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Help solving the Initial Value Problem

  1. Apr 1, 2012 #1
    1. The problem statement, all variables and given/known data
    y''-4y'+4y=0 , y(1)=1 and y'(1)=1



    3. The attempt at a solution

    Auxiliary equation: r2-4r+4=0

    I tried factoring 2 different ways:
    (r-2)2=0

    r=2,r=2

    y1=e2t
    y2=y1

    y(t)=c1e2t+c2e2t

    y(1)=c1e2+c2e2=1 ---eq(1)


    y'(t)=2c1e2t+2c2e2t

    ...c2=1/(2e2)-c1 ---eq(2)

    sub eq(2) into eq(1) --> 1/2=1


    method 2:

    r(r-4)=-4

    r=-4, r=0

    y(1)=c1e-4+c2=1 ---eq(1)

    y'(1)=c1=-1/(4e-4)

    sub eq(2) into eq(1) --> c2=5/4

    solution equation:

    y(t)=1/4(-1+5)
    =1..... which doesnt seem correct

    i checked the solutions on wolframalpha and it solves it using the first method, but ends up with:

    y1=c1e2t
    y2=c2e2t *t

    can someone explain to me why the extra t shows up in y2? Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Apr 1, 2012
  2. jcsd
  3. Apr 1, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi middieman147 :smile:

    (i don't understand your second method at all :confused:)
    it isn't an x, it's a t

    when a root b is repeated (in the auxiliary equation), the two independent solutions are ebt and tebt

    if it's repeated n times, the n independent solutions are ebt tebt … tn-1ebt
     
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