Question

(a) For what values of k does the function y = cos(kt) satisfy the differential equation 81y'' = −4y? (Enter your answers as a comma-separated list.) k = Correct: Your answer is correct. (b) For those values of k, verify that every member of the family of functions y = A sin(kt) + B cos(kt) is also a solution. y = A sin(kt) + B cos(kt) ⇒ y' = Ak cos(kt) − Bk sin(kt) ⇒ y'' = −Ak2 sin(kt) − Bk2 cos(kt).

Answer 1

Answer:

$k = \frac{2}{9}, k=\frac{-2}{9}$

Step-by-step explanation:

The first case is a special case of the second one, so we will solve the question for the second case first.

Consider $y = A\sin(kt) + B\cos(kt)$. Using the properties of derivatives and the derivatives of trigonometric functions we get that

$y' = A\cdot k \cos(kt)- B \cdot k \sin(kt) = k (A\cos(kt)-B\sin(kt))$

$y'' = k(-A\cdot k \sin(kt)-B\cdot k \cos(kt)) = -k^2(y)$

We have the equation $81y''=-4y$. Note that since $y'' = -k^2y$then we have the equation

$-k^2 81y=-4y$,

which implies that $k^2 = \frac{4}{81}$. Then, $k=\pm\frac{2}{9}$

Note that in this case, the value of k doesn't depend on the values of A and B. So, it applies to every value of A and B. The first case is included, since it is the case in which A=0 and B=1.