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The initial value problem of the given system y 1 ′ = − y 1 + 4 y 2 , y 2 ′ = 3 y 1 − 2 y 2 , y 1 ( 0 ) = 3 , y 2 ( 0 ) = 4 by using the Laplace transform. The solution of the given initial value problem is y 1 ( t ) = − e − 5 t + 4 e 2 t and y 2 ( t ) = e − 5 t + 3 e 2 t _ . The given initial value problem is: y 1 ′ = − y 1 + 4 y 2 , y 2 ′ = 3 y 1 − 2 y 2 , y 1 ( 0 ) = 3 , y 2 ( 0 ) = 4 Apply the Laplace transform on both sides of the given system. L ( y 1 ′ ) = − L ( y 1 ) + 4 L ( y 2 ) L ( y 2 ′ ) = 3 L ( y 1 ) − 2 L ( y 2 ) The Laplace differential of transform is L [ y 1 ′ ( t ) ] = s L ( y ( t ) ) − y ( 0 ) , L ( y ) = Y . s Y 1 − y 1 ( 0 ) = − Y 1 + 4 Y 2 s Y 2 − y 2 ( 0 ) = 3 Y 1 − 2 Y 2 Substitute the initial value of y 1 ( 0 ) = 3 and y 2 ( 0 ) = 4 in the above equations: s Y 1 − 3 = − Y 1 + 4 Y 2 ( 1 + s ) Y 1 − 4 Y 2 = 3 s Y 2 − 4 = 3 Y 1 − 2 Y 2 − 3 Y 1 + ( s + 2 ) Y 2 = 4 Solve the above equations by using Cramer’s’ rule. Y 1 = | 3 − 4 4 s + 2 | | 1 + s − 4 − 3 s + 2 | Y 2 = | 1 + s 3 − 3 4 | | 1 + s − 4 − 3 s + 2 | Y 1 = 3 s + 6 + 16 s + 2 + s 2 + 2 s − 12 Y 2 = 4 + 4 s + 9 s + 2 + s 2 + 2 s − 12 On further simplification, the following is obtained: Y 1 ( s ) = 3 s + 22 s 2 + 3 s − 10 = 3 s + 22 ( s + 5 ) ( s − 2 ) Y 2 ( s ) = 4 s + 13 s 2 + 3 s − 10 = 4 s + 13 ( s + 5 ) ( s − 2 ) To convert Y 1 ( s ) into partial fractions, suppose The following. Y 1 ( s ) = 3 s + 22 ( s + 5 ) ( s − 2 ) = A s + 5 + B s − 2 3 s + 22 = A ( s − 2 ) + B ( s + 5 ) Substitute s = 2 in 3 s + 22 = A ( s − 2 ) + B ( s + 5 ) . 3 ( 2 ) + 22 = A ( 2 − 2 ) + B ( 2 + 5 ) 6 + 22 = 7 B B = 28 7 B = 4 Substitute s = − 5 in 3 s + 22 = A ( s − 2 ) + B ( s + 5 ) . 3 ( − 5 ) + 22 = A ( − 5 − 2 ) + B ( − 5 + 5 ) − 18 + 22 = − 7 A A = − 7 7 A = − 1 Therefore, the partial decomposition of Y 1 ( s ) is, Y 1 ( s ) = − 1 s + 5 + 4 s − 2 To convert Y 2 ( s ) into partial fractions, suppose the following. Y 2 ( s ) = 4 s + 13 ( s + 5 ) ( s − 2 ) = A s + 5 + B s − 2 4 s + 13 = A ( s − 2 ) + B ( s + 5 ) Substitute s = 2 in 4 s + 13 = A ( s − 2 ) + B ( s + 5 ) . 4 ( 2 ) + 13 = A ( 2 − 2 ) + B ( 2 + 5 ) 8 + 13 = 7 B B = 21 7 B = 3 Substitute s = − 5 in 4 s + 13 = A ( s − 2 ) + B ( s + 5 ) . 4 ( − 5 ) + 13 = A ( − 5 − 2 ) + B ( − 5 + 5 ) − 20 + 13 = − 7 A A = 7 7 A = 1 Therefore, the partial decomposition of Y 2 ( s ) is, Y 2 ( s ) = 1 s + 5 + 3 s − 2 Take the inverse transformations for Y 1 and Y 2 and use L − 1 ( 1 s − a ) = e a t , to get: y 1 ( t ) = − L − 1 ( 1 s + 5 ) + 4 L − 1 ( 1 s − 2 ) = − e − 5 t + 4 e 2 t y 2 ( t ) = L − 1 ( 1 s + 5 ) + 3 L − 1 ( 1 s − 2 ) = e − 5 t + 3 e 2 t Hence, the solution of the given initial value problem is y 1 ( t ) = − e − 5 t + 4 e 2 t and y 2 ( t ) = e − 5 t + 3 e 2 t _ .

