I have the following equation:

$\displaystyle r''-\frac{2r'^2}{r}-r=0$

I know that the solution is of the form:

$\displaystyle r(\theta)=c_2\sec(c_1+\theta)$

... but only because I used WolframAlpha......

I am not really sure how else to solve this. Any hints, or explicit solutions would be much appreciated.

  • 1
    $\begingroup$ The substitution $u=r'$ gives $r''={du\over d\theta}={du\over dr}{dr\over d\theta}=u{du\over dr}$. Your equation then becomes $u{du\over dr}-{2u^2\over r}-r=0 $, or ${du\over dr}-{2\over r}u={r\over u}$. This is a Bernoulli equation. The substitution $q=u^{2}$ will reduce it to a linear equation. $\endgroup$ – David Mitra Feb 6 '13 at 2:27

You must have done something wrong. That is not a solution of this differential equation.

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  • $\begingroup$ Thanks! Typo... that should be squared $\endgroup$ – Squirtle Feb 6 '13 at 2:19
  • 1
    $\begingroup$ OK, now it's right. $\endgroup$ – Robert Israel Feb 6 '13 at 2:20

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