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I think your questions can be answered by looking at a source that covers singular differential equations of this type, such as R. Gérard and H. Tahara, Singular nonlinear partial differential equatio …

Assuming that $A(t)$ is analytic at $t=0$, getting a power series solution $M(t)$ at $t=0$ of $M'(t) = A(t)M(t)$ is trivial, of course. You just write $A(t) = A_0 + A_1 t + \cdots$ and $M(t) = I_n + …

This isn't much of an answer, but I thought that I'd put down a few observations here that you may find helpful. First of all, if $b=0$, then you have a first integral, in that the ratio $x^{3c}/y^ …

The answer is 'no, the first equation does not imply the second when $n=2$'. The reason is that when $n=2$, the equation is essentially equivalent to requiring that the off-diagonal $n$-by-$n$ block o …

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, …

You are asking about a very classical problem. The Picard-Vessiot theory was developed to show that, in a certain well-defined sense, there is no `closed form' solution to problems of this kind. Yo …

This is a standard geometric formula, disguised because of the old-fashioned notation and the mixture of the PDE with the formula for exponentiation in the orthogonal group. Rewrite it this way: S …

The answer is basically 'no', there is no 'elementary method' involving elementary operations and quadrature (i.e., finding antiderivatives of known holomorphic functions) that will give you a solutio …

N.B.: I checked your calculations a couple of times, and I got a different sign in the Charpit equations: $$ -\frac{x}{N^2}dp=-\frac{1}{x^2 y p}dx=\frac{1}{2\Omega x^2 y^2} dy $$ (Notice the minus si …

When you write the solution, you must have some other conditions in mind, since one generally does not have unique solutions. For example $\zeta = 0$ and $\zeta = 2 w$ both satisfy your equation, so …

Your conjecture is true, and it can be proved by making a few observations. First, for each $\alpha\in\mathbb{R}$, let $V_\alpha$ be the vector space of germs of smooth functions $f$ at the origin th …

There's an obvious analytic answer: $$ v(x,t) = E\qquad u(x,t) = a E^5 (l-x) $$ Is that what you wanted?

The general solution of your equations in a simply connected domain on which $r_2\not=0$ and $r_1\not=\pm1$ is $$ \beta = \frac12 + \frac1{{(r_1}^2{-}1)}\, \left(\frac{\partial a}{\partial\theta_1}+b( …

Depending on what you know about the coefficients $c_{j_0j_1j_2j_3}(\lambda)$, I think that it's not as hopeless as all that. First of all, such a curve would have to lie in the common zero locus $Z\ …

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 …