My mnemonic is to remember a simple special case, and heuristically rederive the equations of motion from that. In the case of wave-related equations, you can obtain most of them by just inventing differential equations that describe different properties of plane waves.
Properties of a plane wave
The simplest solution to a wave equation is the plane wave $\Psi(x,t) = \exp(i\mathbf{k}\cdot\mathbf{x}-i\omega t)$. The time-derivative of this is:
$$\frac{\partial\Psi}{\partial t} = -i\omega\Psi$$
Insert the de Broglie relation $E = \hbar \omega$, and solve for $E$:
$$E = i\hbar\frac{\partial}{\partial t}$$
Then try differentiating the plane wave $\Psi$ with respect to position $\mathbf{x}$:
$$\nabla \Psi = i\mathbf{k}\Psi$$
Now insert the de Broglie relation $\mathbf{p} = \hbar\mathbf{k}$ and solve for $\mathbf{p}$:
$$\mathbf{p} = -i\hbar\nabla$$
We now have two equations $E = i\hbar\partial_t$ and $\mathbf{p} = -i\hbar\nabla$ relating spacetime derivatives of a wave function to physical observables.
Schroedinger equation (energy)
Remember that the Schroedinger equation connects the time-derivative of a wave function to it's energy. Just right-multiply our equation for $E$ above with a wave function $\Psi$, and we get the Schroedinger equation:$^\dagger$
$$E\Psi = i\hbar \frac{\partial\Psi}{\partial t}$$
To get the familiar form of the equation for a single particle in a potential $V(\mathbf{x})$, just remember that classically we have $E = \mathbf{p}^2/2m + V(\mathbf{x})$, use $\mathbf{p} = -i\hbar\nabla$ from above, and insert it into the Schroedinger equation:
$$\left[-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{x})\right]\Psi = i\hbar \frac{\partial\Psi}{\partial t}$$
Helmholtz equation (momentum)
The Helmholtz equation relates the momentum of a wave to its spatial derivative. To obtain it, just square the relation $\mathbf{p} = -i\hbar\nabla$, and right-multiply the result by a wave $\Psi$:
$$\mathbf{p}^2\Psi = -\hbar^2\nabla^2\Psi$$
To obtain the conventional form, divide the equation by $\hbar^2$, and use the de Broglie relation $\mathbf{p} = \hbar\mathbf{k}$:
$$\left[\nabla^2 + \mathbf{k}^2\right]\Psi = 0$$
Wave equation (energy-momentum)
The energy and momentum of a free massless particle is related by $E^2 = \mathbf{p}^2c^2$. Insert $E = i\hbar\partial_t$ and $\mathbf{p} = -i\hbar\nabla$, and we get:
$$-\hbar^2 \frac{\partial^2}{\partial t^2} = -c^2\hbar^2\nabla^2$$
Divide this by $\hbar^2c^2$ and right-multiply by a wave function, and you've got the wave equation:
$$\nabla^2\Psi = \frac{1}{c^2}\frac{\partial^2\Psi}{\partial t^2}$$
Using the relativistic notation $\partial^2 = \partial_t^2/c^2-\nabla^2$, we may write the equation more compactly as:
$$\partial^2\Psi = 0$$
Klein-Gordon equation (energy-momentum)
To get a wave equation for a free massive particle, we just start from the relativistic expression $E^2 = (\mathbf{p}c)^2 + (mc^2)^2$, and insert the relations $\mathbf{p} = -i\hbar\nabla$ and $E = i\hbar\partial_t$ that we got above:
$$-\hbar^2\frac{\partial^2}{\partial t^2} = -\hbar^2c^2\nabla^2 + m^2c^4$$
Then you right-multiply this by a wave $\Psi$, and you've got your equation of motion. To get the conventional form, insert $\partial^2 = \partial_t^2/c^2 - \nabla^2$ and rewrite it:
$$\partial^2\Psi = -\left(\frac{mc}{\hbar}\right)^2\Psi$$
Where does it all come from?
Although the heuristic derivations above consider only the special case of a plane wave, the relations $E \sim \partial/\partial t$ and $\mathbf{p} \sim \nabla$ have quite deep origins. Technically, we say that the Hamiltonian is the generator of time translations, and momentum is the generator of space translations. If you have studied classical mechanics, this is related to Noether's theorem, which connects the time-translation invariance of a theory to conservation of energy, and space-translation invariance to conservation of momentum. (Another important quantity is the angular momentum, the generator of rotations, which is related to rotational invariance of the theory.)
$^\dagger$ We usually refer to the operator as $H$ and the eigenvalues as $E$ in quantum mechanics, so just rename $E$ to $H$ to get the standard form of the Schroedinger equation.
