How Do You Derive u=p/ρ₀c₀ and ρ=p/c₀² from 1D Wave Equations?

[enc08]

Given
The 1D wave equations
$$p_{x}'' - (1/c_{0}^2)p_{t}'' = 0$$
$$u_{x}'' - (1/c_{0}^2)u_{t}'' = 0$$
$$ρ_{x}'' - (1/c_{0}^2)ρ_{t}'' = 0$$

and linearised continuity and momentum equations
$$ρ_{t}' = -ρ_{0}u_{x}'$$, $$ρ_{0}u_{t}'=-p_{x}$$

how may one derive the following two equations?

$$u=p/ρ_{0}c_{0}$$, $$ρ=p/c_{0}^2$$

My notes jump from the first equation to the last two.

Thanks for any input.

 

[enc08]

Bump.

 

[haruspex]

$$ρ_{0}u_{t}'=-p_{x}$$

That should read
$$ρ_{0}u_{t}'=-p'_{x}$$ yes?
I'll use the subscripts only, discarding the ', doubling the subscript for 2nd derivative.
Don't know whether this helps, but investigating u_xt gives c₀²p_xx = c₀²ρ_tt = p_tt = ρ_xx.

 

[enc08]

Hi,

That's right, a single/double prime denotes a single/double derivative.

Could you suggest how to obtain $$u = p/\rho_{0}c_{0}$$?

Thanks.