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Solution to 2nd order ODE using the D operator method with 2 trig terms on RHS

  1. Aug 9, 2012 #1

    I have the DE

    y'' -2y' + 3y = xsin(x) + 2cosh(2x)

    Using the D operator as D = [itex]\frac{dy}{dx}[/itex] this becomes

    (D2 -2D +3)y = xsin(x) + 2cosh(2x)

    so yp = [itex]\frac{1}{p(D^2)}[/itex] operating on xsin(x) + 2cosh(2x)

    (i think)

    So i know if this was say [itex]\frac{1}{p(D^2)}[/itex] operating on sin(x)
    i.e (D2 -2D +3)y = sin(x)
    then yp = 1/(D2 -2D +3) * sin(x)

    and you would substitute D2 = -([itex]\alpha[/itex])2 and proceed to solve.(where alpha is the coefficient of x in the argument of sin, here it is 1)

    My question is, I have a term on the RHS which is polynomial times trig; xsin(x) and a trig term which can be treated as an exponential; 2cosh(2x)

    I know I can split this into
    yp = 1/(D2 -2D +3) * xsin(x) + yp = 1/(D2 -2D +3) * 2cosh(2x)

    But I'm unsure of the 'rules' to use here, i.e for a single trig term you swapped D2
    for -([itex]\alpha[/itex])2
    But what do you do for a poly times a trig and for the cosh function?

    Any points would be much appreciated, if all else fails I will have to resort to the method of undetermined coefficients
  2. jcsd
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