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Any idea how to solve this equation:

$ \ddot \sigma - p e^\sigma - q e^{2\sigma} =0 ~~~~~~~ (1)$

Or

$ \frac{d^2 \sigma}{dt^2} - p e^\sigma - q e^{2\sigma} =0 $

Where p and q are constants.

Edit

Here's what I have tried:

Let $\sigma = log ~ r$, then: $ \dot \sigma= \frac{\dot r}{ r} , ~ \text{and}~ \ddot\sigma= \frac{\ddot r}{ r} - \frac{\dot r^2}{r^2}$. Sub in (1)

$ \frac{\ddot r}{ r} - \frac{\dot r^2}{r^2} - r^2 q - r p =0 ~~~~~~ (2)$

Now to solve (2), will I use something like $ r = e^{\lambda t}$ again ?

Then (2) becomes:

$ \lambda^2 - \lambda^2 - e^{\lambda t} q - p =0 ~~~~ (3) $

$ \therefore \lambda = \frac{1}{t}~ log~ \frac{p}{q} $, or

$ r = \frac{p}{q} $ and

$ \sigma = log \frac{p}{q} $. This solution can not be, cause it means $ \dot \sigma = \ddot \sigma =0!!! $

Have I missed something??

Thanks.

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    $\begingroup$ substituting $\sigma=\log\rho$ gets rid of the exponentials $\endgroup$ – Federico May 14 at 10:55
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    $\begingroup$ Multiply by $\dot{\sigma}$ and use the reverse product rule to get $$\frac{1}{2} \dot{\sigma}^{2} - p e^{\sigma} - \frac{q}{2} e^{2 \sigma} = 0$$ then separate and integrate. $\endgroup$ – mattos May 14 at 12:10
  • $\begingroup$ @Federico, hey, please may you look at the edit now. I think I still can't get the right solution of $ \sigma$. Thanks $\endgroup$ – Dr. phy May 14 at 12:40
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$$\sigma''=pe^{\sigma}+qe^{2\sigma}\Big{|}\times\sigma' \implies \sigma'\sigma''=\sigma'(pe^{\sigma}+qe^{2\sigma})$$ $$\implies \frac{1}{2}\frac{d}{dt}(\sigma')^{2}=\frac{d}{dt}(pe^{\sigma}+qe^{2\sigma}/2)\implies (\sigma')^{2}=pe^{\sigma}+\frac{q}{2}e^{2\sigma}+r$$ $$\implies t+s=\pm\int^{\sigma}d\sigma'\frac{1}{\sqrt{pe^{\sigma}+\frac{q}{2}e^{2\sigma}+r}}$$ $$=\frac{\sigma-\log[2r+pe^{\sigma}+2\sqrt{r(r+pe^{\sigma}+be^{2\sigma})}]}{\sqrt{r}}$$ Here $r, s$ are integration constants.

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