Answer:
a.![k=\pm\frac{5}{2}](https://tex.z-dn.net/?f=k%3D%5Cpm%5Cfrac%7B5%7D%7B2%7D)
Step-by-step explanation:
We are given that a solution
y= cos (kt) satisfied the differential equation
![4y''=-25y](https://tex.z-dn.net/?f=4y%27%27%3D-25y)
We have to find the value of k
a.y=coskt
Differentiate w.r.t x
Then we get
![y'=-k sinkt](https://tex.z-dn.net/?f=y%27%3D-k%20sinkt%20)
![\frac{d(cos ax)}{dx}=-asin ax](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28cos%20ax%29%7D%7Bdx%7D%3D-asin%20ax)
Again differentiate w.r.t x
(![\frac{d(sinax)}{dx}=a cos ax](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28sinax%29%7D%7Bdx%7D%3Da%20cos%20ax)
Substitute the value in given differential equation
![-4 k^2 coskt=-25 coskt](https://tex.z-dn.net/?f=-4%20k%5E2%20coskt%3D-25%20coskt)
coskt cancel on both sides then we get
![4k^2=25](https://tex.z-dn.net/?f=4k%5E2%3D25)
![k^2=\frac{25}{4}](https://tex.z-dn.net/?f=%20k%5E2%3D%5Cfrac%7B25%7D%7B4%7D)
a.![k=\sqrt{\frac{25}{4}}=\pm\frac{5}{2}](https://tex.z-dn.net/?f=k%3D%5Csqrt%7B%5Cfrac%7B25%7D%7B4%7D%7D%3D%5Cpm%5Cfrac%7B5%7D%7B2%7D)
b.We have to show that y=A sin kt + B cos kt is a solution to given differential equation for k=![\pm\frac{5}{2}](https://tex.z-dn.net/?f=%5Cpm%5Cfrac%7B5%7D%7B2%7D)
Substitute the values of k
Then we get
![y=A cos \frac{5}{2} kt+ B sin \frac{5}{2} kt](https://tex.z-dn.net/?f=y%3DA%20cos%20%5Cfrac%7B5%7D%7B2%7D%20kt%2B%20B%20sin%20%5Cfrac%7B5%7D%7B2%7D%20kt%20)
Differentiate w.r.t x
![y'=-\frac{5}{2} A sin \frac{5}{2}t+ \frac{5}{2} B cos \frac{5}{2}t](https://tex.z-dn.net/?f=y%27%3D-%5Cfrac%7B5%7D%7B2%7D%20A%20sin%20%5Cfrac%7B5%7D%7B2%7Dt%2B%20%5Cfrac%7B5%7D%7B2%7D%20B%20cos%20%5Cfrac%7B5%7D%7B2%7Dt)
Again differentiate w.r.t x
Then we get
![y''=-\frac{25}{4}A cos \frac{5}{2} t+\frac{25}{4} B sin\frac{5}{2} t](https://tex.z-dn.net/?f=y%27%27%3D-%5Cfrac%7B25%7D%7B4%7DA%20cos%20%5Cfrac%7B5%7D%7B2%7D%20t%2B%5Cfrac%7B25%7D%7B4%7D%20B%20sin%5Cfrac%7B5%7D%7B2%7D%20t)
Substitute the value of y'' and y in given differential equation
![-25 A cos \frac{5}{2} t +25 B sin \frac{5}{2} t=-25(A cos \frac{5}{2} t+ B sin \frac{5}{2} t)](https://tex.z-dn.net/?f=-25%20A%20cos%20%5Cfrac%7B5%7D%7B2%7D%20t%20%2B25%20B%20sin%20%5Cfrac%7B5%7D%7B2%7D%20t%3D-25%28A%20cos%20%5Cfrac%7B5%7D%7B2%7D%20t%2B%20B%20sin%20%5Cfrac%7B5%7D%7B2%7D%20t%29%20)
![-25 A cos \frac{5}{2} t +25 B sin \frac{5}{2} t=-25 y](https://tex.z-dn.net/?f=-25%20A%20cos%20%5Cfrac%7B5%7D%7B2%7D%20t%20%2B25%20B%20sin%20%5Cfrac%7B5%7D%7B2%7D%20t%3D-25%20y)
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\frac{5}{2}](https://tex.z-dn.net/?f=%5Cpm%5Cfrac%7B5%7D%7B2%7D)
Where A and B are constants