solve system of 1st order nonlinear ODEs

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

• yes e and a are both dependend on t Commented Apr 6, 2022 at 14:32
• You can't use separation of variables here, there is only one independent variable Commented Apr 6, 2022 at 14:38
• 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. Commented Apr 6, 2022 at 14:43
• 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 ? Commented Apr 6, 2022 at 14:54
• @trynerror: express $\frac{\dot a}{a}$ from the first equation and plug it into second. Should be able to take the resulting integral. Commented Apr 6, 2022 at 14:58

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