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One cannot prove existence in the generality that you have stated it. For example, if $g(p) = e^p$, there is no solution when $N>1$ and $f$ is differentiable. To see this, observe that, if there wer …

There is yet another direction in which one could take the question, one that makes more contact with the linear and affine structures that one observes in the standard treatment of linear equations ( …

Yes, there is a generalization that covers this case, and much more general second order systems. For example, you can consult L. P. Eisenhart's 1927 book Non-Riemannian Geometry, where he develops t …

Here is a revised and somewhat expanded version of my answer, with a preparatory 'toy version' to help orient the reader. A simple warmup problem: Before discussing a quantitative variant of the Sch …

Building on Noam's suggestion, you could try using the inherent symmetry of the problem: Set $\sigma_1 = x^2 + y^2 + z^2$, $\sigma_2 = x^2y^2+y^2z^2+z^2x^2$, and $\sigma_3 = x^2y^2z^2$. Then your co …

For your specific question, note that the domain you describe $\mathbb{D}$, consisting of those $z = x+iy$ for which $y > 1/(1+x^2)$, when regarded as a domain in the extended complex plane, $\mathbb{ …

I wasn't able to find a source to cite for the correct formulae, but it turns out that it's not that hard to work out the answer directly using differential geometry. First, a little notation: Let $u …

You are asking a very classical question, which is the question of sufficient conditions for a minimum in the calculus of variations. Quite a lot is known about conditions on Lagrangians that ensure …

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\ …

Here's a counterexample to the OP's literal question: Consider the following four symmetric (in fact, diagonal) $3$-by-$3$ matrices: $A_1 = \mathrm{diag}(0,0,0)$, $A_2 = \mathrm{diag}(\frac12,3,3)$, …

I don't know what you mean by 'without calculation'. I don't think you'll get a simpler explanation than simply solving the equation: If $$ \frac{d^3\bigl((y'')^{-2/3}\bigr)}{dx^3} = 0, $$ then $$ y …

You can solve it parametrically as follows: Write $$ x(t) = \frac{at}{\sqrt{1+at^2}}+\frac{bt}{\sqrt{1+bt^2}} $$ and $$ y(t) = c - \frac{\sqrt{1+at^2}+\sqrt{1+bt^2}}{\sqrt{1+at^2}\sqrt{1+bt^2}} $$ w …

Actually, I now think that the easiest method is to do this: Write $k=\sin z$, so that $|k|<1$, and make the substitution $x = \arcsin(k\sin\theta)$, where $0\le \theta\le \frac\pi2$. The integral b …

You probably need to specify more conditions, as the ones you give are too loose to be interesting. For example, consider the divergence-free vector field in the $xy$-plane given by $$ V = f(x)\,\fra …

You might find it useful to make a change of variables to reduce the equation to a more familiar form. For example, if we assume, as we may, that $a$ and $b$ are not equal, then we can substitute $y …