How Does Energy Density Change with Scale Factor in Cosmology?

[cosmoshadow]

Homework Statement

The fluid equation in cosmology is given as:

$$\dot{\epsilon}$$ + 3*($$\dot{a}$$/a)*($$\epsilon$$+P) = 0

Where $$\epsilon$$ is the energy density and a(t) is a scale factor.

Using the equation of state, P = w*$$\epsilon$$, show how $$\epsilon$$ change with a(t).

Homework Equations

$$\dot{\epsilon}$$ + 3*($$\dot{a}$$/a)*($$\epsilon$$+P) = 0
P = w*$$\epsilon$$

The Attempt at a Solution

I can solve for the equation to the point where I re-arrange it to look like this:

$$\dot{\epsilon}$$/$$\epsilon$$ = -3*(1+w)*($$\dot{a}$$/a)

I do not know how to proceed from here. I know that this equation is supposed to end up like this,

$$\epsilon_w(a)$$ = $$\epsilon_w,0$$*a^-3*(1+w)

but I do not know how to get to this point. Can someone assist me please?

 

[cosmoshadow]

can someone take a look at this? I'm pretty sure its a simple operation that I'm failing to realize.

 

[cosmoshadow]

bump?

 

[scottie_000]

You have your equation
$$\frac{\dot\epsilon}{\epsilon} = -3(w+1)\frac{\dot a}{a}$$
From here you can eliminate the time-dependence
$$\frac{d\epsilon}{\epsilon} = -3(w+1)\frac{da}{a}$$
and this is a differential equation involving just $$\epsilon$$ and $$a$$ you can solve by integrating both sides