(a) For what values of k does the function y k = = cos(kt) satisfy the differential equation 49y” = -16y? (Enter your answers as a comma-separated list.) (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. We begin by calculating the following. y = A sin(kt) + B cos(kt) ⇒ y' = Ak cos(kt) – Bk sin(kt) = y" = Note that the given differential equation 49y" = -16y is equivalent to 49y" + 16y = Now, substituting the expressions for y and y” above and simplifying, we have = LHS 49y" + 16y = 49( [ = -49 + 16(A sin(kt) + B cos(kt)) 49Bk2 cos(kt) + 16A sin(kt) + 16B cos(kt) + (16 − 49k²) B cos(kt) = (16-49k2) = 0 since for all value of k found above, k² =

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Question
(a) For what values of k does the function y
k =
=
cos(kt) satisfy the differential equation 49y” = -16y? (Enter your answers as a comma-separated list.)
(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.
We begin by calculating the following.
y = A sin(kt) + B cos(kt) ⇒ y' = Ak cos(kt) – Bk sin(kt) = y" =
Note that the given differential equation 49y" = -16y is equivalent to 49y" + 16y =
Now, substituting the expressions for y and y” above and simplifying, we have
=
LHS 49y" + 16y =
49( [
=
-49
+ 16(A sin(kt) + B cos(kt))
49Bk2 cos(kt) + 16A sin(kt) + 16B cos(kt)
+ (16 − 49k²) B cos(kt)
=
(16-49k2)
= 0
since for all value of k found above, k² =
Transcribed Image Text:(a) For what values of k does the function y k = = cos(kt) satisfy the differential equation 49y” = -16y? (Enter your answers as a comma-separated list.) (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. We begin by calculating the following. y = A sin(kt) + B cos(kt) ⇒ y' = Ak cos(kt) – Bk sin(kt) = y" = Note that the given differential equation 49y" = -16y is equivalent to 49y" + 16y = Now, substituting the expressions for y and y” above and simplifying, we have = LHS 49y" + 16y = 49( [ = -49 + 16(A sin(kt) + B cos(kt)) 49Bk2 cos(kt) + 16A sin(kt) + 16B cos(kt) + (16 − 49k²) B cos(kt) = (16-49k2) = 0 since for all value of k found above, k² =
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