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Math Advanced Math Custom Kreyszig: Advanced Engineering Mathematics

The value of current i ( t ) shown in the given circuit with the assumption that L = 1 H , C = 0.25 F using Laplace transforms. The value of the current, i ( t ) is i ( t ) = 75 − 75 cos 2 t − 50 t 2 + ( 50 t 2 − 75 + 25 cos ( 2 t − 2 ) − 50 sin ( 2 t − 2 ) ) _ . Definition used: Laplace transform of derivatives: “The first and second derivatives of f ( t ) satisfy ℒ ( f ′ ) = s ℒ ( f ) − f ( 0 ) ⋯ ( 1 ) ℒ ( f ″ ) = s 2 ℒ ( f ) − s f ( 0 ) − f ′ ( 0 ) ⋯ ( 2 ) Formula (1) holds if f ( t ) is continuous for all t ≥ 0 and satisfies the growth restriction (2) and f ′ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 . Similarly (2) holds if f ( t ) and f ′ ( t ) are continuous for all t ≥ 0 and satisfy the growth restriction and f ″ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 .” Unit step function or Heaviside Function: The unit step function or the Heaviside function denoted by u ( t − a ) is defined as follows. u ( t − a ) = { 0 , if t < a 1 , if t > a } Second shifting theorem If f ( t ) has the transform F ( s ) , then the shifted function f ˜ ( t ) = f ( t − a ) u ( t − a ) = { 0 if t < a f ( t − a ) if t > a } has the transform e − a s F ( s ) . That is, if ℒ ( f ( t ) ) = F ( s ) then, ℒ ( f ( t − a ) u ( t − a ) ) = e − a s F ( s ) ⋯ ( 1 ) Calculation: Consider the given conditions as follows. L = 1 H , C = 0.25 F Also, the value of voltage is given as shown below. V = { 200 ( t − 1 3 t 3 ) , 0 < t < 1 0 , t > 1 } It is known that the equation of voltage in an LC circuit is L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) . Substitute the value of voltage in the above equation as shown below. L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = { 200 ( t − 1 3 t 3 ) , 0 < t < 1 0 , t > 1 } Now, differentiate the above equation as follows. L i ″ ( t ) + 1 C i ( t ) = v ′ ( t ) = { 200 ( 1 − t 2 ) , 0 < t < 1 0 , t > 1 } Use the definition of the Heaviside step function rewrite the equation of voltage as shown below. v = 200 ( 1 − t 2 ) ( 1 − u ( t − 1 ) ) = 200 ( 1 − t 2 ) − 200 ( 1 − t ) ( t + 1 ) u ( t − 1 ) = 200 ( 1 − t 2 ) + 200 ( t − 1 ) ( ( t − 1 ) + 2 ) u ( t − 1 ) = 200 ( 1 − t 2 ) + 200 ( ( t − 1 ) 2 + 2 ( t − 1 ) ) u ( t − 1 ) Also, note that the initial current and charge is zero and ℒ ( i ( t ) ) = I ( s ) . Substitute the values L = 1 H , C = 0.25 F and the equation of voltage in the equation L i ″ ( t ) + 1 C i ( t ) = v ′ ( t ) as shown below. i ″ ( t ) + 1 0.25 i ( t ) = 200 ( 1 − t 2 ) + 200 ( ( t − 1 ) 2 + 2 ( t − 1 ) ) u ( t − 1 ) , i ( 0 ) = 0 , i ′ ( 0 ) = 0 Take Laplace transform on both sides of the above equation; apply the second shifting theorem as shown below. ℒ ( i ″ ( t ) + 1 0.25 i ( t ) ) = ℒ ( 200 ( 1 − t 2 ) + 200 ( ( t − 1 ) 2 + 2 ( t − 1 ) ) u ( t − 1 ) ) ℒ ( i ″ ( t ) ) + 4 ℒ ( i ( t ) ) = ℒ ( 200 ( 1 − t 2 ) ) + 200 ℒ ( ( ( t − 1 ) 2 + 2 ( t − 1 ) ) u ( t − 1 ) ) s 2 I ( s ) + 4 I ( s ) = 200 s − 400 s 3 + 200 ( 2 s 3 + 2 s 2 ) e − s ( s 2 + 4 ) I ( s ) = 200 ( s 2 − 2 s 3 ) + 400 ( s + 1 s 3 ) e − s I ( s ) = 200 ( s 2 − 2 s 3 ( s 2 + 4 ) ) + 400 ( s + 1 s 3 ( s 2 + 4 ) ) e − s On further simplification of the above equation the following is obtained. I ( s ) = 200 ( s 2 − 2 s 3 ( s 2 + 4 ) ) + 400 ( s + 1 s 3 ( s 2 + 4 ) ) e − s = 200 I 1 ( s ) + 400 e − s I 2 ( s ) Apply partial fractions in the term I 1 ( s ) = s 2 − 2 s 3 ( s 2 + 4 ) as follows. s 2 − 2 s 3 ( s 2 + 4 ) = A s + B s 2 + C s 3 + D s + E s 2 + 4 s 2 − 2 s 3 ( s 2 + 4 ) = A ( s 2 ) ( s 2 + 4 ) + B s ( s 2 + 4 ) + C ( s 2 + 4 ) + ( D s + E ) s 3 s 3 ( s 2 + 4 ) s 2 − 2 = A ( s 2 ) ( s 2 + 4 ) + B s ( s 2 + 4 ) + C ( s 2 + 4 ) + ( D s + E ) s 3 1 = A s 4 + 4 A s 2 + B s 3 + 4 s B + C s 2 + 4 C + D s 4 + E s 3 Equate the coefficient of s 4 on both sides as shown below. 