Skip to main content

Math Advanced Math ADVANCED ENGINEERING MATHEMATICS

The value of the function f ( t ) if the Laplace transform of f is L ( f ) = 5.5 ( s + 1.5 ) ( s − 4 ) . The function f ( t ) is f ( t ) = e 4 t − e − 1.5 t _ . Given that, the Laplace transform of f is L ( f ) = 5.5 ( s + 1.5 ) ( s − 4 ) . Consider the expression 5.5 ( s + 1.5 ) ( s − 4 ) . Use partial fraction to reduce the expression 5.5 ( s + 1.5 ) ( s − 4 ) as follows. 5.5 ( s + 1.5 ) ( s − 4 ) = A ( s + 1.5 ) + B ( s − 4 ) 5.5 ( s + 1.5 ) ( s − 4 ) = A ( s − 4 ) + B ( s + 1.5 ) ( s + 1.5 ) ( s − 4 ) 5.5 = A ( s − 4 ) + B ( s + 1.5 ) ⋯ ⋯ ( 1 ) Substitute s = 4 in (1) and obtain B as follows. 5.5 = A ( 4 − 4 ) + B ( 4 + 1.5 ) 5.5 = B ( 5.5 ) B = 1 Substitute s = − 1.5 in (1) and obtain A as follows. 5.5 = A ( − 1.5 − 4 ) + B ( − 1.5 + 1.5 ) 5.5 = − A ( 5.5 ) A = − 1 Thus, the expression 5.5 ( s + 1.5 ) ( s − 4 ) becomes − 1 ( s + 1.5 ) + 1 ( s − 4 ) . Now the Laplace transform of f can be written as, L ( f ) = 1 ( s − 4 ) − 1 ( s + 1.5 ) . Take inverse transform on both sides of L ( f ) = 1 ( s − 4 ) − 1 ( s + 1.5 ) as follows. L − 1 { L ( f ) } = L − 1 { 1 ( s − 4 ) − 1 ( s + 1.5 ) } f ( t ) = L − 1 { 1 ( s − 4 ) } − L − 1 { 1 ( s + 1.5 ) } = e 4 t − e − 1.5 t ( ∵ L − 1 { 1 ( s − a ) } = e a t ) Thus, the function f ( t ) is f ( t ) = e 4 t − e − 1.5 t _ . The value of the function f ( t ) if the Laplace transform of f is L ( f ) = ω s 2 ( s 2 + ω 2 ) . The function f ( t ) is f ( t ) = 1 ω 2 ( ω t − sin ω t ) _ . Given that, the Laplace transform of f is L ( f ) = ω s 2 ( s 2 + ω 2 ) . Consider the expression ω s 2 ( s 2 + ω 2 ) . Use partial fraction to reduce the expression ω s 2 ( s 2 + ω 2 ) as follows. ω s 2 ( s 2 + ω 2 ) = A s + B s 2 + C s + D s 2 + ω 2 ω s 2 ( s 2 + ω 2 ) = A s ( s 2 + ω 2 ) + B ( s 2 + ω 2 ) + ( C s + D ) s 2 s 2 ( s 2 + ω 2 ) ω = A s ( s 2 + ω 2 ) + B ( s 2 + ω 2 ) + ( C s + D ) s 2 ⋯ ⋯ ( 1 ) Substitute s = 0 in (1) and obtain B as follows. ω = A s ( s 2 + ω 2 ) + B ( s 2 + ω 2 ) + ( C s + D ) s 2 ω = A ( 0 ) ( 0 2 + ω 2 ) + B ( 0 2 + ω 2 ) + ( C ( 0 ) + D ) ( 0 ) 2 ω = B ω 2 B = 1 ω Equate s 2 coefficient and obtain D as follows. B + D = 0 1 ω + D = 0 D = − 1 ω Substitute s = 1 in (1) and obtain the equation as follows. ω = A s ( s 2 + ω 2 ) + B ( s 2 + ω 2 ) + ( C s + D ) s 2 ω = A ( 1 ) ( 1 2 + ω 2 ) + B ( 1 2 + ω 2 ) + ( C ( 1 ) + D ) ( 1 ) 2 A + A ω 2 + B + B ω 2 + C + D = ω A + A ω 2 + 1 ω + ω + C − 1 ω = ω Simplify further, A + A ω 2 + ω + C = ω A + C + A ω 2 = 0 C = − ( A + A ω 2 ) Substitute s = − 1 in (1) and obtain the equation as follows. ω = A s ( s 2 + ω 2 ) + B ( s 2 + ω 2 ) + ( C s + D ) s 2 ω = A ( − 1 ) ( ( − 1 ) 2 + ω 2 ) + B ( ( − 1 ) 2 + ω 2 ) + ( C ( − 