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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:




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


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

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

    method 2:


    r=-4, r=0

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


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

    solution equation:

    =1..... which doesnt seem correct

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

    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


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    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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