Question

For what values of k does the function y = cos(kt) satisfy the differential equation 9y'' = −100y? (enter your answers as a comma-separated list.)

Answer 1

Answer:

-10/3, 10/3

Step-by-step explanation:

(In this answer I will use y' to denote the derivative of y with respect to t. You shouldn't normally do this because y' normally means the derivative of y with respect to x but I'll be a bit messy for this case)

First calculate the derivatives:

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

Then plug the derivtes y'' and y into the equation:

$-9k^2\cos(kt) = -100\cos(kt)$

Solve the equation for k:

$100\cos(kt) - 9k^2\cos(kt) = 0 \\\\\Rightarrow \cos(kt)(100-9k^2) = 0$

So then we have that $y=\cos(kt)$ satisfies the differential equation when $\cos(kt) = 0$ or when $100-9k^2=0$ (or both). The solutions to these equations are:

$\left \{ {{\cos(kt)=0 \Rightarrow k=\frac{n\pi}{2t}} \atop {100-9k^2 = 0 \Rightarrow k= \pm \sqrt{\frac{100}{9}}=\pm \frac{10}{3}}} \right.$

I understand that looks a bit complicated and I doubt you would have to give your answers in terms of t so if it asks for a separated list of answers I would go for:

k = -10/3, 10/3.