1 ) + D ) ( − 1 ) 2 − A − A ω 2 + B + B ω 2 − C + D = ω − A − A ω 2 + 1 ω + ω − C − 1 ω = ω Simplify further, − A − A ω 2 + ω − C = ω − A − A ω 2 − ( − ( A + A ω 2 ) ) = 0 ( ∵ C = − ( A + A ω 2 ) ) A ( − 1 − ω 2 + 1 + ω 2 ) = 0 A = 0 The value of A = 0 then the value of C = 0 . Thus, the expression ω s 2 ( s 2 + ω 2 ) becomes 1 ω s 2 − 1 ω ( s 2 + ω 2 ) . Now the Laplace transform of f can be written as, L ( f ) = 1 ω s 2 − 1 ω ( s 2 + ω 2 ) . Take inverse transform on both sides of L ( f ) = 1 ω s 2 − 1 ω ( s 2 + ω 2 ) as follows. L − 1 { L ( f ) } = L − 1 { 1 ω s 2 − 1 ω ( s 2 + ω 2 ) } f ( t ) = L − 1 { 1 ω s 2 } − L − 1 { 1 ω ( s 2 + ω 2 ) } = 1 ω t − 1 ω [ sin ω t ω ] ( ∵ L − 1 { 1 ( s 2 + a 2 ) } = sin a t a ) = 1 ω 2 ( ω t − sin ω t ) Thus, the function f ( t ) is f ( t ) = 1 ω 2 ( ω t − sin ω t ) _ . The value of the function f ( t ) if the Laplace transform of f is L ( f ) = 40.5 s ( s 2 − 9 ) . The function f ( t ) is f ( t ) = 4.5 ( cosh 3 t − 1 ) _ . Given that, the Laplace transform of f is L ( f ) = 40.5 s ( s 2 − 9 ) . Consider the expression 40.5 s ( s 2 − 9 ) . Use partial fraction to reduce the expression 40.5 s ( s 2 − 9 ) as follows. 40.5 s ( s 2 − 9 ) = A s + B s − 3 + C s + 3 40.5 s ( s 2 − 9 ) = A ( s + 3 ) ( s − 3 ) + B s ( s + 3 ) + C s ( s − 3 ) s ( s 2 − 9 ) 40.5 = A ( s + 3 ) ( s − 3 ) + B s ( s + 3 ) + C s ( s − 3 ) ⋯ ⋯ ( 1 ) Substitute s = 0 in (1) and obtain A as follows. 40.5 = A ( s + 3 ) ( s − 3 ) + B s ( s + 3 ) + C s ( s − 3 ) 40.5 = A ( 0 + 3 ) ( 0 − 3 ) + B ( 0 ) ( 0 + 3 ) + C ( 0 ) ( 0 − 3 ) 40.5 = − 9 A + 0 + 0 A = − 4.5 Substitute s = 3 in (1) and obtain B as follows. 40.5 = A ( s + 3 ) ( s − 3 ) + B s ( s + 3 ) + C s ( s − 3 ) 40.5 = A ( 3 + 3 ) ( 3 − 3 ) + B ( 3 ) ( 3 + 3 ) + C ( 3 ) ( 3 − 3 ) 40.5 = 0 + 18 B + 0 B = 2.25 Substitute s = − 3 in (1) and obtain C as follows. 40.5 = A ( s + 3 ) ( s − 3 ) + B s ( s + 3 ) + C s ( s − 3 ) 40.5 = A ( − 3 + 3 ) ( − 3 − 3 ) + B ( − 3 ) ( − 3 + 3 ) + C ( − 3 ) ( − 3 − 3 ) 40.5 = 0 + 18 C C = 2.25 Thus, the expression 40.5 s ( s 2 − 9 ) becomes − 4.5 s + 2.25 s − 3 + 2.25 s + 3 . Now the Laplace transform of f can be written as, L ( f ) = − 4.5 s + 2.25 s − 3 + 2.25 s + 3 . Take inverse transform on both sides of L ( f ) = − 4.5 s + 2.25 s − 3 + 2.25 s + 3 as follows. L − 1 { L ( f ) } = L − 1 { − 4.5 s + 2.25 s − 3 + 2.25 s + 3 } f ( t ) = L − 1 { − 4.5 s } + L − 1 { 2.25 s − 3 } + L − 1 { 2.25 s + 3 } = − 4.5 + 2.25 e 3 t + 2.25 e − 3 t = 4.5 ( − 1 + e 3 t 2 + e − 3 t 2 ) Simplify further as follows. f ( t ) = 4.5 ( − 1 + e 3 t 2 + e − 3 t 2 ) = 4.5 ( − 1 + ( e 3 t 2 + e − 3 t 2 ) ) = 4.5 ( − 1 + ( cosh 3 t ) ) = 4.5 ( cosh 3 t − 1 ) Thus, the function f ( t ) is f ( t ) = 4.5 ( cosh 3 t − 1 ) _ .

