For what nonzero values of k does the function y=sin(kt) satisfy the differential equation y'' + 9y = 0?
Do you know how to get started? First just find the derivatives of y.
$\displaystyle y'=kcos(kt)$
$\displaystyle y''=-k^2sin(kt)$
So just substitute into the differential equation:
$\displaystyle -k^2sin(kt)+9sin(kt)=0$
so $\displaystyle k=3$ is one:
$\displaystyle sin(kt)=0$ is the other possible solution. Just find the k