Published by : Politeknik Sultan Azlan Shah Behrang Stesen, Behrang 35950 Perak. Tel :05-4544431 Faks:05-4544993 Email : http://www.psas.edu.my First published 2022 All rights reserved. No part of this publication may be reproduced stored in a retrievel system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without premission of Sultan Azlan Shah Polytechnics. PERPUSTAKAAN NEGARA MALAYSIA Numerical Method and Laplace Transform e-ISBN 978-967-2409-80-9
PREFACE E-book Numerical Method and Laplace Transform specially written for all polytechnic students and also for those who wish to pursue studies in science and engineering. Following the latest syllabus prescribed by Department of Polytechnic Education and Community College Education (JPPKK), this revise, updated e-book presents relevant functions and concise manner. This book consists of two chapters: Numerical Method and Laplace Transform. Each chapter also includes knowledge assessment to prepare them for their understanding. We acknowledge that there are elements to this e-book that can be improved upon which we will strive to continually address in coming editions. We would like to thank everybody involved in the development of this e-book and it is our hope that this book would serve its purpose in helping students gain better understanding of the subject. SITI ZUBAIDAH SAHARUDIN SITI JANARIAH JANTAN NORHASLIZA ABDULLAH
TABLE OF CONTENT 1.0 NUMERICAL METHOD 1.1 CONSTRUCT SOLUTIONS OF LINEAR EQUATIONS …………………………………………………... 1 1.1.1 APPLY GAUSSIAN ELIMINATION METHOD ………………………………………………… 1 1.1.2 APPLY LU DECOMPOSITION USING DOOLITTLE METHOD ……………………….… 4 1.1.3 APPLY LU DECOMPOSITION USING CROUT METHOD ………………………………… 9 SELF ASSESSMENT 1A …………………………………………………..…………………………………………………… 15 1.2 CONSTRUCT SOLUTIONS OF POLYNOMIAL EQUATION ……………………………………..….. 16 1.2.1 APPLY FIXED POINT ITERATION METHOD ……………………………………………..…. 16 1.2.2 APPLY NEWTON-RAPHSON METHOD .…………………………………………….……….. 21 SELF ASSESSMENT 1B ……………………………………………………………………………………………………….. 24 2.0 LAPLACE TRANSFORM 2.1 APPLY DEFINITION OF LAPLACE TRANSFORM ………………………………………………………. 26 2.2 APPLY LAPLACE TRANSFORM PROPERTIES …………………………………………………………… 29 2.3 APPLY INVERSE LAPLACE TRANSFORM …………………………………………………………….…… 32 2.3.1 SOLVE DIFFERENTIAL EQUATIONS USING LAPLACE TRANSFORM …………..… 37 SELF ASSESSMENT 2A ………………………………………………………………………………………………………. 42
1.0 NUMERICAL METHOD 1.1 CONSTRUCT SOLUTIONS OF LINEAR EQUATIONS i. 3 methods to find the solution for 3 variable simultaneous linear equations are Gaussian Elimination Method, Doolittle Method and Crout Method. ii. 3 linear equations with 3 variables: + + = + + = + + = 1.1.1 APPLY GAUSSIAN ELIMINATION METHOD Gaussian Elimination is a systematic application of elementary row operations to a system of linear equations to convert the system to upper triangular form. Once the coefficient matrix is in upper triangular form, we use back substitution to find a solution. Gaussian Elimination Steps: 1. Write the augmented matrix for the system of linear equations. 2. Use elementary row operations on the augmented matrix [A|b] to transform A into upper triangular form. 3. Use back substitution to find the solution of the problem. At the end of this topic, student should be able to: Solve a solution of linear equations using: Gaussian Elimination method LU Decomposition using Doolittle Method LU Decomposition using Crout method Solve a solution of polynomial equation using: The Fixed-Point Iteration Method The Newton-Raphson Method 1