The value of the function f ( t ) if the Laplace transform of f is L ( f ) = 5.5 ( s + 1.5 ) ( s − 4 ) . The function f ( t ) is f ( t ) = e 4 t − e − 1.5 t _ . Given that, the Laplace transform of f is L ( f ) = 5.5 ( s + 1.5 ) ( s − 4 ) . Consider the expression 5.5 ( s + 1.5 ) ( s − 4 ) . Use partial fraction to reduce the expression 5.5 ( s + 1.5 ) ( s − 4 ) as follows. 5.5 ( s + 1.5 ) ( s − 4 ) = A ( s + 1.5 ) + B ( s − 4 ) 5.5 ( s + 1.5 ) ( s − 4 ) = A ( s − 4 ) + B ( s + 1.5 ) ( s + 1.5 ) ( s − 4 ) 5.5 = A ( s − 4 ) + B ( s + 1.5 ) ⋯ ⋯ ( 1 ) Substitute s = 4 in (1) and obtain B as follows. 5.5 = A ( 4 − 4 ) + B ( 4 + 1.5 ) 5.5 = B ( 5.5 ) B = 1 Substitute s = − 1.5 in (1) and obtain A as follows. 5.5 = A ( − 1.5 − 4 ) + B ( − 1.5 + 1.5 ) 5.5 = − A ( 5.5 ) A = − 1 Thus, the expression 5.5 ( s + 1.5 ) ( s − 4 ) becomes − 1 ( s + 1.5 ) + 1 ( s − 4 ) . Now the Laplace transform of f can be written as, L ( f ) = 1 ( s − 4 ) − 1 ( s + 1.5 ) . Take inverse transform on both sides of L ( f ) = 1 ( s − 4 ) − 1 ( s + 1.5 ) as follows. L − 1 { L ( f ) } = L − 1 { 1 ( s − 4 ) − 1 ( s + 1.5 ) } f ( t ) = L − 1 { 1 ( s − 4 ) } − L − 1 { 1 ( s + 1.5 ) } = e 4 t − e − 1.5 t ( ∵ L − 1 { 1 ( s − a ) } = e a t ) Thus, the function f ( t ) is f ( t ) = e 4 t − e − 1.5 t _ . The value of the function f ( t ) if the Laplace transform of f is L ( f ) = ω s 2 ( s 2 + ω 2 ) . The function f ( t ) is f ( t ) = 1 ω 2 ( ω t − sin ω t ) _ . Given that, the Laplace transform of f is L ( f ) = ω s 2 ( s 2 + ω 2 ) . Consider the expression ω s 2 ( s 2 + ω 2 ) . Use partial fraction to reduce the expression ω s 2 ( s 2 + ω 2 ) as follows. ω s 2 ( s 2 + ω 2 ) = A s + B s 2 + C s + D s 2 + ω 2 ω s 2 ( s 2 + ω 2 ) = A s ( s 2 + ω 2 ) + B ( s 2 + ω 2 ) + ( C s + D ) s 2 s 2 ( s 2 + ω 2 ) ω = A s ( s 2 + ω 2 ) + B ( s 2 + ω 2 ) + ( C s + D ) s 2 ⋯ ⋯ ( 1 ) Substitute s = 0 in (1) and obtain B as follows. ω = A s ( s 2 + ω 2 ) + B ( s 2 + ω 2 ) + ( C s + D ) s 2 ω = A ( 0 ) ( 0 2 + ω 2 ) + B ( 0 2 + ω 2 ) + ( C ( 0 ) + D ) ( 0 ) 2 ω = B ω 2 B = 1 ω Equate s 2 coefficient and obtain D as follows. B + D = 0 1 ω + D = 0 D = − 1 ω Substitute s = 1 in (1) and obtain the equation as follows. ω = A s ( s 2 + ω 2 ) + B ( s 2 + ω 2 ) + ( C s + D ) s 2 ω = A ( 1 ) ( 1 2 + ω 2 ) + B ( 1 2 + ω 2 ) + ( C ( 1 ) + D ) ( 1 ) 2 A + A ω 2 + B + B ω 2 + C + D = ω A + A ω 2 + 1 ω + ω + C − 1 ω = ω Simplify further, A + A ω 2 + ω + C = ω A + C + A ω 2 = 0 C = − ( A + A ω 2 ) Substitute s = − 1 in (1) and obtain the equation as follows. ω = A s ( s 2 + ω 2 ) + B ( s 2 + ω 2 ) + ( C s + D ) s 2 ω = A ( − 1 ) ( ( − 1 ) 2 + ω 2 ) + B ( ( − 1 ) 2 + ω 2 ) + ( C ( − 1 ) + D ) ( − 1 ) 2 − A − A ω 2 + B + B ω 2 − C + D = ω − A − A ω 2 + 1 ω + ω − C − 1 ω = ω Simplify further, − A − A ω 2 + ω − C = ω − A − A ω 2 − ( − ( A + A ω 2 ) ) = 0 ( ∵ C = − ( A + A ω 2 ) ) A ( − 1 − ω 2 + 1 + ω 2 ) = 0 A = 0 The value of A = 0 then the value of C = 0 . Thus, the expression ω s 2 ( s 2 + ω 2 ) becomes 1 ω s 2 − 1 ω ( s 2 + ω 2 ) . Now the Laplace transform of f can be written as, L ( f ) = 1 ω s 2 − 1 ω ( s 2 + ω 2 ) . Take inverse transform on both sides of L ( f ) = 1 ω s 2 − 1 ω ( s 2 + ω 2 ) as follows. L − 1 { L ( f ) } = L − 1 { 1 ω s 2 − 1 ω ( s 2 + ω 2 ) } f ( t ) = L − 1 { 1 ω s 2 } − L − 1 { 1 ω ( s 2 + ω 2 ) } = 1 ω t − 1 ω [ sin ω t ω ] ( ∵ L − 1 { 1 ( s 2 + a 2 ) } = sin a t a ) = 1 ω 2 ( ω t − sin ω t ) Thus, the function f ( t ) is f ( t ) = 1 ω 2 ( ω t − sin ω t ) _ . The value of the function f ( t ) if the Laplace transform of f is L ( f ) = 40.5 s ( s 2 − 9 ) . The function f ( t ) is f ( t ) = 4.5 ( cosh 3 t − 1 ) _ . Given that, the Laplace transform of f is L ( f ) = 40.5 s ( s 2 − 9 ) . Consider the expression 40.5 s ( s 2 − 9 ) . Use partial fraction to reduce the expression 40.5 s ( s 2 − 9 ) as follows. 40.5 s ( s 2 − 9 ) = A s + B s − 3 + C s + 3 40.5 s ( s 2 − 9 ) = A ( s + 3 ) ( s − 3 ) + B s ( s + 3 ) + C s ( s − 3 ) s ( s 2 − 9 ) 40.5 = A ( s + 3 ) ( s − 3 ) + B s ( s + 3 ) + C s ( s − 3 ) ⋯ ⋯ ( 1 ) Substitute s = 0 in (1) and obtain A as follows. 40.5 = A ( s + 3 ) ( s − 3 ) + B s ( s + 3 ) + C s ( s − 3 ) 40.5 = A ( 0 + 3 ) ( 0 − 3 ) + B ( 0 ) ( 0 + 3 ) + C ( 0 ) ( 0 − 3 ) 40.5 = − 9 A + 0 + 0 A = − 4.5 Substitute s = 3 in (1) and obtain B as follows. 40.5 = A ( s + 3 ) ( s − 3 ) + B s ( s + 3 ) + C s ( s − 3 ) 40.5 = A ( 3 + 3 ) ( 3 − 3 ) + B ( 3 ) ( 3 + 3 ) + C ( 3 ) ( 3 − 3 ) 40.5 = 0 + 18 B + 0 B = 2.25 Substitute s = − 3 in (1) and obtain C as follows. 40.5 = A ( s + 3 ) ( s − 3 ) + B s ( s + 3 ) + C s ( s − 3 ) 40.5 = A ( − 3 + 3 ) ( − 3 − 3 ) + B ( − 3 ) ( − 3 + 3 ) + C ( − 3 ) ( − 3 − 3 ) 40.5 = 0 + 18 C C = 2.25 Thus, the expression 40.5 s ( s 2 − 9 ) becomes − 4.5 s + 2.25 s − 3 + 2.25 s + 3 . Now the Laplace transform of f can be written as, L ( f ) = − 4.5 s + 2.25 s − 3 + 2.25 s + 3 . Take inverse transform on both sides of L ( f ) = − 4.5 s + 2.25 s − 3 + 2.25 s + 3 as follows. L − 1 { L ( f ) } = L − 1 { − 4.5 s + 2.25 s − 3 + 2.25 s + 3 } f ( t ) = L − 1 { − 4.5 s } + L − 1 { 2.25 s − 3 } + L − 1 { 2.25 s + 3 } = − 4.5 + 2.25 e 3 t + 2.25 e − 3 t = 4.5 ( − 1 + e 3 t 2 + e − 3 t 2 ) Simplify further as follows. f ( t ) = 4.5 ( − 1 + e 3 t 2 + e − 3 t 2 ) = 4.5 ( − 1 + ( e 3 t 2 + e − 3 t 2 ) ) = 4.5 ( − 1 + ( cosh 3 t ) ) = 4.5 ( cosh 3 t − 1 ) Thus, the function f ( t ) is f ( t ) = 4.5 ( cosh 3 t − 1 ) _ .