The initial value problem of the given system y 1 ′ = − y 1 + 4 y 2 , y 2 ′ = 3 y 1 − 2 y 2 , y 1 ( 0 ) = 3 , y 2 ( 0 ) = 4 by using the Laplace transform. The solution of the given initial value problem is y 1 ( t ) = − e − 5 t + 4 e 2 t and y 2 ( t ) = e − 5 t + 3 e 2 t _ . The given initial value problem is: y 1 ′ = − y 1 + 4 y 2 , y 2 ′ = 3 y 1 − 2 y 2 , y 1 ( 0 ) = 3 , y 2 ( 0 ) = 4 Apply the Laplace transform on both sides of the given system. L ( y 1 ′ ) = − L ( y 1 ) + 4 L ( y 2 ) L ( y 2 ′ ) = 3 L ( y 1 ) − 2 L ( y 2 ) The Laplace differential of transform is L [ y 1 ′ ( t ) ] = s L ( y ( t ) ) − y ( 0 ) , L ( y ) = Y . s Y 1 − y 1 ( 0 ) = − Y 1 + 4 Y 2 s Y 2 − y 2 ( 0 ) = 3 Y 1 − 2 Y 2 Substitute the initial value of y 1 ( 0 ) = 3 and y 2 ( 0 ) = 4 in the above equations: s Y 1 − 3 = − Y 1 + 4 Y 2 ( 1 + s ) Y 1 − 4 Y 2 = 3 s Y 2 − 4 = 3 Y 1 − 2 Y 2 − 3 Y 1 + ( s + 2 ) Y 2 = 4 Solve the above equations by using Cramer’s’ rule. Y 1 = | 3 − 4 4 s + 2 | | 1 + s − 4 − 3 s + 2 | Y 2 = | 1 + s 3 − 3 4 | | 1 + s − 4 − 3 s + 2 | Y 1 = 3 s + 6 + 16 s + 2 + s 2 + 2 s − 12 Y 2 = 4 + 4 s + 9 s + 2 + s 2 + 2 s − 12 On further simplification, the following is obtained: Y 1 ( s ) = 3 s + 22 s 2 + 3 s − 10 = 3 s + 22 ( s + 5 ) ( s − 2 ) Y 2 ( s ) = 4 s + 13 s 2 + 3 s − 10 = 4 s + 13 ( s + 5 ) ( s − 2 ) To convert Y 1 ( s ) into partial fractions, suppose The following. Y 1 ( s ) = 3 s + 22 ( s + 5 ) ( s − 2 ) = A s + 5 + B s − 2 3 s + 22 = A ( s − 2 ) + B ( s + 5 ) Substitute s = 2 in 3 s + 22 = A ( s − 2 ) + B ( s + 5 ) . 3 ( 2 ) + 22 = A ( 2 − 2 ) + B ( 2 + 5 ) 6 + 22 = 7 B B = 28 7 B = 4 Substitute s = − 5 in 3 s + 22 = A ( s − 2 ) + B ( s + 5 ) . 3 ( − 5 ) + 22 = A ( − 5 − 2 ) + B ( − 5 + 5 ) − 18 + 22 = − 7 A A = − 7 7 A = − 1 Therefore, the partial decomposition of Y 1 ( s ) is, Y 1 ( s ) = − 1 s + 5 + 4 s − 2 To convert Y 2 ( s ) into partial fractions, suppose the following. Y 2 ( s ) = 4 s + 13 ( s + 5 ) ( s − 2 ) = A s + 5 + B s − 2 4 s + 13 = A ( s − 2 ) + B ( s + 5 ) Substitute s = 2 in 4 s + 13 = A ( s − 2 ) + B ( s + 5 ) . 4 ( 2 ) + 13 = A ( 2 − 2 ) + B ( 2 + 5 ) 8 + 13 = 7 B B = 21 7 B = 3 Substitute s = − 5 in 4 s + 13 = A ( s − 2 ) + B ( s + 5 ) . 4 ( − 5 ) + 13 = A ( − 5 − 2 ) + B ( − 5 + 5 ) − 20 + 13 = − 7 A A = 7 7 A = 1 Therefore, the partial decomposition of Y 2 ( s ) is, Y 2 ( s ) = 1 s + 5 + 3 s − 2 Take the inverse transformations for Y 1 and Y 2 and use L − 1 ( 1 s − a ) = e a t , to get: y 1 ( t ) = − L − 1 ( 1 s + 5 ) + 4 L − 1 ( 1 s − 2 ) = − e − 5 t + 4 e 2 t y 2 ( t ) = L − 1 ( 1 s + 5 ) + 3 L − 1 ( 1 s − 2 ) = e − 5 t + 3 e 2 t Hence, the solution of the given initial value problem is y 1 ( t ) = − e − 5 t + 4 e 2 t and y 2 ( t ) = e − 5 t + 3 e 2 t _ .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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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

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