anonymous
  • anonymous
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?
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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IrishBoy123
  • IrishBoy123
you 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
  • IrishBoy123
but the mere fact you have a restorative force [-kx] means that you have simple harmonic motion and so the solution is sinusoid
anonymous
  • anonymous
I'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?

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IrishBoy123
  • IrishBoy123
maybe 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
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anonymous
  • anonymous
Well, now I have understood what my mechanics book meant with "analytically difficult"!!!!! Thank you IrishBoy, you've been very kind!
IrishBoy123
  • IrishBoy123
and thank you for this insight! i feel very humble again :p
beginnersmind
  • beginnersmind
Is 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
  • beginnersmind
Silly me, it should be y'' + P(y) = 0
anonymous
  • anonymous
I 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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