The action potential satisfy a 2nd order ODE according to the Hodgkin-Huxley model. See equation (30) of their seminal paper [1]. This equation has no closed form solution, but can be solved numerically.
[1]. HODGKIN AL, HUXLEY AF. A quantitative description of membrane current and its...
Nonlinear dynamics and chaos by Steven Strogatz is a classic and often the book used in introductory courses on the subject. It's exceptional well written and easily digestible. More advanced treatments of the subject depends on the direction you want to go in. Chaos is a big field with a lot of...
I would expect the equation to be
$$ \bigg(\frac{dR}{dt}\bigg)^2 = \frac{GM}{R} $$
if it where to describe the gravitational collapse of a non-relativistic star, as a fluid parcel located at distance ##R## is gravitational bound (the Viral theorem). But I might be wrong, I haven't really studied...
No worries! It was me that read your original post in a hurry. Your differential equation in ##y^\prime## belong to a notorious difficult class of ODE's called Abel's nonlinear ODE's of the fist-kind. I haven't had the change nor time to study this class of ODE's, so I'm afraid that I can't...
Your ODE is a second-order linear equation with constant coefficients. It is rather straightforward to solve, simply observe that you can write it in the following form
$$ \big(e^{kt}y^\prime\big)^\prime = - ae^{kt}.$$
Now you simply have to integrate twice.
Let ##c## be a rational number (possibly irrational), then there exist a sequence ##\{q_n\}_{n\in\mathbb{N}}## of purely rational numbers that converges to ##c##. Therefore (assuming ##f## is continuous)
$$
\begin{align*}
f(cx) &= \lim_{n\rightarrow\infty}f(q_nx) \\
&=...
The function ##f(x) = \mathrm{max}(x,0)## (i.e. the positive part) is also a solution to Cauchy as well and though it is continuous it isn't differentiable at ##x=0## and thus not analytic. Check for yourself if also satisfy the original functional equation.
I believe that ##f(x) = ax## for ##a\in\{0,1\}## are the entire family of continuous solutions (consisting of elementary functions) to the original functional equation. As any continuous solution also has to be a solution to the Cauchy's functional equation. However there exist non-continuous...
From ##f(x^2)=xf(x)## conclude that ##f## is an odd function i.e. ##f(-x)=-f(x)##. If we take ##(x,y) = (-x,y)## then we get that
$$ f(-xf(y)+f(-x)) + yf(y) = f(-x) + yf(y-x) $$
or, by using the odd property of ##f##, that
$$ -f(xf(y)+f(x)) + yf(y) = -f(x) + yf(y-x). $$
Now, add this equation...
Sure! However call ##f(x) = 2020##. Then, by an identical argument, we instead get that ##f(2020) = 1/2020##, which clearly contradicts the condition ##f(2020) = 2019##.
I agree. If the function exist then its graph is intersting. It clearly isn't monotone.
Not sure if this is going to be of any help. But doesn't the condition ##f(2020)=2019## seem to be contradictory? I mean, let ##y=f(x)## for some arbitrary chosen ##x\in\mathbb{R}##. Then
$$yf(y) = 1,$$
according to the functional equation. Or equivalently that
$$ f(x) = \frac{1}{x}.$$
This...
Your PDE is linear, you should therefore try to look for a solution by separation of variables; i.e. assume that the solution is of the form ##T(t,x) = u(t)v(x)## and derive a space-independent equation for ##u(t)## and a time-independent equation for ##v(x)##.
Remember that ##v(x)## inherit...
Legendre polynomials are orthogonal but not orthonormal over the interval ##[-1,1]##. Thus, you shouldn't expect your orthonormal basis to be identical to the Legendre polynomials.
NB. If you are trying to construct and orthonormal set ##\{p_0,p_1,p_2\}## of polynomials over the interval...
Your series can be put on the form of a telescope series and thereby summed up.
$$
\begin{align*}
\sum_{n=1}^\infty\frac{n^2+3n+1}{n^4+2n^3+n^2} &= \sum_{n=1}^\infty\frac{n^2+3n+1}{n^2(n+1)^2} \\
&= \lim_{N\rightarrow\infty}\sum_{n=1}^N\bigg\lbrace\frac{2n^2+3n}{(n+1)^2} -...
I assume (since you are not precisely specifying it) that ##B(0,r)\subseteq\mathbb{R}^n## is the closed ball of radius ##r>0## centered at ##0##.
You are given that ##0\in B(0,1)## is a fixed point of ##f##. This immediately tells us that there exist a point in ##x\in B(0,1)## for which ##f(x)...
Here is an example of how you would go about computing a definite integral of ##f(x) = x^2## from "first-principle" without utilizing the fundamental theorem of calculus. I will only show the calculations for the right evaluated Riemann sum, as the calculations for the left evaluated Riemann sum...
There are two tricks to solve this ODE. The first is to notice that
$$(y^{\prime\prime})^2 = k^2\big[1+(y^\prime)^2\big]$$
is nothing more than a 1st-order ODE in ##y^\prime##. Thus, let ##p = y^\prime## (and thus ##p^\prime = y^{\prime\prime}##) and the original ODE becomes
$$(p^\prime)^2 =...
It is possible to express ##y## in terms of ##x##. From your calculations you have that
$$y^{-2}e^{y^2} = e^{x^2},$$
this it great. Rearrange the equation to look like
$$-y^2e^{-y^2} = -e^{-x^2}.$$
This is an equation of the form ##we^w = z## and can thus be solved using the Lambert ##W##...
Begin by deviding both side of
$$\frac{d}{d\theta}\bigg(\sin\theta\frac{d\Theta}{d\theta}\bigg) = - l(l+1)\sin\theta\,\Theta$$
by ##\sin\theta## to get
$$\frac{1}{\sin\theta}\frac{d}{d\theta}\bigg(\sin\theta\frac{d\Theta}{d\theta}\bigg) + l(l+1)\Theta = 0.$$
Now, introduce the new variable...
Apply l'Hospital's rule to deduce that
\begin{align*}
\lim_{m\rightarrow\infty}\frac{m\pi \pm 2}{4m\pi} &= \frac{1}{4}
\end{align*}
and conclude that the probability for finding the particle in the leftmost quarter of the box is ##1/4## at high energies (i.e. large ##m##).
Sure! However, as you probably know, an antiderivative to ##f(x) = x\sin(x)## is ##F(x) = \sin(x) - x\cos(x)##. Thus, the original improper integral is
\begin{align}
I &= \int_0^\infty x\sin(x)\,dx \\ &= \lim_{a\rightarrow\infty}\Big[\sin(x) - x\cos(x)\big].
\end{align}
This limit doesn't exist...
How sure are you that the equation reads $$\ddot{\xi} = -b\xi + \cos(\omega t)(a-c\xi^2)$$ and not instead $$\dot{\xi} = -b\xi + \cos(\omega t)(a-c\xi^2).$$ I only ask because the latter is an Ricatti equation and thus exactly solvable.
If you only have calculated the pertubation series upto...
There are an infinite number of trivial solutions to this ODE.
Supose ##f:\Omega\rightarrow\mathbb{C}## is a holomorphic function, then ##\big(y(x), z(x)\big) = \big(f(x),\pm f(x)\big)## are trivial solutions to your original ODE $$(y^\prime)^2 - (z^\prime)^2 + 2m^2(y^2 - z^2) = 0.$$
So to...