0 = A + D A = − D Now, equate the constant terms on both sides as follows. − 2 = 4 C − 1 2 = C Equate the coefficient of s 3 on both sides as shown below. 0 = B + E B = − E Equate the coefficient of s on both sides as shown below. 0 = 4 B 0 = B Therefore, the value of E is 0 . Equate the coefficient of s 2 on both sides as shown below. 1 = 4 A + C Substitute the value of C in the above equation and solve for A as follows. 1 = 4 A + C 1 = 4 A − 1 2 3 2 = 4 A 3 8 = A Therefore, the value of D is − 3 8 . Thus, the value of I 1 ( s ) is as follows. I 1 ( s ) = s 2 − 2 s 3 ( s 2 + 4 ) = A s + B s 2 + C s 3 + D s + E s 2 + 4 = 3 8 s − 1 2 s 3 − 3 s 8 ( s 2 + 4 ) Apply partial fractions in the term I 2 ( s ) = s + 1 s 3 ( s 2 + 4 ) as follows. s + 1 s 3 ( s 2 + 4 ) = A s + B s 2 + C s 3 + D s + E s 2 + 4 s + 1 s 3 ( s 2 + 4 ) = A s 2 ( s 2 + 4 ) + B s ( s 2 + 4 ) + C ( s 2 + 4 ) + ( D s + E ) ( s 3 ) s 3 ( s 2 + 4 ) s + 1 = A s 2 ( s 2 + 4 ) + B s ( s 2 + 4 ) + C ( s 2 + 4 ) + ( D s + E ) ( s 3 ) s + 1 = A s 4 + 4 A s 2 + B s 3 + 4 B s + C s 2 + 4 C + D s 4 + E s 3 Equate the coefficient of s 4 on both sides as shown below. 0 = A + D A = − D Now, equate the constant terms on both sides as follows. 1 = 4 C 1 4 = C Equate the coefficient of s 3 on both sides as shown below. 0 = B + E B = − E Equate the coefficient of s on both sides as shown below. 1 = 4 B 1 4 = B Therefore, the value of E is − 1 4 . Equate the coefficient of s 2 on both sides as shown below. 0 = 4 A + C − 1 4 = 4 A − 1 16 = A Therefore, the value of E is 1 16 . Thus, the value of I 2 ( s ) is as follows. I 2 ( s ) = s + 1 s 3 ( s 2 + 4 ) = − 1 16 s + 1 4 s 2 + 1 4 s 3 + s 16 ( s 2 + 4 ) − 1 4 ( s 2 + 4 ) Thus, the value of I ( s ) is obtained as follows. I ( s ) = 200 I 1 ( s ) + 400 e − s I 2 ( s ) = 200 ( 3 8 s − 1 2 s 3 − 3 s 8 ( s 2 + 4 ) ) + 400 e − s ( − 1 16 s + 1 4 s 2 + 1 4 s 3 + s 16 ( s 2 + 4 ) − 1 4 ( s 2 + 4 ) ) Now, take inverse Laplace transform on both sides as shown below. ℒ − 1 ( I ( s ) ) = 200 ℒ − 1 ( 3 8 s − 1 2 s 3 − 3 s 8 ( s 2 + 4 ) ) + 400 ℒ − 1 ( e − s ( − 1 16 s + 1 4 s 2 + 1 4 s 3 + s 16 ( s 2 + 4 ) − 1 4 ( s 2 + 4 ) ) ) = ( 200 ℒ − 1 ( 3 8 s ) − 200 ℒ − 1 ( 200 2 s 3 ) − 200 ℒ − 1 ( 3 s 8 ( s 2 + 4 ) ) + 400 ℒ − 1 ( − e − s ( 16 s ) ) + 400 ℒ − 1 ( e − s 4 s 2 ) + 400 ℒ − 1 ( e − s 4 s 3 ) + 400 ℒ − 1 ( e − s s 16 ( s 2 + 4 ) ) − 400 ℒ − 1 ( e − s 4 ( s 2 + 4 ) ) ) = ( ℒ − 1 ( 75 s e − s ) − ℒ − 1 ( 100 s 3 e − s ) − ℒ − 1 ( 75 s s 2 + 4 e − s ) + ℒ − 1 ( 100 s 3 e − s ) + ℒ − 1 ( 25 s s 2 + 4 e − s ) − ℒ − 1 ( 100 s 2 + 4 e − s ) + ℒ − 1 ( 100 s 2 e − s ) − ℒ − 1 ( 25 s e − s ) ) On further simplifications, i ( t ) = 75 − 75 cos 2 t − 50 t 2 + ( 50 ( t − 1 ) 2 + 25 cos 2 ( t − 1 ) − 50 sin 2 ( t − 1 ) + 100 ( ( t − 1 ) + 25 ) u ( t − 1 ) ) = 75 − 75 cos 2 t − 50 t 2 + ( 50 t 2 − 75 + 25 cos ( 2 t − 2 ) − 50 sin ( 2 t − 2 ) ) Thus, the value of i(t) is i ( t ) = 75 − 75 cos 2 t − 50 t 2 + ( 50 t 2 − 75 + 25 cos ( 2 t − 2 ) − 50 sin ( 2 t − 2 ) ) _ .

