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Differential equation problem

  1. Apr 11, 2004 #1
    The last example on my homework assignment this week is this: Solve the following differential equation.

    [tex]y'' + 2y' + y = e^{-x}[/tex]

    I started by solving it like it was y'' + 2y' +y = 0 (Instert y = e^ax and so on) and got the following equation and solution:

    [tex]e^{ax}(a^2 + 2a + 1) = 0 => a = -1[/tex]

    [tex]y'' + 2y' + y = 0, y = e^{-x}(k_1x + k_2)[/tex]

    Since the right side of the original equation is of the form Me^hx, and h is a solution to the equation above (a = -1), I tried inserting f(x) = Axe^(-x).

    From that I got that f'(x) = Ae^(-x) - Axe^(-x) and f''(x) = Axe^(-x) -2Ae^(-x).

    But when I insert that into the original equation everything cancels out and I'm left with e^(-x) = 0. Since that never applies I'm inclined to think that there are no solutions to the equation. What do you think?
     
  2. jcsd
  3. Apr 11, 2004 #2

    arildno

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    You might try a particular solution of the form f(x)=Ax^(2)e^(-x)
     
  4. Apr 11, 2004 #3
    That doesn't seem to work, I get that A should equal both 0 and 1/2. I've also tried Ae^-x, which didn't work either.
     
  5. Apr 11, 2004 #4

    arildno

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    f(x)=Ax^(2)e^(-x)
    f'(x)=2Axe^(-x)-Ax^(2)e^(-x)
    f''(x)=Ax^(2)e^(-x)-4Axe^(-x)+2Ae^(-x)
    f''+2f'+f=2Ae^(-x)-->A=1/2
     
  6. Apr 11, 2004 #5
    Hmm... I must have made some mistake. Well, thanks a lot for the help.
     
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