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(a) For what values of k does the function y = cos(kt) satisfy the differential equation 25y" = -36y? (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 25y" = -36y is equivalent to 25y" + 36y= LHS = 25y" + 36y= 25(-Ak² sin(kt) - Bk² cos(kt)) + 36( = -25AK² sin(kt) - 25Bk² cos(kt) + = (36 - 25k²)A sin(kt) + - since k² = cos(kt) . Thus, sin(kt) + 36B cos(kt)

(a) For what values of k does the function y = cos(kt) satisfy the differential equation 25y" = -36y? (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 25y" = -36y is equivalent to 25y" + 36y= LHS = 25y" + 36y= 25(-Ak² sin(kt) - Bk² cos(kt)) + 36( = -25AK² sin(kt) - 25Bk² cos(kt) + = (36 - 25k²)A sin(kt) + - since k² = cos(kt) . Thus, sin(kt) + 36B cos(kt)

Calculus: Early Transcendentals
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 25y" = -36y? (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" -Ak² sin(kt) - Bk² cos(kt). y = A sin(kt) + B cos(kt) → y' = Ak cos(kt) - Bk sin(kt) The given differential equation 25y" = -36y is equivalent to 25y" + 36y= LHS = 25y" + 36y= 25(-Ak² sin(kt) - Bk² cos(kt)) + 36( = -25Ak² sin(kt) - 25Bk² cos(kt) + ( = (36 - 25k²)A sin(kt) + since k² = cos(kt) . Thus, sin(kt) + 36B cos(kt)
Transcribed Image Text:(a) For what values of k does the function y = cos(kt) satisfy the differential equation 25y" = -36y? (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" -Ak² sin(kt) - Bk² cos(kt). y = A sin(kt) + B cos(kt) → y' = Ak cos(kt) - Bk sin(kt) The given differential equation 25y" = -36y is equivalent to 25y" + 36y= LHS = 25y" + 36y= 25(-Ak² sin(kt) - Bk² cos(kt)) + 36( = -25Ak² sin(kt) - 25Bk² cos(kt) + ( = (36 - 25k²)A sin(kt) + since k² = cos(kt) . Thus, sin(kt) + 36B cos(kt)
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