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Math Advanced Math ADV.ENG.MATH (LL) W/WILEYPLUS BUNDLE

The value of current i ( t ) shown in the given circuit with the assumption that R = 2 Ω , L = 1 H , C = 0.1 F using Laplace transforms. The value of the current, i ( t ) is i ( t ) = { 27 cos t + 6 sin t − ( 27 cos 3 t + 11 sin 3 t ) e − t , 0 < t < 2 π ( 27 cos 3 t + 11 sin 3 t ) ( e − ( t − 2 π ) − e − t ) , t > 2 π } _ . 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. R = 2 Ω , L = 1 H , C = 0.1 F Also, the value of voltage is given as shown below. V = { 255 sin t , 0 < t < 2 π 0 , t > 2 π } It is known that the equation of voltage in an LC circuit is R i ( t ) + 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. R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = { 255 sin t , 0 < t < 2 π 0 , t > 2 π } Use the definition of the Heaviside step function rewrite the equation of voltage as shown below. v ( t ) = 255 sin t ( 1 − u ( t − 2 π ) ) = 255 sin t − 255 sin ( t − 2 π ) u ( t − 2 π ) Also, note that the initial current and charge is zero and L ( i ( t ) ) = I ( s ) . Substitute the values R = 2 Ω , L = 1 H , C = 0.1 F and the equation of voltage in the equation R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) L i ″ ( t ) + 1 C i ( t ) = v ′ ( t ) as shown below. 2 i ( t ) + i ′ ( t ) + 1 0.1 ∫ 0 t i ( τ ) d τ = 255 sin t − 255 sin ( t − 2 π ) u ( t − 2 π ) , 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 ( 2 i ( t ) + i ′ ( t ) + 1 0.1 ∫ 0 t i ( τ ) d τ ) = L ( 255 sin t − 255 sin ( t − 2 π ) u ( t − 2 π ) ) L ( i ′ ( t ) ) + 2 L ( i ( t ) ) + 10 L ( ∫ 0 t i ( τ ) d τ ) = L ( 255 sin t ) − L ( 255 sin ( t − 2 π ) u ( t − 2 π ) ) s I ( s ) + 2 I ( s ) + 10 I ( s ) s = 255 s 2 + 1 − 255 s 2 + 1 e − 2 π s s 2 + 2 s + 10 s I ( s ) = 255 ( 1 s 2 + 1 − e − 2 π s s 2 + 1 ) I ( s ) = 255 s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) ( 1 − e − 2 s ) Apply partial fractions on the term s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) as shown below. s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) = A s + B s 2 + 1 + C s + D s 2 + 2 s + 10 s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) = ( A s + B ) ( s 2 + 2 s + 10 ) + ( C s + D ) ( s 2 + 1 ) ( s 2 + 1 ) ( s 2 + 2 s + 10 ) s = ( A s + B ) ( s 2 + 2 s + 10 ) + ( C s + D ) ( s 2 + 1 ) s = A s 3 + 2 A s 2 + 10 A s + B s 2 + 2 B s + 10 B + C s 3 + C s + D s 2 + D Equate the coefficient of s 3 on both sides as shown below. 0 = A + C A = − C Now, equate the coefficient of s 2 on both sides as follows. 0 = 2 A + B + D Equate the coefficient of s on both sides as shown below. 1 = 10 A + 2 B + C Now, equate the constant terms as follows. 0 = 10 B + D − 10 B = D Substitute the value − 10 B = D in the equation 0 = 2 A + B + D as follows. 