Differential equations solveable independently of coordinate system? Looking from a physics viewpoint ODEs tend to look very differently when setting up the problem in different coordinate systems. For instance the Laplacian in spherical coordinates involves way more terms than the Laplacian in cartesian coordinates. 
By using a certain coordinate system, we can exploit symmetries of our problem, which makes the ODE easier. However I was wondering whether one could analytically solve a ODE in every coordinate system, if it is analytically solvable in one coordinate system? Or in other words: Can a ODE be analytically solvable in one, but not analytically solvable in another coordinate system?
Thanks in advance 
 A: Perhaps my casual understanding of Lie theory can lend some perspective here. The question you pose "Given a differential operator, under what conditions is the characteristic of having analytic solutions invariant?" (which I think is a very good question to ask) can be pondered in tandem with the question of antiquity "Given a polynomial, under what conditions are the roots expressible in terms of arithmatic operations on the base field of rationals?". 
The second question and its analogue in differential equations "Given a differential operator, under what conditions are the solutions (its kernel) expressible in terms of arithmatic combinations of the base field of fractions of analytic functions?", have answers in Galois and Lie theory, respectively. The first question has a very complete and satisfactory answer. The second has partial answers. 
For a given polynomial, the question of whether it has solutions of the type specified above is invariantly positive or negative under ring automorphisms, which can be thought of as coordinate transformations of the polynomial ring. My casual understanding of Lie theory would lead me to believe that there is a notion under which transformations between different coordinate systems of analytic functions can be understood in terms of automorphisms in the right context, and that the question you pose would have the same result as its analogous result for polynomials -- namely what you would expect that yes, a closed-form in one analytic coordinate system means a closed-form in another.
A: Caveats: I'd never heard the acronym DGL, and there seem to be a couple of plausible interpretations. I assume you mean some sort of (probably differential) equation defined by an operator acting on (scalar-valued) smooth functions, with "smooth" meaning "infinitely differentiable" (or "class $\mathscr{C}^\infty$") for simplicity.
If I understand the context and spirit of the question, analytic solvability is independent of coordinates, thanks to the definition of "expressing an operator in a new coordinate system".
In math-y language, let $U$ and $V$ be open sets in $\mathbf{R}^n$, and $\varphi:U \to V$ a diffeomorphism. If $f$ is a smooth function on $V$, then $F = f\circ \varphi$ is a smooth function on $U$. (This notation is used systematically below.)
In physics-y language, coordinates on $U$ are the new (!) variables, coordinates on $V$ are original variables, and $\varphi$ is a change of coordinates. Precomposition with $\varphi$ "transfers" functions from $V$ to $U$, or "$F$ is $f$ in the new coordinate system".
For example, we might have $U = (0, \infty) \times (-\pi, \pi)$ (in the $(r, \theta)$-plane), $V = \mathbf{R}^2 \setminus (-\infty, 0]$, and $\varphi(r, \theta) = (r\cos\theta, r\sin\theta) = (x, y)$ the polar coordinates mapping. If (say) $f(x, y) = x^2 + y^2$, then $F(r,\theta) = (f\circ\varphi)(r, \theta) = r^2$ is "$f$ in polar coordinates".
If $T:\mathscr{C}^\infty(V)\to\mathscr{C}^\infty(V)$ is an operator, define the induced operator $\varphi^*T:\mathscr{C}^\infty(U)\to\mathscr{C}^\infty(U)$ by
$$
(\varphi^*T)(F) = \bigl(T(F\circ\varphi^{-1})\bigr)\circ\varphi.
$$
In words, express $F$ in the original coordinates ($f = F\circ\varphi^{-1}$ is a smooth function on $V$), apply $T$ (to obtain a smooth function $Tf$ on $V$), and express the function $Tf$ in the new coordinate system (precompose with $\varphi$).
Suppose the equation $Tf = 0$ can be solved analytically ("in the original coordinate system"). A tiny bit of map-chasing shows that $F = f\circ\varphi$ is a solution of $(\varphi^*T)(F) = 0$ ("in the new coordinate system"). Indeed, since $f = F\circ\varphi^{-1}$, we have
$$
(\varphi^*T)(F) = (Tf)\circ\varphi = 0\circ\varphi = 0.
$$
Again, the heart of this argument is the definition of $\varphi^*T$, which causes "application of $T$" to intertwine with "changing coordinates".
A: Integral solutions of differential equations, don't they count as coordinate free solutions?Plenty examples of these are found in e.g. Fluid Dynamics. Here is a picture from an old textbook, as an illustration of how it works.
First define a closed control volume as indicated in the picture (dotted line).
Assumptions are: incompressible fluid, uniform velocity distribution over $A_1$ and $A_2$, friction along $A_3$ negligible.
Known quantities: pressure $p_1$, areas $A_1$, $A_2$, velocity $V_1$. Then $V_2$ follows from integrating the continuity (partial differential) equation over the control volume, giving: $$V_2 A_2 = V_1 A_1$$
Next find $(p_2 - p_1)$ from integrating the momentum (partial differential) equation in $x$-direction, if written as in the book: $$ \iint_A \rho u (\vec{v}\cdot\vec{n}) df = - \iint_A p \cos(n,x) df $$
Giving: $$ - \rho V_1^2 A_1 + \rho V_2^2 A_2 = p_1 A_2 - p_2 A_2 $$
Combining this with the continuity equation gives the unknown pressure $p_2$ as:
$$ p_2 = p_1 + \rho V_2 (V_1 - V_2) $$
There are plenty examples as well with the integral (instead of the differential) formulations of Electromagnetism.
