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The answer is 'no'. Making the substitution $$ x = \frac{(t-1)(t-5)(t^2+2t+5)}{16t^2}, $$ one finds $$ {\textstyle\sqrt{x+\sqrt{x+\sqrt{x+1}}}\,\mathrm{d}x} = \frac{(t^2-2t+5)(t^2-5)\sqrt{t^4{-}2t^2 …

I'm adding a separate answer for the general question that the OP asked, which settles the question in the negative for all $n>2$ (and gives an alternate proof for $n=3$ to the one I gave above). Rec …

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 …

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 …

There exist $F$ for which there is no global solution $f$ to the above equation. Here is how you can construct an example: First, regard $F$ as a vector field on $\mathbb{R}^3$ and consider its dual …

There is a straightforward way to deduce necessary conditions for a space curve to lie on an ellipsoid, and it's really a matter of calculation to make these conditions explicit in terms of the curvat …

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 …

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) $ …

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 …

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 …

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 …

Well, right away, you can see that the answer is 'no', in general. Consider the round $n$-sphere $S^n$ with its standard metric. When $n>1$, it has no nonzero harmonic $1$-forms, but it has nontrivi …

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 …

There is a counterexample with $d=1$ and $m=n=2$. Here is one way to construct such an example: Let $g:\mathbb{R}\to\mathbb{R}$ be a smooth function such that $g'(t)>0$ for $t\not=0$ and $g^{(k)}(0) …