Exact solution of scalar field cosmology with exponential potential Considering a homogenious and isotropic universe described by the Robertson-Walker metric:
$$ ds^2= dt^2-a^2(t)\left[\frac{dr^2}{1-kr^2}+r^2(d\theta^2+\sin^2(\theta)d \phi ^2)\right]$$
and also assuming that this universe is dominated by a scalar field $\phi (t)$ and a potential of the form $ V(\phi)=V_0e^{- \lambda \phi}$ we can obtain the equations:
The $00$ component of Einstein´s equation is the Friedmann equation:
$$ H^2 + \frac{k}{a^2}= \frac{8 \pi G}{3}\left[\frac{1}{2} \dot\phi^2+V(\phi)\right]$$
where $ H(t)=\frac{\dot a(t)}{a(t)} $ is the Hubble parameter and the equation for the field $\phi$ is:
$$  \ddot\phi+3H\dot\phi-\lambda V_0e^{-\lambda \phi}=0  $$
Solving those 2 equations, considering ($8 \pi G=1$) and ($k=0$), yields the following results:
$$ a(t)=a_0 t^{P} \space \space \space, \space\space\space P=\frac{2}{\lambda^2}\space\space\space\space\space\space (1)$$
$$ \phi=\phi_0 + \frac{2}{\lambda}\ln \left( \frac{t}{t_0}\right)\space\space\space\space\space\space\space\space (2)$$
I tried to obtain them but I can´t get to those solutions $(1)$ and $(2)$. Can someone show me how to do it?
 A: So we can apply the slow-roll condition/approximation. It is given as
$$\dot{\phi}/2 \ll V(\phi)$$ When you take the derivative of both sides with respect to $\phi$, you get another identity.
$$\ddot{\phi} \ll \frac{dV(\phi)}{d\phi}$$
Now, when you apply these conditions to the above equations you end up with
$$3H\dot{\phi} = \lambda V(\phi)~~~~(1)$$
$$3H^2 = V(\phi)~~~~(2)$$
From now on its algebra actually (which I failed at first). But let me do a couple more steps. So at this point, by inserting one in another, you'll obtain two equations
$$\dot{\phi} = \lambda H~~~~(3)$$
$$\dot{\phi} = \frac{\lambda \sqrt{3V(\phi)}}{3}~~~~(4)$$
After this, by using $H = \frac{da/dt}{a}$ in equation $(3)$
So we can apply the slow-roll condition/approximation. It is given as
$$\dot{\phi}/2 \ll V(\phi)$$ When you take the derivative of both sides with respect to $\phi$, you get another identity.
$$\ddot{\phi} \ll \frac{dV(\phi)}{d\phi}$$
Now, when you apply these conditions to the above equations you end up with
$$3H\dot{\phi} = \lambda V(\phi)~~~~(1)$$
$$3H^2 = V(\phi)~~~~(2)$$
From now on its algebra actually (which I failed at first). But let me do a couple more steps. So at this point, by inserting one in another, you'll obtain two equations
$$\dot{\phi} = \lambda H~~~~(3)$$
$$\dot{\phi} = \frac{\lambda \sqrt{3V(\phi)}}{3}~~~~(4)$$
After this, by using $H = \frac{da/dt}{a}$ in equation $(3)$.
Let me do first $(3)$
So we have
$$\frac{d\phi}{dt} = \lambda\frac{da}{dta}$$
$dt$ cancels so we are left with
$$d\phi =\lambda\frac{da}{a}$$
taking integral of both sides
$$\int_{\phi_0}^{\phi}d\phi = \int_{a_0}^a\lambda\frac{da}{a}$$
$$\phi - \phi_0 = \lambda [ln(a) - ln(a_0)] $$
Thus,
$$a(\phi) =a_0 e^{(\phi-\phi_0) / \lambda}~~~~(5)$$
Let us look $(4)$ to find $\phi(t)$
$$\dot{\phi} = \frac{\lambda \sqrt{3V_0e^{-\lambda \phi}}}{3}$$
Let me call $C = \frac{\lambda \sqrt{3V_0}}{3}$
so we obtain,
$$\frac{d\phi}{dt} = Ce^{(-\lambda \phi)/2}$$
$$\frac{d\phi}{e^{(-\lambda \phi)/2}} = Cdt$$
taking integral of both sides,
$$\frac{2}{\lambda}[e^{(\lambda \phi)/2}] = Ct$$
$$e^{(\lambda \phi)/2} = \frac{\lambda}{2}C(t-t_0) + e^{(\lambda \phi_0)/2} $$
Take the ln of both side
$$\frac{\lambda}{2}\phi = ln(C\frac{\lambda}{2}) + ln(t)$$
so
$$\phi = \phi_0 + \frac{2}{\lambda}ln(t)$$
now put this into (5). I know that the equations that you have wrote and this is not exactly the same but this is all I can do