The value of current i ( t ) shown in the given circuit with the assumption that L = 1 H , C = 0.25 F using Laplace transforms. The value of the current, i ( t ) is i ( t ) = 75 − 75 cos 2 t − 50 t 2 + ( 50 t 2 − 75 + 25 cos ( 2 t − 2 ) − 50 sin ( 2 t − 2 ) ) _ . Definition used: Laplace transform of derivatives: “The first and second derivatives of f ( t ) satisfy ℒ ( f ′ ) = s ℒ ( f ) − f ( 0 ) ⋯ ( 1 ) ℒ ( f ″ ) = s 2 ℒ ( f ) − s f ( 0 ) − f ′ ( 0 ) ⋯ ( 2 ) Formula (1) holds if f ( t ) is continuous for all t ≥ 0 and satisfies the growth restriction (2) and f ′ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 . Similarly (2) holds if f ( t ) and f ′ ( t ) are continuous for all t ≥ 0 and satisfy the growth restriction and f ″ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 .” Unit step function or Heaviside Function: The unit step function or the Heaviside function denoted by u ( t − a ) is defined as follows. u ( t − a ) = { 0 , if t < a 1 , if t > a } Second shifting theorem If f ( t ) has the transform F ( s ) , then the shifted function f ˜ ( t ) = f ( t − a ) u ( t − a ) = { 0 if t < a f ( t − a ) if t > a } has the transform e − a s F ( s ) . That is, if ℒ ( f ( t ) ) = F ( s ) then, ℒ ( f ( t − a ) u ( t − a ) ) = e − a s F ( s ) ⋯ ( 1 ) Calculation: Consider the given conditions as follows. L = 1 H , C = 0.25 F Also, the value of voltage is given as shown below. V = { 200 ( t − 1 3 t 3 ) , 0 < t < 1 0 , t > 1 } It is known that the equation of voltage in an LC circuit is L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) . Substitute the value of voltage in the above equation as shown below. L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = { 200 ( t − 1 3 t 3 ) , 0 < t < 1 0 , t > 1 } Now, differentiate the above equation as follows. L i ″ ( t ) + 1 C i ( t ) = v ′ ( t ) = { 200 ( 1 − t 2 ) , 0 < t < 1 0 , t > 1 } Use the definition of the Heaviside step function rewrite the equation of voltage as shown below. v = 200 ( 1 − t 2 ) ( 1 − u ( t − 1 ) ) = 200 ( 1 − t 2 ) − 200 ( 1 − t ) ( t + 1 ) u ( t − 1 ) = 200 ( 1 − t 2 ) + 200 ( t − 1 ) ( ( t − 1 ) + 2 ) u ( t − 1 ) = 200 ( 1 − t 2 ) + 200 ( ( t − 1 ) 2 + 2 ( t − 1 ) ) u ( t − 1 ) Also, note that the initial current and charge is zero and ℒ ( i ( t ) ) = I ( s ) . Substitute the values L = 1 H , C = 0.25 F and the equation of voltage in the equation L