ADVANCED ENGINEERING MATHEMATICS

ADVANCED ENGINEERING MATHEMATICS

10th Edition

ISBN: 9781119664697

Author: Kreyszig

Publisher: WILEY

See similar textbooks

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

Blurred answer

Chapter 6.5, Problem 25P

Chapter 6.6, Problem 1P

Students have asked these similar questions

Q4 3 Points Consider the matrices A = B = C = 2 3 View them as elements of the vector space M2x2 of all 2 x 2 matrices. Determine if C is an element of Span{A, B}. Show work to justify your answer. (Hint: You may want to remind yourself what the definition of Span{A, B} is.) Please select file(s) Select file(s) Save Answer

Q2 29 2 Points Q2.1 1 Point [120] Let A 120 and v = 001 v is an element of Col A. v is not an element of Col A. Save Answer Q2.2 1 Point [12 0 Let A 120 and w= [001] w is an element of Nul A. Ow is not an element of Nul A. Save Answer 1 Select the statement that is true. 0 Select the statement that is true.

Q3 3 Points In the vector space V = P₂ of all polynomials of degree at most 2, consider the elements Pi(t)=1+t, P2(t) = 1-t, Ps(t)=1+2t+t². Determine if it is possible to write the polynomial q(t) = t² as a linear combination of Pi(t), p2(t), ps(t). If possible, show which linear combination of p₁(t), p2(t), ps(t) is equal to q(t) . If not possible, explain why. Show all of your work. Please select file(s) Select file(s) Save Answer

Chapter 6 Solutions

ADVANCED ENGINEERING MATHEMATICS

Ch. 6.1 - Prob. 1P Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....

Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Find the transform. Show the details of your work....Ch. 6.1 - Table 6.1. Convert this table to a table for...Ch. 6.1 - Using in Prob. 10, find , where f1(t) = 0 if t ≦...Ch. 6.1 - Table 6.1. Derive formula 6 from formulas 9 and...Ch. 6.1 - Nonexistence. Show that does not satisfy a...Ch. 6.1 - Nonexistence. Give simple examples of functions...Ch. 6.1 - Existence. Show that . [Use (30) in App. 3.1.]...Ch. 6.1 - Change of scale. If and c is any positive...Ch. 6.1 - Inverse transform. Prove that is linear. Hint:...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - Given F(s) = ℒ(f), find f(t). a, b, L, n are...Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.1 - In Probs. 33–36 find the transform. In Probs....Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the IVPs by the Laplace transform. If...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Solve the shifted data IVPs by the Laplace...Ch. 6.2 - Using (1) or (2), find if f(t) equals: t cos 4t Ch. 6.2 - Using (1) or (2), find if f(t) equals: te−at Ch. 6.2 - Using (1) or (2), find if f(t) equals: cos2 2t Ch. 6.2 - Using (1) or (2), find if f(t) equals: sin2 ωt Ch. 6.2 - Using (1) or (2), find if f(t) equals: sin4 t....Ch. 6.2 - Using (1) or (2), find if f(t) equals: cosh2 t Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.2 - INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3,...Ch. 6.3 - Report on Shifting Theorems. Explain and compare...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Sketch or graph the given function, which is...Ch. 6.3 - Find and sketch or graph f(t) if equals e−3s/(s −...Ch. 6.3 - Prob. 13P Ch. 6.3 - Prob. 14P Ch. 6.3 - Find and sketch or graph f(t) if equals e−3s/s4 Ch. 6.3 - Prob. 16P Ch. 6.3 - Prob. 17P Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Prob. 20P Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Prob. 23P Ch. 6.3 - Using the Laplace transform and showing the...Ch. 6.3 - Prob. 25P Ch. 6.3 - Prob. 26P Ch. 6.3 - Prob. 27P Ch. 6.3 - Prob. 28P Ch. 6.3 - Prob. 29P Ch. 6.3 - Prob. 30P Ch. 6.3 - Prob. 31P Ch. 6.3 - Prob. 32P Ch. 6.3 - Prob. 33P Ch. 6.3 - Prob. 34P Ch. 6.3 - Prob. 35P Ch. 6.3 - Prob. 36P Ch. 6.3 - Prob. 37P Ch. 6.3 - Prob. 38P Ch. 6.3 - Prob. 39P Ch. 6.3 - Prob. 40P Ch. 6.4 - Prob. 3P Ch. 6.4 - Prob. 4P Ch. 6.4 - Prob. 5P Ch. 6.4 - Prob. 6P Ch. 6.4 - Prob. 7P Ch. 6.4 - Prob. 8P Ch. 6.4 - Prob. 9P Ch. 6.4 - Prob. 11P Ch. 6.4 - Prob. 15P Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: 2. Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: 3. Ch. 6.5 - CONVOLUTIONS BY INTEGRATION Find: 4. Ch. 6.5 - Prob. 5P Ch. 6.5 - Prob. 6P Ch. 6.5 - Prob. 7P Ch. 6.5 - Prob. 8P Ch. 6.5 - Prob. 9P Ch. 6.5 - Prob. 10P Ch. 6.5 - Prob. 11P Ch. 6.5 - Prob. 12P Ch. 6.5 - Prob. 13P Ch. 6.5 - Prob. 14P Ch. 6.5 - CAS EXPERIMENT. Variation of a Parameter. (a)...Ch. 6.5 - Prob. 17P Ch. 6.5 - Prob. 18P Ch. 6.5 - Prob. 19P Ch. 6.5 - Prob. 20P Ch. 6.5 - Prob. 21P Ch. 6.5 - Prob. 22P Ch. 6.5 - Prob. 23P Ch. 6.5 - Prob. 24P Ch. 6.5 - Prob. 25P Ch. 6.5 - Prob. 26P Ch. 6.6 - Prob. 2P Ch. 6.6 - Prob. 3P Ch. 6.6 - Prob. 4P Ch. 6.6 - Prob. 5P Ch. 6.6 - Prob. 6P Ch. 6.6 - Prob. 7P Ch. 6.6 - Prob. 8P Ch. 6.6 - Prob. 9P Ch. 6.6 - Prob. 10P Ch. 6.6 - Prob. 11P Ch. 6.6 - Prob. 14P Ch. 6.6 - Prob. 15P Ch. 6.6 - Prob. 16P Ch. 6.6 - Prob. 17P Ch. 6.6 - Prob. 18P Ch. 6.6 - Prob. 19P Ch. 6.6 - Prob. 20P Ch. 6.7 - Prob. 2P Ch. 6.7 - Prob. 3P Ch. 6.7 - Prob. 4P Ch. 6.7 - Prob. 5P Ch. 6.7 - Prob. 6P Ch. 6.7 - Prob. 7P Ch. 6.7 - Prob. 8P Ch. 6.7 - Prob. 9P Ch. 6.7 - Prob. 10P Ch. 6.7 - Prob. 11P Ch. 6.7 - Prob. 12P Ch. 6.7 - Prob. 13P Ch. 6.7 - Prob. 14P Ch. 6.7 - Prob. 15P Ch. 6.7 - Prob. 16P Ch. 6.7 - Prob. 19P Ch. 6.7 - Prob. 20P Ch. 6 - Prob. 1RQ Ch. 6 - Prob. 2RQ Ch. 6 - Prob. 3RQ Ch. 6 - Prob. 4RQ Ch. 6 - Prob. 5RQ Ch. 6 - When and how do you use the unit step function and...Ch. 6 - If you know f(t) = ℒ−1{F(s)}, how would you find...Ch. 6 - Explain the use of the two shifting theorems from...Ch. 6 - Prob. 9RQ Ch. 6 - Prob. 10RQ Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the transform, indicating the method used and...Ch. 6 - Find the inverse transform, indicating the method...Ch. 6 - Prob. 21RQ Ch. 6 - Prob. 22RQ Ch. 6 - Prob. 23RQ Ch. 6 - Prob. 24RQ Ch. 6 - Prob. 25RQ Ch. 6 - Prob. 26RQ Ch. 6 - Prob. 27RQ Ch. 6 - Prob. 28RQ Ch. 6 - Prob. 29RQ Ch. 6 - Prob. 30RQ Ch. 6 - Prob. 31RQ Ch. 6 - Prob. 32RQ Ch. 6 - Prob. 33RQ Ch. 6 - Prob. 34RQ Ch. 6 - Prob. 35RQ Ch. 6 - Prob. 36RQ Ch. 6 - Prob. 37RQ Ch. 6 - Prob. 38RQ Ch. 6 - Prob. 39RQ Ch. 6 - Prob. 40RQ Ch. 6 - Prob. 41RQ Ch. 6 - Prob. 42RQ Ch. 6 - Prob. 43RQ Ch. 6 - Prob. 44RQ Ch. 6 - Prob. 45RQ