Solve the linear system by Gauss elimination method. 1 + 52 = 7 −21 − 72 = −5 SOLUTION: Convert the system of equations above into an augmented matrix: ( − − | − ) Keep R1 (eliminated R2 by adding with R1) 2R1+ R2 R2 () + (−) = () + (−) = () + (−) = Use back substitution to find the solution of the problem = , so = Substitute 2 = 3 1 + 52 = 7 1 + 5(3) = 7 1 = −8 x1 8, x2 3 EXAMPLE 1 R1 R2 2
Solve the linear system by Gauss elimination method. − 3 + = 4 2 − 8 + 8 = −2 −6 + 3 − 15 = 9 SOLUTION: Convert the system of equations above into an augmented matrix: ( − − − − | − ) Keep R1 and R3 (eliminated R2 by adding with R1) -2R1+ R2 R2 −() + = −(−) + (−) = − −() + = −() + (−) = − Keep R1 and R2 (eliminated R3 by adding with R1) 6R1 + R3 R3 () + (−) = (−) + = − () + (−) = − () + = R2 R2 Keep R1 and R3 (eliminated R3 by adding with R2) -15R2 + R3 R3 () + = (−) + (−) = () + (−) = − (−) + = EXAMPLE 2 R1 R2 R3 ( 1 −3 1 − −6 3 −15 | 4 − 9 ) ( 1 −3 1 0 −2 6 − − | 4 −10 ) ( 1 −3 1 − 0 −15 −9 | 4 − 33 ) ( 1 −3 1 0 −1 3 − | 4 −5 ) 3
Use back substitution to find the solution of the problem. −54 = 108 = −2 Substitute = −2 − + 3 = −5 − + 3(−2) = −5 , = −1 Substitute = −2 and = −1 − 3 + = 4 − 3(−1) − 2 = 4 = 3 x 3, y 1, z 1 1.1.2 APPLY LU DECOMPOSITION USING DOOLITTLE METHOD A modification of the elimination method called LU decomposition. The technique will rewrite the matrix as the product of two matrices, A = LU. The technique breaks the matrix into a product of two matrices, L and U. L is a lower triangular matrix and U is an upper triangular matrix. Using LU decomposition, [] = [][] = []([]) = Solve []() = and then solve []() = Step for the solution to the linear system = : 1. Construct the matrices L and U, if possible 2. Solve = for using forward substitution 3. Solve = for using back substitution In the lower triangular matrix, all elements above the diagonal are zero, in the upper triangular matrix; all the elements below the diagonal are zero 4
Doolittle method (L has one on the diagonal) [ 1 0 0 21 1 0 31 32 0 ][ 11 12 13 0 22 23 0 0 33 ] = [ 11 12 13 21 22 23 31 32 33 ] Determine an LU-Decomposition for the following linear system by using Doolittle method + 2 −3 = 5 4 − 3 =7 5 = 5 SOLUTION: Step 1 ( 1 2 −3 0 4 −3 0 0 5 ) ( ) = ( 5 7 5 ) Step 2 = × ( − − ) = ( ) × ( ) ( − − ) = ( + + + + + ) Column 1 Column 2 Column 3 11 = 1 12 = 2 13 = −3 2111 = 0 21(1) = 0 21 = 0 2112 + 22 = 4 (0)(2) + 22 = 4 22 = 4 2131 + 23 = −3 (0)(−3) + 23 = −3 23 = −3 3111 = 0 31(1) = 0 31 = 0 3112 + 3222 = 0 (0)(2) + 32(4) = 0 32 = 0 3113 + 3223 + 33 = 5 (0)(−3) + (0)(−3) + 33 = 5 33 = 5 = EXAMPLE 3 A X B 5
= ( ) × ( − − ) Step 3 L Y = B ( ) ( ) = ( ) 1 = 5 , 2 = 7, 3 = 5 Step 4 U X = Y ( 1 2 −3 0 4 −3 0 0 5 ) ( ) Use back substitution to find the solution of the problem 5 = 5 = 1 Substitute = 1 4 + (−3) = 7 4 + (−3(1)) = 7 4 − 3 = 7 = 2.5 Substitute = −2.5, = 1 + 2 − 3 = 5 + 2(2.5) − 3(1) = 5 = 3 We find that: = 3, = 2.5 and = 1 6
Solve the simultaneous equation below by using the Doolittle -LU Decomposition Method to find the value of s, t and u + 3 + 3 = 4 2 − 3 − 2 = 2 3 + + 2 = 5 SOLUTION: Step 1 ( 1 3 3 2 −3 −2 3 1 2 ) ( ) = ( 4 2 5 ) Step 2 = × ( − − ) = ( ) × ( ) ( − − ) = ( + + + + + ) Column 1 Column 2 Column 3 11 = 1 12 = 3 13 = 3 2111 = 2 21(1) = 2 21 = 2 2112 + 22 = −3 (2)(3) + 22 = −3 22 = −9 2131 + 23 = −2 (2)(3) + 23 = −2 23 = −8 = () = = + = () + (−) = = + + = () + ( ) (−) + = = = = ( ) × ( − − ) EXAMPLE 4 A X B 7
Step 3 L Y = B ( ) ( ) = ( ) = + = () + = = − + + = () + (−) + = = − Step 4 U X = Y ( − − ) ( ) = ( − − ) Use back substitution to find the solution of the problem = − = − ⁄ ⁄ = − Substitute = −15 −9 − 8 = −6 −9 − 8(−15) = −6 −9 = −6 – 120 = −126 −9 = 14 8
