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Differential equation

  1. Apr 14, 2005 #1
    diff. eq.

    --------------------------------------------------------------------------------

    solve the following promblem
    Y^(,,,)-3y^(,,)+31y^(,)-37y=0
    i let y = e^kx
    y'= ke^kx
    y''=k^2e^kx
    y'''=k^3e^kx

    so i got this
    k^3e^kx-3(k^2e^kx)+31(ke^kx)-37(e^kx)=0
    e^kx(k^3-3K2+31k-37)=0

    so,

    (k^3-3K2+31k-37)=0

    now i have to find (K)


    how should i solve for (k) from this equation k^3-3K2+31k-37=0
    can i use synthetic division if yes how should i use it or which other method can i use

    is this right

    i solve the k by synthetic division
    5 1 1 -17 -65
    5 30 65
    1 6 13 0

    so the factor is (k-5) (K^2+6k+13)
    then i use this equation
    (-b+-squrt(b^2-(4ac)))/2a

    and got k = -3 +- 2i

    and my fianl answer is
    y=c1e^5x+c2e^(-3x)cos(2x)+c3e^(-3x)sin(2x)

    sorry this is wrong info i type in
    please dont do this problem
     
    Last edited: Apr 14, 2005
  2. jcsd
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