0 = 2 A + B + D 0 = 2 A + B − 10 B 0 = 2 A − 9 B 2 A = 9 B A = 9 2 B Therefore, the value of C is − 9 2 B . Substitute the values A = 9 2 B , C = − 9 2 B in the equation 1 = 10 A + 2 B + C and solve for B as follows. 1 = 10 A + 2 B + C 1 = 90 2 B + 2 B − 9 2 B 1 = 45 B + 2 B − 9 2 B 1 = 47 B − 9 2 B 1 = 94 B − 9 B 2 On further simplifications, 1 = 85 2 B 2 85 = B Therefore, obtain the value of A as shown below. A = 9 2 ⋅ ( 2 85 ) = 9 85 Similarly, obtain the value of C as follows. C = − 9 2 B = − 9 2 ( 2 85 ) = − 9 85 Obtain the value of D as shown below. − 10 B = D − 10 ⋅ 2 85 = D − 20 85 = D − 4 17 = D Substitute the values of A , B , C and D in the equation s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) = A s + B s 2 + 1 + C s + D s 2 + 2 s + 10 as follows. s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) = A s + B s 2 + 1 + C s + D s 2 + 2 s + 10 = 9 85 s + 2 85 s 2 + 1 + − 9 85 s − 4 17 s 2 + 2 s + 10 = 9 s + 2 85 ( s 2 + 1 ) − 9 s + 20 85 ( s 2 + 2 s + 10 ) Thus, the value of I ( s ) is obtained as follows. I ( s ) = 255 ( 9 s + 2 85 ( s 2 + 1 ) − 9 s + 20 85 ( s 2 + 2 s + 10 ) ) − 255 e − 2 s ( 9 s + 2 85 ( s 2 + 1 ) − 9 s + 20 85 ( s 2 + 2 s + 10 ) ) = ( − 27 s s 2 + 1 + 6 s 2 + 1 − 27 s s 2 + 2 s + 10 − 60 s 2 + 2 s + 10 − 27 s s 2 + 1 ⋅ e − 2 s − 6 s 2 + 1 ⋅ e − 2 s + 27 s s 2 + 2 s + 10 ⋅ e − 2 s + 60 s 2 + 2 s + 10 ⋅ e − 2 s ) = ( − 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 − 33 ( s + 1 ) 2 + 3 2 + 27 s s 2 + 1 + 6 s + 1 + 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 ⋅ e − 2 s + 33 ( s + 1 ) 2 + 3 2 ⋅ e − 2 s − 27 s s 2 + 1 ⋅ e − 2 s − 6 s + 1 ⋅ e − 2 s ) Now, take inverse Laplace transform on both sides as shown below. L − 1 ( I ( s ) ) = L − 1 ( − 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 − 33 ( s + 1 ) 2 + 3 2 + 27 s s 2 + 1 + 6 s + 1 + 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 ⋅ e − 2 s + 33 ( s + 1 ) 2 + 3 2 ⋅ e − 2 s − 27 s s 2 + 1 ⋅ e − 2 s − 6 s + 1 ⋅ e − 2 s ) i ( t ) = ( L − 1 ( − 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 ) − L − 1 ( 33 ( s + 1 ) 2 + 3 2 ) + L − 1 ( 27 s s 2 + 1 ) + L − 1 ( 6 s + 1 ) + L − 1 ( 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 ⋅ e − 2 s ) + L − 1 ( 33 ( s + 1 ) 2 + 3 2 ⋅ e − 2 s ) − L − 1 ( 27 s s 2 + 1 ⋅ e − 2 s ) − L − 1 ( 6 s + 1 ⋅ e − 2 s ) ) i ( t ) = ( − 27 cos 3 t e − t − 11 sin 3 t e − t + 27 cos t + 6 sin t + ( 27 cos 3 ( t − 2 π ) e − ( t − 2 π ) + 11 sin 3 ( t − 2 π ) e − ( t − 2 π ) − 27 cos ( t − 2 π − 6 sin ( t − 2 π ) ) ) u ( t − 2 π ) ) Thus, the value of i(t) is i ( t ) = { 27 cos t + 6 sin t − ( 27 cos 3 t + 11 sin 3 t ) e − t , 0 < t < 2 π ( 27 cos 3 t + 11 sin 3 t ) ( e − ( t − 2 π ) − e − t ) , t > 2 π } _ .