bstitute = 14, = −15 + 3 + 3 = 4 + 3(14) + 3(−15) = 4 = 4 − 42 + 45 = 7 We find that: = 7, = 14 and = −15 1.1.3 APPLY LU DECOMPOSITION USING CROUT METHOD The Crout matrix decomposition algorithm differs slightly from the Doolittle method. Doolittle's method returns a unit lower triangular matrix and an upper triangular matrix, while the Crout method returns a lower triangular matrix and a unit upper triangular matrix. Crout method (U has one on the diagonal) [ 11 0 0 21 22 0 31 32 33 ][ 1 12 13 0 1 23 0 0 1 ] = [ 11 12 13 21 22 23 31 32 33 ] Determine an LU-Decomposition for the following linear system by using Crout’s method − + + = + = − + = − EXAMPLE 5 9
SOLUTION: Step 1 ( −1 2 1 0 8 6 −2 0 5 ) ( ) = ( 0 10 −11 ) Step 2 = × ( − − ) = ( ) × ( ) ( − − ) = ( + + + + + ) Column 1 Column 2 Column 3 11 = −1 1112 = 2 (−1)12 = 2 12 = −2 1113 = 1 (−1)13 = 1 13 = −1 21 = 0 2112 + 22 = 8 (0)(−2) + 22 = 8 22 = 8 2113 + 2223 = 6 (0)(−1) + (8)23 = 6 = 31 = −2 3112 + 32 = 0 (−2)(−2) + 32 = 0 32 = −4 + + = (−)(−) + (−) ( ) + = = = = ( − − − ) × ( − − ) Step 3 L Y = B ( − − − ) ( ) = ( − ) A X B 10
Substitute 1 = 0, 2 = 10 8 −21 − 42 + 63 = −11 −2(0) − 4 ( 10 8 ) + 63 = −11 3 = −1 Step 4 U X = Y ( − − )( ) = ( − ) = −1 Substitute = −1, + 3 4 = 5 4 + 3 4 = 5 4 = 2 Substitute = −1, = 2 − 2 − = 0 − 2(2) − (−1) = 0 = 3 We find that: = 3, = 2 and = −1 Solve the following linear system by using Crout’s method 2 − 4 + = 4 6 + 2 − = 10 −2 + 6 − 2 = −6 EXAMPLE 6 11
SOLUTION: Step 1 ( 2 −4 1 6 2 −1 −2 6 −2 ) ( ) = ( 4 10 −6 ) Step 2 = × ( − − − − ) = ( ) × ( ) ( − − − − ) = ( + + + + + ) Column 1 Column 2 Column 3 11 = 2 1112 = −4 (2)12 = −4 12 = −2 1113 = 1 (2)13 = 1 = 21 = 6 2112 + 22 = 2 (6)(−2) + 22 = 2 22 = 14 2113 + 2223 = −1 ( ) + = − = − 31 = −2 3112 + 32 = 6 (−2)(−2) + 32 = 6 32 = 2 3113 + 3223 + 33 = −2 − ( ) + (− ) + = − = − = = ( − − ) × ( − − ) A X B 12
Step 3 L Y = B ( − − ) ( ) = ( − ) = = Substitute = + = ( ) + = = − = − = − Substitute = , = − − + − = − − ( ) + ( − ) − = − − − − = − = Step 4 U X = Y ( − − ) ( ) = ( − ) = 4 13
Substitute = , − = − − () = − = Substitute = , = − + = − () + () = = We find that: = 2, = 1 and = 4 14
SELF ASSESSMENT 1A (LINEAR EQUATIONS) Determine the value of , and from the linear equation below by using: i. Doolittle Method ii. Crout Method QUESTION 1 − 3 + = 4 2 − 8 + 8 = −2 −6 + 3 + 15 = 9 QUESTION 2 3 + = 2 − − = 3 2 + 2 = 2 QUESTION 3 4 + 4 − 3 = 7 5 + 7 − 13 = −9 + 2 − 5 = −6 QUESTION 4 2 + 3 − = 5 4 + 4 − 3 = 3 −2 + 3 − = 1 ANSWER SELF ASSESSMENT 1A (LINEAR EQUATIONS) 1. = −17, = − 17 2 , = − 9 2 2. = 4, = −10, = 11 3. = −1, = 5, = 3 4. = 1, = 2, = 3 15
A nonlinear system of equations is a set of equations where one or more terms have a variable of degree two or higher and/or there is a product of variables in one of the equations. Most real-life physical systems are non-linear systems, such as the weather. If you've ever tried to find a root of a complicated function algebraically, you may have had some difficulty. Using some basic concepts of calculus, we have ways of numerically evaluating roots of complicated functions. There are several methods for finding the roots such as Fixed-Point Iteration Method and Newton Raphson Method. 1.2.1 APPLY FIXED POINT ITERATION METHOD Fixed Point Iteration can be used in attempting to solve any equation than can be written in than form = (). We use +1 = () to generate a sequence [] that hopefully converges to a root. Fixed Point Iteration Step 1. Find the appropriate iteration functions if it is not given in the question. 2. Assume the first approximate (0) to the root of the question (if not given) 3. Substitute the first approximation into the iteration function to get the second approximation. Repeat until converge or until we have the same value repeating itself ( +1 − ≤ 0). How to find the first approximation if not given: 1. Assume the value of and substitute into the function () = 3 3 + 2 − 4 If = 0, = 3(0 3 ) + 2(0) − 4 = −4 If = 1, = 3(1 3 ) + 2(1) − 4 = 1 ∗Make sure the answer for function is positive and negative 2. Find the first approximation ° = 0+1 2 = 0.5 16 1.2 CONSTRUCT SOLUTIONS OF POLYNOMIAL EQUATION
