Please help this is too hard for me
2 answers:
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Answer:
Check attachment for complete question
Step-by-step explanation:
Given that,
y=Coskt
We are looking for value of k, that satisfies 4y''=-25y
Let find y' and y''
y=Coskt
y'=-kSinkt
y''=-k²Coskt
Then, applying this 4y'"=-25y
4(-k²Coskt)=-25Coskt
-4k²Coskt=-25Coskt
Divide through by Coskt and we assume Coskt is not equal to zero
-4k²=-25
k²=-25/-4
k²=25/4
Then, k=√(25/4)
k= ± 5/2
b. Let assume we want to use this
y=ASinkt+BCoskt
Since k= ± 5/2
y=A•Sin(±5/2t)+ B •Cos(±5/2t)
y'=±5/2ACos(±5/2t)-±5/2BSin(±5/2t)
y''=-25/4ASin(±5/2t)-25/4BCos(±5/2t
Then, inserting this to our equation given to check if it a solution to y=ASinkt+BCoskt
4y''=-25y
For 4y''
4(-25/4ASin(±5/2t)-25/4BCos(±5/2t))
-25A•Sin(±5/2t)-25B•Cos(±5/2t).
Then,
-25y
-25(A•Sin(±5/2t)+ B •Cos(±5/2t))
-25A•Sin(±5/2t) - 25B •Cos(±5/2t)
Then, we notice that, 4y'' is equal to -25y, then we can say that y=Coskt is a solution to y=ASinkt+BCoskt
