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2nd ODE [Particular Integral]

  1. May 25, 2009 #1

    Air

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    When finding the particular integral, I understand that for:

    [tex]y''+y'+y=e^{2x}[/tex], One would choose [tex]y=Ae^{2x}[/tex]
    [tex]y''+y'+y=x[/tex], One would choose [tex]y=Ax+B[/tex]

    But what am I supposed to choose if I have: [tex]y''+y'+y=xe^{2x}[/tex]?
     
  2. jcsd
  3. May 25, 2009 #2

    Well, you know that in case of:

    Dy = exp(p x)

    where D is a linear differential operator,

    you would choose y = A exp(px)

    You also know that a solution of:

    Dy = f(x) + g(x)

    can be obtained by solving

    Dy = f(x)

    and

    Dy =g(x)

    separately and adding up the solutions. Linearitity of D implies that this will work.

    Then, in case of:

    Dy = x exp(px)

    you could simply consider the factor x to be the derivative w.r.t. p. So, if you simply solve:

    Dy = exp(px)

    by putting y = A(p)exp(px), then the linearity of D implies that the solution of

    Dy = [exp[(p+epsilon)x] - exp(px)]/epsilon

    is given by

    [A(p+epsilon)exp[(p+epsilon)x] - A(p)exp(px)]/epsilon

    So, taking the derivative of A(p)exp(px) w.r.t. p will do. By Leibnitz's rule, you see that this amounts to putting

    y = (A + B x)exp(px)
     
  4. May 25, 2009 #3
    [itex] (A + Bx) e ^{2x} [/tex]

    for [itex] x^{2} e^{2x}[/tex] use [itex] (A + Bx +Cx^{2})e^{2x} [/itex]

    and so on
     
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