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Differential equation

  1. Oct 22, 2006 #1
    Find the solution of the given inital value problem.
    y"-2y'+y=te^t+4 y(0)=1 y'(0)=1

    r^2-2r+1=0
    (r-1)^2
    so r= -1
    my homogenous solution is y_h(t)=c_1e^-t+c_2te^-t

    I have no idea where to begin for the particular solution. I know a need a guess, and I would think it would be Ate^t, but I really don't know.
    Can someone please help me out?
     
  2. jcsd
  3. Oct 22, 2006 #2

    arildno

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    Is r=-1?
    Why?
     
  4. Oct 22, 2006 #3
    Oh sorry, I meant r= 1
    then my homogenous solution would be c_1e^t +c_2te^t
     
  5. Oct 22, 2006 #4

    HallsofIvy

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    For a right hand side of the form
    te^t+4
    You should immediately think y= (At+ B)e^t+ C since when you have t^n, you might need lower powers of t.

    Then, noticing that e^t and te^t are already solutions to the homogeneous equation, think of multiplying the e^t solution by the next power of t, t^2. That is, try y= (At^3+ Bt^2)e^t + C.
     
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