The value of current i ( t ) shown in the given circuit with the assumption that R = 2 Ω , L = 1 H , C = 0.1 F using Laplace transforms. The value of the current, i ( t ) is i ( t ) = { 27 cos t + 6 sin t − ( 27 cos 3 t + 11 sin 3 t ) e − t , 0 < t < 2 π ( 27 cos 3 t + 11 sin 3 t ) ( e − ( t − 2 π ) − e − t ) , t > 2 π } _ . 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. R = 2 Ω , L = 1 H , C = 0.1 F Also, the value of voltage is given as shown below. V = { 255 sin t , 0 < t < 2 π 0 , t > 2 π } It is known that the equation of voltage in an LC circuit is R i ( t ) + 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. R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = { 255 sin t , 0 < t < 2 π 0 , t > 2 π } Use the definition of the Heaviside step function rewrite the equation of voltage as shown below. v ( t ) = 255 sin t ( 1 − u ( t − 2 π ) ) = 255 sin t − 255 sin ( t − 2 π ) u ( t − 2 π ) Also, note that the initial current and charge is zero and L ( i ( t ) ) = I ( s ) . Substitute the values R = 2 Ω , L = 1 H , C = 0.1 F and the equation of voltage in the equation R i ( t ) + L i ′ ( t ) + 1 C ∫ 0 t i ( τ ) d τ = V ( t ) L i ″ ( t ) + 1 C i ( t ) = v ′ ( t ) as shown below. 2 i ( t ) + i ′ ( t ) + 1 0.1 ∫ 0 t i ( τ ) d τ = 255 sin t − 255 sin ( t − 2 π ) u ( t − 2 π ) , 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 ( 2 i ( t ) + i ′ ( t ) + 1 0.1 ∫ 0 t i ( τ ) d τ ) = L ( 255 sin t − 255 sin ( t − 2 π ) u ( t − 2 π ) ) L ( i ′ ( t ) ) + 2 L ( i ( t ) ) + 10 L ( ∫ 0 t i ( τ ) d τ ) = L ( 255 sin t ) − L ( 255 sin ( t − 2 π ) u ( t − 2 π ) ) s I ( s ) + 2 I ( s ) + 10 I ( s ) s = 255 s 2 + 1 − 255 s 2 + 1 e − 2 π s s 2 + 2 s + 10 s I ( s ) = 255 ( 1 s 2 + 1 − e − 2 π s s 2 + 1 ) I ( s ) = 255 s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) ( 1 − e − 2 s ) Apply partial fractions on the term s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) as shown below. s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) = A s + B s 2 + 1 + C s + D s 2 + 2 s + 10 s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) = ( A s + B ) ( s 2 + 2 s + 10 ) + ( C s + D ) ( s 2 + 1 ) ( s 2 + 1 ) ( s 2 + 2 s + 10 ) s = ( A s + B ) ( s 2 + 2 s + 10 ) + ( C s + D ) ( s 2 + 1 ) s = A s 3 + 2 A s 2 + 10 A s + B s 2 + 2 B s + 10 B + C s 3 + C s + D s 2 + D Equate the coefficient of s 3 on both sides as shown below. 0 = A + C A = − C Now, equate the coefficient of s 2 on both sides as follows. 0 = 2 A + B + D Equate the coefficient of s on both sides as shown below. 1 = 10 A + 2 B + C Now, equate the constant terms as follows. 0 = 10 B + D − 10 B = D Substitute the value − 10 B = D in the equation 0 = 2 A + B + D as follows. 0 = 2 A + B + D 0 = 2 A + B − 10 B 0 = 2 A − 9 B 2 A = 9 B A = 9 2 B Therefore, the value of C is − 9 2 B . Substitute the values A = 9 2 B , C = − 9 2 B in the equation 1 = 10 A + 2 B + C and solve for B as follows. 