(b). y" + 3y' + 2y = sin(2t — 6)u3(t); y(0) = 1, y'(0) = 0 Note that sin(2t - 6) = sin(2(t − 3)) is the shifted version of sin(2t) s2Ys3sY3+2Y: 2 -3s (s²+3s+2)Y= = 3 2 $2 +4 = e $2 +4 -3s e +s+3 Let 1 Y = 2e-3s 1 (s²+4)(s+1)(s+2) (s²+4)(s+1)(s+2) S+3 (s + 1)(s+2) (A+B)s+2A+B A s+1 B Cs+ D Cs + D + S+2 s² +4 + s² + 3s+2 s² +4 (A + B)s³ + (4A + 4B)s + (2A + B)s² + 8A+4B + Cs³ + 3Cs² + 2Cs + Ds² + 3Ds + 2D = (s²+4)(s+1)(s+2) (A+B+C)s³ + (2A + B + 3C + D)s² + (4A + 4B + 2C+ 3D)s + 8A+4B + 2D (s²+4)(s+1)(s+2) Therefore, to make the numerator 1, we need A+B+C 0 2A+B+3C + D = 0 4A 4B2C + 3D = 0 8A4B2D = 1 = Combining the first and third equation we have C fourth equation we have 12C + 2D = −1, and thus D : and 2A + B = 11, so A = }} and B = −1}. For the second term we have == S+3 A B = (s + 1)(s+2) s+1 S+2 so A+B = 1 and 2A + B = 3. Therefore, A = 2, B = = D. Combining the second and 1 20' 40° C = - 3. Now A+B = 3 40 (A+B)s+2A + B (s+ 1)(s+2) −1.
(b). y" + 3y' + 2y = sin(2t — 6)u3(t); y(0) = 1, y'(0) = 0 Note that sin(2t - 6) = sin(2(t − 3)) is the shifted version of sin(2t) s2Ys3sY3+2Y: 2 -3s (s²+3s+2)Y= = 3 2 $2 +4 = e $2 +4 -3s e +s+3 Let 1 Y = 2e-3s 1 (s²+4)(s+1)(s+2) (s²+4)(s+1)(s+2) S+3 (s + 1)(s+2) (A+B)s+2A+B A s+1 B Cs+ D Cs + D + S+2 s² +4 + s² + 3s+2 s² +4 (A + B)s³ + (4A + 4B)s + (2A + B)s² + 8A+4B + Cs³ + 3Cs² + 2Cs + Ds² + 3Ds + 2D = (s²+4)(s+1)(s+2) (A+B+C)s³ + (2A + B + 3C + D)s² + (4A + 4B + 2C+ 3D)s + 8A+4B + 2D (s²+4)(s+1)(s+2) Therefore, to make the numerator 1, we need A+B+C 0 2A+B+3C + D = 0 4A 4B2C + 3D = 0 8A4B2D = 1 = Combining the first and third equation we have C fourth equation we have 12C + 2D = −1, and thus D : and 2A + B = 11, so A = }} and B = −1}. For the second term we have == S+3 A B = (s + 1)(s+2) s+1 S+2 so A+B = 1 and 2A + B = 3. Therefore, A = 2, B = = D. Combining the second and 1 20' 40° C = - 3. Now A+B = 3 40 (A+B)s+2A + B (s+ 1)(s+2) −1.
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
Can you explain how all A B C D terms are calculated
Transcribed Image Text:(b). y" + 3y' + 2y = sin(2t — 6)u3(t); y(0) = 1, y'(0) = 0
Note that sin(2t - 6) = sin(2(t − 3)) is the shifted version of sin(2t)
s2Ys3sY3+2Y:
2
-3s
(s²+3s+2)Y=
=
3
2
$2 +4
=
e
$2 +4
-3s
e
+s+3
Let
1
Y = 2e-3s
1
(s²+4)(s+1)(s+2)
(s²+4)(s+1)(s+2)
S+3
(s + 1)(s+2)
(A+B)s+2A+B
A
s+1
B
Cs+ D
Cs + D
+
S+2 s² +4
+
s² + 3s+2
s² +4
(A + B)s³ + (4A + 4B)s + (2A + B)s² + 8A+4B + Cs³ + 3Cs² + 2Cs + Ds² + 3Ds + 2D
=
(s²+4)(s+1)(s+2)
(A+B+C)s³ + (2A + B + 3C + D)s² + (4A + 4B + 2C+ 3D)s + 8A+4B + 2D
(s²+4)(s+1)(s+2)
Therefore, to make the numerator 1, we need
A+B+C 0
2A+B+3C + D = 0
4A 4B2C + 3D = 0
8A4B2D = 1
=
Combining the first and third equation we have C
fourth equation we have 12C + 2D = −1, and thus D :
and 2A + B = 11, so A = }} and B = −1}.
For the second term we have
==
S+3
A
B
=
(s + 1)(s+2)
s+1
S+2
so A+B = 1 and 2A + B = 3. Therefore, A
= 2, B =
=
D. Combining the second and
1
20'
40°
C = - 3. Now A+B =
3
40
(A+B)s+2A + B
(s+ 1)(s+2)
−1.
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