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Verify that the expression: 1 C(t,T) = 1 e-b(T-t) 1 + ab -a(T-t) a(a – b) b(a – b) is the solution to the differential equation: 1 — е-а(Т-t) C:(t, T) + - bC(t, T) = 0

Verify that the expression: 1 C(t,T) = 1 e-b(T-t) 1 + ab -a(T-t) a(a – b) b(a – b) is the solution to the differential equation: 1 — е-а(Т-t) C:(t, T) + - bC(t, T) = 0

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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Verify that the expression: 1 -a(T-t) 1 1 -6(T-t) C(t,T) = a(a – b) b(a – b)* ab - is the solution to the differential equation: 1 —е а(Т-t) C:(t,T)+ - bC(t,T) = 0 - a
Transcribed Image Text:Verify that the expression: 1 -a(T-t) 1 1 -6(T-t) C(t,T) = a(a – b) b(a – b)* ab - is the solution to the differential equation: 1 —е а(Т-t) C:(t,T)+ - bC(t,T) = 0 - a
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