Mathematical Physics 02 - Carl Bender

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Recommends user3002473 · 2014-10 · Mathematics Stack Exchange · on “Factoring differential equations”

@JasonKnapp Here's a video demonstrating it (it's not the original text that I got the equation from, but I think it's similar in context). [this video] Specifically, the professor writes it on the board at around the 17th minute.

Recommends Ivan Burbano · 2022-03 · Mathematics Stack Exchange · on “The certainty behind the spanning set of solution for a second order linear homogeneous ODE with constant coefficients”

There is a very neat way of approaching this that leaves absolutely no doubt of all of the solutions. I don't have a lot of time so I will be schematic about this. In any case, I think everything is found in this lecture [this video] First, as shown in the lecture, you can use an integrating factor to reduce to the case $a=0$ . In that case, the remaining differential equation can be factored as $$\left(\frac{d}{dx}+i\sqrt{b}\right)\left(\frac{d}{dx}-i\sqrt{b}\right)y=0.$$ Setting $$Y=\left(\frac{d}{dx}-i\sqrt{b}\right)y,$$ you are left with a simple first order equation $$\left(\frac{d}{dx}+i\sqrt{b}\right)Y=0,$$ which can be rewritten as $$\frac{d\ln(Y)}{dx}=-i\sqrt{b}.$$ At which point is clea

Cites bolbteppa · 2013-07 · Mathematics Stack Exchange · on “Perturbation Theory on Finite Domains”

In this video [this video] (from 27.00 - 50.00, which you don't need to watch!) a guy shows how you can solve the general second order ode $y'' + P(x)y = 0$ using perturbation theory. However he points out that the domain must be finite in order for this to work, I'm wondering how you would phrase a question like this or how you would know when you are working with an infinite domain etc...? If I was just given $y'' + x^2y = 0$ & asked to solve it using perturbation theory I wouldn't know if I was getting into trouble about finite or infinite domains, how do I re-phrase the question so that it makes sense? Thanks for your time.

Cites Spine Feast · 2013-12 · Mathematics Stack Exchange · on “$(D^2 -1)y=0 \iff (D-\tanh x)(D+ \tanh x)y=0$ where $D \equiv \frac{d}{dx}$?!”

In this [this video] lecture, the professor explains that factoring differential operators if different from factoring numbers, and provides this example. I have no idea where it comes from and indeed why it is true. Could someone provide an explanation for the fact that: $$(D^2 -1)y=0 \iff (D-\tanh x)(D+ \tanh x)y=0$$

Cites Johannes Maria Frank · 2020-06 · Mathematics Stack Exchange · on “Multiplying out a quadratic equation of differential factors acting on a function”

I am having trouble understanding a multiplying out of a quadratic equation with differential factors: from the following video [this video] : $$ (D + A(x))(D + B(x))y(x) = 0\\ (D^2 + AD + AB + B^{'} + BD)y = 0 \\ $$ What I don't get is his remark on the last multiplication two terms. It looks like he is using the multiplication rule for differentiation, but in the end, the multiplied out parenthesis is again acting on y, so for me it looks like a double use. I see that $B^{'} = DB$ but I still do not get why two terms are produced?