Answer
ω=±4
Work Step by Step
We are given that $y = A sin( ωt)$ satisfies the differential equation $y''+16y=0$.
Therefore,
$(A sin( ωt))''+16(A sin( ωt))=0$
$-A ω^2sin( ωt)+16(A sin( ωt))=0$
$(-ω^2+16)A sin( ωt))=0$
Therefore,
$-ω^2+16=0$
or ω=±4
