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

$49y=-36y?$

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

$k=$

$(b)$

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

$y=$ A $sin(kt)+Bcos(kt)$

$is$ also a solution.

$We$ begin $by$ calculating the following.

$y=$ A $sin(kt)+Bcos(kt)y=Akcos(kt)-Bksin(kt)y=$

Note that the given differential equation

$49y=-36y$

$is$ equivalent $to$

$49y+36y=$

$.$

Now, substituting the expressions for $y$ and

$y$

above and simplifying, $we$ have

$LHS=49y+36y=49$

$+36(Asin(kt)+Bcos(kt))=-49$

$-49Bk2cos(kt)+36$A $sin(kt)+36Bcos(kt)=(36-49k2)$

$+(36-49k2)Bcos(kt)=0$

since for all value $ofk$ found above,

$k2=$

$.$