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Utilizing Partial Fractions (Cont.) When using Partial Fraction, we can use the decomposition to make the algebraic equation suitable to inverse these terms into differential equations. Normally, using algebraic tactic can solve this problem. However, using partial fraction is important because there will be equation that will be tedious and difficult when the same tactic. As such Y(s) = Algebraic tactics 3+1 + 2 (s + 2)² (s+2)+ (s+ 2)² (s+2)4 (s+2)-1 2 + 1 1 = 2 + 5+2 (s + 2)² (s+2)4 Decomposition s+1 2 Y(s) = + (s + 2)² (s+2)4 s+1 (s+2)² A B + (s+2) (s+2)² s+1 A(s + 2) + B 1 = 2A + B 1 = 2A+ (-1) 2 = 2A A 1 s+1 1 1 = (s+2)² (s+2) (s+2)² If s=0 0+1 A(0+2)+B 12A + B If s=-2 -2+1=A(-2+2)+ BB = -1

Utilizing Partial Fractions (Cont.) When using Partial Fraction, we can use the decomposition to make the algebraic equation suitable to inverse these terms into differential equations. Normally, using algebraic tactic can solve this problem. However, using partial fraction is important because there will be equation that will be tedious and difficult when the same tactic. As such Y(s) = Algebraic tactics 3+1 + 2 (s + 2)² (s+2)+ (s+ 2)² (s+2)4 (s+2)-1 2 + 1 1 = 2 + 5+2 (s + 2)² (s+2)4 Decomposition s+1 2 Y(s) = + (s + 2)² (s+2)4 s+1 (s+2)² A B + (s+2) (s+2)² s+1 A(s + 2) + B 1 = 2A + B 1 = 2A+ (-1) 2 = 2A A 1 s+1 1 1 = (s+2)² (s+2) (s+2)² If s=0 0+1 A(0+2)+B 12A + B If s=-2 -2+1=A(-2+2)+ BB = -1

College Algebra
College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
ChapterP: Prerequisites
SectionP.4: Factoring Polynomials
Problem 84E: The rate of change of an autocatalytic chemical reaction is kQxkx2 where Q is the amount of the...
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Utilizing Partial Fractions (Cont.) When using Partial Fraction, we can use the decomposition to make the algebraic equation suitable to inverse these terms into differential equations. Normally, using algebraic tactic can solve this problem. However, using partial fraction is important because there will be equation that will be tedious and difficult when the same tactic. As such Y(s) = Algebraic tactics 3+1 + 2 (s + 2)² (s+2)+ (s+ 2)² (s+2)4 (s+2)-1 2 + 1 1 = 2 + 5+2 (s + 2)² (s+2)4 Decomposition s+1 2 Y(s) = + (s + 2)² (s+2)4 s+1 (s+2)² A B + (s+2) (s+2)² s+1 A(s + 2) + B 1 = 2A + B 1 = 2A+ (-1) 2 = 2A A 1 s+1 1 1 = (s+2)² (s+2) (s+2)² If s=0 0+1 A(0+2)+B 12A + B If s=-2 -2+1=A(-2+2)+ BB = -1
Transcribed Image Text:Utilizing Partial Fractions (Cont.) When using Partial Fraction, we can use the decomposition to make the algebraic equation suitable to inverse these terms into differential equations. Normally, using algebraic tactic can solve this problem. However, using partial fraction is important because there will be equation that will be tedious and difficult when the same tactic. As such Y(s) = Algebraic tactics 3+1 + 2 (s + 2)² (s+2)+ (s+ 2)² (s+2)4 (s+2)-1 2 + 1 1 = 2 + 5+2 (s + 2)² (s+2)4 Decomposition s+1 2 Y(s) = + (s + 2)² (s+2)4 s+1 (s+2)² A B + (s+2) (s+2)² s+1 A(s + 2) + B 1 = 2A + B 1 = 2A+ (-1) 2 = 2A A 1 s+1 1 1 = (s+2)² (s+2) (s+2)² If s=0 0+1 A(0+2)+B 12A + B If s=-2 -2+1=A(-2+2)+ BB = -1
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