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anonymous

  • one year ago

2. degree diff. equation? I have honestly forgot more or less everything about 2. degree diff. equations. How do I solve this equation? (posted below) Note that I am not after "the answer". I am after the steps taken towards getting an answer - so that I may (hopefully) recall how to do it

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  1. anonymous
    • one year ago
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  2. Adi3
    • one year ago
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    What do you think??

  3. Adi3
    • one year ago
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    @Ivanskodje

  4. anonymous
    • one year ago
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    I think: Kx^2 + 9.81 - which has no real roots so it will be imaginary Then I think I need to find another something and put it into an ... "general"-somethingsomething

  5. Adi3
    • one year ago
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    @IrishBoy123 is that right??

  6. anonymous
    • one year ago
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    I meant x^2 + 9.81 (or PHI or whatever symbol)

  7. Adi3
    • one year ago
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    bump your question

  8. Adi3
    • one year ago
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    @IrishBoy123

  9. anonymous
    • one year ago
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    x1 = root(9.81) * i x2 = -root(9.81) * i Although I have no idea what to do with this now. I think there was a formula or something somewhere.

  10. anonymous
    • one year ago
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    I wonder if this has something to do with Laplace

  11. IrishBoy123
    • one year ago
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    let's take a different example as @Ivan wants a general approach and not the answer say \(y'' - 3y' + 2y = 0\) if we use the differential operator \(D = \dfrac{d}{dx}\) to simplify the algebra we have \((D^2 - 3 D + 2)y = 0\) \((D-2)(D-1)y = 0\) taking the first root: \((D-1)y = 0 \implies y'-y = 0\) so \(y' = y\) \(\dfrac{dy}{y} = dx \) \(\ln y = x + c\) \(y = e^{x+c} = Ae^x\) by the same logic \((D-2)y = 0 \implies y'-2y = 0\) and gives \(y = Be^{2x}\) to the solution is \(y = A e^x + B e^{2x}\)

  12. IrishBoy123
    • one year ago
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    @Ivanskodje , if you follow that, we have a way to solve the one you have posted. just pattern match, some complex exponentials will crop up but we can deal with that... of course we can also use laplace transform if you like. normally you'd save laplace for initial value problems...but we can do it if you are familiar with laplace

  13. anonymous
    • one year ago
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    D = d/dt - However, isn't y'' d^2y / dt^2 ? I mean... or ... \[y ^{''} = \frac{ d^2y }{ dt^2 } -> D=\frac{ d }{ dt } \] We set D to be d/dt, how can y'' be D^2 if the d below isn't ^2 as well? Or is the dt^2 = (dt)^2 ?

  14. Adi3
    • one year ago
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    can some one medal me pls.

  15. IrishBoy123
    • one year ago
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    \(D^2 \implies \dfrac{d}{dx}\left( \dfrac{d}{dx} \right) = \dfrac{d^2}{dx^2}\) it's an algebraic shortcut the alternative is to assume straight out that the solutions to the equation that i gave as an example come in the form \(y = A x^{\ x} + B x^{2 x}\) ie having found the roots of the underlying equation. i personally think that less gratifying but that is called the method of undetermined coefficients you can find loads of it here: http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

  16. anonymous
    • one year ago
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    Although shou... oooh! "dx" below is like one unit? Not d*x right?

  17. IrishBoy123
    • one year ago
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    \[D^2 (y) = \dfrac{d^2 y}{dx^2}\]

  18. IrishBoy123
    • one year ago
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    or \(D^2 (\theta) = \dfrac{d^2 \theta}{dt^2}\), if \(\theta\) and \(t\) are the dependent and independent variable, as seems to be the case in your DE.

  19. Adi3
    • one year ago
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    thank you for giving me a medal.

  20. Adi3
    • one year ago
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    appreciate it.

  21. Adi3
    • one year ago
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    :(

  22. Adi3
    • one year ago
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    Thank you, i have 49 C now

  23. Adi3
    • one year ago
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    1 more medal for 50.

  24. Adi3
    • one year ago
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    1 more medal for 50.

  25. anonymous
    • one year ago
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    I am trying to understand the e^(x+c) = Ae^x - How do we figure it is equal to Ae^x? Formula?

