STEP 1: To evaluate #y''+16y=0#, we first need to find #y''# given that #y=sin(kt)#
#y=sin(kt)#
#y'=cos(kt)*k=kcos(kt)#
#y''=-ksin(kt)*k=-k^2sin(kt)#
STEP 2: Now we can evaluate #y''+16y=0#
Remember, we just found #y''=-k^2sin(kt)# and the original problem told us that #y=sin(kt)#
So: #y''+16y=0 -->#
#-k^2sin(kt) + 16(sin(kt)) = 0#
#k^2sin(kt) - 16(sin(kt)) = 0#
#sin(kt) (k^2 - 16) = 0#
STEP 3: Solve
#sin(kt) (k^2 - 16) = 0#
#sin(kt) = 0# OR #k^2-16=0#
STEP 3(a): Find where #sin(kt) = 0#
#sin(kt) = 0# when #kt = 0# OR #kt = pi# OR #kt=2pi# ...
More generally, when #kt = 0 + pi*n#
Now, remember that k is a constant but t is a variable!
Let's first look at where #kt=0#. #kt=0# when #k=0#. So that's one solution!
Now let's look at where #kt = pi#. There is no value k such that for all values of t, #kt=pi#. So we have no solutions here.
STEP 3(b): Find where #k^2=16#
#k^2=16# when #k=4# or #k=-4#
STEP 4: Putting it all together
Our final answer: when #k=-4,0,4#
