Differential Equations Translation of Laplace Transforms and Partial-Fractions., guaranteed an verified 100% Pass

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Translation of Laplace Transforms and Partial Fractions


𝑃(𝑠)
Suppose 𝑅 (𝑠) = , where 𝑃(𝑠) and 𝑄(𝑠) are polynomials, and the degree
𝑄(𝑠)
of 𝑃(𝑠) is less than 𝑄(𝑠).



The portion of the partial fraction decomposition of 𝑅(𝑠) corresponding to a
linear factor (𝑠 − 𝑎) of multiplicity 𝑛 is a sum of terms of the form:
𝐴1 2𝐴 𝑛 𝐴
+ (𝑠−𝑎) 2 + ⋯ + (𝑠−𝑎)𝑛
𝑠−𝑎

where 𝐴1 , 𝐴2 , … 𝐴𝑛 are constants.


Ex. Find the form of the partial fraction expansion of
1
𝑅 (𝑠) = .
(𝑠2 −1)(𝑠+1)



1 1 1
= (𝑠−1)(𝑠+1)(𝑠+1) = (𝑠−1)(𝑠+1)2
(𝑠 2 −1)(𝑠+1)
1 𝐴 𝐵 𝐶
(𝑠−1)(𝑠+1)2
= 𝑠−1 + 𝑠+1 + (𝑠+1)2 .


The portion of the partial fraction decomposition corresponding to the irreducible
quadratic factor (𝑠 − 𝑎)2 + 𝑏 2 of multiplicity 𝑛 is a sum if 𝑛 partial fractions of
the form:


𝐴1 𝑠+𝐵1 𝐴 𝑠+𝐵 𝐴 𝑠+𝐵
(𝑠−𝑎)2 +𝑏 2
+ ((𝑠−𝑎2 )2 +𝑏2 2)2 + ⋯ + ((𝑠−𝑎𝑛 )2 +𝑏𝑛2 )𝑛 .

, 2


Ex. Find the form of the partial fraction expansion for
1
𝑅 (𝑠) = 2.
(𝑠−1)2 (𝑠2 +1)



1 𝐴 𝐵 𝐶𝑠+𝐷 𝐸𝑠+𝐹
(𝑠−1)2 (𝑠 2 +1)2
= 𝑠−1 + (𝑠−1)2 + 𝑠2 +1 + (𝑠2 +1)2 .


Note: any quadratic polynomial with real coefficients can always put in the form
of (𝑠 − 𝑎)2 ± 𝑏 2 by completing the square.



Ex. Put 𝑠 2 + 8𝑠 + 40 in the form (𝑠 − 𝑎)2 ± 𝑏 2 by completing the square.



𝑠 2 + 8𝑠 + 40 = 𝑠 2 + 8𝑠 + 16 − 16 + 40
= (𝑠 + 4)2 + 24.


Theorem: If 𝐹 (𝑠) = ℒ(𝑓 (𝑡)) exists for 𝑠 > 𝑐, then ℒ(𝑒 𝑎𝑡 𝑓 (𝑡)) exists
for 𝑠 > (𝑎 + 𝑐) and ℒ(𝑒 𝑎𝑡 𝑓(𝑡)) = 𝐹(𝑠 − 𝑎), or equivalently,
ℒ −1 (𝐹 (𝑠 − 𝑎)) = 𝑒 𝑎𝑡 𝑓(𝑡). Thus, the translation 𝑠 → (𝑠 − 𝑎)
in the transform corresponds to multiplication of the original
function by 𝑒 𝑎𝑡 .



Proof:
∞ ∞
ℒ(𝑒 𝑎𝑡 𝑓(𝑡)) = ∫0 𝑒 −𝑠𝑡 𝑒 𝑎𝑡 𝑓(𝑡)𝑑𝑡 = ∫0 𝑒 −(𝑠−𝑎)𝑡 𝑓(𝑡)𝑑𝑡 = 𝐹 (𝑠 − 𝑎).