Determine L ¹{F}. F(s) = -s²-2s+5 (s+ 2)² (s+3)
Chapter12: Sequences, Series And Binomial Theorem
Section12.4: Binomial Theorem
Problem 12.78TI: Find the coefficient of the x4 term of (x+2)7.
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Where did I go wrong? I've attached the question and my work. I used partial fraction decomposition method, then used the laplace formula I thought was relevant.
Did I use the wrong Laplace formula? Not sure why my answer is incorrect.
Transcribed Image Text:Determine L ¹{F}.
F(s) =
-s²-2s+5
(s+ 2)² (s+3)
![= A + B
(2+2) (a+1)
+ C
(a+1)
-
- 2² -20 + 5 = A(2+2) (241) + B(212) 4(2+3)
((02)² (ats)
(A+)
(2+)
= A(a+1) (2+)) + B(2+2)(2+3) + C(242(2+2)
[1] = A (at + 5/2+6) + B(2²+S2+b) + C (2²+42+4)
⑥ F(2) == n-d+s
(ad)² (2+3)
· - 22 = 2² (A+B+C) · -20=2 (SA+SB+4C)
-1 = A+B+ㄴ
1-7-8--1
(72) = v
A = -8 61 + 116
3
10
15
A = -128
15
- 128-11126
15
a +2
at
e
J"{F}=-128 C
15
A=-37 +2 C +3A+3-1-2=25-SB-100
20 3
4
6
A=-29 + 2C+3A
3
15
4
-1=20-29
43
15
A =& L + 116
13
IS
To
17
= A2² + SA2+ 6A + B2² +5B~+6B + Co² +4L2L4L
= 2* (A+B+C) + 2 (SA+SB + 4K) + 6A+6B+42
9
• S = 6A+B+C
-2 = SA+SB+C 16A = S-6B-4C
A
9
-1 -2=5 S-B-2C+SB+UL
ㅅ
0-1) 2
9
3.
to
+37 = + SB+ 10 C
6
SB = 37
6
B =
- 12e
at
B=37-2 C
1 30
B=37
30
17
+
1
9
19
-
~~ 아~
2+3
이
21 -0-21 host
도
10C
3
-st
+ ble
A S
37
20
=동-(끎-04-0
6
3
A=-2-4C
S
:)
C=-3 A-3
ㅗ
10
4
(=-3-8 (+ 116
43
15
(=26-29
+2
C = 61
10
S
19-79=2
10
1
ह
10
USING 오{으룩 (2):1
ह
01
2-6
at
ㄱ =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48ae96ba-8296-4f48-84d9-a65ba8f4727e%2F0717d0ff-0713-4e61-9519-a27040b2e05f%2Fpbwyt4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:= A + B
(2+2) (a+1)
+ C
(a+1)
-
- 2² -20 + 5 = A(2+2) (241) + B(212) 4(2+3)
((02)² (ats)
(A+)
(2+)
= A(a+1) (2+)) + B(2+2)(2+3) + C(242(2+2)
[1] = A (at + 5/2+6) + B(2²+S2+b) + C (2²+42+4)
⑥ F(2) == n-d+s
(ad)² (2+3)
· - 22 = 2² (A+B+C) · -20=2 (SA+SB+4C)
-1 = A+B+ㄴ
1-7-8--1
(72) = v
A = -8 61 + 116
3
10
15
A = -128
15
- 128-11126
15
a +2
at
e
J"{F}=-128 C
15
A=-37 +2 C +3A+3-1-2=25-SB-100
20 3
4
6
A=-29 + 2C+3A
3
15
4
-1=20-29
43
15
A =& L + 116
13
IS
To
17
= A2² + SA2+ 6A + B2² +5B~+6B + Co² +4L2L4L
= 2* (A+B+C) + 2 (SA+SB + 4K) + 6A+6B+42
9
• S = 6A+B+C
-2 = SA+SB+C 16A = S-6B-4C
A
9
-1 -2=5 S-B-2C+SB+UL
ㅅ
0-1) 2
9
3.
to
+37 = + SB+ 10 C
6
SB = 37
6
B =
- 12e
at
B=37-2 C
1 30
B=37
30
17
+
1
9
19
-
~~ 아~
2+3
이
21 -0-21 host
도
10C
3
-st
+ ble
A S
37
20
=동-(끎-04-0
6
3
A=-2-4C
S
:)
C=-3 A-3
ㅗ
10
4
(=-3-8 (+ 116
43
15
(=26-29
+2
C = 61
10
S
19-79=2
10
1
ह
10
USING 오{으룩 (2):1
ह
01
2-6
at
ㄱ =
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