1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 2nd ODE [Particular Integral]

  1. May 25, 2009 #1


    User Avatar

    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)


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook