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inkyvoyd

  • 3 years ago

Solve the general second order linear ODE.

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  1. inkyvoyd
    • 3 years ago
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    @nincompoop , I'm waiting.

  2. experimentX
    • 3 years ago
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    write the DE in terms of operator, as you assumed the linearity, factorize it ... let (first factored operator) y = u (second factor) u = your input solve the second for u, then use it to solve for first.

  3. inkyvoyd
    • 3 years ago
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    y''+p(x)y'+q(x)y=0 I'll start with homogenous first? And find two linearly independent solutions?

  4. experimentX
    • 3 years ago
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    i don't think you can generally solve for the systems containing p(x) or q(x) even if there is specific values for p(x) and q(x), they are not so easily solvable Eg. http://mathworld.wolfram.com/HermiteDifferentialEquation.html

  5. inkyvoyd
    • 3 years ago
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    Mm - I should've clarified and used constant coefficients. >.<

  6. inkyvoyd
    • 3 years ago
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    http://mathworld.wolfram.com/Second-OrderOrdinaryDifferentialEquation.html wait, lemme read this >.>

  7. experimentX
    • 3 years ago
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    yep!! if you had constant coefficients, you can factor up the operator as (D-a)(D-b)y = f(x) let (D-b)y = u then it becomes (D-a)u = f(x) <--- this is first order linear, solve for 'u' let u = g(x) be it's solution then, you have (D-a)y = g(x) <--- again linear in y. The e^(kx) are characteristics of linear equations. There are better methods, but this is only for concept/

  8. inkyvoyd
    • 3 years ago
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    @experimentX , what i never understood is why can one factor the operator? is it only because you can relate it to the characterisitic equation?

  9. experimentX
    • 3 years ago
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    depending on those values 'a' and 'b', you either get decaying type or harmonic or damped harmonic solution.

  10. experimentX
    • 3 years ago
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    this is because the operators are commutative. (D-a)[(D-b)y(x)] = (D-b)[(D-a)y(x)]

  11. experimentX
    • 3 years ago
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    at least for constant coefficients .. since there is no 'x' factor to differentiate.

  12. inkyvoyd
    • 3 years ago
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    @nincompoop , I don't need your copy pasting it's lagging my post.

  13. inkyvoyd
    • 3 years ago
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    I see @experimentX - so in case of so in other cases you can't simply factor the operator?

  14. experimentX
    • 3 years ago
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    No ... check this out (D + a)(x^2D + b) y(x)

  15. inkyvoyd
    • 3 years ago
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    ?

  16. experimentX
    • 3 years ago
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    (D + a)(x^2D + b) y(x) =/= (x^2D + b)(D + a) y(x)

  17. inkyvoyd
    • 3 years ago
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    ohhh

  18. inkyvoyd
    • 3 years ago
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    Actually I had originially asked this question to make @nincompoop stop being so annoying - I just realized that it has practical applications in RLC circuits though x.x

  19. experimentX
    • 3 years ago
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    there are lot's of application ... usually where you get sinusoidal periodic solutions. like all sorts of pendulums, RLC, etc ,etc

  20. inkyvoyd
    • 3 years ago
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    Hmm I'ma close this and open a new question, actually more of a physics question. I'm guessing it'll pop up a first order ODE, I just want to formulate one I guess.

  21. experimentX
    • 3 years ago
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    sure

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