Problem: Determine the particular solution of y"–3y'+2y=4e²"; 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)=: s-2 4 (4) F (s)[s² – 3s +2]+3s – 5–9=- -2 (5) F(s)[s² – 3s +2]=1-3s(s– 2)+14(s–2) s-2 4- 3s? – 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". (s-2)(s² – 3s+2)|
Problem: Determine the particular solution of y"–3y'+2y=4e²"; 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)=: s-2 4 (4) F (s)[s² – 3s +2]+3s – 5–9=- -2 (5) F(s)[s² – 3s +2]=1-3s(s– 2)+14(s–2) s-2 4- 3s? – 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". (s-2)(s² – 3s+2)|
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
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=4e²"; 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)=:
s-2
4
(4) F (s)[s² – 3s +2]+3s – 5–9=-
-2
(5) F(s)[s² – 3s +2]=1-3s(s– 2)+14(s–2)
s-2
4- 3s? – 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'.
(s– 2)(s² – 3s+2)|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F98c2a3a6-d360-4784-bfed-6dde17ec371f%2F882d5160-9cc2-40ce-b28b-348be6214b05%2F7vx3xvk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem:
Determine the particular solution of y"–3y'+2y=4e²"; 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)=:
s-2
4
(4) F (s)[s² – 3s +2]+3s – 5–9=-
-2
(5) F(s)[s² – 3s +2]=1-3s(s– 2)+14(s–2)
s-2
4- 3s? – 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'.
(s– 2)(s² – 3s+2)|

Transcribed Image Text:-3s + 20s – 24
(8)
(s– 2)(s– 2)(s–1)
A. B
+
S-2
s-2
§-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';
-2' (s– 2)
(12) y(1)=4e²ª +4te²' – 7e' = 4e' +4te' – 7
%3D
-
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