Fluids pressure and Perturbation Suppose the fluid pressure is given by $p = p_0 +c^2(ρ−ρ_0)$, where $p_0, ρ_0, $and  c are positive constants. Consider small perturbations to a uniform rest state with density $ρ_0$, so that ρ = $ρ_0 + ρ'$. Show that ρ' and the velocity potential φ satisfy the wave equations
$$ \frac{∂^2φ}{∂_t^2}−c^2∇^2φ=0$$
$$\frac{∂^2\rho'}{∂t^2}−c^2∇^2ρ'=0,$$ and find a relation between φ and ρ′.
My idea is to use the Euler equation $\frac{∂\rho}{∂t}$+$\nabla \cdot (\rho \mathbf u)=0$ and $\rho\frac {D\mathbf u}{Dt}=-\nabla p$ But how to use the relationship between p and $\rho$
Plug in to the equation I have:
first equation$$\frac{∂\rho'}{∂t}+\nabla \bullet (\rho'\mathbf u)=0$$
second equation$$(\rho_0+\rho')(\frac{∂\mathbf u}{∂t}+(\mathbf u\bullet\nabla)\mathbf u)=-\nabla c^2\rho'$$ Then how should I continue?
 A: This is a simple application of perturbation theory to the Euler equations. So, you will have
$$
\rho=\rho_0+\epsilon\rho'
$$
and
$$
{\bf u}={\bf u}_0-\epsilon\nabla\phi.
$$
I have introduced $\epsilon$ as a bookeeping and I will set it to 1 to the end of the computation. Now, from the continuity equation
$$
  \epsilon\frac{\partial\rho'}{\partial t}+\nabla[(\rho_0+\epsilon\rho')({\bf u}_0-\epsilon\nabla\phi)]=0
$$
Then, expand and take all the derivative of $\rho_0$ and ${\bf u}_0$ to 0. You will get
$$
   \epsilon\frac{\partial\rho'}{\partial t}+
   \nabla\cdot(\rho_0{\bf u}_0)+
   {\epsilon}{\bf u}_0\cdot\nabla\rho'-\epsilon\rho_0\Delta_2\phi+O(\epsilon^2)=0.
$$
From the other equation you will get
$$
   -\epsilon\nabla\frac{\partial\phi}{\partial t}
-\epsilon{\bf u}_o\cdot\nabla(\nabla\phi)=-c^2\frac{\nabla\rho'}{\rho_0}+O(\epsilon^2).
$$
where I used the equation $p=p_0+c^2(\rho-\rho_0)$ on the right hand side. Now, ${\bf u}_0$ is a constant vector and we can take it to be 0 without loss of generality (we are studying sound waves in an otherwise static fluid) and so, we have the first order approximation
$$
\frac{\partial\phi}{\partial t}=c^2\frac{\rho'}{\rho_0}
$$
where I have chosen an integration contribution being 0. If you put this in the continuity equation at the same order you will get
$$
  \frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}-\Delta_2\phi=0
$$
as required. You see that several approximations have been done in different stages of this derivation. To get the equation for the density, just derive the continuity equation with respect to time and use the equation between $\rho'$ and $\phi$ and you are done.
ADDED ON OP'S REQUEST: To derive the second equation I have to consider
$$
   \epsilon\frac{\partial\rho'}{\partial t}+
   \nabla\cdot(\rho_0{\bf u}_0)+
   {\epsilon}{\bf u}_0\cdot\nabla\rho'-\epsilon\rho_0\Delta_2\phi+O(\epsilon^2)=0.
$$
that becomes
$$
   \epsilon\frac{\partial\rho'}{\partial t}+\epsilon\rho_0\Delta_2\phi+O(\epsilon^2)=0.
$$
Now, from the other Euler equation is
$$
\frac{\partial\phi}{\partial t}=c^2\frac{\rho'}{\rho_0}. \qquad   {\bf (1)}
$$
Then, I take the derivative of the first yielding at order $\epsilon$
$$
   \frac{\partial^2\rho'}{\partial t^2}+\rho_0\Delta_2\frac{\partial\phi}{\partial t}=0.
$$
and using eq.(1) one has
$$
  \frac{1}{c^2}\frac{\partial^2\rho'}{\partial t^2}-\Delta_2\rho'=0.
$$
