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Remark: I had a little time to write a draft of my notes on the proofs of the claims I make below and have posted it on my home webpage here. (It would have made a very long post on MO, so I decided …

Well the standard techniques would take advantage of the fact that the equation doesn't explicitly involve the independent variable $x$ to integrate the equation once, thereby leading to the conservat …

There is probably no single proof that would provide a rigorous justification of the OP's principle in all cases. Moreover, without specifying more clearly what is meant by a 'natural map', the princ …

Your question is a bit vague, but let me try the following statement, which might be the kind of answer you are looking for: If $M$ is a manifold and $S\to M$ is a vector bundle over $M$ endowed with …

As for asking about whether the symmetries of this equation would help you solve it, here are a few remarks that you may (or may not) find useful: I assume that you want to consider what are usually …

Of course, Igor's answer points the way to working out the answer the OP wanted, but it may not be clear, even after you have got the eigenvalues, what the corresponding eigenfunctions are, or that th …

I would guess that you are just seeing the effects of a dynamical system with a hyperbolic fixed point. Consider the matrix equation $$ A'(x) = \begin{pmatrix} 0 & 1\\ \cos(x) & 0\end{pmatrix} A(x) $ …

Jeanne's calculations give the right answer, i.e., that the solutions depend on two arbitrary functions of 2 variables. It turns out, though, that, with the right choice of variables, one can reduc …

There's a straightforward abstract answer that you may not like, but, because it clarifies your question and explains a uniform way to answer similar questions, I'll sketch it here. First, consider a …

What you are looking for nowadays goes by the name of the Rumin complex and is defined on any contact manifold. Moreover, there is a vast generalization of this that sometimes goes by the name of 'th …

In fact, using the moving frame, it is easy explicitly to solve the equations and get the formula for the slope $\tan\bigl(\theta(s)\bigr)$ as a function of arc-length along the curve. However, one s …

If you mean a (real) analytical solution with $y(0)=0$, then the answer is 'no'. If you write this as the problem of looking for integral curves of $\omega = (2x-x^2y)\ dx + y\ dy$ in the $xy$-plane, …

Edited on May 2, 2020: The OP pointed out that I had not addressed a special case (namely $C=1$ below), so I am amending my answer to address this and reorganizing so that the $C=1$ case gets addresse …

Actually, no matter what $A$ is, there will be nonzero solutions that will converge to zero, so you can't prove unboundedness, even though it may be true that the 'generic' solution is unbounded. Her …

As Ben's argument suggests, the proof that, if the flow of $X$ preserves $J$ then the flow of $JX$ preserves $J$ does depend on the integrability of $J$. As a concrete example of an almost-complex ma …