Find k such that y=sin(kt) satisfies y’’+16y=0?

1 Answer
Feb 20, 2018

k=-4,0,4

Explanation:

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