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Y(s) = Utilizing Partial Fractions (Cont.) When using partial fractions, we can use decomposition to make the algebraic equation suitable for inversely converting these terms into differential equations. Normally, using algebraic tactics can solve this problem. However, using partial fractions is essential because there will be equations that will be tedious and difficult when using the same tactic. As such, we take this equation to break it up into simpler pieces that match known formulas. Decomposition = = Algebraic tactics s+1 2 Y(s) = + (s+2)2 (s+2)4 (s+2)-1 2 + (s + 2)² (s+2)4 1 2 - + 5+2 (s + 2)² (s+2)4 s+1 2 s+1 A B + (s + 2)² (s + 2)4 + (s+2)2 (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)+ B12A + B If s=-2 -2+1 A(-2+2)+ BB = -1

Y(s) = Utilizing Partial Fractions (Cont.) When using partial fractions, we can use decomposition to make the algebraic equation suitable for inversely converting these terms into differential equations. Normally, using algebraic tactics can solve this problem. However, using partial fractions is essential because there will be equations that will be tedious and difficult when using the same tactic. As such, we take this equation to break it up into simpler pieces that match known formulas. Decomposition = = Algebraic tactics s+1 2 Y(s) = + (s+2)2 (s+2)4 (s+2)-1 2 + (s + 2)² (s+2)4 1 2 - + 5+2 (s + 2)² (s+2)4 s+1 2 s+1 A B + (s + 2)² (s + 2)4 + (s+2)2 (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)+ B12A + B If s=-2 -2+1 A(-2+2)+ BB = -1

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section: Chapter Questions
Problem 18T
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I need help in explaining, as I want to make sure that I describe the partial fractions decomposition accurately. 

Y(s) = Utilizing Partial Fractions (Cont.) When using partial fractions, we can use decomposition to make the algebraic equation suitable for inversely converting these terms into differential equations. Normally, using algebraic tactics can solve this problem. However, using partial fractions is essential because there will be equations that will be tedious and difficult when using the same tactic. As such, we take this equation to break it up into simpler pieces that match known formulas. Decomposition = = Algebraic tactics s+1 2 Y(s) = + (s+2)2 (s+2)4 (s+2)-1 2 + (s + 2)² (s+2)4 1 2 - + 5+2 (s + 2)² (s+2)4 s+1 2 s+1 A B + (s + 2)² (s + 2)4 + (s+2)2 (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)+ B12A + B If s=-2 -2+1 A(-2+2)+ BB = -1
Transcribed Image Text:Y(s) = Utilizing Partial Fractions (Cont.) When using partial fractions, we can use decomposition to make the algebraic equation suitable for inversely converting these terms into differential equations. Normally, using algebraic tactics can solve this problem. However, using partial fractions is essential because there will be equations that will be tedious and difficult when using the same tactic. As such, we take this equation to break it up into simpler pieces that match known formulas. Decomposition = = Algebraic tactics s+1 2 Y(s) = + (s+2)2 (s+2)4 (s+2)-1 2 + (s + 2)² (s+2)4 1 2 - + 5+2 (s + 2)² (s+2)4 s+1 2 s+1 A B + (s + 2)² (s + 2)4 + (s+2)2 (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)+ B12A + B If s=-2 -2+1 A(-2+2)+ BB = -1
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