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The solution of initial value problem y ″ + 4 y = δ ( t − π ) where y ( 0 ) = 8 , y ′ ( 0 ) = 0 . The solution of the initial value problem y ″ + 4 y = δ ( t − π ) is y ( t ) = { 8 cos 2 t + 1 2 sin 2 t if t > π 8 cos 2 t if t < π _ . Result used: Laplace transformation derivative: Let f ′ and f ″ be continuous for all t ≥ 0 and satisfy the growth restriction and let f ′ and f ″ be piecewise continuous on every finite interval on the semi axis t ≥ 0 . Then the transform of f ′ and f ″ satisfies, L ( f ″ ) = s 2 L { f } − s f ( 0 ) − f ′ ( 0 ) L ( f ′ ) = s L { f } − f ( 0 ) . Formula used: Laplace transform for the impulse function is L { δ ( t − a ) } = e − a s . Shifted function: If f ( t ) has the transform F ( s ) then L − 1 { e − a s F ( s ) } = f ( t − a ) u ( t − a ) , here u ( t − a ) is an unit step function and it is defined as, u ( t − a ) = { 0 if t < a 1 if t > a . Calculation: Given that, the initial value differential equation is y ″ + 4 y = δ ( t − π ) . Take Laplace transformation on both sides of the equation y ″ + 4 y = δ ( t − π ) as follows. L { y ″ + 4 y } = L { δ ( t − π ) } L { y ″ } + 4 L { y } = L { δ ( t − π ) } Use Laplace transform derivative and substitute the initial value y ( 0 ) = 8 , y ′ ( 0 ) = 0 as follows. L { y ″ } + 4 L { y } = L { δ ( t − π ) } { s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) } + 4 Y ( s ) = e − π s s 2 Y ( s ) − 8 s − 0 + 4 Y ( s ) = e − π s Y ( s ) ( s 2 + 4 ) − 8 s = e − π s Simplify further as follows. Y ( s ) ( s 2 + 4 ) − 8 s = e − π s Y ( s ) ( s 2 + 4 ) = 8 s + e − π s Y ( s ) = 8 s + e − π s ( s 2 + 4 ) Y ( s ) = 8 s ( s 2 + 4 ) + e − π s ( s 2 + 4 ) Now take inverse Laplace transformation on both sides of above function as follows. L − 1 ( Y ( s ) ) = L − 1 { 8 s ( s 2 + 4 ) + e − π s ( s 2 + 4 ) } y ( t ) = 8 L − 1 { s ( s 2 + 4 ) } + L − 1 { e − π s ( s 2 + 4 ) } ⋯ ⋯ ( 1 ) Here, the inverse Laplace transform of the function s ( s 2 − 4 ) is computed as follows. L − 1 ( s ( s 2 + 4 ) ) = L − 1 [ s ( s 2 + 4 ) ] = L − 1 [ s ( s 2 + 2 2 ) ] = cos 2 t The inverse Laplace transform of the function 1 ( s 2 + 4 ) is computed as follows. L − 1 ( 1 ( s 2 + 4 ) ) = L − 1 ( 1 ( s 2 + 2 2 ) ) = 1 2 sin 2 t The inverse Laplace transform of the function e − π s ( s 2 + 4 ) is computed as follows. L − 1 ( e − π s ( s 2 + 4 ) ) = L − 1 ( e − π s G ( s ) ) = u ( t − π ) ⋅ g ( t − π ) = u ( t − π ) ⋅ { 1 2 sin 2 ( t − π ) } = ( 1 2 sin 2 t ) u ( t − π ) Substitute all value in the equation (1) and simplify as follows. y ( t ) = 8 L − 1 { s ( s 2 + 4 ) } + L − 1 { e − π s ( s 2 + 4 ) } = 8 cos 2 t + ( 1 2 sin 2 t ) u ( t − π ) = { 8 cos 2 t + 1 2 sin 2 t if t > π 8 cos 2 t if t < π Therefore, the solution of the initial value problem y ″ + 4 y = δ ( t − π ) is y ( t ) = { 8 cos 2 t + 1 2 sin 2 t if t > π 8 cos 2 t if t < π _ . Draw the graph of the function y ( t ) = { 8 cos 2 t + 1 2 sin 2 t if t > π 8 cos 2 t if t < π as shown below in Figure 1 From Figure 1, it is observed that the graph of the function y ( t ) up to π is y ( t ) = 8 cos 2 t and the graph of the function y ( t ) at ( π , ∞ ) is y ( t ) = 8 cos 2 t + 1 2 sin 2 t .

