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General solution to 2nd order ODE

  1. May 25, 2008 #1
    Find the general solution of the ordinary differential equation.

    y'' - 7y'+ 6y = 2e^(3t) + te^(t)

    First i found GS(H) by lettings y = e^(cx)

    and got GS(H) = Ae^(6t) + Be^(t)

    i then found PS(IH) y'' - 7y'+ 6y = 2e^(3t) by letting y = ae^(3t)

    and got PS(IH) = -(1/3)e^(3t)

    Now my problem
    I am trying to find the other PS(IH) y'' - 7y'+ 6y = te^(t)
    do i do this by letting y = (a+bt)e^(t) ??
     
  2. jcsd
  3. May 25, 2008 #2

    Defennder

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    This is what I have from my notes:

    [​IMG]
     
  4. May 25, 2008 #3

    HallsofIvy

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    Normally, you would try that but since et is already a solution to the homogeneous equation, this will not work. (Surely you have already tried and discovered that?) The standard step for this situation is to multiply by t: try y= (at+ bt2et.
     
  5. May 25, 2008 #4
    you say try y= (at+ bt2et.

    is that bracket suppose to be there? im not sure if thats (at+bt^2)e^t or at+b(t^2).e^t
     
    Last edited: May 25, 2008
  6. May 25, 2008 #5
    I used y = (at+bt^2)e^t
    y' = (at+bt^2)e^t + (a+2bt)e^t
    y'' = (at+bt^2)e^t + (a+2bt)e^t + (a+2bt)e^t + 2be^t

    i substituted those in and equated coefficients and got the result a = o and b = 1/10

    have i done this right ?
     
  7. May 26, 2008 #6

    Defennder

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    I didn't get that. For me, neither a nor b is zero. Why don't you check this by substituting it into the modified DE to see if it satifies the equation?
     
  8. May 26, 2008 #7

    HallsofIvy

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    = (at+ bt^2)e^t+ 2(a+ 2bt)e^t+ 2be^t



    You want y'' - 7y'+ 6y = 2e^(3t) + te^(t) and you've already done the "2e^(3t)" part.

    y'' = (at+ bt^2)e^t+ 2(a+ 2bt)e^t+ 2be^t
    -7y'= -7(at+bt^2)e^t - 7(a+ 2bt)e^t
    +6y= 6(at+ bt^2)e^t

    so y"- 7y'+ 6y= -5(a+ 2bt)e^t+ 2be^t= (-5a+ 2b)e^t+ -10bte^t= te^t. Equating "like coefficients, -10b= 1 so b= -1/10, not 1/10.
     
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