How is the Schroedinger equation a wave equation? Wave equations take the form:
$$\frac{ \partial^2 f}  {\partial t^2} = c^2 \nabla ^2f$$
But the Schroedinger equation takes the form:
$$i \hbar \frac{ \partial f}  {\partial t} = - \frac{\hbar ^2}{2m}\nabla ^2f + U(x) f$$
The partials with respect to time are not the same order. How can Schroedinger's equation be regarded as a wave equation?  And why are interference patterns (e.g in the double-slit experiment) so similar for water waves and quantum wavefunctions?
 A: As stressed in other answers and comments, the common point between these equations is that their solutions are "waves". It is the reason why the physics they describe (eg interference patterns) is similar.
Tentatively, I would define a "wavelike" equation as


*

*a linear PDE

*whose space of spacially bounded* solutions admits a (pseudo-)basis of the form
$$e^{i \vec{k}.\vec{x} - i \omega_{\alpha}(\vec{k})t}, \vec{k} \in \mathbb{R}^n, \alpha \in \left\{1,\dots,r\right\}$$ with $\omega_1(\vec{k}),\dots,\omega_r(\vec{k})$ real-valued (aka. dispersion relation).
For example, in 1+1 dimension, these are going to be the PDE of the form $$\sum_{p,q} A_{p,q} \partial_x^p \partial_t^q \psi = 0$$ such that, for all $k \in \mathbb{R}$ the polynomial $$Q_k(\omega) := \sum_{p,q} (i)^{p+q} A_{p,q} k^p \omega^q$$ only admits real roots. In this sense this is reminiscent of the hyperbolic vs parabolic classification detailed in @DaniH answer, but without giving a special role to 2nd order derivatives.
Note that with such a definition the free Schrödinger equation would qualify as wavelike, but not the one with a potential (and rightly so I think, as the physics of, say, the quantum harmonic oscillator is quite different, with bound states etc). Nor would the heat equation $\partial_t \psi -c \partial_x^2 \psi = 0$: the '$i$' in the Schrödinger equation matters!
* Such equations will also often admit evanescent waves solutions corresponding to imaginary $\vec{k}$.
A: This answer elaborates on my comment to David Z's answer.  I think his definition of a wave equation is excessively broad, because it includes every translationally invariant PDE and every value of $v$.  For simplicity, let's specialize to linear PDE's in one spatial dimension.  A general order-$N$ such equation takes the form
$$\sum_{n=0}^N \sum_{\mu_1, \dots, \mu_n \in \{t, x\}} c_{\mu_1, \dots, \mu_n} \partial_{\mu_1} \dots \partial_{\mu_n}\, f(x, t) = 0.$$
To simplify the notation, we'll let $\{ \mu \}$ denote $\mu_1, \dots, \mu_n$, so that 
$$\sum_{n=0}^N \sum_{\{\mu\}} c_{\{\mu\}} \partial_{\mu_1} \dots \partial_{\mu_n}\, f(x, t) = 0.$$
Let's make the ansatz that $f$ only depends on $\xi := x - v t$.  Then $\partial_x f(\xi) = f'(\xi)$ and $\partial_t f(\xi) = -v\, f'(\xi)$.  If we define $a_{\{\mu\}} \in \mathbb{N}$ to simply count the number of indices $\mu_i \in \{\mu\}$ that equal $t$, then the PDE becomes
$$\sum_{n=0}^N f^{(n)}(\xi) \sum_{\{\mu\}} c_{\{\mu\}} (-v)^{a_{\{\mu\}}} = 0.$$
Defining $c'_n := \sum \limits_{\{\mu\}} c_{\{\mu\}} (-v)^{a_{\{\mu\}}}$, we get the ordinary differential equation with constant coefficients
$$\sum_{n=0}^N c'_n\ f^{(n)}(\xi) = 0.$$
Now as usual, we can make the ansatz $f(\xi) = e^{i z \xi}$ and find that the differential equation is satisfied as long as $z$ is a root of the characteristic polynomial $\sum \limits_{n=0} c_n' (iz)^n$.  So our completely arbitrary translationally invariant linear PDE will have "wave-like solutions" traveling at every possibly velocity!
A: Actually, a wave equation is any equation that admits wave-like solutions, which take the form $f(\vec{x} \pm \vec{v}t)$. The equation $\frac{\partial^2 f}{\partial t^2} = c^2\nabla^2 f$, despite being called "the wave equation," is not the only equation that does this.
If you plug the wave solution into the Schroedinger equation for constant potential, using $\xi = x - vt$
$$\begin{align}
-i\hbar\frac{\partial}{\partial t}f(\xi) &= \biggl(-\frac{\hbar^2}{2m}\nabla^2 + U\biggr) f(\xi) \\
i\hbar vf'(\xi) &= -\frac{\hbar^2}{2m}f''(\xi) + Uf(\xi) \\
\end{align}$$
This clearly depends only on $\xi$, not $x$ or $t$ individually, which shows that you can find wave-like solutions. They wind up looking like $e^{ik\xi}$.
A: While it is not very technical in nature it is worth going back to the first definition of what a wave is (that is, the one you use before learning that "a wave is a solution to a wave-equation"). The wording I use in the introductory classes is

