F(s) = 28²+218+61 K₁ K2 K3 (8+4)³ + 8+4 (8+4)2 + d (8+4)³ ds (8 + 4) ³ 28²+218+61 = (8+4)³ K₁ d K₂ 3. d K3 − — [(s + 4)³ ù ] + £; [(s + 4) ³¸ Ka³ ] + 1; [(s + 4)³ _ *,, ds d ds s+4 ds (8+4)2 ds = ± [K₁ (8 + 4)²] + [K2 (s + 4)] + [Ks] d ds ds (s+r)³
F(s) = 28²+218+61 K₁ K2 K3 (8+4)³ + 8+4 (8+4)2 + d (8+4)³ ds (8 + 4) ³ 28²+218+61 = (8+4)³ K₁ d K₂ 3. d K3 − — [(s + 4)³ ù ] + £; [(s + 4) ³¸ Ka³ ] + 1; [(s + 4)³ _ *,, ds d ds s+4 ds (8+4)2 ds = ± [K₁ (8 + 4)²] + [K2 (s + 4)] + [Ks] d ds ds (s+r)³
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
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What is the partial-farction expansion of the rational function F(s) if it has repeated real roots and a solution of a sum of three different fractions?
![F(s) = 28²+218+61
K₁
K2
K3
(8+4)³
+
8+4 (8+4)2
+
d
(8+4)³
ds
(8 + 4) ³ 28²+218+61
=
(8+4)³
K₁
d
K₂
3.
d
K3
− — [(s + 4)³ ù ] + £; [(s + 4) ³¸ Ka³ ] + 1; [(s + 4)³ _ *,,
ds
d
ds
s+4
ds
(8+4)2 ds
= ± [K₁ (8 + 4)²] + [K2 (s + 4)] + [Ks]
d
ds
ds
(s+r)³](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d5b5049-d8dd-402a-aa3b-0cfd97dc82be%2Ffe15199c-823e-45d2-98ef-933c93ac2e6a%2Fgelh78b_processed.png&w=3840&q=75)
Transcribed Image Text:F(s) = 28²+218+61
K₁
K2
K3
(8+4)³
+
8+4 (8+4)2
+
d
(8+4)³
ds
(8 + 4) ³ 28²+218+61
=
(8+4)³
K₁
d
K₂
3.
d
K3
− — [(s + 4)³ ù ] + £; [(s + 4) ³¸ Ka³ ] + 1; [(s + 4)³ _ *,,
ds
d
ds
s+4
ds
(8+4)2 ds
= ± [K₁ (8 + 4)²] + [K2 (s + 4)] + [Ks]
d
ds
ds
(s+r)³
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