# Math Help - Differential equation

1. ## Differential equation

For what nonzero values of k does the function y=sin(kt) satisfy the differential equation y'' + 9y = 0?

2. Originally Posted by jamessmith
For what nonzero values of k does the function y=sin(kt) satisfy the differential equation y'' + 9y = 0?

$y=\sin(kt)$

$

y'=k\cos(kt)
$

$

y''= -k^2\sin(kt)
$

$y''+9y=0$

$
-k^2\sin(kt)+9\sin(kt)=0
$

And so $\sin(kt)(-k^2+9)=0$

And you can solve that for k right?

3. Originally Posted by jamessmith
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.

$y'=kcos(kt)$

$y''=-k^2sin(kt)$

So just substitute into the differential equation:

$-k^2sin(kt)+9sin(kt)=0$

so $k=3$ is one:

$sin(kt)=0$ is the other possible solution. Just find the k