Non-linear coupled differential equations I was trying to solve these coupled differential equations but can´t quite get to the solution. The differential equations are:
$$ H^2= \frac{1}{3}\left[\frac{1}{2} \dot\phi^2+V(\phi)\right] \space\space\space\space\space(1)$$
$$  \ddot\phi+3H\dot\phi-\lambda V_0e^{-\lambda \phi}=0  \space\space\space\space\space\space (2)$$
where $ H(t)=\frac{\dot a(t)}{a(t)} $ , $\lambda$ and $V_0$ are constants and the dot notation represents $\space \dot a= \frac{da}{dt}$. The respective solutions of $a(t)$ and $\phi(t)$ are the following:
$$ a(t)=a_0 t^{P} \space \space \space, \space\space\space P=\frac{2}{\lambda^2}\space\space\space\space\space\space (3)$$
$$ \phi(t) =\phi_0 + \frac{2}{\lambda}\ln \left( \frac{t}{t_0}\right)\space\space\space\space\space\space\space\space (4)$$
I was told to do a power law solution for $a(t)$ and $\psi(t)$, given that $\psi(t)=e^{- \lambda \phi (t)}$ , but making that substitution $ \psi(t)$ I got an equation that seems even worse:
$$ -\ddot \psi \psi^2 \left(\frac{1}{\lambda} \right) + \left( \frac{\dot \psi}{\lambda}- \frac{3H \psi}{\lambda}  \right) \dot \psi = \lambda V_0 \psi^3$$
I tried using aproximations too but to no luck. Any help would be appreciated.
 A: This question is not answerable without knowing $V(\phi)$. In your similar post on Physics SE, you mentioned that $V(\phi)=V_0e^{-\lambda\phi}$ so I will assume that.
The two equations to be solved are
$$\left(\frac{\dot a}{a}\right)^2=\frac13\left(\frac12\dot\phi^2+V_0e^{-\lambda\phi}\right)\tag{1a}$$
and
$$\ddot\phi+3\left(\frac{\dot a}{a}\right)\dot\phi-\lambda V_0e^{-\lambda\phi}=0.\tag{1b}$$
The presence of the exponential $e^{-\lambda\phi}$ complicates these equations so change variables from $\phi$ to
$$\psi\equiv e^{-\lambda\phi}.\tag2$$
In terms of $\psi$ we have
$$\phi=-\frac{1}{\lambda}\ln\psi\tag{3a}$$
$$\dot\phi=-\frac{1}{\lambda}\frac{\dot\psi}{\psi}\tag{3b}$$
$$\ddot\phi=-\frac{1}{\lambda}\left[\frac{\ddot\psi}{\psi}-\left(\frac{\dot\psi}{\psi}\right)^2\right]\tag{3c}$$
so equations (1) become
$$3\left(\frac{\dot a}{a}\right)^2=\frac{1}{2\lambda^2}\left(\frac{\dot\psi}{\psi}\right)^2+V_0\psi\tag{4a}$$
and
$$-\frac{1}{\lambda}\left[\frac{\ddot\psi}{\psi}-\left(\frac{\dot\psi}{\psi}\right)^2\right]-\frac{3}{\lambda}\left(\frac{\dot a}{a}\right)\left(\frac{\dot\psi}{\psi}\right)-\lambda V_0\psi=0.\tag{4b}$$
These look complicated, but let's see if they have a simple power-law solution of the form
$$a=a_0t^p\tag{5a}$$
$$\psi=\psi_0t^q\tag{5b}.$$
It seems reasonable to try this because all the terms will be just powers of $t$ and maybe we can make those powers work out to satisfy the equations. We have
$$\frac{\dot a}{a}=\frac{p}{t}\tag{6a}$$
$$\frac{\dot\psi}{\psi}=\frac{q}{t}\tag{6b}$$
$$\frac{\ddot\psi}{\psi}=\frac{q(q-1)}{t^2}\tag{6c}$$
so equations (4) become
$$\frac{3p^2}{t^2}-\frac{q^2}{2\lambda^2t^2}=V_0\psi_0t^q\tag{7a}$$
$$\frac{q}{\lambda t^2}-\frac{3pq}{\lambda t^2}=\lambda V_0\psi_0t^q\tag{7b}.$$
Now look at the dependence of each term on $t$. These equations can only be satisfied if
$$q=-2\tag{8}$$
and, with this value of $q$, equations (7) become
$$3p^2-\frac{2}{\lambda^2}=V_0\psi_0\tag{9a}$$
$$-\frac{2}{\lambda}+\frac{6p}{\lambda}=\lambda V_0\psi_0\tag{9b}.$$
We can eliminate $V_0\psi_0$ by taking (9a) and subtracting (9b) divided by $\lambda$ from it. This gives
$$3p^2-\frac{6p}{\lambda^2}=0\tag{10a}$$
or
$$p=\frac{2}{\lambda^2}.\tag{10b}$$
So we have our solution!
$$a(t)=a_0t^{2/\lambda^2}\tag{11a}$$
$$\phi(t)=-\frac{1}{\lambda}\ln(\psi_0t^{-2})=-\frac{1}{\lambda}\ln\psi_0+\frac{2}{\lambda}\ln t\tag{11b}.$$
This is equivalent to the form you gave,
$$\phi=\phi_0+\frac{2}{\lambda}\ln\frac{t}{t_0}\tag{12}$$
if we take
$$\phi_0=\frac{2}{\lambda}\ln{t_0}-\frac{1}{\lambda}\ln{\psi_0}.\tag{13}$$
