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

1. Jan 14, 2006

mr_coffee

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

Any help would be great!
thanks!

2. Jan 14, 2006

qbert

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...

3. Jan 14, 2006

mr_coffee

Thanks again qbert!! worked out great! good old quadratic formula!
low = 6, high = 10.

4. Jan 15, 2006

leright

: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