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Solving DE- which approach and more importantly - why

  1. Apr 18, 2013 #1
    My end goal is to solve for [itex]G(m)[/itex] in terms of the other functions, but first I have to solve the DE :

    [itex]\frac{d}{dm}[F(m) G(m)] = (\frac{d}{dm}F(m)) D(m)[/itex].

    What I've done is to say (using integration by parts)

    [itex]F(m) G(m) = F(m)D(m) - \int F(m) (\frac{d}{dm}D(m)) dm[/itex].

    This is one method I tried. Another approach would be a separation of variables, which I tried as...

    [itex]\frac{d}{dm}[F(m) G(m)] = \frac{dF}{dm} G + F \frac{dG}{dm} = \frac{dF}{dm} D [/itex]

    which leads to...

    [itex] \ln |F| = \int \frac{dG}{D - G} dm [/itex] and I believe that the RHS cannot be evaluated without specifying functions [itex]D[/itex] and [itex]G[/itex] (if the RHS can be evaluated without specification - please let me know...and please show me the way).

    Help on understanding the correct method would be extremely appreciated.

    Thanks in advance.
  2. jcsd
  3. Apr 18, 2013 #2

    Simon Bridge

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    If you just need an expression for G in terms of F and D, then the first one looks more like it gets you there.

    I noticed that $$\renewcommand{\dd}[2]{\frac{d #1}{d #2}}
    \renewcommand{\dm}[1]{\frac{d #1}{dm}}

    G\dm{F}+ F\dm{G} = D\dm{F}$$ rearranges to $$\dm{G}= \frac{D-G}{F}\dm{F}$$ ... integrate both sides wrt m ... but it's early and I havn't had my coffee yet...
  4. Apr 19, 2013 #3
    Hi Simon, thanks for your reply.

    Thanks for your suggestion, but I think I'll stick with the integration by parts method, since I have convenient expressions for [itex]F(m) (\frac{d}{dm}D(m))[/itex] that I will need to integrate.

    Once again, thanks for the help.
  5. Apr 20, 2013 #4
    Hi !
    it's a separable EDO :

    Attached Files:

  6. Apr 20, 2013 #5

    Simon Bridge

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    Yeah - that's where I was going ;)
    (Once I'd had my coffee...)
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