(a) For what values of k does the function y = cos(kt) satisfy the differential equation 81y'' = −4y? (Enter your answers as a c

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(a) For what values of k does the function y = cos(kt) satisfy the differential equation 81y'' = −4y? (Enter your answers as a c

omma-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).

Mathematics

1 answer:

tester [3.9K] · Answer 1 · 2026-03-02T02:01:39+03:00

Answer:

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

Step-by-step explanation:

The initial scenario is a specific case of the subsequent one, so we will address the second case first.

Consider $y = A\sin(kt) + B\cos(kt)$ . Through the utilization of derivatives and trigonometric function properties, it is determined that

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

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

The equation is represented as $81y''=-4y$ . It's important to note that since $y'' = -k^2y$ it leads to the equation

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

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

It's notable that in this instance, the value of k is independent of A and B. Thus, it applies universally to any values of A and B. The first scenario is included since it corresponds to A=0 and B=1.