Question

Find all values of k for which the function y=sin(kt) satisfies the differential equation y′′+19y=0. separate your answers by commas.

Answer 1

The given function is

$y=sin(kt)$

Differentiating

$y'=kcos(kt)$

Again differentiating

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

Substituting the values of y '' and y in

$y''+19y=0$

We will get

$-k^2 sin(kt)+19sin(kt)=0 \\ sin(kt) (19-k^2)=0 sin(kt) =0, 19-k^2=0 \\ kt = \pi n , k =+- \sqrt{19} \\ k = \frac{ \pi n}{t} , - \sqrt{17}, \sqrt{17}$