Use Fixed Point Iteration Method to determine an approximation of the equation () = 2 − − 1 (Let 0 = 2 and give your answer correct to 4 decimal places) SOLUTION: Write the equation in the form of x. () = 2 − − 1 () = 0 2 = + 1 = √ + 1 Label left side as + and right side with +1 = √ + 1 EXAMPLE 7 How to find the x as a subject formula: - To choose the , we have to differentiate first either take the or 2 : |()| = ( + 1) 1 2 ⁄ | ′ ()| = 1 2 ( + 1) −1 2 ⁄ | ′ ()| = 1 2( + 1) 1 2 ⁄ ° = 2 , = 1 2(2+1) 1 2 ⁄ = 0.287 < 1 |()| = 2 − 1 | ′ ()| = 2 ° = 2 , = 2(2) = 4 > 1 So, we choose the value of x less than 1, () = ( + 1) 1 2 ⁄ 17
Repeat until converge Let ° = = √ + = . = √. + = . = √. + = . = √. + = . = √. + = . = √. + = . = √. + = . = √. + = . − = From , we have the same value repeating itself so that the answer is 1.6181 (correct to 4 decimal places) Find the root for 2 = 1 and give your answer correct to 4 decimal places. Given 0 = 0.5. SOLUTION: Write the equation in the form of x 2 = 1 = 1 2 Label left side as + and right side with +1 = 1 2 EXAMPLE 8 18
Repeat until converge 0 = 0.5 1 = 1 2 0.5 = 0.3033 2 = 1 2 0.3033 = 0.3692 3 = 1 2 0.3692 = 0.3456 4 = 1 2 0.3456 = 0.3539 5 = 1 2 0.3539 = 0.3510 6 = 1 2 0.3510 = 0.3520 7 = 1 2 0.3520 = 0.3516 8 = 1 2 0.3516 = 0.3518 9 = 1 2 0.3518 = 0.3517 10 = 1 2 0.3517 = 0.3517 From 10, we have the same value repeating itself so that the answer is 0.3517 Given the equation + ln − 2 = 0. Find the real root of the equation correct to 3 decimal places if the value of x between 1.5 and 1.6 SOLUTION: Write the equation in the form of x + ln − 2 = 0 = 2 − ln EXAMPLE 9 19
Find the first iteration = 1.5 and = 1.6 ° = . + . = . = − (. ) = . = − (. ) = . = − (. ) = . = − (. ) = . = − (. ) = . = − (. ) = . = − (. ) = . From , we have the same value repeating itself so that the answer is 1.557(correct to 3 decimal places) The equation for sin + 3 − 4 = 0 has a real root between = 1 and = 2. Show that +1 = 4− 3 is the appropriate iterative function for the equation. Hence, find the real root of the equation correct to 3 decimal places. SOLUTION: sin + 3 − 4 = 0 3 = 4 − sin = 4−sin 3 , must be change in radian mode ° = + = . 1 = 4 − sin 1.5 3 = 1.0008 2 = 4 − sin 1.0008 3 = 1.0527 3 = 4 − sin 1.0527 3 = 1.0437 EXAMPLE 10 20
4 = 4 − sin 1.0437 3 = 1.0452 5 = 4 − sin 1.0452 3 = 1.045 6 = 4 − sin 1.045 3 = 1.045 So, the real root is 1.045 (correct to 3 decimal places) 1.2.2 APPLY NEWTON-RAPHSON METHOD A method for finding successively better approximations to the roots (or zeroes) of a real- valued function. The Newton-Raphson method in one variable: Given a function ƒ(x) and its derivative ƒ '(x), we begin with a first guess ° for a root of the function. Provided the function is reasonably well-behaved a better approximation 1 is 1 = ° − (°) ′(°) Geometrically, 1 is the intersection point of the tangent line to the graph of f, with the xaxis. The process is repeated until a sufficiently accurate value is reached: +1 = − () ′() Use Newton - Raphson Method to find the root of the equation () = 3 + 2 2 − 1, starting with 0 = 0.5. Give your answer correct to three decimal places. SOLUTION: () = 3 + 2 2 − 1 ′ () = 3 2 + 4 Substitute in Newton Raphson formula, +1 = − () ′() EXAMPLE 11 21
° = 0.5 1 = 0.5 − (0.5) 3 + 2(0.5) 2 − 1 3(0.5) 2 + 4(0.5) = 0.6364 2 = 0.6364 − (0.6364) 3 + 2(0.6364) 2 − 1 3(0.6364) 2 + 4(0.6364) = 0.6183 3 = 0.6183 − (0.6183) 3 + 2(0.6183) 2 − 1 3(0.6183) 2 + 4(0.6183) = 0.6180 4 = 0.6180 − (0.6180) 3 + 2(0.6180) 2 − 1 3(0.6180) 2 + 4(0.6180) = 0.6180 From 4, we have the same value repeating itself so that the answer is 0.618(correct to 3 decimal places) Find the root for 3 − 6 + 4 = 0 and give your answer correct to 4 decimal places. Given 0 = 1. SOLUTION: () = 3 − 6 + 4 ′ () = 3 2 − 6 Substitute in Newton Raphson formula, +1 = − () ′() ° = 1 1 = 1 − 1 3 − 6(1) + 4 3(1) 2 − 6 = 0.6667 2 = 0.6667 − 0.66673 − 6(0.6667) + 4 3(0.6667) 2 − 6 = 0.7302 3 = 0.7302 − 0.73023 − 6(0.7302) + 4 3(0.7302) 2 − 6 = 0.7320 4 = 0.7320 − 0.73203 − 6(0.7320) + 4 3(0.7320) 2 − 6 = 0.7321 EXAMPLE 12 22
5 = 0.7321 − 0.73213 − 6(0.7321) + 4 3(0.7321) 2 − 6 = 0.7321 From 5, we have the same value repeating itself so that the answer is 0.7321 (correct to 4 decimal places) Solve the equation () = − 2 + ln by using Newton Raphson Method. Give your answer correct to 4 decimal places with initial guess of 0 = 0.5 SOLUTION: () = − 2 + ln ′ () = + 1 Substitute in Newton Raphson formula, +1 = − () ′() ° = 0.5 1 = 0.5 − 0.5 − 2 + ln 0.5 0.5 + 1 0.5 = 1.2310 2 = 1.2310 − 1.2310 − 2 + ln 1.2310 1.2310 + 1 1.2310 = 1.5406 3 = 1.5406 − 1.5406 − 2 + ln 1.5406 1.5406 + 1 1.5406 = 1.5571 4 = 1.5571 − 1.5571 − 2 + ln 1.5571 1.5571 + 1 1.5571 = 1.5571 From 4, we have the same value repeating itself so that the answer is 1.5571 (correct to 4 decimal places) Use Newton - Raphson Method to find the root of the equation EXAMPLE 13 EXAMPLE 14 23
() = − 1 − sin . Give your answer correct to four decimal places. SOLUTION: () = − 1 − ′ () = 1 − cos - The value of must be change in radian mode Find 0, when 0 is not given. Assume the value of and make sure the value of () positive & negative value using the equation () = − 1 − 0 1 3 () -1 -0.8415 1.8589 ° = + = . 1 = 1.5 − 1.5 − 1 − 1.5 1 − cos 1.5 = 2.0350 2 = 2.0350 − 2.0350 − 1 − 2.0350 1 − cos 2.0350 = 1.9377 3 = 1.9377 − 1.9377 − 1 − 1.9377 1 − cos 1.9377 = 1.9346 4 = 1.9346 − 1.9346 − 1 − 1.9346 1 − cos 1.9346 = 1.9346 From x4, we have the same value repeating itself so that the answer is 1.9346 (correct to 4 decimal places) SELF ASSESSMENT 1B (POLYNOMIAL EQUATIONS) 1. Determine an approximate of the equation () = 2 − 5 + 1 using Fixed Point Iteration method. (Given ° = 1, and give your answer correct to 3 decimal places) 2. Find the real root of the equation 2 − 4 sin = 0 using Fixed Point Iteration method correct to 3 decimal places. (Given the ° = 1) 24
3. Find the real root of the equation 2 = cos + 3 using Fixed Point Iteration method correct to 3 decimal places. 4. Find the real root for 3 − 5 − 16 − 0 and give your answer correct to 4 decimal places. Given 0 = 3.5 by using Fixed Point Iteration Method. 5. Solve the equation of 3 + 3 2 − 2 = 0 using Newton-Raphson Method, state your answer to 3 decimal places with an initial guess ° = 1 6. Determine a real root of () = 2 − 4 using Newton-Raphson Method. Give your answer correct to 2 decimal places 7. Solve the equation − 4 = 0 using Newton-Raphson Method. Give your answer correct to 4 decimal places if the value of x between 0 and 1 8. Solve the equation of 5 2 + 11 − 17 = 0 using Newton-Raphson Method, state your answer to 4 decimal places with an initial guess ° = 1 9. Find the real root of the equation cos + 3 − 2 = 0 using Newton- Raphson method correct to 4 decimal places. (Given the initial guess, ° = 0.1) ANSWER SELF ASSESSMENT 1B (POLYNOMIAL EQUATIONS) 1. 0.209 2. 1.934 3. 1.524 4. 3.1698 5. 0.732 6. 2.00 7. 0.3574 8. 1.0471 9. 1.5236 25
