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Homework Help: Finding a fundamental set of solutions for a 2nd order differential equation

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    64y''+144y'=0

    y1(0)=1 y'1(0)=0
    and
    y2(0)=0 and y'2(0)=1

    2. Relevant equations
    y1=c1*e^(r1*t) + c2*e^(r2*t)


    3. The attempt at a solution

    I start by finding the characteristic equation:
    64r^2+144r=0
    r1=-9/4 and r2=0

    y1=c1e(r1*t) + c2e(r2*t)

    so I get
    y1=c1e^(-9/4 *t) + c2e^(0*t)

    e^(0*t) = 1 will always = 0, which gives
    y1=c1e^(-9/4 *t) + c2(1)

    so I suppose I am asking if I started this wrong or if not, because I need y'1.

    With these values I would have:
    y'1=(-9/4)c1e^(-9/4 *t) + 0(c2)
    ?? because the derivative of 1 is zero

    Is this correct or have I gone about the problem in the wrong way?
     
  2. jcsd
  3. Oct 12, 2012 #2

    fluidistic

    User Avatar
    Gold Member

    This looks like correct to me.
     
  4. Oct 12, 2012 #3
    Okay, so when I go from there
    y=c1*e^(-9/4 *t) + c2
    y'=(-9/4)*e^(-9/4 *t)*c1 + c2??

    Which leads me to
    1=c1+c2
    0=(-9/4)c1+c2
    for y(0)=1 y'(0)=0

    c1=(1-c2)
    0=(-9/4)(1-c2) + c2
    9/4=(13/4)*c2
    c2=9/13
    c1=4/13

    but this isn't correct according to webwork, so what is the mistake? is it supposed to be


    y=c1*e^(-9/4 *t) + c2
    y'=(-9/4)*e^(-9/4 *t)*c1
    giving
    1=c1+c2
    0=(-9/4)c1
    c1=(1-c2)
    0=(-9/4)(1-c2)
    c2=9/4
    c1=-5/4

    but that isn't right either?

    I'm just stumped on this, and it is probably a rather silly mistake somewhere in my algebra or something...
     
  5. Oct 12, 2012 #4

    fluidistic

    User Avatar
    Gold Member

    As you noted the derivative of a constant is 0 so this is wrong.
    You might want to redo the arithmetics in the red part. :)
    Other than this, you're doing fine.
     
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