i ″ ( t ) + 1 C i ( t ) = v ′ ( t ) as shown below. i ″ ( t ) + 1 0.25 i ( t ) = 200 ( 1 − t 2 ) + 200 ( ( t − 1 ) 2 + 2 ( t − 1 ) ) u ( t − 1 ) , i ( 0 ) = 0 , i ′ ( 0 ) = 0 Take Laplace transform on both sides of the above equation; apply the second shifting theorem as shown below. ℒ ( i ″ ( t ) + 1 0.25 i ( t ) ) = ℒ ( 200 ( 1 − t 2 ) + 200 ( ( t − 1 ) 2 + 2 ( t − 1 ) ) u ( t − 1 ) ) ℒ ( i ″ ( t ) ) + 4 ℒ ( i ( t ) ) = ℒ ( 200 ( 1 − t 2 ) ) + 200 ℒ ( ( ( t − 1 ) 2 + 2 ( t − 1 ) ) u ( t − 1 ) ) s 2 I ( s ) + 4 I ( s ) = 200 s − 400 s 3 + 200 ( 2 s 3 + 2 s 2 ) e − s ( s 2 + 4 ) I ( s ) = 200 ( s 2 − 2 s 3 ) + 400 ( s + 1 s 3 ) e − s I ( s ) = 200 ( s 2 − 2 s 3 ( s 2 + 4 ) ) + 400 ( s + 1 s 3 ( s 2 + 4 ) ) e − s On further simplification of the above equation the following is obtained. I ( s ) = 200 ( s 2 − 2 s 3 ( s 2 + 4 ) ) + 400 ( s + 1 s 3 ( s 2 + 4 ) ) e − s = 200 I 1 ( s ) + 400 e − s I 2 ( s ) Apply partial fractions in the term I 1 ( s ) = s 2 − 2 s 3 ( s 2 + 4 ) as follows. s 2 − 2 s 3 ( s 2 + 4 ) = A s + B s 2 + C s 3 + D s + E s 2 + 4 s 2 − 2 s 3 ( s 2 + 4 ) = A ( s 2 ) ( s 2 + 4 ) + B s ( s 2 + 4 ) + C ( s 2 + 4 ) + ( D s + E ) s 3 s 3 ( s 2 + 4 ) s 2 − 2 = A ( s 2 ) ( s 2 + 4 ) + B s ( s 2 + 4 ) + C ( s 2 + 4 ) + ( D s + E ) s 3 1 = A s 4 + 4 A s 2 + B s 3 + 4 s B + C s 2 + 4 C + D s 4 + E s 3 Equate the coefficient of s 4 on both sides as shown below. 0 = A + D A = − D Now, equate the constant terms on both sides as follows. − 2 = 4 C − 1 2 = C Equate the coefficient of s 3 on both sides as shown below. 0 = B + E B = − E Equate the coefficient of s on both sides as shown below. 0 = 4 B 0 = B Therefore, the value of E is 0 . Equate the coefficient of s 2 on both sides as shown below. 1 = 4 A + C Substitute the value of C in the above equation and solve for A as follows. 1 = 4 A + C 1 = 4 A − 1 2 3 2 = 4 A 3 8 = A Therefore, the value of D is − 3 8 . Thus, the value of I 1 ( s ) is as follows. I 1 ( s ) = s 2 − 2 s 3 ( s 2 + 4 ) = A s + B s 2 + C s 3 + D s + E s 2 + 4 = 3 8 s − 1 2 s 3 − 3 s 8 ( s 2 + 4 ) Apply partial fractions in the term I 2 ( s ) = s + 1 s 3 ( s 2 + 4 ) as follows. s + 1 s 3 ( s 2 + 4 ) = A s + B s 2 + C s 3 + D s + E s 2 + 4 s + 1 s 3 ( s 2 + 4 ) = A s 2 ( s 2 + 4 ) + B s ( s 2 + 4 ) + C ( s 2 + 4 ) + ( D s + E ) ( s 3 ) s 3 ( s 2 + 4 ) s + 1 = A s 2 ( s 2 + 4 ) + B s ( s 2 + 4 ) + C ( s 2 + 4 ) + ( D s + E ) ( s 3 ) s + 1 = A s 4 + 4 A s 2 + B s 3 + 4 B s + C s 2 + 4 C + D s 4 + E s 3 Equate the coefficient of s 4 on both sides as shown below. 