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"-3y¹+2y} = L{4e²"} 4 (2)[s²F (s) — sy (0) — y'(0)] — 3[sF (s) — y'(0)] +2F(s) = ¸±₂ — S-2 4 (3) s²F (s)-s(-3)-5-3sF(s)+3(-3)+2F(s) = S-2 4 (4) F (s)[s² − 3s+2]+3s−5−9= S-2 4-3s(s-2)+14(s—2 (5) F (s)[s² − 3s +2] = S-2 4-3s²-6s+14s-28 (6) F (s) = -3s² +20s-24 (s-2)(s²-3s+2) (S-2) (s²-3s+2) -3s² +20s-24 (7) y(t)=L¹ (s-2)(s²-3s+2) -3s² +20s-24 A B C (8) + + (s-2)(s-2)(s-1) S-2 S-2 s-1 (9)-3s² +20s-24 = A(s−2)(s−1)+B(s−1)+C(s−2)² (10) Let s=-2 → B=4 ; Let s=1 → C=7; Let s=0⇒ A=4 4 4 7 (11) y(t)=L¹ + s-2 (S-2)² S- (12) y(t)=4e²¹ +4te²¹ - 7e' = 4e' +4te' - 7
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"-3y¹+2y} = L{4e²"} 4 (2)[s²F (s) — sy (0) — y'(0)] — 3[sF (s) — y'(0)] +2F(s) = ¸±₂ — S-2 4 (3) s²F (s)-s(-3)-5-3sF(s)+3(-3)+2F(s) = S-2 4 (4) F (s)[s² − 3s+2]+3s−5−9= S-2 4-3s(s-2)+14(s—2 (5) F (s)[s² − 3s +2] = S-2 4-3s²-6s+14s-28 (6) F (s) = -3s² +20s-24 (s-2)(s²-3s+2) (S-2) (s²-3s+2) -3s² +20s-24 (7) y(t)=L¹ (s-2)(s²-3s+2) -3s² +20s-24 A B C (8) + + (s-2)(s-2)(s-1) S-2 S-2 s-1 (9)-3s² +20s-24 = A(s−2)(s−1)+B(s−1)+C(s−2)² (10) Let s=-2 → B=4 ; Let s=1 → C=7; Let s=0⇒ A=4 4 4 7 (11) y(t)=L¹ + s-2 (S-2)² S- (12) y(t)=4e²¹ +4te²¹ - 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
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Determine the particular solution of the given using Laplace and Inverse Laplace transform.
Just two or three answer on the questions please? Thank you.
![Problem:
Determine the particular solution of y"-3y'+2y=4e²; y(0)=-3, y'(0) = 5 using Laplace and
Inverse Laplace Transforms.
Solution:
(1) Z{y"−3y¹+2y} = L{4e²"}
4
(2)[s²F (s) — sy(0) — y'(0)] — 3[sF (s) —y'(0)]+2F(s)= ¸±₂
— —
S-2
4
(3) s²F
(s)-s(-3)-5-3sF(s)+3(-3)+2F(s) =
S-2
4
(4) F (s)[s² − 3s+2]+3s−5−9=
S-2
4-3s(s-2)+14(s—2)
(5) F (s)[s² − 3s +2] =
S-2
4-3s²-6s+14s-28
(6) F (s) =
-3s² +20s-24
(s-2)(s²-3s+2)
(S-2) (s²-3s+2)
-3s² +20s-24
(7) y(t)=L¹
(s-2)(s²-3s+2)
-3s² +20s-24
A
B
C
(8)
+
+
(s-2)(s-2)(s-1) S-2 S-2 s-1
(9)−3s² +20s-24 = A(s−2)(s−1)+B(s−1)+C(s−2)²
(10) Let s=-2 → B=4 ; Let s=1 → C=7; Let s=0⇒ A=4
4
4
7
(11) y(t)=L¹ +
s-2 (S-2)² S-
(12) y(t)=4e²¹ +4te²¹ - 7e' = 4e' +4te' - 7](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F60b35b95-6c2d-4498-99a9-c50ea06a6fb5%2F44d372c2-e47e-4c1e-8f14-8de31115821c%2Fmgb691q_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) Z{y"−3y¹+2y} = L{4e²"}
4
(2)[s²F (s) — sy(0) — y'(0)] — 3[sF (s) —y'(0)]+2F(s)= ¸±₂
— —
S-2
4
(3) s²F
(s)-s(-3)-5-3sF(s)+3(-3)+2F(s) =
S-2
4
(4) F (s)[s² − 3s+2]+3s−5−9=
S-2
4-3s(s-2)+14(s—2)
(5) F (s)[s² − 3s +2] =
S-2
4-3s²-6s+14s-28
(6) F (s) =
-3s² +20s-24
(s-2)(s²-3s+2)
(S-2) (s²-3s+2)
-3s² +20s-24
(7) y(t)=L¹
(s-2)(s²-3s+2)
-3s² +20s-24
A
B
C
(8)
+
+
(s-2)(s-2)(s-1) S-2 S-2 s-1
(9)−3s² +20s-24 = A(s−2)(s−1)+B(s−1)+C(s−2)²
(10) Let s=-2 → B=4 ; Let s=1 → C=7; Let s=0⇒ A=4
4
4
7
(11) y(t)=L¹ +
s-2 (S-2)² S-
(12) y(t)=4e²¹ +4te²¹ - 7e' = 4e' +4te' - 7
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