Verify every number of a family of functions is also a solution

[Aerospace93]

Homework Statement

(a)For what values of k does the function y=coskt satisfy the differential equation 4y''=-25y?
(b) For those values of k, verify that every number of the family of functions y=Asinkt+Bcoskt is also a solution.

The Attempt at a Solution

(a) y=coskt, y'=-ksinkt, y''=-k²coskt.
4(-k²coskt)=-25coskt; 4k²=25, so k=+/-5/2
(b) y= Asinkt+Bcoskt; y'= +/-Akcoskt+/-BKcoskt; y''= +-Ak²sinkt+/-bk²coskt
Im not sure how to verify this...

 

[Ray Vickson]

Homework Statement

(a)For what values of k does the function y=coskt satisfy the differential equation 4y''=-25y?
(b) For those values of k, verify that every number of the family of functions y=Asinkt+Bcoskt is also a solution.

The Attempt at a Solution

(a) y=coskt, y'=-ksinkt, y''=-k²coskt.
4(-k²coskt)=-25coskt; 4k²=25, so k=+/-5/2
(b) y= Asinkt+Bcoskt; y'= +/-Akcoskt+/-BKcoskt; y''= +-Ak²sinkt+/-bk²coskt
Im not sure how to verify this...

Well, do you have 4y'' = -25y or not?

 

[Aerospace93]

I believe you would if y''=-Ak²sinkt-Bk²coskt?

 

[HallsofIvy]

Did you miss where the problem says "for those values of k"? You are asked to show that y= A sin((2/5)t)+ B cos((2/5)t) and y= Asin((-2/5)t)+ Bcos((-2/5)t) both satisfy $4d^2y/dx^2= -25y[/quote] for any A and B.$

 

[Aerospace93]

ok so (for a positive k) y= Asinkt+Bcoskt; y'= -Akcoskt+BKsinkt; y''= -Ak²sinkt-bk²coskt
so,
-4[A(5/2)²sin(5/2)t-B(5/2)²cos(5/2)t= -25[Asin(5/2)t+Bcos(5/2)t
-25Asin(5/2)t-25Bcos/5/2)t= RHS