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Second Order DE

  1. Nov 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Not really a homework problem, but more of a project I am working on.

    Solve the following differential equation (if possible):

    [tex]
    C=\frac{\phi ''}{\phi (\phi ' ^2 +1)^2}
    [/tex]

    Where C is a constant, and [itex]\phi [/itex] is a function.

    2. Relevant equations

    See above.

    3. The attempt at a solution

    First, note that
    [tex]
    \left ( \frac{1}{1+\phi ' ^2} \right ) '=\frac{-2\phi ' \phi ''}{(1+\phi ' )^2}
    [/tex]

    So the original equation can be rewritten:

    [tex]
    C=\frac{\phi ''}{\phi (\phi ' ^2 +1)^2}=\frac{1}{-2\phi \phi '} \left ( \frac{-2\phi ' \phi ''}{(1+\phi ')^2} \right )=\frac{1}{-2 \phi \phi '} \left ( \frac{1}{1+\phi ' ^2} \right ) '
    [/tex]

    So

    [tex]
    -2C\phi \phi '=\left ( \frac{1}{1+\phi ' ^2} \right ) '
    [/itex]

    Integrating both sides with respect to the independent variable (call it v)

    [tex]
    \int -2C\phi \phi ' \, dv=\int \left ( \frac{1}{1+\phi ' ^2} \right ) ' dv
    [/tex]

    By the fundamental theorem of calculus, we have
    [tex]
    -C\phi ^2+C_2=\left ( \frac{1}{1+\phi ' ^2} \right ) + C_3 [/tex]
    [tex] -C\phi ^2=\left ( \frac{1}{1+\phi ' ^2} \right ) + C_4
    [/tex]

    Rewriting:
    [tex]\phi ' = \sqrt{\frac{1-C\phi ^2 + C_4}{C\phi ^2+C_4}}[/tex]

    I have tried trig substitution to no avail, so I went to look at the slope field on my TI 89, but nothing shows up. It isn't a technical problem, so I was wondering what was going on. Does anyone see anyway I can make further progress? Also, how can I approximate solutions (or just look at them)?
     
  2. jcsd
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