To determine To solve: The initial value problem 9 y ″ − 6 y ′ + y = 0 , y ( 0 ) = 3 , y ′ ( 0 ) = 1 using Laplace transforms. Explanation Formula used: The Laplace transform of the function e a t is 1 s − a and the inverse Laplace transform of the function 1 s − a is e a t . Theorem used: Laplace transform of derivatives: The transforms of the first and second derivatives of f ( t ) satisfies the following conditions. L ( f ′ ) = s L ( f ) − f ( 0 ) ⋯ ( 1 ) L ( f ″ ) = s 2 L ( f ) − s f ( 0 ) − f ′ ( 0 ) ⋯ ( 2 ) Where (1) holds if f ( t ) is continuous for all t ≥ 0 and f ′ ( t ) is piecewise continuous on every finite interval on the semi axis t ≥ 0 . Similarly, (2) holds if f and f ′ are continuous for all t ≥ 0 and satisfy the growth restriction and f ″ is piecewise continuous on every finite interval on the semi axis t ≥ 0 . Procedure used: “Consider an initial value problem y ″ + a y ′ + b y = r ( t ) , y ( 0 ) = K 0 , y ′ ( 0 ) = K 1 ⋯ ( 3 ) where a and b are constants. Step 1: Set up the subsidiary equation: This is an algebraic equation for the transform of Y = L ( y ) which is obtained by transforming (3) in terms of (1) and (2) as ( s 2 Y − s y ( 0 ) − y ′ ( 0 ) ) + a ( s Y − y ( 0 ) ) + b Y = R ( s ) where R ( s ) = L ( r ) . Collect the Y terms and obtain the subsidiary equation as ( s 2 + a s + b ) Y = ( s + a ) y ( 0 ) + y ′ ( 0 ) R ( s ) . Step 2: Obtain the solution of the subsidiary equation by algebra. That is, divide by s 2 + a s + b and use the transfer function as shown below. Q ( s ) = 1 s 2 + a s + b = 1 ( s + 1 2 a ) 2 + b − 1 4 a 2 Step 3: Simplify the above equation and obtain the solution as follows. Y ( s ) = ( ( s + a ) y ( 0 ) + y ′ ( 0 ) ) Q ( s ) + R ( s ) Q ( s )

10th Edition

ISBN: 9781119096023

Author: Kreyszig

Publisher: WILEY

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Chapter 6.3, Problem 18P

To determine

To solve: The initial value problem 9y″−6y′+y=0, y(0)=3,y′(0)=1 using Laplace transforms.

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Chapter 6.3, Problem 17P

Chapter 6.3, Problem 19P

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