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Finding general soln of differential eqn (zero under the root)

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data

    9y'' + 6y' + y = 0

    r^2 + 6r + 1

    ( -6 +/- root 36 - 36 ) / 18

    lamda = -6/18 = -(2/9)
    mu = 0

    (this is the part im unsure about, because if i end up with a zero under the root, then does this make the final equation into this:

    y = c_1*cos(0*x) + c_2*sin(0*x)

    thanks for any help.
    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 26, 2010 #2


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    Homework Helper
    Gold Member

    First, [itex]-\frac{6}{18}=-\frac{1}{3}\neq-\frac{2}{9}[/tex].

    Second, when you have a double root, [itex]\lambda[/itex], you look for a solution of the form [itex]y(t)=C_1e^{\lambda t}+C_2 t e^{\lambda t}[/itex]
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