...transformation called the Laplace transform. It is very effective in the study of initial value problem involving linear differential equation with constant coefficient. Laplace transform was first introduced by a French mathematician called Pierre Simon Marquis de Laplace about 1780’s. This method associated with the isolation of the original problem that is function ƒ(t) of a real variable and some function ƒ(s) of a complex variable so that the ordinary differential equation for the function ƒ(t) is transformed into an algebraic equation for ƒ(s) which in most cases can readily be solved. The solution of the original differential equation can be arrived at by obtaining the inverse transformation. The transformation and its inverse can be derived by consulting already prepared table of transform. This method is particularly useful in the solution of differential equation and has more application in various fields of technology e.g. electrical network, mechanical vibrations, structural problems, control systems. Meanwhile in this research work, I shall look into the Laplace transform, the properties of the Laplace transform and the use of this technique in solving delay differential equation will be looked into. 1.2 Statement of the Problem There are so many engineering and other related problems that can be expressed in the form of ordinary differential equations. But such problems cannot easily be solved using the elementary method of solution. In such cases, the Laplace transform...
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...coefficients & method of variation of parameters. Discuss & compare the advantages and disadvantages of each method. Illustrate your findings with examples. ” I am thankful to Ms Manreet Singh who guided me during the difficult moments ,I faced during the completion of this project & I am to thankful to librarian sir. Yours faithfully, Anirban Sarkar. Abstract, Method of undetermined coefficients, otherwise known as the Lucky Guess Method, is an approach to finding a particular solution to certain inhomogeneous ordinary and recurrence relations. It is closely related to the annihilator method, but instead of using a...
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...DSMHT 201: Applied Differential Equation (credit 02) 1. Ordinary differential equations and their solutions: Initial value problems, Boundary value problems, Basic existence and uniqueness theorems (statement and illustration only). 2. Solution of first order equations: separable equations and equations reducible to this form, Linear equations, exact equations, Special integrating factors, Substitutions and transformations. 3. Solution of higher order linear differential equations: Solution space of homogeneous linear equations. Fundamental solutions of homogeneous systems. Reduction of order. Homogeneous linear equations with constant coefficients. Non homogeneous equations. 4. Method of undermined coefficients. Variations of parameters. Cauchy-Euler differential equations. 5. Systems of differential equations, linear system, Fundamental matrix, Solutions of linear systems with constant coefficients. References: 1. D.G. Zill, A First Course in Differentil Equations with Applications. 2. F. Braur and J.A. Nohel, Differential Equations. 3. S.L. Ross, Differential Equation. DSMHT 202: Geological and Hydro-meteorological Hazards (credit 02) 1. Introduction to Geological and Hydro-meteorological Hazards. 2. Geological Hazards Earthquakes (causes, types and effects of Earthquakes), Tsunami & Seiches. (naming, Tsunami Generation, Velocity and height, coastal effects and vulnerability), Volcanoes (origin & types of volcanic hazards...
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...Engineering and Science subjects. Here our intention is to make the students acquainted with the concept of basic topics from Mathematics, which they need to pursue their Engineering degree in different disciplines. Course Contents: Module I: Differential Calculus Successive differentiation, Leibnitz’s theorem (without proof), Mean value theorem, Taylor’s theorem (proof), Remainder terms, Asymptote & Curvature, Partial derivatives, Chain rule, Differentiation of Implicit functions, Exact differentials, Tangents and Normals, Maxima, Approximations, Differentiation under integral sign, Jacobians and transformations of coordinates. Module II: Integral Calculus Fundamental theorems, Reduction formulae, Properties of definite integrals, Applications to length, area, volume, surface of revolution, improper integrals, Multiple Integrals-Double integrals, Applications to areas, volumes. Module III: Ordinary Differential Equations Formation of ODEs, Definition of order, degree & solutions, ODE of first order : Method of separation of variables, homogeneous and non homogeneous equations, Exactness & integrating factors, Linear equations & Bernoulli equations, General linear ODE of nth order, Solution of homogeneous equations, Operator method, Method of undetermined coefficients, Solution of simple simultaneous ODE. Module IV: Vector Calculus Scalar and Vector Field, Derivative of a Vector, Gradient, Directional...
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...1.0 INTRODUCTION Differential Equation In Motion A differential equation is a mathematical equation that relates some functions of one or more variables with its derivatives. Besides that, differential equations differ from ordinary equations of mathematics in that in addition to variables and constants they also contain derivatives of one or more of the variables involved. The history of differential equations was independently invented by English physicist Isaac Newton (1642-1727) and German mathematician Gottfried Leibniz (1646-1716). Both of them credited initial development. Further improvements were seen in the work of other mathematicians, namely, Leonhard Euler (1713-1765), Alexis-Claude Clairaut (1713-1765) and Sophus Lie (1842-1899). Moreover, differential equations play a prominent role in many aspects. This is because there are many real life applications of differential equations. For example, differential equations describe the way electrical currents move through circuits in engineering. In chemistry, they describe the rates of chemical reactions. 2.0 METHODOLOGY 2.1 Newton Second Law of Motion If an object of mass m is moving with acceleration a and being acted on with force F then Newton’s Second Law tells us, F=ma. Newton Second Law is used to analyze the motion of the object. 2.2 Separable Variable The differential equation can be solved by using the method of separation of variables. For example, a differential equation like this, Nydydx=M(x)...
