calculation. Identify the line where the error is found. Explain why it is wrong and provide a correction for that particular line. If there is no error found in the solution, just reply with NO ERROR. Problem: Determine the particular solution of y–3y'+2y - 4e; y(0)--3, y'(0) - 5 using Laplace and Inverse Laplace Transforms. Solution: (1)L{y"- 3y'+ 2y} = L{4e*} (2)[s*F(s)= sy(0)– y'(0)]-3[sF(s)– y'(0)]+2F(s)- 4 S-2 4 (3)s°F(s)– s(-3)–5– 3sF(s)+3(-3)+2F(s) =- 8-2 (4) F(s)[s* – 3s +2]+3s =5-9= 4 +3s – 5-9= S-2 (5) F()[s* – 3s + 2] = - 4- 3s(s – 2)+14(s–2) s-2 4- 3.s – 6s + 14s- 28 - 3s? + 20s – 24 (6) F (s) =- (s– 2)(s² – 3s +2) (s – 2)(s² – 3s +2) -3s? + 20s – 24 (7)y(1)=L' :- 2)(s* s-2)(s² – 3s+2) –3s² +20s - 24 (8) (s- 2)(s– 2)(s– 1) s-2 's-2 A B S-1 (9)– 3s² +20s – 24 = A(s–2)(s–1)+B(s– 1)+C(s–2)* (10) Let s=-2 B=4; Let s=1C=7;Let s=0- A=4 4 4 7 (11)y(1)= L', 2 – 2)' (12) y(1)=4e²" +4te²" – 7e' = 4e' +4te' – 7 :-
calculation. Identify the line where the error is found. Explain why it is wrong and provide a correction for that particular line. If there is no error found in the solution, just reply with NO ERROR. Problem: Determine the particular solution of y–3y'+2y - 4e; y(0)--3, y'(0) - 5 using Laplace and Inverse Laplace Transforms. Solution: (1)L{y"- 3y'+ 2y} = L{4e*} (2)[s*F(s)= sy(0)– y'(0)]-3[sF(s)– y'(0)]+2F(s)- 4 S-2 4 (3)s°F(s)– s(-3)–5– 3sF(s)+3(-3)+2F(s) =- 8-2 (4) F(s)[s* – 3s +2]+3s =5-9= 4 +3s – 5-9= S-2 (5) F()[s* – 3s + 2] = - 4- 3s(s – 2)+14(s–2) s-2 4- 3.s – 6s + 14s- 28 - 3s? + 20s – 24 (6) F (s) =- (s– 2)(s² – 3s +2) (s – 2)(s² – 3s +2) -3s? + 20s – 24 (7)y(1)=L' :- 2)(s* s-2)(s² – 3s+2) –3s² +20s - 24 (8) (s- 2)(s– 2)(s– 1) s-2 's-2 A B S-1 (9)– 3s² +20s – 24 = A(s–2)(s–1)+B(s– 1)+C(s–2)* (10) Let s=-2 B=4; Let s=1C=7;Let s=0- A=4 4 4 7 (11)y(1)= L', 2 – 2)' (12) y(1)=4e²" +4te²" – 7e' = 4e' +4te' – 7 :-
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
100%
![Spot the error/errors (if there is any) in the following
calculation. Identify the line where the error is found.
Explain why it is wrong and provide a correction for
that particular line. If there is no error found in the
solution, just reply with NO ERROR.
Problem:
Determine the particular solution of y'–3y'+2y=4e; y(0) – –3, y'(0) – 5 using Laplace and
Inverse Laplace Transforms.
Solution:
(1) L{v"- 3y'+ 2y} = L{4e"}
(2)[s*F(s)– sy (0)– y'(0)] -3[sF(s)– y'(0))+2F(s)=
4
S-2
4
(3)s°F(s)– s(-3)–5– 3sF(s)+3(-3)+2F(s)=-
s-2
(4) F(»\[s° – 3s +2]+3s – 5-9
4
3s – 5– 9=
§-2
-2)
(5) F(s)[s² – 3s +2] = 4–3s(s – 2)+14(s-
§-2
4- 3s – 6s+14.s – 28
-3s² + 20s – 24
(6) F (s)=
(s– 2)(s² – 3s +2)
(s– 2)(s² – 3s +2)
-3s2 + 20s – 24
(7) y(1)=L
|(s–2)(s² – 3s +2)|
- 3s? +20s – 24
(8)
(s– 2)(s– 2)(s–1) s-2
A
В
C
§-2
s-1
(9)– 3s² +20s – 24 = A(s– 2)(s–1)+B(s – 1)+C(s– 2)*
(10) Let s=-2→ B=4 ; Let s=1→ C=7;Let s=0 –→ A=4
4
4
7
(11) y(1)= L'
s–2' (s–2)
S-1
(12) y(1)=4e²' +4te²' – 7e' = 4e' +4te' – 7](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F80abc339-0ae5-421a-9ee5-7adb2109b1d6%2Ff246ae2f-9f3e-4a3d-afc0-8c0a83e4b375%2Fjvsd32d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Spot the error/errors (if there is any) in the following
calculation. Identify the line where the error is found.
Explain why it is wrong and provide a correction for
that particular line. If there is no error found in the
solution, just reply with NO ERROR.
Problem:
Determine the particular solution of y'–3y'+2y=4e; y(0) – –3, y'(0) – 5 using Laplace and
Inverse Laplace Transforms.
Solution:
(1) L{v"- 3y'+ 2y} = L{4e"}
(2)[s*F(s)– sy (0)– y'(0)] -3[sF(s)– y'(0))+2F(s)=
4
S-2
4
(3)s°F(s)– s(-3)–5– 3sF(s)+3(-3)+2F(s)=-
s-2
(4) F(»\[s° – 3s +2]+3s – 5-9
4
3s – 5– 9=
§-2
-2)
(5) F(s)[s² – 3s +2] = 4–3s(s – 2)+14(s-
§-2
4- 3s – 6s+14.s – 28
-3s² + 20s – 24
(6) F (s)=
(s– 2)(s² – 3s +2)
(s– 2)(s² – 3s +2)
-3s2 + 20s – 24
(7) y(1)=L
|(s–2)(s² – 3s +2)|
- 3s? +20s – 24
(8)
(s– 2)(s– 2)(s–1) s-2
A
В
C
§-2
s-1
(9)– 3s² +20s – 24 = A(s– 2)(s–1)+B(s – 1)+C(s– 2)*
(10) Let s=-2→ B=4 ; Let s=1→ C=7;Let s=0 –→ A=4
4
4
7
(11) y(1)= L'
s–2' (s–2)
S-1
(12) y(1)=4e²' +4te²' – 7e' = 4e' +4te' – 7
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