\mu s}\int_{-\infty }^{s}f\left ( {-s}' \right )e^{-\mu {s}'}d{s}'[/tex]
    [tex]g\left ( s \right )=-f\left ( -s \right )+ 2\mu e^{\mu s}\int_{-\infty }^{s}f\left ( {-s}' \right )e^{-\mu {s}'}d{s}'[/tex]
    Swapping limits and changing 's' for '-s' gives
    [tex]g\left ( s \right )=-f\left ( -s \right )+ 2\mu e^{\mu s}\int_{-s}^{\infty }f\left ( {s}' \right )e^{\mu {s}'}d{s}'[/tex]
    This is giving a sign swap in the final solutioin that is having my head spin. Can't see the route of this. Can anyone spot the mistake?
     
  2. jcsd
  3. Apr 27, 2014 #2

    HallsofIvy

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    Staff Emeritus
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    Swapping [itex]\int_{-s}^\infty ds'[/itex] for [itex]\int_{-\infty}^s ds'[/itex] is equivalent to making the substitution [itex]u= -s'[/itex]. Then [itex]du= -ds'[/itex]. When [itex]s'= -\infty[/itex], [itex]u= \infty[/itex] and when [itex]s'= s[/itex], [itex]u= -s[/itex].

    So [tex]\int_{-s}^\infty f(-s')e^{-\mu s'}ds'= \int_u^{-\infty}f(u)e^{\mu u}(-du)= \int_{-\infty}^u f(u)e^{\mu u}du[/tex]
     
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