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Differentiating y= Acosαx + Bsinαx and finding Constants

  1. Dec 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Don't really know what to do with this one, any help welcome

    If A, B and α are constants, independent of x, and

    y= Acosαx + Bsinαx

    Evaluate the derivates
    y'=dy/dx
    y^n=d^2y/dx^2

    Determine values of A, B and α such that y satisfies
    y^n+y=0 and y(0)=0, y'(0)=1


    2. Relevant equations



    3. The attempt at a solution

    I've tried to differentiate y
    y'=-αAsinαx + αBcosαx
    y'=αBcosαx - αAsinαx

    y^n=-ααBsinαx + ααAcosαx
    y^n=ααAcosαx - ααBsinαx
     
  2. jcsd
  3. Dec 4, 2009 #2

    LCKurtz

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    You might find it easier to see if you used standard notation and parentheses. Use y'' for second derivative and use exponents like a2 instead of aa. (Use the X2 icon for superscripts.

    Rewriting what you have for the second derivative:

    y'' = a2Acos(αx) - α2Bsin(αx)

    Now can you figure out a so that

    y''+y=0

    And once you have that what do the two conditions

    y(0)=0, y'(0)=1

    tell you about A and B?
     
  4. Dec 5, 2009 #3
    But are my y' and y'' values correct because their wouldn't be any point in starting part 2 of the question if they are not
     
  5. Dec 5, 2009 #4

    LCKurtz

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    You have a sign mistake in y'' which will become obvious when you try to make y'' + y = 0.
     
  6. Dec 5, 2009 #5
    y''=-ααBsinαx - ααAcosαx then, yes?
     
  7. Dec 5, 2009 #6

    LCKurtz

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    Yes. Did you understand about the X2 button so you can write a2 instead of aa?
     
  8. Dec 6, 2009 #7
    y+yn=0

    2Bsinαx - α2Acosαx + Acosαx + Bsinαx=0

    Acosαx + Bsinαx=α2Bsinαx + α2Acosαx

    Acosαx(1 - α2)=Bsinαx(α2 - 1)

    so α=1?

    b) 0=Acosα(0) + Bsinα(0)
    A=0

    c) 1=αBcosα(0) - αAsinα(0)
    B=1
     
    Last edited: Dec 7, 2009
  9. Dec 6, 2009 #8

    LCKurtz

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    Right here, if you collect terms on cos and sin you can see what a has to be.

    Is a = 1 the only solution to a2=1?

    That's right if a = 1 is the only value for a that works.
     
  10. Dec 6, 2009 #9
    My bad, a=1 or -1
    B=1 or -1
    A=0

    Thank you very much for the help
     
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