a moving disturbance

where the 'disturbance' is allowed to be in any measurable quantity, and simple means that the quantity is seen to vary from its equilibrium value and then return to that value.
The surprising thing is not how general that expression is, but that it is necessary to use something that general to cover all the basic cases: waves on strings, surface waves on liquids: sound and light.
And by that definition Schrödinger's equation is used to describe the moving variation of various observables, so arguably qualifies.
There is room to quibble—the wave-function itself is not an observable, and even the distributions of values that can be observed are often statistical in nature—but I've always been comfortable with this approach.
A: Both are types of wave equations because the solutions behave as you expect for "waves". However, mathematically speaking they are partial differential equations (PDE) which are not of the same type (so you expect that the class of solutions, given some boundary conditions, will present different behaviour). The constraints on the eigenvalues of the linear operator are also particular to each of the types of PDE. Generally, a second order partial differential equation in two variables can be written as
$$A \partial_x^2 u + B \partial_x \partial_y u + C \partial_y^2 u + \text{lower order terms} = 0 $$
The wave equation in one dimension you quote is a simple form for a hyperbolic PDE satisfying $B^2 - 4AC > 0$.
The Schrödinger equation is a parabolic PDE in which we have $B^2 - 4AC = 0$ since $B=C=0$. It can be mapped to the heat equation.
A: In the technical sense, the Schrödinger equation is not a wave equation (it is not a hyperbolic PDE). In a more heuristic sense, though, one may regard it as one because it exhibits some of the characteristics of typical wave-equations. In particular, the most important property shared with wave-equations is the Huygens principle. For example, this principle is behind the double slit experiment.
If you want to read about this principle and the Schrödinger equation, see Huygens' principle, the free Schrodinger particle and the quantum anti-centrifugal force and Huygens’ Principle as Universal Model of Propagation. See also this Math.OF post for more details about the HP and hyperbolic PDE's.
A: As Joe points out in his answer to a duplicate, the Schrodinger equation for a free particle is a variant on the slowly-varying envelope approximation of the wave equation, but I think his answer misses some subtleties.
Take a general solution $f(x)$ to the wave equation $\partial^2 f = 0$ (we use Lorentz-covariant notation and the -+++ sign convention).  Imagine decomposing $f$ into a single plane wave modulated by an envelope function $\psi(x)$: $f(x) = \psi(x)\, e^{i k \cdot x}$, where the four-vector $k$ is null.  The wave equation then becomes
$$(\partial^\mu + 2 i k^\mu) \partial_\mu \psi = ({\bf \nabla} + 2 i\, {\bf k}) \cdot {\bf \nabla} \psi + \frac{1}{c^2} (-\partial_t + 2 i \omega) \partial_t \psi= 0,$$
where $c$ is the wave velocity.
If there exists a Lorentz frame in which $|{\bf k} \cdot {\bf \nabla} \psi| \ll |{\bf \nabla} \cdot {\bf \nabla} \psi|$ and $|\partial_t \dot{\psi}| \ll \omega |\dot{\psi}|$, then in that frame the middle two terms can be neglected, and we are left with
$$i \partial_t \psi = -\frac{c^2}{2 \omega} \nabla^2 \psi,$$
which is the Schrodinger equation for a free particle of mass $m = \hbar \omega / c^2$.
$|\partial_t \dot{\psi}| \ll \omega |\dot{\psi}|$ means that the envelope function's time derivative $\dot{\psi}$ is changing much more slowly than the plane wave is oscillating (i.e. many plane wave oscillations occur in the time $|\dot{\psi} / \partial_t \dot{\psi}|$ that it takes for $\dot{\psi}$ to change significantly) - hence the name "slowly-varying envelope approximation."  The physical interpretation of $|{\bf k} \cdot {\bf \nabla} \psi| \ll |{\bf \nabla} \cdot {\bf \nabla} \psi|$ is much less clear and I don't have a great intuition for it, but it seems to basically imply that if we take the direction of wave propagation to be $\hat{{\bf z}}$, then $\partial_z \psi$ changes very quickly in space along the direction of wave propagation (i.e. you only need to travel a small fraction of a wavelength $\lambda$ before $\partial_z \psi$ changes significantly).  This is a rather strange limit, because clearly it doesn't really make sense to think of $\psi$ as an "envelope" if it changes over a length scale much shorter than the wavelength of the wave that it's supposed to be enveloping.  Frankly, I'm not even sure if this limit is compatible with the other limit $|\partial_t \dot{\psi}| \ll \omega |\dot{\psi}|$.  I would welcome anyone's thoughts on how to interpret this limit.
