Question

Find all values of k for which the function y=sin(kt) satisfies the differential equation y′′+16y=0. Separate your answers by commas. isn't the answer just ±4?

Answer 1

Answer:

0, 4, -4 and they may want you to mention formally all the kt multiples of $\pi$.

Step-by-step explanation:

Let's do the second derivative of the function: $y(t)=sin(k\,t)$

$y'(t) =k\,cos(k\,t)\\y"(t)=-k^2\,sin(kt)$

So now we want:

$y"+16\,y'=0\\-k^2\,sin(kt)+\,16\,sin(kt)=0\\sin(kt)\,(16-k^2)=0$

Then we have to include the zeros of the binomial ($16-k^2$) which as you say are +4 and -4, and also the zeros of $sin(kt)$, which include all those values of

$kt=0\,,\,\pi\,\,,\,2\pi\, ,\,etc.$

So an extra one that they may want you to include is k = 0