Yes, even without rotation, you would have tides with bulges. The usual approach is to use the shallow water equations and linearize them to get the Laplace tidal equations. In spherical coordinates $(\theta,\phi)$ for a perfectly spherical Earth covered with a small layer of water:
$$
\begin{align}
\partial_t h &= - \frac H{R\sin\theta}\left(\partial_\theta(\sin\theta u_\theta) +\partial_\phi u_\phi\right) \\
\partial_tu_\theta &= -\frac gR\partial_\theta h \\
\partial_tu_\phi &= -\frac g{R\sin\theta}\partial_\phi h.
\end{align}
$$
$h$ is the height displacement (bulges will be positive $h$ values), $u$ is the 2D tangential velocity field. $H=3.7\cdot10^3m$ is the mean ocean depth, $R=6.4\cdot10^6m$ the radius of the Earth, and $g$ is the surface gravity, the usual $9.8m/s^2$. Naturally, $H\ll R$.
Under the influence of an external potential $U$ you need to add the additional terms:
$$
\begin{align}
\partial_t h &= - \frac H{R\sin\theta}\left(\partial_\theta(\sin\theta u_\theta) +\partial_\phi u_\phi\right) \\
\partial_tu_\theta &= -\frac1R\partial_\theta (gh+U) \\
\partial_tu_\phi &= -\frac1{R\sin\theta}\partial_\phi (gh+U).
\end{align}
$$
To solve the linear response, decompose the forcing into harmonic components by Fourier transform, then expand in spherical scalar/vector harmonics:
$$
h = e^{i\omega t}\sum_{l=0}^\infty\sum_{m=-l}^lh_{lm}Y_{lm} \quad U = e^{i\omega t}\sum_{l=0}^\infty\sum_{m=-l}^lU_{lm}Y_{lm} \\
u = e^{i\omega t}\sum_{l=0}^\infty\sum_{m=-l}^lu_{lm}^{(1)}\Psi_{lm}+u_{lm}^{(2)}\Phi_{lm}.
$$
The equations are now algebraic:
$$
\begin{align}
i\omega h_{lm} &= \frac HRl(l+1)u_{lm}^{(1)} \\
i\omega u_{lm}^{(1)} &= -\frac1R(gh_{lm}+U_{lm}) \\
i\omega u_{lm}^{(2)} &= 0
\end{align}
$$
and easily solvable:
$$
\begin{align}
h_{lm} &= -\frac{\omega_l^2}{\omega_l^2-\omega^2}U_{lm} & u_{lm}^{(1)} &= -\frac{i\omega}{\omega_l^2-\omega^2}\frac{U_{lm}}R \\
u_{lm}^{(2)} &= 0 & \omega_l &= \frac{\sqrt{l(l+1)gH}}R
\end{align}
$$
with $\omega_l$ the resonant frequency of the mode. In particular, for $\omega=0$ you recover the Newton's static theory of tides:
$$
h = -U \quad u = 0.
$$
Physically, the water piles up at the minima of external potential and follow them quasi-statically as they rotate.
In your setup, $U$ is only due to the Moon. The simplest model is to assume that the Moon executes uniform circular motion, centered on the Earth. Since it is far away, the tidal forces is approximated by its quadrupole component. Furthermore, aligning the spherical coordinates so that $\theta=\pi/2$ is the plane of the Moon's orbit, you get:
$$
U = e^{-i2\omega t}U_{2,-2}Y_{2,-2}+U_{2,0}Y_{2,0}+e^{i2\omega t}U_{2,2}Y_{2,2}
$$
with $\omega$ the angular velocity of the Moon's revolution and the amplitudes ($D=3.8\cdot10^8m$ the Earth-Moon distance, $g_m = \frac{GM_m}{D^2} = g\frac{M_m R^2}{M_eD^2} = 3.4\cdot10^{-6}g$ the gravity due to the Moon on Earth):
$$
U_{2,0} = -\frac{g_mR^2}{D}\sqrt{\frac\pi5} \quad U_{2,2} = U_{2,2} = \frac{g_mR^2}{D}\sqrt{\frac{6\pi}5} \\
U = \frac{g_mR^2}{D}\left(-\frac34\sin^2\theta\cos[2(\phi-\omega t)]+\frac{3\cos^2\theta-1}4\right)
$$
The amplitude of the potential's variations is due to the $|m|=2$ terms:
$$
\Delta U = \frac{3g_mR^2}{4D}.
$$
You therefore only have a quadrupole response in the height field, so two bulges and tide amplitude:
$$
\Delta h = \underbrace{\frac{\omega_2^2}{\omega_2^2-4\omega^2}}_{dynamic}\underbrace{\frac{3g_mR^2}{4gD}}_{static}
$$
In this simple model, you therefore just need to compare $\omega$ and $\omega_2$. If $\omega<\omega_2$, then you get you get a qualitatively similar result as in the static case. Only as $\omega$ increases to $\omega_2$, the tides are dynamically amplified by dynamics, diverging at the resonant frequency $\omega_2$. For $\omega>\omega_2$, past the resonance, you have a reversal in the sign (dips rather than bulges) and at very high frequency you have essentially no tides.
Plugging the numbers, the static theory predicts tides of the order of the centimetre: $\Delta U/g \sim 10^{-2}m$. For the dynamic theory, use $\omega = 7.27\cdot 10^{-5} rad/s$ for a period of $1$ day and $\omega_2 = 7.30\cdot 10^{-5} rad/s\sim\omega$ of also roughly a day. Therefore, the $m=2$ harmonic is past the resonance with an attenuation of $-1/3$. You therefore still have tides of the order of the centimetre, but in opposite phase. Geometrically, it predicts the bulges to be a right angle to the Earth-Moon line.
Naturally, this model is too simplistic for modelling actual tides. Rotation needs to be taken into account, which couples the harmonics. It is not negligible, it has the same order of magnitude as the driving frequency and the resonant frequency. Furthermore, to really get large tides as observed on Earth, you need to look at how the dynamics interacts with the topography.
Hope this helps.
