Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
3. Solve Equations Use the Laplace transform to solve the given initial value problem. (a). y" — y' — 6y = 4-u2(t); y(0) = 1, y'(0) = −1 s²Y – sy(0) – y' (0) — sY + y(0) – 6Y - (s²-s-6)Y 1 s(s − 3) (s + 2) (A+B)s 3A - 4 -2s e = S 4 -2s e - +s- S S -2s e + s(s-3)(s+2) S-2 (s - 3) (s + 2) (A+B)s²+2(A + B)s - 3As - 6A + Cs² - 3Cs s(s - 3) (s +2) (A+B+C)s² + (-A + 2B - 3C)s – 6A s(s - 3) (s +2) - For numerator being 1, we have A + B + C = 0, -A + 2B - 3C = 0 and -6A: Y(s) = = 4. Note that A B C + + = = S S- 3 S+2 s(s - 3) s+2 = 1. Then A = = B = 15' C= = 1 10° For the last term of Y, S-2 (s - 3) (s + 2) A B = - S 3 S+2 As 2A+Bs - 3B (s - 3) (s + 2) Therefore, A + B = 1 and 2A – 3B = −2, so A = ½, B = ½½. - (A+B)s+2A - 3B (s − 3)(s+2) In summary, 11 1 1 1 1 1 1 Y(s) = (- + 6 s 15 s - 3 2 y(t) 4 +(· + 3 15 y(t) + 7 1 + 2 + 15° 5 5² + -2t 10 s +2 4 +u2(t)(* + 5) (4 − e−2s) + - 5s-3 -2t e +u2(t)( (1½½ 1 e3(t−2) 4 1 5s+2 1 - e -2(1–2)) 6 15 10 1 1 -e³ (t−2) 15 10° Te-2(1–2))

3. Solve Equations Use the Laplace transform to solve the given initial value problem. (a). y" — y' — 6y = 4-u2(t); y(0) = 1, y'(0) = −1 s²Y – sy(0) – y' (0) — sY + y(0) – 6Y - (s²-s-6)Y 1 s(s − 3) (s + 2) (A+B)s 3A - 4 -2s e = S 4 -2s e - +s- S S -2s e + s(s-3)(s+2) S-2 (s - 3) (s + 2) (A+B)s²+2(A + B)s - 3As - 6A + Cs² - 3Cs s(s - 3) (s +2) (A+B+C)s² + (-A + 2B - 3C)s – 6A s(s - 3) (s +2) - For numerator being 1, we have A + B + C = 0, -A + 2B - 3C = 0 and -6A: Y(s) = = 4. Note that A B C + + = = S S- 3 S+2 s(s - 3) s+2 = 1. Then A = = B = 15' C= = 1 10° For the last term of Y, S-2 (s - 3) (s + 2) A B = - S 3 S+2 As 2A+Bs - 3B (s - 3) (s + 2) Therefore, A + B = 1 and 2A – 3B = −2, so A = ½, B = ½½. - (A+B)s+2A - 3B (s − 3)(s+2) In summary, 11 1 1 1 1 1 1 Y(s) = (- + 6 s 15 s - 3 2 y(t) 4 +(· + 3 15 y(t) + 7 1 + 2 + 15° 5 5² + -2t 10 s +2 4 +u2(t)(* + 5) (4 − e−2s) + - 5s-3 -2t e +u2(t)( (1½½ 1 e3(t−2) 4 1 5s+2 1 - e -2(1–2)) 6 15 10 1 1 -e³ (t−2) 15 10° Te-2(1–2))

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question

Can you explain how they calcualted the A B C values in more depth

3. Solve Equations Use the Laplace transform to solve the given initial value problem. (a). y" — y' — 6y = 4-u2(t); y(0) = 1, y'(0) = −1 s²Y – sy(0) – y' (0) — sY + y(0) – 6Y - (s²-s-6)Y 1 s(s − 3) (s + 2) (A+B)s 3A - 4 -2s e = S 4 -2s e - +s- S S -2s e + s(s-3)(s+2) S-2 (s - 3) (s + 2) (A+B)s²+2(A + B)s - 3As - 6A + Cs² - 3Cs s(s - 3) (s +2) (A+B+C)s² + (-A + 2B - 3C)s – 6A s(s - 3) (s +2) - For numerator being 1, we have A + B + C = 0, -A + 2B - 3C = 0 and -6A: Y(s) = = 4. Note that A B C + + = = S S- 3 S+2 s(s - 3) s+2 = 1. Then A = = B = 15' C= = 1 10° For the last term of Y, S-2 (s - 3) (s + 2) A B = - S 3 S+2 As 2A+Bs - 3B (s - 3) (s + 2) Therefore, A + B = 1 and 2A – 3B = −2, so A = ½, B = ½½. - (A+B)s+2A - 3B (s − 3)(s+2) In summary, 11 1 1 1 1 1 1 Y(s) = (- + 6 s 15 s - 3 2 y(t) 4 +(· + 3 15 y(t) + 7 1 + 2 + 15° 5 5² + -2t 10 s +2 4 +u2(t)(* + 5) (4 − e−2s) + - 5s-3 -2t e +u2(t)( (1½½ 1 e3(t−2) 4 1 5s+2 1 - e -2(1–2)) 6 15 10 1 1 -e³ (t−2) 15 10° Te-2(1–2))
Transcribed Image Text:3. Solve Equations Use the Laplace transform to solve the given initial value problem. (a). y" — y' — 6y = 4-u2(t); y(0) = 1, y'(0) = −1 s²Y – sy(0) – y' (0) — sY + y(0) – 6Y - (s²-s-6)Y 1 s(s − 3) (s + 2) (A+B)s 3A - 4 -2s e = S 4 -2s e - +s- S S -2s e + s(s-3)(s+2) S-2 (s - 3) (s + 2) (A+B)s²+2(A + B)s - 3As - 6A + Cs² - 3Cs s(s - 3) (s +2) (A+B+C)s² + (-A + 2B - 3C)s – 6A s(s - 3) (s +2) - For numerator being 1, we have A + B + C = 0, -A + 2B - 3C = 0 and -6A: Y(s) = = 4. Note that A B C + + = = S S- 3 S+2 s(s - 3) s+2 = 1. Then A = = B = 15' C= = 1 10° For the last term of Y, S-2 (s - 3) (s + 2) A B = - S 3 S+2 As 2A+Bs - 3B (s - 3) (s + 2) Therefore, A + B = 1 and 2A – 3B = −2, so A = ½, B = ½½. - (A+B)s+2A - 3B (s − 3)(s+2) In summary, 11 1 1 1 1 1 1 Y(s) = (- + 6 s 15 s - 3 2 y(t) 4 +(· + 3 15 y(t) + 7 1 + 2 + 15° 5 5² + -2t 10 s +2 4 +u2(t)(* + 5) (4 − e−2s) + - 5s-3 -2t e +u2(t)( (1½½ 1 e3(t−2) 4 1 5s+2 1 - e -2(1–2)) 6 15 10 1 1 -e³ (t−2) 15 10° Te-2(1–2))
Expert Solution
steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS