To determine To solve: The initial value problem y ″ − 4 y ′ + 3 y = 6 t − 8 , y ( 0 ) = 0 , y ′ ( 0 ) = 0 by using Laplace transform. Explanation Formula used: 1. 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 . 2. The Laplace transform of the function t is 1 s 2 and the inverse Laplace transform of the function 1 s 2 is 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 ) . Step 4: Apply the given initial conditions y ( 0 ) = y ′ ( 0 ) = 0 and obtain Y = R Q . Therefore, the value of Q is obtained as Q = Y R = L ( Output ) L ( input ) . Step 5: Take the inverse of Y to obtain y = L − 1 ( Y ) . Calculation: The given initial value problem is y ″ − 4 y ′ + 3 y = 6 t − 8 , y ( 0 ) = 0 , y ′ ( 0 ) = 0 . Use the above mentioned procedure and solve the given initial value problem as shown below. Suppose that Y = L ( y ) . Substitute the value of L ( y ′ ) from (1) in the given initial value problem and obtain the subsidiary equation as shown below. s 2 Y − 4 s Y + 3 Y = 6 s 2 − 8 s ( s 2 − 4 s + 3 ) Y = 6 − 8 s s 2 Y = 6 − 8 s ( s 2 − 4 s + 3 ) ( s 2 ) Y = 6 − 8 s ( s − 3 ) ( s − 1 ) s 2 Now, apply partial fractions in the right hand side of the above equation and simplify as shown below. 6 − 8 s ( s − 3 ) ( s − 1 ) s 2 = A s − 3 + B s − 1 + C s 2 + D s 6 − 8 s ( s − 3 ) ( s − 1 ) s 2 = A ( s − 1 ) s 2 + B ( s − 3 ) s 2 + C ( s − 3 ) ( s − 1 ) + D s ( s − 3 ) ( s − 1 ) ( s − 3 ) ( s − 1 ) s 2 Cancel like terms on both sides as follows

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 6.2, Problem 9P

To determine

To solve: The initial value problem y″−4y′+3y=6t−8, y(0)=0,y′(0)=0 by using Laplace transform.

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Advanced Engineering Mathematics

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