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I'm trying to solve this second order differential equation, but it seems my solution isn’t accurate, since i could not find the correct $\sigma$ in $$ \ddot \sigma - p e^\sigma - q e^{2\sigma} =0\qquad\qquad(1)$$ or $$\frac{d^2 \sigma}{dt^2} - p e^\sigma - q e^{2\sigma} =0$$ where $p$ and $q$ are constants. So any help is appreciated.

Here's what I have tried:

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

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

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

Then (2) becomes:

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

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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    $\begingroup$ This is not linear, forget about homogeneous. You cannot use the ansatz $r=\exp(\lambda t)$ $\endgroup$ – Shubham Johri May 15 at 15:30
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You can try to reduce the order of the DE: $$\ddot \sigma - p e^\sigma - q e^{2\sigma} =0$$ $$2\dot\sigma\ddot \sigma - 2\dot\sigma p e^\sigma -2\dot \sigma q e^{2\sigma} =0$$ Integrate: $$\dot \sigma^2- 2p e^\sigma -q e^{2\sigma} =C_1$$ $$\dot \sigma=\pm \sqrt { 2p e^\sigma +q e^{2\sigma} +C_1}$$ $$\int \dfrac {d\sigma}{ \sqrt { 2p e^\sigma +q e^{2\sigma} +C_1}}=\pm\int dt$$

Then substitute $e^{\sigma}=u$.

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  • $\begingroup$ Then I will have again, $\frac{\dot u^2}{u^2} -2 p u + 2 q u =0$ it's nonlinear DE as “shubham mentioned before, so I think I will not use the exponential again to proceed in solving. $\endgroup$ – Dr. phy May 16 at 17:35
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    $\begingroup$ No take the root on both side then integrate with substitution method@Dr.phy $\endgroup$ – MtGlasser May 16 at 17:39
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    $\begingroup$ I added more lines check my answer @Dr.phy $\endgroup$ – MtGlasser May 16 at 17:43
  • $\begingroup$ Fine, thanks for your answer @Aryadeva. $\endgroup$ – Dr. phy May 16 at 18:09
  • $\begingroup$ You're welcome @Dr.phy $\endgroup$ – MtGlasser May 16 at 18:15

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