(b) For those values of k, verify that every member of the family of functions y = A sin(kt) + B cos(kt) is also a solution. y =

Answer:

Refer to the attached document for the complete question

Step-by-step explanation:

Given that,

y=Cos(kt)

We are looking for the value of k that satisfies 4y''=-25y

Let’s find y' and y''

y=Cos(kt)

y'=-kSin(kt)

y''=-k²Cos(kt)

Substituting this back into 4y''=-25y

4(-k²Cos(kt))=-25Cos(kt)

-4k²Cos(kt)=-25Cos(kt)

Dividing through by Cos(kt) under the assumption that Cos(kt) is not zero

-4k²=-25

k²=-25/-4

k²=25/4

Thus, k=√(25/4)

k= ± 5/2

b. Assuming we need to use this

y=ASin(kt)+BCos(kt)

Since k= ± 5/2

y=A•Sin(±5/2t)+ B•Cos(±5/2t)

y'=(±5/2)ACos(±5/2t)-(±5/2)BSin(±5/2t)

y''=-25/4ASin(±5/2t)-25/4BCos(±5/2t

Then, substituting this into our equation can check if it holds true for y=ASin(kt)+BCos(kt)

4y''=-25y

For 4y''

4(-25/4ASin(±5/2t)-25/4BCos(±5/2t))

-25A•Sin(±5/2t)-25B•Cos(±5/2t.

We see that, 4y'' is indeed equal to -25y, confirming that y=Cos(kt) resolves into y=ASin(kt)+BCos(kt)