(a) For what values of k does the function y = cos(kt) satisfy the differential equation 100y" = -49y? (Enter your answers as a comma-separated list.) k = (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). The given differential equation 100y" = -49y Is equivalent to 100y" + 49y = Thus, LHS = 100y" + 49y= 100(-Ak2 sin(kt) – Bk2 cos(kt)) + = -100AK2 sin(kt) – 100BK2 cos(kt) + sin(kt) + 49B cos(kt) = (49 - 100k2)A sin(kt) + since 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...
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(a) For what values of k does the function y = cos(kt) satisfy the differential equation 100y" = -49y? (Enter your answers as a comma-separated list.)
k =
(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).
The given differential equation 100y" = -49y Is equivalent to 100y" + 49y =
Thus,
LHS = 100y" + 49y= 100(-Ak2 sin(kt) – Bk2 cos(kt)) +
= -100AK2 sin(kt) – 100BK2 cos(kt) +
sin(kt) + 49B cos(kt)
= (49 - 100k2)A sin(kt) +
since k2 =
Transcribed Image Text:(a) For what values of k does the function y = cos(kt) satisfy the differential equation 100y" = -49y? (Enter your answers as a comma-separated list.) k = (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). The given differential equation 100y" = -49y Is equivalent to 100y" + 49y = Thus, LHS = 100y" + 49y= 100(-Ak2 sin(kt) – Bk2 cos(kt)) + = -100AK2 sin(kt) – 100BK2 cos(kt) + sin(kt) + 49B cos(kt) = (49 - 100k2)A sin(kt) + since k2 =
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