The solution of initial value problem y ″ + 4 y = δ ( t − π ) where y ( 0 ) = 8 , y ′ ( 0 ) = 0 . The solution of the initial value problem y ″ + 4 y = δ ( t − π ) is y ( t ) = { 8 cos 2 t + 1 2 sin 2 t if t > π 8 cos 2 t if t < π _ . Result used: Laplace transformation derivative: Let f ′ and f ″ be continuous for all t ≥ 0 and satisfy the growth restriction and let f ′ and f ″ be piecewise continuous on every finite interval on the semi axis t ≥ 0 . Then the transform of f ′ and f ″ satisfies, L ( f ″ ) = s 2 L { f } − s f ( 0 ) − f ′ ( 0 ) L ( f ′ ) = s L { f } − f ( 0 ) . Formula used: Laplace transform for the impulse function is L { δ ( t − a ) } = e − a s . Shifted function: If f ( t ) has the transform F ( s ) then L − 1 { e − a s F ( s ) } = f ( t − a ) u ( t − a ) , here u ( t − a ) is an unit step function and it is defined as, u ( t − a ) = { 0 if t < a 1 if t > a . Calculation: Given that, the initial value differential equation is y ″ + 4 y = δ ( t − π ) . Take Laplace transformation on both sides of the equation y ″ + 4 y = δ ( t − π ) as follows. L { y ″ + 4 y } = L { δ ( t − π ) } L { y ″ } + 4 L { y } = L { δ ( t − π ) } Use Laplace transform derivative and substitute the initial value y ( 0 ) = 8 , y ′ ( 0 ) = 0 as follows. L { y ″ } + 4 L { y } = L { δ ( t − π ) } { s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) } + 4 Y ( s ) = e − π s s 2 Y ( s ) − 8 s − 0 + 4 Y ( s ) = e − π s Y ( s ) ( s 2 + 4 ) − 8 s = e − π s Simplify further as follows. Y ( s ) ( s 2 + 4 ) − 8 s = e − π s Y ( s ) ( s 2 + 4 ) = 8 s + e − π s Y ( s ) = 8 s + e − π s ( s 2 + 4 ) Y ( s ) = 8 s ( s 2 + 4 ) + e − π s ( s 2 + 4 ) Now take inverse Laplace transformation on both sides of above function as follows. L − 1 ( Y ( s ) ) = L − 1 { 8 s ( s 2 + 4 ) + e − π s ( s 2 + 4 ) } y ( t ) = 8 L − 1 { s ( s 2 + 4 ) } + L − 1 { e − π s ( s 2 + 4 ) } ⋯ ⋯ ( 1 ) Here, the inverse Laplace transform of the function s ( s 2 − 4 ) is computed as follows. L − 1 ( s ( s 2 + 4 ) ) = L − 1 [ s ( s 2 + 4 ) ] = L − 1 [ s ( s 2 + 2 2 ) ] = cos 2 t The inverse Laplace transform of the function 1 ( s 2 + 4 ) is computed as follows. L − 1 ( 1 ( s 2 + 4 ) ) = L − 1 ( 1 ( s 2 + 2 2 ) ) = 1 2 sin 2 t The inverse Laplace transform of the function e − π s ( s 2 + 4 ) is computed as follows. L − 1 ( e − π s ( s 2 + 4 ) ) = L − 1 ( e − π s G ( s ) ) = u ( t − π ) ⋅ g ( t − π ) = u ( t − π ) ⋅ { 1 2 sin 2 ( t − π ) } = ( 1 2 sin 2 t ) u ( t − π ) Substitute all value in the equation (1) and simplify as follows. y ( t ) = 8 L − 1 { s ( s 2 + 4 ) } + L − 1 { e − π s ( s 2 + 4 ) } = 8 cos 2 t + ( 1 2 sin 2 t ) u ( t − π ) = { 8 cos 2 t + 1 2 sin 2 t if t > π 8 cos 2 t if t < π Therefore, the solution of the initial value problem y ″ + 4 y = δ ( t − π ) is y ( t ) = { 8 cos 2 t + 1 2 sin 2 t if t > π 8 cos 2 t if t < π _ . Draw the graph of the function y ( t ) = { 8 cos 2 t + 1 2 sin 2 t if t > π 8 cos 2 t if t < π as shown below in Figure 1 From Figure 1, it is observed that the graph of the function y ( t ) up to π is y ( t ) = 8 cos 2 t and the graph of the function y ( t ) at ( π , ∞ ) is y ( t ) = 8 cos 2 t + 1 2 sin 2 t .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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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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