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Diff Eq problem, finding the small value and large value of k?

  1. Jan 14, 2006 #1
    Hello everyone i'm lost on how they want me to approach this:
    Find the two values of k for which
    y(x) = e^{kx}
    is a solution of the differential equation
    y'' - 16 y' + 60 y = 0.

    smaller value =?
    larger value = ?

    I did the following:
    y(x) = e^(kx);
    y' = ke^(kx);
    y'' = k^2e^(kx);

    (k^2e^(kx)) - 16(ke^(kx)) + 60(e^(kx)) = 0;
    is there a sysematic way to solve this problem rather then just trying to randomly guess numbers? i tried the randomly guessing k values and it isn't working out :bugeye:

    Any help would be great!
  2. jcsd
  3. Jan 14, 2006 #2
    so you have:
    (k^2e^(kx)) - 16(ke^(kx)) + 60(e^(kx)) = 0
    or slightly rewriten
    (k^2 - 16k + 60) e^(kx) = 0
    then since e^(kx) is never 0

    k^2 - 16k + 60 = 0

    I believe there might be some sort of systematic
    way to figure out what k is. i seem to remember
    some sort of an algebraic formula...
  4. Jan 14, 2006 #3
    Thanks again qbert!! worked out great! good old quadratic formula!
    low = 6, high = 10.
  5. Jan 15, 2006 #4
    :rofl: good one. Yeah, just solve for k. Once you take the derivatives and plug them in. the rest is just algebra.
    Last edited: Jan 15, 2006
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