Question

A. For what values of k does the function y = cos(kt) satisfy the differential equation 25y'' = −16y?

Answer 1

Answer:

Step-by-step explanation:

Given y = coskt

y' = -ksinkt

y'' = -k²coskt

Substitute this y'' into the expression 25y'' = −16y

25(-k²coskt) = -16(coskt)

25k²coskt = 16(coskt)

25k² = 16

k² = 16/25

k = ±√16/25

k = ±4/5

b) from the DE 25y'' = −16y

Rearrange

25y''+16y = 0

Expressing using auxiliary equation

25m² + 16 = 0

25m² = -16

m² = -16/25

m = ±4/5 I

m = 0+4/5 I

Since the auxiliary root is complex number

The solution to the DE will be expressed as;

y = Asinmt + Bsinmt

Since k = m

y = Asinkt+Bsinkt where A and B are constants