Extension of trigonometric functions (like Bessel functions) I'm physics student learning mathematical physics, now especially Bessel functions.
I learned that the Bessel functions and that family are derived from the ODE
$$
y''+y'/x+(1-\nu^2/x^2)y=0,
$$
and $J_\nu(x)$, $Y_\nu(x)$, $H_\nu^{(1)}(x)$, $H_2^{(2)}(x)$, $I_\nu(x)$, $K_\nu(x)$ have similar properties to $\cos(x)$, $\sin(x)$, $e^{ix}$, $e^{-ix}$, $\cosh(x)$, $\sinh(x)$. (I'm using the notation in the Arfken's Mathematical methods for physicists.)
So I naturally began to wonder if we can also generalize the ODE for trigonometric function $y''+y=0$, to depend on certain index like $\nu$, so we can generate special functions which depend on the index, and which becomes trigonometric function for certain index.  But I'm not sure where should I put the index $\nu$ to make family of functions. If I set, like, $y''+\nu^2y=0$, then solution is just $\sin(\nu x)$, which is basically same as $\sin(x)$. Wait... but maybe we can do this way, $y''+(1-\nu^2/x^2)y=0$?
Is there a extension of trigonometric functions in this(or similar) way? Are they just not practical functions? Or is there a reason that trigonometric functions are special so they should be unique?
 A: The Bessel functions are the generalization of sine and cosine across multiple dimensionalities. Classically, they're the generalization for $d=2$, but it turns out that they can be mapped onto the general case, too.
Specifically, the Laplacian in $d$-dimensions and hyperspherical coordinates is given by
$$\nabla^2 = \frac{1}{r^{d-1}}\frac{\partial}{\partial r} \left(r^{d-1}\frac{\partial}{\partial r}\right) + \frac{1}{r^2} \nabla^2_{S_d}, \tag1$$
where $\nabla^2_{S_d}$ is the Laplacian on the surface of a $d$-dimensional unit sphere. Now, let's look at the eigenvalue problem for the Laplacian, $\nabla^2 f = -k^2f$, which can be solved using separation of variables. The angular part is solved with Gegenbauer polynomials of sines/cosines of the angular variables and has eigenvalue $-\ell(\ell+d-2)$. The radial part then becomes the ODE
$$\frac{\partial^2R}{\partial r^2} + \frac{d-1}{r} \frac{\partial R}{\partial r} - \frac{\ell(\ell+d-2)}{r^2}R = -k^2R, \tag2$$
where $\ell$ is an integer (see Equation 2 from this paper). 
Equation 2 becomes Bessel's differential equation if you make the substitution $R(r)=g(r) / r^{d/2-1}$, explicitly
$$g'' + \frac{ g'}{r} + \left(k^2 - \frac{\left[\frac{d}{2}- 1 + \ell\right]^2}{r^2}\right)g = 0, \tag3$$
which goes the rest of the way with the substitutions $y=g$, $x=kr$, and $\nu=\frac{d}{2}- 1 + \ell$. 
For $d=2$ you get back to Bessel's equation. For $d=3$ you get the spherical Bessel's equation. For $d=1$ you can use $\ell=0,1$ to produce cosine and sine (the rotation symmetry there is just discrete parity).
I find this generalization so useful that I call the solution to these equations hyperspherical Bessel functions and defined a shorthand notation for them
$$X_{[d]_\pm\ell}(r) \equiv \frac{X_{\pm(\ell + d/2-1)}(r)}{r^{d/2-1}}, \tag4$$
where $X$ is any of the Bessel/Hankel function symbols, modified or otherwise: $J$, $Y$, $I$, $K$, $H^{(1)}$, or $H^{(2)}$.
It is possible to generalize even further, and you get into special functions that are so general I've never actually seen them used outside of Mathematica spitting them out from time to time: hypergeometric functions, generalized hypergeometric functions, the Meijer's $G$-function, etc.
A: There isn't really a single unique possible extension, because there's a myriad ways in which you might deform the trigonometric functions, and it comes down to the human subjective judgement on whether the deformation is useful enough to be classified as a special function or not.
However, there is a pretty clear answer in the Jacobi elliptic functions $\operatorname{sn}(x,k)$, $\operatorname{cn}(x,k)$, and related objects; these are explained in depth in Wikipedia and the Digital Library of Mathematical Functions. Since you explicitly mentioned the second-order ODE, I'll quote one relevant differential equation
$$
\frac{{\mathrm{d}}^{2}}{{\mathrm{d}z}^{2}}\operatorname{sn}\left(z,k\right)
=
-(1+k^{2})\operatorname{sn}\left(z,k\right) 
+2k^{2}{\operatorname{sn}^{3}}\left(z,k\right),
$$
(with more found here), but that isn't really the cleanest way to obtain that family in a natural way; instead, the first place they tend to pop up is as inverses of the elliptic integrals.
