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anonymous
 one year ago
I was studying the differential equation of the pendulum, you know, y'' + k sin (y) = 0, with k constant, and I know it can be approssimated, for little y angles, to y'' + ky = 0. My question is: how do I solve the equation without estimating sin (y) = y?
anonymous
 one year ago
I was studying the differential equation of the pendulum, you know, y'' + k sin (y) = 0, with k constant, and I know it can be approssimated, for little y angles, to y'' + ky = 0. My question is: how do I solve the equation without estimating sin (y) = y?

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3you can note that \(y'' + ky = 0\) \((D^2 + k) y = 0\) where D is the differential operator \(D = \pm i \sqrt{k}\) so characteristic equation \(y = A e^{i \ \sqrt{k}x} + B e^{i \ sqrt{k}x}\) and try the suggested solution to see if it works in this differential equation and expanding that solution out gives you sinusoids

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3but the mere fact you have a restorative force [kx] means that you have simple harmonic motion and so the solution is sinusoid

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry, what do you mean with "expanding"? Taylor? And there's a thing I don't understand, did you do that approximation, sin (y) = y, in the first equation you wrote?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3maybe there is some ambiguity there yes, without the linearisation you have to solve something else \(y'' + k \sin y = 0\) and that's way above my pay grade!! here's why :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, now I have understood what my mechanics book meant with "analytically difficult"!!!!! Thank you IrishBoy, you've been very kind!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3and thank you for this insight! i feel very humble again :p

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0Is the equation y'' + P(x) = 0, solvable for a general polynomial P(x)? It looks like it should be, but the answer wolfram gives me for y'' + k(x(1/6)*x^3) = 0 is giberrish.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0Silly me, it should be y'' + P(y) = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I had the same idea, expanding with Taylor, it looks like a Bernoulli equation, even though it's not a first order equation, so I'll try to divide and change y variables. Let's see if something sensed can come out of this!
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