Find the two values of k for which yxekx is a solution of th

Find the two values of \"k\" for which y(x)=e^(kx) is a solution of the differential equation y\'\'-16y\'+60y=0.

Smaller Value=
Larger Value=

if this is a solution then

y\' = ke^(kx) and y\"= k²e^(kx) plug them in the differential equation

k2ekx -16kekx + 60ekx =0 then factor out e^kx

(k2 - 16k + 60)ekx =0 that means

(k2 - 16k + 60) = 0

k=6 and k = 10

$Find the two values of \$