2.0 LAPLACE TRANSFORM In mathematics, the Laplace transform, named after its inventor Pierre-Simon Laplace (/ləˈplɑːs/), is an integral transform that converts a function of a real variable (often time) to a function of a complex variable (complex frequency). The transform has many applications in science and engineering because it is a tool for solving differential equations. It transforms differential equations into algebraic equations and convolution into multiplication. 2.1 APPLY DEFINITION OF LAPLACE TRANSFORM Definition of Laplace transform At the end of this topic, student should be able to: Apply definition of Laplace Transform Apply Laplace Transform Apply Inverse Laplace Transform ℒ{()} = ∫ − ∞ 0 () = () 26
Calculate F() if given () = 8 by using definition of Laplace Transform. SOLUTION: () = ∫ − ∞ 0 (8) = 8[ − − ] 0 ∞ = 8[ ∞ − − 0 − ] = 8[0 + 1 ] = 8 Calculate F() if given () = −4by using definition of Laplace Transform SOLUTION: () = ∫ ∞ − 0 ( −4 ) = ∫ ∞ −(+4) 0 = [ −(+4) −(+4) ] 0 ∞ = [ ∞ −(+4) − 0 −(+4) ] = [0 + +4 ] = + EXAMPLE 1 EXAMPLE 2 27
2.2 APPLY LAPLACE TRANSFORM PROPERTIES SOLVING LAPLACE TRANSFORM BY USING TABLE OF LAPLACE Find the Laplace Transform of the function ℒ {cos (3)} SOLUTION: ℒ {cos (3)} = 2+3 2 = 2+9 SCAN QR CODE FOR THE TABLE OF LAPLACE EXAMPLE 3 Refer table no 10 cos = ( 2 + 2) 29
APPLY LAPLACE TRANSFORM PROPERTIES Linearity Find the Laplace Transform of the function ℒ{5 3 − −3 } SOLUTION: ℒ{5 3 − −3 } = 5 ( 3! 4 ) − 1 +3 = 30 4 − 1 +3 Find the Laplace Transform of the function () = −3 5 + 5 SOLUTION: ℒ{ −3 5 + 5 } = (+) + + = (+) + + ℒ{1 () + 2()} = ℒ{ℒ{()}} + ℒ{()} EXAMPLE 4 Refer table no 3 and no 5 = ! +1 − = 1 + EXAMPLE 5 Refer table no 1 and no 13 = − sin = ( + ) 2 + 2 30
First Shifting theorem The first shifting theorem provides a convenient way of calculating the Laplace transform of functions that are of the form () = −() where a is a constant and g is a given function. By using First Shifting theorem, find { −} SOLUTION: ℒ{e −3t} = 1 +3 ℒ{2} = 2 2+2 2 = 2 2+4 ℒ{ −3 2} = 2 (+3) 2+4 Multiplication by By multiplication by t, find ℒ { 2 } SOLUTION: { } = (−) {() } = (−) { ) } EXAMPLE 6 { ()} = (−) {()} EXAMPLE 7 31
= () ( ( − )) = () ( − (−) ) = () [ ( − ) () − ()( − )() ( − ) ] = ( − ) 2.3 APPLY INVERSE LAPLACE TRANSFORM SOLVE THE INVERSE LAPLACE TRANSFORM BY USING: a. Simple Function If {()} = () , () Inverse Laplace Transform for () defined as − {()} = () Where − is called the inverse Laplace transform operator. Obtain inverse Laplace transform of the following: a) b) c) 2 +5 d) 3 2−6 e) 8 2−9 f) 1 4 2+9 SOLUTION: a) −1 { 4 } = 4 −1 { 1 } = 4 EXAMPLE 8 32
b) −1 { 12 3 } = 6 −1 { 2! 2+1 } = 6 2 c) 2 +5 −1 { 2 +5 } = 2 −1 { 1 +5 } = 2 −5 d) 3 2−6 −1 { 3 2−6 } = −1 { 3 2(−3) } = 3 2 −1 { 1 (−3) } = 3 2 3 e) 8 2−9 −1 { 8 2−9 } = −1 { 8 3 ∙ 3 2−3 2 } = 8 3 −1 { 3 2−3 2 } = 8 3 sinh 3 f) 1 4 2+9 −1 { 1 4 2+9 } = −1 { 1 4 ∙ 1 2+ 9 4 } = 1 4 −1 { 2 3 ∙ 3 2 2+( 3 2 ) 2 } = 1 6 sin 3 2 Find the inverse Laplace transform of the following: a) + + b) + − SOLUTION: a)+ + −1 { 3+2 2+4 } = −1 { 3 2+4 + 2 2+4 } = −1 {3 ∙ 2+2 2 + 2 2+2 2 } −1 { 3+2 2+4 } = 3 cos 2 + sin 2 b)+ − −1 { 7+3 2−9 } = −1 { 7 2−9 + 3 2−9 } = −1 {7 ∙ 2−3 2 + 3 2−3 2 } −1 { 7+3 2−9 } = 7 cosh 3 + sinh 3 EXAMPLE 9 33
