For what values of k does the function y $=$ cos$($kt$)$ satisfy the differential equation

$49$y$=64$y$?$

$($Enter your answers as a comma$-$separated list.$)$

k $=$

$($b$)$

For those values of k$,$ verify that every member of the family of functions

y $=$ A sin$($kt$)+$ B cos$($kt$)$

is also a solution.

We begin by calculating the following.

y $=$ A sin$($kt$)+$ B cos$($kt$)$ y$=$ Ak cos$($kt$)$ Bk sin$($kt$)$ y$=$

Note that the given differential equation

$49$y$=64$y

is equivalent to

$49$y$+64$y $=$

$.$

Now, substituting the expressions for y and

y$$

above and simplifying, we have

LHS $=49$y$+64$y$=49$

$+64($A sin$($kt$)+$ B cos$($kt$))=49$

$49$Bk$2$ cos$($kt$)+64$A sin$($kt$)+64$B cos$($kt$)=(6449$k$2)$

$+(6449$k$2)$ B cos$($kt$)=0$

since for all value of k found above,

k$2$