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.