  26. anonymous
    • one year ago
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    Because e^c is a constant - so A? I think I got it

  27. IrishBoy123
    • one year ago
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    yes!

  28. anonymous
    • one year ago
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    On the second part; (D−2)y=0 - Where did the B come from? I got to e^2t. Do we just assume it is multiplied with some unknown constant? There is a formular that is like y = Ae^(n_1)x + Be^(n_2)x .... hmm

  29. anonymous
    • one year ago
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    n_1 = 1, n_2 = 2

  30. anonymous
    • one year ago
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    so Ae^x + Be^2x

  31. anonymous
    • one year ago
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    Applies when n_1 is not equal to n_2, but are real

  32. IrishBoy123
    • one year ago
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    \( (D−2)y=0 \implies y' = 2y\) \(\dfrac{dy}{y} = 2 dx\) \(\ln y = 2x + c\) \(y = e^{2x+c} = Be^{2x} \)

  33. anonymous
    • one year ago
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    oh of course! Forgot the C... never forget the C!

  34. IrishBoy123
    • one year ago
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    yes you are seeing the pattern emerge.

  35. anonymous
    • one year ago
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    Now, when going back to the task at hand; I see that because we get an unreal number1 and number2 - there is a different way to progress? According to a book; the two numbers I get, somehow becomes n_1 = a+ib and n_2 = a-ib. However, when I solve for O, I get root(9.81) * i and -root(9.81)*i. How do that translate?

  36. anonymous
    • one year ago
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    Like this;

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  37. anonymous
    • one year ago
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    The one to the right is the "format" I need in order to use the second formula where I use sin and cos

  38. IrishBoy123
    • one year ago
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    well, leave the actual numbers out of it until we solve. we have \(\ddot \theta + \omega^2 \theta = 0\) where \(\omega^2 = \dfrac{g}{R}\) [that \(\omega^2 \) substitution (??),....., just go with it and you'll see why it's handy] \((D^2+\omega^2)\theta = 0\) \(D = \pm i \omega\) thus as per above pattern, we have \(\theta = A e^{i \omega \; t} + B e^{-i \omega\; t}\) so we have the complex solution you mention but the real part is zero. we are left with \(i \omega\) in the solution

  39. IrishBoy123
    • one year ago
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    we then switch that to sin and cos, when you are ready....

  40. anonymous
    • one year ago
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    Ah, much easier! Yes - I got to Ae^wit + Be^-wit How do we proceed ?

  41. IrishBoy123
    • one year ago
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    \(\theta = A e^{i \omega \; t} + B e^{-i \omega\; t}\) \(= A ( \cos \omega t + i \sin \omega t) + B ( \cos (-\omega t) + i \sin (-\omega t)) \) \(= A ( \cos \omega t + i \sin \omega t) + B ( \cos \omega t - i \sin \omega t) \) \( = (A + B) \cos \omega t + i( A - B ) \sin \omega t \) \(= C \cos \sqrt{\dfrac{g}{R}} t + D \sin \sqrt{\dfrac{g}{R}} t \) where \(C, D = const.\)

  42. anonymous
    • one year ago
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    That looks very familiar - I just need to find a reference to the convertion between e^(...) to cos(...)+isin(...) Working on it now

  43. IrishBoy123
    • one year ago
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    eulers formule \[e^{ix}=\cos x+i\sin x\] https://en.wikipedia.org/wiki/Euler%27s_formula

  44. anonymous
    • one year ago
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    I just need to find a reference to it in the formula book - and put the pieces together. I find something similar but not quite the same - working on it

  45. anonymous
    • one year ago
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    This is the formula I have to work with; \[\lambda_1 = \alpha +i \beta, \lambda_2 = \alpha -i \beta\] \[y = e ^{\alpha x} (A \sin \beta x + B \cos \beta x)\] I assume this makes the alpha 0 and beta w ? Or would this be the wrong direction?

  46. anonymous
    • one year ago
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    Apparently the answer is supposed to be

  47. IrishBoy123
    • one year ago
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    no here \(\alpha = 0\) so you have \(y =1 \cdot (A \sin \beta x + B \cos \beta x)\) and \(\beta = \omega = \sqrt{\dfrac{g}{R}}\)

  48. IrishBoy123
    • one year ago
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    yes, you also can use a angle formula to express it as a cosine or a sine with a phase shift \(\phi\). so \(\sin (\omega t + \phi) = \sin \omega t \cos \phi + \cos \omega t \sin \phi\) again you can pattern match back into the solution above these are all ways of expressing the solution, you take your pick.

  49. IrishBoy123
    • one year ago
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    so\[A \sin (\omega t + \phi) = A [\sin \omega t \cos \phi + \cos \omega t \sin \phi] \\= B \sin \omega t + C \cos \omega t \]

  50. anonymous
    • one year ago
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    Processing information... ! Still working on it

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