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

1. Feb 18, 2013

### Aerospace93

1. The problem statement, all variables and given/known data
(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.

3. 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...

2. Feb 18, 2013

### Ray Vickson

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

3. Feb 18, 2013

### Aerospace93

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

4. Feb 18, 2013

### HallsofIvy

Staff Emeritus
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 [itex]4d^2y/dx^2= -25y[/quote] for any A and B.

5. Feb 18, 2013

### 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

Last edited: Feb 18, 2013