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Y''(x) + A sin(y(x)) - B = 0; A,B : positive, real

  1. Mar 8, 2012 #1
    Hello there,

    I have no idea how to solve the differential equation

    y''(x) + A sin(y(x)) - B = 0 ,

    where A and B are positive real numbers. I do also have initial conditions: y(0) = 0 and y'(0) = 0.

    I would be grateful for any help.
  2. jcsd
  3. Mar 8, 2012 #2
    y''+A sin(y)-B=0
    y''y'+A sin(y)y'-B y'=0
    y'²/2-A cos(y)-B y =C
    y'²=2A cos(y)+2B y +2C
    dy/dx =y' = (+or-)sqrt(2A cos(y)+2B y+2C)
    dx=(+or-) dy/sqrt(2A cos(y)+2B y+2C)
    x =(+or-) integal (dy/sqrt(2A cos(y)+2B y+2C)) +c
    cannot be expressed with a finit number of usual functions;
  4. Mar 8, 2012 #3
    Thank you very much.
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