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Particular solution for dif eq

  1. Feb 18, 2006 #1
    [tex]-3y''-2y'+y=-t^2+2t+2e^{-4t}[/tex]

    i am to find the particular solution to this.

    i started with the non-exponential:
    [tex]y=At^2+Bt[/tex]
    [tex]y'=2At+B[/tex]
    [tex]y''=2A[/tex]

    [tex](-6A-2B)+(-4A+B)t+A(t^2)[/tex]
    [tex]-6A-2B=0,-4A+B=2,A(t^2)=-1[/tex]
    [tex]A=-1, B=3[/tex]

    i started with the exponential:
    [tex]y=Ce^{-4y}[/tex]
    [tex]y'=-4Ce^{-4t}[/tex]
    [tex]y''=16Ce^{-4t}[/tex]

    [tex]-48Ce^{-4t}+8Ce^{-4t}+Ce^{-4t}=2Ce^{-4t}[/tex]
    [tex]C=-\frac{2}{39}[/tex]

    so I think the particular solution is:
    [tex]y_p=-t^2+3t- \frac{2}{39}e^{-4t}[/tex]

    but this is wrong, not sure why, any ideas?
     
  2. jcsd
  3. Feb 18, 2006 #2

    Fermat

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    you have -4A + B = 2, but that is inconsistent with A = -1 and B = 3.

    What have you missed?

    (Check a few lines before)
     
  4. Feb 18, 2006 #3
    if A=-1, then b=-2? but for the other equation it is 3? i dont understand this
     
  5. Feb 18, 2006 #4

    Fermat

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    You forgot the C!
     
  6. Feb 18, 2006 #5
    couldnt I do it seperated? solve for A and B, and then solve for C would be the same as solving for A B and C at the same time right?
     
  7. Feb 18, 2006 #6

    Fermat

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    y=At² + Bt + C
     
  8. Feb 18, 2006 #7

    Fermat

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    My apologies Urban...
    I only read your post as far as the ommission of C from the 1st eqn.
    I just now saw the C from the 2nd part you did.
    I must have been confusing you.
    Sorry about that.
     
  9. Feb 18, 2006 #8
    it's no problem Fermat, any ideas is to what I have done wrong? since B can be both equal to -2 and 3, i am not sure what to do.
     
  10. Feb 18, 2006 #9

    Fermat

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    do what you did before , but with y=At² + Bt + C.

    y=At² + Bt + C
    y' = 2At + B
    y'' = 2A

    -3y'' - 2y + y = -t² + t
    -6A -4At - 2B + At² + Bt + C = -t² +t

    etc.
     
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