Explanation Procedure used: 1. Apply Laplace transform to each term of the given equation. 2. By using the integrating factor, find the solution of the first-order linear equation as given below. a). The first order linear equation y ′ + P y = Q . b). The integrating factor is given as, I F = e ∫ P ( t ) d t . c). The solution of first-order linear equation is given as, y ′ × I F = ∫ ( Q × I F ) d s . 3. Apply boundary conditions to get the solution. Calculation: Given that, the differentiation equation y ″ + 3 t y ′ − 6 y = 0 . Take Laplace transformation on both sides as follows. L { y ″ + 3 t y ′ − 6 y } = 0 L { y ″ } + 3 L { t y ′ } − 6 { y } = 0 ⋯ ⋯ ( 1 ) Here, the Laplace transform of y ″ is, L { y ″ } = s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) . The Laplace transform of y ′ is, L { t y ′ } = − d d s [ s Y ( s ) − y ( 0 ) ] . Substitute − d d s [ s Y ( s ) − y ( 0 ) ] for L { t y ′ } , s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) for L { y ″ } , Y ( s ) for { y } and 1 s for { 1 } in Equation (1) as, s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) − 3 d d s [ s Y ( s ) − y ( 0 ) ] − 6 Y ( s ) = 0 ⋯ ⋯ ( 2 ) . Substitute 1 for y ( 0 ) and 0 for y ′ ( 0 ) in Equation (2) and simplify as follows. s 2 Y ( s ) − s ( 0 ) − ( 1 ) − 3 d d s [ s Y ( s ) − ( 0 ) ] − 6 Y ( s ) = 0 s 2 Y ( s ) − s − 3 d d s [ s Y ( s ) ] − 6 Y ( s ) = 0 s 2 Y ( s ) − 3 d d s [ s Y ( s ) ] − 6 Y ( s ) = s s 2 Y ( s ) − 3 Y ( s ) − 3 s Y ′ ( s ) − 6 Y ( s ) = s Simplify further, − ( 9 − s 2 ) Y ( s ) − 3 s Y ′ ( s ) = s 3 s Y ′ ( s ) + ( 9 − s 2 ) Y ( s ) = − s Y ′ ( s ) + ( 3 s − s 3 ) Y ( s ) = − 1 3 ⋯ ⋯ ( 3 ) The integration factor for Equation (3) is computed as follows

5th Edition

ISBN: 9780128149485

Author: Abell, Martha L. L.

Publisher: Elsevier Science

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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 18E

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The solution of the equation y″+3ty′−6y=0 where y(0)=1 and y′(0)=0.

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Chapter 8.3, Problem 17E

Chapter 8.3, Problem 19E

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Chapter 8 Solutions

Introductory Differential Equations

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The solution of the equation y ″ + 3 t y ′ − 6 y = 0 where y ( 0 ) = 1 and y ′ ( 0 ) = 0 .

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The solution of the equation y″+3ty′−6y=0 where y(0)=1 and y′(0)=0.

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Chapter 8 Solutions