0 = A + D A = − D Now, equate the constant terms on both sides as follows. 1 = 4 C 1 4 = C Equate the coefficient of s 3 on both sides as shown below. 0 = B + E B = − E Equate the coefficient of s on both sides as shown below. 1 = 4 B 1 4 = B Therefore, the value of E is − 1 4 . Equate the coefficient of s 2 on both sides as shown below. 0 = 4 A + C − 1 4 = 4 A − 1 16 = A Therefore, the value of E is 1 16 . Thus, the value of I 2 ( s ) is as follows. I 2 ( s ) = s + 1 s 3 ( s 2 + 4 ) = − 1 16 s + 1 4 s 2 + 1 4 s 3 + s 16 ( s 2 + 4 ) − 1 4 ( s 2 + 4 ) Thus, the value of I ( s ) is obtained as follows. I ( s ) = 200 I 1 ( s ) + 400 e − s I 2 ( s ) = 200 ( 3 8 s − 1 2 s 3 − 3 s 8 ( s 2 + 4 ) ) + 400 e − s ( − 1 16 s + 1 4 s 2 + 1 4 s 3 + s 16 ( s 2 + 4 ) − 1 4 ( s 2 + 4 ) ) Now, take inverse Laplace transform on both sides as shown below. ℒ − 1 ( I ( s ) ) = 200 ℒ − 1 ( 3 8 s − 1 2 s 3 − 3 s 8 ( s 2 + 4 ) ) + 400 ℒ − 1 ( e − s ( − 1 16 s + 1 4 s 2 + 1 4 s 3 + s 16 ( s 2 + 4 ) − 1 4 ( s 2 + 4 ) ) ) = ( 200 ℒ − 1 ( 3 8 s ) − 200 ℒ − 1 ( 200 2 s 3 ) − 200 ℒ − 1 ( 3 s 8 ( s 2 + 4 ) ) + 400 ℒ − 1 ( − e − s ( 16 s ) ) + 400 ℒ − 1 ( e − s 4 s 2 ) + 400 ℒ − 1 ( e − s 4 s 3 ) + 400 ℒ − 1 ( e − s s 16 ( s 2 + 4 ) ) − 400 ℒ − 1 ( e − s 4 ( s 2 + 4 ) ) ) = ( ℒ − 1 ( 75 s e − s ) − ℒ − 1 ( 100 s 3 e − s ) − ℒ − 1 ( 75 s s 2 + 4 e − s ) + ℒ − 1 ( 100 s 3 e − s ) + ℒ − 1 ( 25 s s 2 + 4 e − s ) − ℒ − 1 ( 100 s 2 + 4 e − s ) + ℒ − 1 ( 100 s 2 e − s ) − ℒ − 1 ( 25 s e − s ) ) On further simplifications, i ( t ) = 75 − 75 cos 2 t − 50 t 2 + ( 50 ( t − 1 ) 2 + 25 cos 2 ( t − 1 ) − 50 sin 2 ( t − 1 ) + 100 ( ( t − 1 ) + 25 ) u ( t − 1 ) ) = 75 − 75 cos 2 t − 50 t 2 + ( 50 t 2 − 75 + 25 cos ( 2 t − 2 ) − 50 sin ( 2 t − 2 ) ) Thus, the value of i(t) is i ( t ) = 75 − 75 cos 2 t − 50 t 2 + ( 50 t 2 − 75 + 25 cos ( 2 t − 2 ) − 50 sin ( 2 t − 2 ) ) _ .

Custom Kreyszig: Advanced Engineering Mathematics

Custom Kreyszig: Advanced Engineering Mathematics

10th Edition

ISBN: 9781119166856

Author: Kreyszig

Publisher: JOHN WILEY+SONS INC.CUSTOM

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Custom Kreyszig: 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

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