Question

) For what values of k does the function y = cos(kt) satisfy the differential equation 4y 00 = −25y? (1) (b) For those values of k that you found in part (a), verify that every function of the form y = A sin(kt) + B cos(kt) is also a solution to the differential equation (1). (Here, A and B are constants.)

Answer 1

a.
`k=\pm(5)/(2)`

Explanation:

We are given that a solution

y= cos (kt) satisfied the differential equation

`4y''=-25y`

We have to find the value of k

a.y=coskt

Differentiate w.r.t x

Then we get

`y'=-k sinkt`

`(d(cos ax))/(dx)=-asin ax`

Again differentiate w.r.t x

`y''=-k^2 cos kt` (
`(d(sinax))/(dx)=a cos ax`

Substitute the value in given differential equation

`-4 k^2 coskt=-25 coskt`

coskt cancel on both sides then we get

`4k^2=25`

`k^2=(25)/(4)`

a.
`k=\sqrt{(25)/(4)}=\pm(5)/(2)`

b.We have to show that y=A sin kt + B cos kt is a solution to given differential equation for k=
`\pm(5)/(2)`

Substitute the values of k

Then we get

`y=A cos (5)/(2) kt+ B sin (5)/(2) kt`

Differentiate w.r.t x

`y'=-(5)/(2) A sin (5)/(2)t+ (5)/(2) B cos (5)/(2)t`

Again differentiate w.r.t x

Then we get

`y''=-(25)/(4)A cos (5)/(2) t+(25)/(4) B sin(5)/(2) t`

Substitute the value of y'' and y in given differential equation

`-25 A cos (5)/(2) t +25 B sin (5)/(2) t=-25(A cos (5)/(2) t+ B sin (5)/(2) t)`

`-25 A cos (5)/(2) t +25 B sin (5)/(2) t=-25 y`

LHS=RHS

Hence, every function of the form

y=A cos kt +B sin kt is a s solution of given differential equation for k=
`\pm(5)/(2)`

Where A and B are constants