Find the inverse Laplace transform of the following: a) (+) b) + (+)+ c) 8+13 2+4−5 SOLUTION: a) (+) −1 { 5 (+2) 3 } = −1 { 5 2 ∙ 2 ! (+2) 2+1 } = 5 2 −1 { 2 (+2) 2+1 } −1 { 5 (+2) 3 } = 5 2 2 −2 b) + (+)+ − { + (+)+} = − { (+)− (+)+} = − { (+) (+)+− (+)+} −1 { +1 (+3) 2+16} = −1 { (+3) (+3) 2+4 2 − 1 2 ∙ 4 (+3) 2+4 2 } = −3 cos 4 − 1 2 −3 sin 4 a) 8+13 2+4−5 For this question, we need to use the technique of completing the square to rewrite the denominator in a form that we could find in the Laplace Transform table. 2 + 4 − 5 = ( + 4 2 ) 2 − 5 − ( 4 2 ) 2 = 0 = ( + 2) 2 − 9 = 0 −1 { 8 + 13 2 + 4 − 5 } = −1 { 8( + 2) − 3 ( + 2) 2 − 9 } = −1 { 8( + 2) ( + 2) 2 − 9 − 3 ( + 2) 2 − 9 } EXAMPLE 10 34
−1 { 8 + 13 2 + 4 − 5 } = −1 { 8( + 2) ( + 2) 2 − 3 2 − 3 ( + 2) 2 − 3 2 } −1 { 8 + 13 2 + 4 − 5 } = 8 −2 cosh 3 − −2 sinh 3 b. Partial Fraction Method Recall from calculus that a rational function of the form () () , where () and () are polynomials with the degree of less than the degree of , has a partial fraction expansion whose form is based on the linear and quadratic factors of (), (We assume the coefficients of the polynomials to be real numbers). Thera are few cases to consider: A linear factor + in the denominator produces a partial fraction + , where is an unknown constant Repeated linear factors ( + ) gives: 1 + + 2 ( + ) 2 Repeated linear factors ( + ) gives 1 + + 2 ( + ) 2 + 3 ( + ) 3 + ⋯ + ( + ) Quadratic factor ( + + ) gives: + 2 + + Find the partial fraction of the following. Hence obtain the inverse Laplace Transform. a) (−) b) 3 2+6 (−1)2 (+2) c) 2 2++1 (+1)( 2+1) EXAMPLE 11 35
SOLUTION: a) (−) By using partial fraction 1 ( − 1) = + − 1 1 = ( − 1) + 1 = − + Coefficient: 1 ∶ 0 = + 0 ∶ 1 = − By using calculator, find the value of and : = −1 = 1 1 ( − 1) = −1 + 1 − 1 Hence, −1 { 1 (−1) } = −1 { −1 } + −1 { 1 −1 } = −1 + b) 3 2+6 (−1)2 (+2) By using partial fraction 3 2+6 (−1) 2 (+2) = (−1) + (−1) 2 + (+2) 3 2 + 6 = ( − 1)( + 2) + ( + 2) + ( − 1) 2 3 2 + 6 = 2 + − 2 + + 2 + 2 − 2 + Coefficient: 2 ∶ 3 = + ∶ 0 = + − 2 0 ∶ 6 = −2 + 2 + By using calculator, find the value of , and : = 1 , = 3 = 2 3 2 + 6 ( − 1) 2 ( + 2) = 1 ( − 1) + 3 ( − 1) 2 + 2 ( + 2) 36
Hence, − { + ( − ) ( + ) } = − { ( − ) } + − { ( − ) } + − { ( + ) } = + 3 + 2 −2 a) 2 2++1 (+1)( 2+1) By using partial fraction 2 2 + + 1 ( + 1)( 2 + 1) = + 1 + + 2 + 1 2 2 + + 1 = ( 2 + 1) + ( + ) ( + 1) 2 2 + + 1 = 2 + + 2 + + + Coefficient: 2 ∶ 2 = + ∶ 1 = + 0 ∶ 1 = + By using calculator, find the value of , and : = 1 , = 1 = 0 2 2 + + 1 ( + 1)( 2 + 1) = 1 + 1 + 2 + 1 Hence, −1 { 2 2 + + 1 ( + 1)( 2 + 1) } = −1 { 1 + 1 } + −1 { 2 + 1 } = −1 + cos 2.3.1 SOLVE DIFFERENTIAL EQUATIONS USING LAPLACE TRANSFORM Our focus is to show how Laplace Transform can be used to solve initial problems for linear differential equations. As we will see, the method of Laplace Transforms leads to the solutions of the initial value problem without first finding a general solution. 37
The basic concept Method of solving linear differential equations for initial value problems by using Laplace Transform can be simplifies as: Differentiation for Laplace Transform: If { ()} = () , so { ′ ()} = () − (0) { ′′ ()} = 2 () − (0) − ′ (0) { ′′′ ()} = 3 () − 2 (0) − ′ (0) − ′′ (0) { ()} = () − −1 (0) − −2 (0) − −1 (0) METHOD OF LAPLACE TRANSFORM To solve initial value problems: a) Take the Laplace Transform of both sides of the equations b) Use the properties of Laplace transform and initial conditions to obtain an equation form the Laplace transform of the solution and then solve this equation for the transform c) Determine the inverse Laplace transform of the solutions by looking it up in the table or by using a suitable method (such as partial fractions) in combination with the table. 38
