Explanation Formula used: Laplace transform of L { y ″ } = s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) and L { y ′ } = s Y ( s ) − y ( 0 ) . Calculation: Given that, the differential equation is, y ″ + 2 y ′ + 4 y = t − e − t . Take Laplace transform on both side y ″ + 2 y ′ + 4 y = t − e − t and simplify as follows. L { y ″ + 2 y ′ + 4 y } = L { t − e − t } L { y ″ } + 2 L { y ′ } + 4 L { y } = L { t } − L { e − t } Substitute s Y ( s ) − y ( 0 ) for L { y ′ } , s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) for L { y ″ } and Y ( s ) for { y } in Equation L { y ″ } + 2 L { y ′ } + 4 L { y } = L { t } − L { e − t } and obtain as, s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) + 2 s Y ( s ) − 2 y ( 0 ) + 4 Y ( s ) = L { t } − L { e − t } . Substitute 1 for y ( 0 ) and − 1 for y ′ ( 0 ) in Equation s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) + 2 s Y ( s ) − 2 y ( 0 ) + 4 Y ( s ) = L { t } − L { e − t } , s 2 Y ( s ) − s ( 1 ) − ( − 1 ) + 2 ( s Y ( s ) − ( 1 ) ) + 4 Y ( s ) = L { t } − L { e − t } s 2 Y ( s ) − s + 1 + 2 s Y ( s ) − 2 + 4 Y ( s ) = 1 s 2 − 1 s + 1 ( s 2 + 2 s + 4 ) Y ( s ) = 1 s 2 − 1 s + 1 + s + 1 Y ( s ) = 1 s 2 ( s 2 + 2 s + 4 ) − 1 ( s + 1 ) ( s 2 + 2 s + 4 ) + s + 1 ( s 2 + 2 s + 4 ) Here, it is observed that the partial fraction of 1 s 2 ( s 2 + 2 s + 4 ) is s 8 ( s 2 + 2 s + 4 ) + 1 4 s 2 − 1 8 s . The partial fraction of 1 ( s + 1 ) ( s 2 + 2 s + 4 ) is − s − 1 3 ( s 2 + 2 s + 4 ) + 1 3 ( s + 1 )

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 8.3, Problem 27E

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The solution of the differential equation y″+2y′+4y=t−e−t with y(0)=1,y′(0)=−1 and graph it.

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Chapter 8.3, Problem 24E

Chapter 8.3, Problem 29E

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Chapter 8 Solutions

Introductory Differential Equations

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The solution of the differential equation y ″ + 2 y ′ + 4 y = t − e − t with y ( 0 ) = 1 , y ′ ( 0 ) = − 1 and graph it.

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The solution of the differential equation y″+2y′+4y=t−e−t with y(0)=1,y′(0)=−1 and graph it.

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Chapter 8 Solutions