University:
University of California, Berkeley
Course:
MATH 51 | Calculus I
Academic year:

2023
Views:

427

Pages:

2
Author:

Clint W.

Section Theorem: If L {f (t)} = F (s) and a is any real number, then L {eatf (t)} = F (s − a). Exercise 1. Determine the Laplace transform of eat sin bt. (Nagle Section 7.3 Example 1) Exercise 2. Evaluate: (a) L { te−6t } (b) L { t10 e−7t }. (#2,4) Class Exercise 1. Evaluate: (a) L { e2t (t − 1)2 } (c) L { e3t (9 − 4t + 10 sin 2t ) }. (#6,8,10) (b) L { e−2t cos 4t } Corollary: L −1 { F (s − a) } = eat f (t). Exercise 3. Determine L −1 { 5 s−6 Exercise 4. Determine L −1 { 5 (s+2)4 Exercise 5. Determine L −1 { 3s+2 s2 +2s+10 Exercise 6. Evaluate: (a) L −1 { − 6s s2 +9 (c) L −1 { (s+1) (s+2)4 3 2s2 +8s+10 }. (Nagle Section 7.4 Example 2) }. (Nagle Section 7.4 Example 3) }. (Nagle Section 7.4 Example 4) 1 (s−1)4 } Class Exercise 2. Evaluate: (a) L −1 { 2 + (b) L −1 { 2s+5 s2 +6s+34 } 1 s2 +2s+5 }. (#12, 14) (b) L −1 { 5s (s−2)2 } }. (#16, 18, 20) Exercise 7. Use the Laplace transform to solve the given initial-value problem. (#22,24) (a) y ′ − y = 1 + tet , y(0) = 0 (b) y ′′ − 4y ′ + 4y = t3 e2t , y(0) = 0, y ′ (0) = 0 Class Exercise 3. Use the Laplace transform to solve the given initial-value problem. (a) y ′′ − 4y ′ + 4y = t3 , y(0) = 1, y ′ (0) = 0 (#26) (b) 2y ′′ + 20y ′ + 51y = 0, y(0) = 2, y ′ (0) = 0 (#28) (c) y ′′ − 2y ′ + 5y = 1 + t, y(0) = 0, y ′ (0) = 4 (#30) Definition: The unit step function U (t − a) is defined to be ( 0, 0 ≤ t < a U (t − a) = 1, t ≥ a Theorem: If F (s) = L { f (t) } and a > 0, then L { f (t − a) U (t − a) } = e−as F (s). Corollary: L −1 { e−as F (s) } = f (t − a)U (t − a). Corollary: L { g(t)U (t − a) } = e−as L { g(t + a) }. Exercise 8. Evaluate: L { e2−t U (t − 2) }. (#38) Class Exercise 4. Evaluate: (a) L { (3t + 1) U (t − 1) } Exercise 9. Evaluate: L −1 { (1+e−2s )2 s+2 (b) L { sin t U (t − π2 ) } (#40, 42) }. (#44) Class Exercise 5. Evaluate: (a) L −1 { se−πs/2 s2 +4 } (b) L −1 { e−2s s2 (s−1) }. (#46, 48) Fact: A general piecewise-defined function of the type ( g(t), 0 ≤ t < a f (t) = h(t), t ≥ a is the same as f (t) = g(t) − g(t)U (t − a) + h(t)U (t − a). Exercise 10. Write the function in terms of unit step functions. Find the Laplace transform of the given function. (#58) ( 0, 0 ≤ t < 3π/2 f (t) = sin t, t ≥ 3π/2 Class Exercise 6. Write each function in terms of unit step functions. Find the Laplace transform of the given function. (#60) ( sin t, 0 ≤ t < 2π f (t) = 0, t ≥ 2π Fact: A general piecewise-defined function of the   0, f (t) = g(t),   0, type 0≤t< a a≤t< b t≥b can be written f (t) = g(t)[U (t − a) − U (t − b)]. Exercise 11. Write each function in terms of unit step functions. Find the Laplace transform of the given function. (#56)   1, 0 ≤ t < 4 f (t) = 0, 4 ≤ t < 5   1, t ≥ 5 Exercise 12. Use the Laplace transform to solve the given initial-value problem. y ′ + y = f (t), y(0) = 0, where ( 1, 0≤t< 1 f (t) = −1, t ≥ 1 (#64) Class Exercise 7. Use the Laplace transform to solve the given initial-value problem. y ′′ + 4y = f (t), y(0) = 0, y ′ (0) = −1 where ( 1, 0 ≤ t < 1 f (t) = 0, t ≥ 1 (#66) Exercise 13. Use the Laplace transform to solve the given initial-value problem. y ′′ − 5y ′ + 6y = U (t − 1), y(0) = 0, y ′ (0) = 1 (#68) Class Exercise 8. Use the Laplace transform to solve the given initial-value problem. y ′′ + 4y ′ + 3y = 1 − U (t − 2) − U (t − 4) + U (t − 6), y(0) = 0, y ′ (0) = 0 (#70) Homework: 3-27 (every 4th), 31, 37-69 (every 4th)