Solve the initial value problem ′ + = , () = SOLUTION: The differential equation in ′ + = cos ( 1 ) is and identity between two functions of . Hence equality holds for the Laplace transform for these functions: { ′ + } = {cos } { ′ } + {} = {cos } Using the linearity property of and the previously computed transform of the logarithmic functions, we can write: { ′ } + {} = 2+1 ( 2 ) Now let () = {} () . From the formulas for the Laplace transform and the initial conditions in (1) , we find { ′ } = () − (0) = () − 0 {} = () Substituting this expression into (2) and solving for () yields () − 0 + () = 2 + 1 ( + 1) () = 2 + 1 () = ( + 1)( 2 + 1) EXAMPLE 12 39
Our remaining task is to compute the inverse Laplace transform of the rational functions (). This was done by using partial fraction expansion, we found () = 1 2 ( + 1) − 2( 2 + 1) + 1 2 ( 2 + 1) Function of () can be obtained by taking the Laplace transform of these equations. So () = 1 2 −− 1 2 cos + 1 2 sin Solve the initial value problem ′′ + 4 ′ − 5 = , (0) = 1 , ′ (0) = 0 SOLUTION: Assume, { ()} = () { ′′} + 4 { ′′} − 5{} = { } { ′′} + 4 { ′′} − 5 () = 1 (−1) 2 (2) { ′′} = 2() − (0) − ′ (0) = 2() − { ′ } = () − (0) = () − 1 Substituting this expression into (2) and solving for () gives [ 2 () − ] − 4 [ () − 1] − 5 () = 1 ( − 1) 2 ( 2 + 4 − 5) () = + 4 + 1 ( − 1) 2 ( + 5)( − 1) = 3 + 2 2 − 7 + 5 ( − 1) 2 () = 3 + 2 2 − 7 + 5 ( + 5)( − 1) 3 EXAMPLE 13 40
The partial fraction expansion for () has the form 3 + 2 2 − 7 + 5 ( + 5)( − 1) 3 = 35 216 ( + 5) + 181 216 ( − 1) − 1 36 ( − 1) 2 + 1 6 ( − 1) 3 () = 35 216 ( + 5) + 181 216 ( − 1) − 1 36 ( − 1) 2 + 1 6 ( − 1) 3 Function of () can be obtained by taking the inverse Laplace transform of this equations. So, () = 35 216 −5 + 181 216 − 1 36 + 1 12 2 41
SELF ASSESSMENT 2A 1. Calculate F() if given () = 9 by using definition of Laplace Transform ANSWER: 9 2. Calculate F() if given () = 5 by using definition of Laplace Transform ANSWER: 1 −5 3. Determine the Laplace Transform below on Laplace Transform Table. i. () = 5 + 4 7 ANSWER: 120 6 + 4 7 ii. () = 3 cos 5 + 4 5 ANSWER: 3 2+25 + 4 −5 iii. ℒ{5 3 − −3 } ANSWER: 30 4 − 1 +3 4. Solve () = −2 cosh 3by using first shift theorem ANSWER: (+2)2−9 5. Solve ℒ{ 3 4} by using first shift theorem ANSWER: −3 (−3) 2+16 6. Show the steps ℒ{ 2 −3 } = 2 (+3)3 [using first shift theorem] 7. Determine the Laplace transform for () = −2 [Using Multiplication by ] ANSWER: 1 (+2)2 42
8. Obtain inverse Laplace Transform of: i. ANSWER: ii. − ANSWER: iii. + ANSWER: − iv.2 + 1 +2 ANSWER: 2 + 2 v. +2 2+9 ANSWER: cos 3 + 2 3 sin 3 vi.2+5 2−4 ANSWER: 2 cosh 2 + 5 2 sinh 2 9. Find the partial fraction of the following. Hence obtain the inverse Laplace transform: i. (−) ANSWER: − ii. ( +) ANSWER: − iii. 2+9+2 (−1)2 (+3) ANSWER: (2 + 3) − −3 iv. 2 2+10 ( 2−2+5) (+1) ANSWER: (3 cos 2 + 4 sin 2) − − 10. Solve the following differential equations: i. ′ + = −2 (0) = 0 ANSWER: () = ( − 1 − ) −2 ii. ′ + 3 = 13 sin 2 (0) = 6 ANSWER: () = 8 −3− 2 cos 2 + 3 sin 2 iii. ′′ + 2 ′ − 3 = , (0) = 2 , ′ (0) = 1 ANSWER: () = − 1 9 (3 + 2) + 2 + 2 9 −3 iv. ′′ − 4 ′ = 4 2 , (0) = 0 , ′ (0) = 5 ANSWER: () = (1 + − −4 ) 2 43
ACKNOWLEDGEMENT First and foremost, thanks to Allah, the Almighty, for His grace can e-book of Electrical Engineering Mathematics completed successfully. Our deepest gratitude to our family and our colleagues of Mathematics, Science and Computer Department for their support, love and encouragement to complete of this project. Finally, this book is dedicated to our students, who struggle and find success through diligence and hard work. They provide us with inspiration and ideas to communicate mathematics effectively. Thank you. Siti Zubaidah Saharudin Siti Janariah Jantan Norhasliza Abdullah 44
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