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I am having trouble finding solutions for the functions e(t) and a(t) for the following system of equations. I have tried solving equation 2 by seperation of variables and got to $e(t) = -1 +\sqrt{1+\frac{c_1}{a^9(t)}}$ while only taking the positive solution. Plugging this into 1 leaves me with an equation that I cant seperate anymore. Also differenting my solution again doesn't fulfill the equation 2. Note: $\dot a = \frac{da}{dt}$ and c1 and k are constants. Other restrictions $a(t)>0$

1.) $\dot a(t)^2 = c_1 e(t)a^2(t)-k$

2.) $\dot e(t)= -3 \frac{\dot a(t)}{a(t)}(\frac{e(t)(e(t)+2)}{e(t)+1})$

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  • $\begingroup$ yes e and a are both dependend on t $\endgroup$
    – trynerror
    Commented Apr 6, 2022 at 14:32
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    $\begingroup$ You can't use separation of variables here, there is only one independent variable $\endgroup$
    – Vasili
    Commented Apr 6, 2022 at 14:38
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    $\begingroup$ Could you also give the original equation that you started from? Perhaps there is a different way. // Not every ODE, even more so for systems, has a symbolic solution. You should be able to give a reason why you would expect a symbolic solution for an equation that does not fall into the usual example classes. $\endgroup$ Commented Apr 6, 2022 at 14:43
  • $\begingroup$ I am trying to solve the 2 Friedmann equations in cosmology the first one is $(\frac{\dot a (t)}{a(t)})^2=c_1 e(t) - \frac{k} {a(t)^2}$ and the second one is $\dot e(t)=-3 \frac{\dot a(t)}{a(t)}(e+p)$. This system has 3 degrees of freedom a(t), e(t) and p(t) with the unusual equation of state (relation between pressure p and energy density e): $p(e)=\frac {e(t)}{e(t)+1}$ this reduces to 2 degrees of freedoms for 2 equations . I plugged my equation of state into the second equation to arrive at my equation 2. Thats where I started. But maybe there is no symbolic solution to this ? $\endgroup$
    – trynerror
    Commented Apr 6, 2022 at 14:54
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    $\begingroup$ @trynerror: express $\frac{\dot a}{a}$ from the first equation and plug it into second. Should be able to take the resulting integral. $\endgroup$
    – Vasili
    Commented Apr 6, 2022 at 14:58

1 Answer 1

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From

$$ \dot e(t)= -3 \frac{\dot a(t)}{a(t)}\left(\frac{e(t)(e(t)+2)}{e(t)+1}\right) $$

we have

$$ \frac{(e(t)+1)\dot e(t)}{e(t)(e(t)+2)}+3\frac{\dot a(t)}{a(t)}=0 $$

and then

$$ \frac 12\ln\left(e(t)(e(t)+2)\right)+3\ln a(t) = c_0 $$

so

$$ e(t)(e(t)+2)a(t)^6 = c_1 $$

etc.

NOTE

$$ \dot u(t)^2=f(u(t)) $$

is separable

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