Knowledge Booster

Background pattern image

Similar questions

SEE MORE QUESTIONS

Recommended textbooks for you

Advanced Engineering Mathematics
Advanced Math
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Wiley, John & Sons, Incorporated
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...
Advanced Math
ISBN:9781118141809
Author:Nathan Klingbeil
Publisher:WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:9781337798310
Author:Peterson, John.
Publisher:Cengage Learning,
Basic Technical Mathematics
Advanced Math
ISBN:9780134437705
Author:Washington
Publisher:PEARSON
Topology
Advanced Math
ISBN:9780134689517
Author:Munkres, James R.
Publisher:Pearson,

Text book image

Advanced Engineering Mathematics

Advanced Math

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Wiley, John & Sons, Incorporated

Text book image

Numerical Methods for Engineers

Advanced Math

ISBN:9780073397924

Author:Steven C. Chapra Dr., Raymond P. Canale

Publisher:McGraw-Hill Education

Text book image

Introductory Mathematics for Engineering Applicat...

Advanced Math

ISBN:9781118141809

Author:Nathan Klingbeil

Publisher:WILEY

Text book image

Mathematics For Machine Technology

Advanced Math

ISBN:9781337798310

Author:Peterson, John.

Publisher:Cengage Learning,

Text book image

Basic Technical Mathematics

Advanced Math

ISBN:9780134437705

Author:Washington

Publisher:PEARSON

Text book image

Advanced Math

ISBN:9780134689517

Author:Munkres, James R.

Publisher:Pearson,

SEE MORE TEXTBOOKS