1 = 10 A + 2 B + C 1 = 90 2 B + 2 B − 9 2 B 1 = 45 B + 2 B − 9 2 B 1 = 47 B − 9 2 B 1 = 94 B − 9 B 2 On further simplifications, 1 = 85 2 B 2 85 = B Therefore, obtain the value of A as shown below. A = 9 2 ⋅ ( 2 85 ) = 9 85 Similarly, obtain the value of C as follows. C = − 9 2 B = − 9 2 ( 2 85 ) = − 9 85 Obtain the value of D as shown below. − 10 B = D − 10 ⋅ 2 85 = D − 20 85 = D − 4 17 = D Substitute the values of A , B , C and D in the equation s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) = A s + B s 2 + 1 + C s + D s 2 + 2 s + 10 as follows. s ( s 2 + 1 ) ( s 2 + 2 s + 10 ) = A s + B s 2 + 1 + C s + D s 2 + 2 s + 10 = 9 85 s + 2 85 s 2 + 1 + − 9 85 s − 4 17 s 2 + 2 s + 10 = 9 s + 2 85 ( s 2 + 1 ) − 9 s + 20 85 ( s 2 + 2 s + 10 ) Thus, the value of I ( s ) is obtained as follows. I ( s ) = 255 ( 9 s + 2 85 ( s 2 + 1 ) − 9 s + 20 85 ( s 2 + 2 s + 10 ) ) − 255 e − 2 s ( 9 s + 2 85 ( s 2 + 1 ) − 9 s + 20 85 ( s 2 + 2 s + 10 ) ) = ( − 27 s s 2 + 1 + 6 s 2 + 1 − 27 s s 2 + 2 s + 10 − 60 s 2 + 2 s + 10 − 27 s s 2 + 1 ⋅ e − 2 s − 6 s 2 + 1 ⋅ e − 2 s + 27 s s 2 + 2 s + 10 ⋅ e − 2 s + 60 s 2 + 2 s + 10 ⋅ e − 2 s ) = ( − 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 − 33 ( s + 1 ) 2 + 3 2 + 27 s s 2 + 1 + 6 s + 1 + 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 ⋅ e − 2 s + 33 ( s + 1 ) 2 + 3 2 ⋅ e − 2 s − 27 s s 2 + 1 ⋅ e − 2 s − 6 s + 1 ⋅ e − 2 s ) Now, take inverse Laplace transform on both sides as shown below. L − 1 ( I ( s ) ) = L − 1 ( − 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 − 33 ( s + 1 ) 2 + 3 2 + 27 s s 2 + 1 + 6 s + 1 + 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 ⋅ e − 2 s + 33 ( s + 1 ) 2 + 3 2 ⋅ e − 2 s − 27 s s 2 + 1 ⋅ e − 2 s − 6 s + 1 ⋅ e − 2 s ) i ( t ) = ( L − 1 ( − 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 ) − L − 1 ( 33 ( s + 1 ) 2 + 3 2 ) + L − 1 ( 27 s s 2 + 1 ) + L − 1 ( 6 s + 1 ) + L − 1 ( 27 ( s + 1 ) ( s + 1 ) 2 + 3 2 ⋅ e − 2 s ) + L − 1 ( 33 ( s + 1 ) 2 + 3 2 ⋅ e − 2 s ) − L − 1 ( 27 s s 2 + 1 ⋅ e − 2 s ) − L − 1 ( 6 s + 1 ⋅ e − 2 s ) ) i ( t ) = ( − 27 cos 3 t e − t − 11 sin 3 t e − t + 27 cos t + 6 sin t + ( 27 cos 3 ( t − 2 π ) e − ( t − 2 π ) + 11 sin 3 ( t − 2 π ) e − ( t − 2 π ) − 27 cos ( t − 2 π − 6 sin ( t − 2 π ) ) ) u ( t − 2 π ) ) Thus, the value of i(t) is i ( t ) = { 27 cos t + 6 sin t − ( 27 cos 3 t + 11 sin 3 t ) e − t , 0 < t < 2 π ( 27 cos 3 t + 11 sin 3 t ) ( e − ( t − 2 π ) − e − t ) , t > 2 π } _ .

ADV.ENG.MATH (LL) W/WILEYPLUS BUNDLE

ADV.ENG.MATH (LL) W/WILEYPLUS BUNDLE

10th Edition

ISBN: 9781119809210

Author: Kreyszig

Publisher: WILEY

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ADV.ENG.MATH (LL) W/WILEYPLUS BUNDLE

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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