Second Order DE

  • Thread starter awetawef
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Homework Statement



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.

Homework Equations



See above.

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)?
 

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