Inserting \(S=S_0e^{mx}\) into the Webster horn equation gives
\begin{align*}
\frac{\partial^2 p }{\partial x^2} + m \frac{\partial p}{\partial x} = \frac{1}{c_0^2}\frac{\partial^2 p}{\partial t^2}\,.
\end{align*}
A solution proportional to \(e^{j(\omega t-kx)}\) is sought, giving
\begin{align*}
-k^2 - jkm &= -\frac{\omega^2}{c_0^2}\\
k^2 + jkm -k_0^2 &=0\\
\implies k &= -j\frac{m}{2} \pm \frac{\omega}{c_0}\sqrt{1-(m/2k_0)^2}
\end{align*}
The phase speed is found by taking the real part of the above and solving for \(\omega/k\):
\begin{align*}
c_\text{ph} = \frac{\omega}{k} = \frac{c_0}{\sqrt{1-(m/2k_0)^2}}
\end{align*}
Writing \(k_0 = \omega/c_0\) gives \(c_\text{ph} = \frac{c_0}{\sqrt{1-(mc_0/2\omega)^2}} = \frac{c_0}{\sqrt{1-(mc_0/4\pi f)^2}}\). The cutoff frequency is identified to be \(f^{(c)}= mc_0/4\pi\), and the phase speed can therefore be written as
\begin{align*}
c_\text{ph} = \frac{c_0}{\sqrt{1-(f^{(c)}/f)^2}}\,.
\end{align*}
By substituting the expression for \(k\) into the form of solution \(e^{j(\omega t-kx)}\), it is seen that the magnitude of the solution grows as
\[|p| \propto e^{-mx/2}.\]
This result could be derived by noting that the power, remains constant as the wave propagates through the horn:
\begin{align*}
W = IS\propto p_\text{rms}^2 S &= \text{constant}\\
\implies p_\text{rms} &\propto 1/\sqrt{S}\\
&\propto 1/\sqrt{e^{mx}}\\
|p|&\propto e^{-mx/2}
\end{align*}