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Diferencial equation

  1. Mar 8, 2014 #1
    How can I demostrate that a solution of d2u/(dθ)^2+u=0 is u=cos(θ-θ0)

    Thanks
     
  2. jcsd
  3. Mar 8, 2014 #2

    LCKurtz

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    Differentiate it twice, add it to the result, and see if you get zero.
     
  4. Mar 8, 2014 #3
    Yes, thanks, but how can I get that result
     
  5. Mar 8, 2014 #4

    LCKurtz

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    Show us what happens when you try it.
     
  6. Mar 8, 2014 #5
    I already differentiate it twice and then I replace it to the equation and I get to zero, I know that cos(θ-θ0) is a solution, but I don t know how to get that solution.
    When I solve the equation I reach other result more complicated with imaginary terms etc...
     
  7. Mar 8, 2014 #6

    LCKurtz

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    To save typing I will use ##x## instead of ##\theta## for the independent variable. When you solved ##u''+u=0## you probably got solutions like ##e^{ix}## and ##e^{-ix}##, so the general solution is ##u = Ae^{ix}+Be^{-ix}##. Using the Euler formulas you can write this equivalently as ##u = C\cos x + D\sin x##. A solution of the form ##\cos(x-x_0)## can be written using the addition formula as ##\cos x \cos x_0 - \sin x \sin x_0##. You can get that from the previous form by letting ##C=\cos x_0,~D= -\sin x_0##.

    You can read about constant coefficient DE's many places on the internet. One such place is:
    http://www.cliffsnotes.com/math/differential-equations/second-order-equations/constant-coefficients
     
  8. Mar 8, 2014 #7
    D is imaginary?
     
  9. Mar 8, 2014 #8

    LCKurtz

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    The constants can be imaginary or complex. But there is a theorem that if the coefficients of the DE are real and the boundary conditions are real, the constants C and D will be real in the {sine,cosine} expression. If you leave the solution in the complex exponential form, the constants A and B will come out complex. So for real coefficients and real boundary conditions, you really just make it complicated if you leave it in the complex exponential form. Use the {sine,cosine} form.
     
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