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Very important

  1. May 21, 2006 #1
    hello genius guys,
    can u people tell me the method of solving this question

    QUESTION:1
    (D^2 + a^2 )y=Sec x
    STEP 1:
    y=sec x/((D+ai)(D-ai))
    STEP2:
    y=1/(d+ai) *\exp ^aix [tex]\int Secax \exp ^-aix dx[/tex]


    plz plz tell me any method of solving this question i have a test tomorrow plz tell me the solution
     
    Last edited: May 22, 2006
  2. jcsd
  3. May 21, 2006 #2

    benorin

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    Homework Helper

    Is D is a differential operator?
     
  4. May 21, 2006 #3
    yes D is a differential operator
     
  5. May 21, 2006 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Benorin, that's a fairly standard notation.

    Mohdfasieh, it looks like you are trying to put it into form for a transform solution. I don't much like those.

    Here's how I would do it: The associated homogeneous d.e. is
    (D2+ a^2)y= 0 which has solutions (for a not 0)
    y(x)= C cos(ax)+ D sin(ax).

    Now find a solution to the entire equation by assuming a solution of the form y(x)= u(x)cos(ax)+ v(x)sin(ax). Differentiating,
    y'= u' cos(ax)- au sin(ax)+ v'sin(ax)+ avcos(ax).

    Since there are, in fact, many such solutions, reduce the search to those for which u' cos(ax)+ v' sin(ax)= 0.

    Now we have y'= -au sin(ax)+ av cos(ax) so y"= -au' sin(ax)- a2 cos(ax)+ av' cos(ax)- a2v sin(ax)

    Putting those into the equation, y"+ a2y= -au' sin(ax)- a2u cos(ax)+ av' cos(ax)- a2v sin(ax)+ a2u cos(ax)+ a2v sin(ax)= -au' sin(ax)+ a v' cos(ax)= sec(ax).

    That, together with u' cos(ax)+ v' sin(ax)= 0 gives to equations to solve for u' and v' which then can (theoretically) be integrated.
     
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