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...METHOD This Program will find the Root of the Equation f(x) = x^3 - 4x -9 ************ By Bisection Method ************ Enter the range in which root lies................. Enter the value for the lower limit - 2 Enter the value for the upper limit - 3 Enter the limit of error allowed - 0.00001 The value of function at 2.000000 is = -9.000000 The value of function at 3.000000 is = 6.000000 Iterations Start now............................ The value of function at 2.500000 is = -3.375000 The value of function at 2.750000 is = 0.796875 The value of function at 2.625000 is = -1.412109 The value of function at 2.687500 is = -0.339111 The value of function at 2.718750 is = 0.220917 The value of function at 2.703125 is = -0.061077 The value of function at 2.710938 is = 0.079423 The value of function at 2.707031 is = 0.009049 The value of function at 2.705078 is = -0.026045 The value of function at 2.706055 is = -0.008506 The value of function at 2.706543 is = 0.000270 The value of function at 2.706299 is = -0.004118 The value of function at 2.706421 is = -0.001924 The value of function at 2.706482 is = -0.000827 The value of function at 2.706512 is = -0.000279 The Approximate root of the above function is = 2.706528 and the value of function is = -0.000004 The number of Iterations are = 16 Output For “C” code of NEWTON – RAPHSON METHOD This Program will find the Root of the Equation f(x) = x.logx - 1.2 ************ By...
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...Lab 0: MATLAB and ODE Solvers | ME 4173 Robot Kinematics | | | | Introduction The following report will display the results and conclusions of an experiment to simulate the output of an inverse pendulum system in MATLAB. The objectives of this experiment were to review MATLAB programming and using MATLAB to simulate ODEs and systems. Objectives * Examine the basics of MATLAB * Use MATLAB to simulate a system * Use ODE solvers to numerically integrate the system over a set time period Apparatus The apparatus used in this experiment was MATLAB. It was used to provide a simulation environment to analyze the inverse pendulum’s motion. Experiments and Results There were six components of this experiment. This experiment was mostly familiarizing with MATLAB. All code used is illustrated in the Appendix – Code. The first part consisted of learning commands within the MATLAB environment. It was a brief overview of how commands work in MATLAB. There was no code used in the part of the experiment. The second part of the experiment examined how arrays were created and used in MATLAB. The first step was to create a matrix. This matrix was then subjected to various commands including eye( ), zeroes( ), and ones( ). Indexing was also used to access various parts of the matrix. Matrix operations such as transpose, inverse, size and length were also shown. Part three of the experiment explained how a script was created and what it was...
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...COORDINATE GEOMETRY. EQUATION OF A STRAIGHT LINE SOLVED EXAMPLES. 1. ( ) ( Solution. ( Now, using the formula ) we have: ( ( ). ) . . ( 2. ) Solution. ( ) ( ) ( ( ) ), ( ) ) ( ) ( ) Solution. ( ) ( ) ( ) ( ) EQUATION OF A CIRCLE. The general equation of a circle is of the form .Where (– √ is the centre of the circle and the radius is: Finding the equation of a circle of a circle given its radius and centre Let ( ) be any point on a circle whose centre is ( circle is given by : ( ) ( ) ) and ( r ( ( ) ) ) 0 If the centre is at the origin ( ( ) ( ) ) then the equation becomes: ‘ the equation of a ) Solved Examples. 1. Find the equation of the circle with centre ( ) and radius Solution. Using ( ) ( , ( ( )- ) we have, ) ( ) ( ) 2. Find the equation of a circle with centre ( ) which passes through the point( Solution. ( ) ( ( ) ) ( ) √ ( ) ( ( ) ( , ) ) (√ ) 3. Find the centre and radius of the circle Solution. ( ) Comparing (1) with the general equation of a circle The radius of the circle is: √ The centre is ( √ √ ) ( ) ) EXERCISE. 1. Find the centre and radius of the circle 2. Find the centre and radius of the circle 3. Find the equation of the circle which passes...
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...THE USE OF IMPULSE ACTIONS AT CHECKING FOR SHORT CIRCUITS IN ROTOR TURBINE. \ Abstract: Unbalanced magnetic attraction of the rotor and its uneven heating is caused by coil short circuit in the rotor winding, which causes its bending. These processes increase the vibration due to the composing reverse speed. The damage of the coil isolation can be caused both by the movement of the windings in the slot and the displacement of insulating spacers between the turns of the rotor coils. Keywords: Rotor, turbine, impulse action, short circuits. 1. Introduction The coil short circuit remained insignificant for a long time unless due to thermal imbalance it caused the shift of the vibration level. For a long time turbogenerators (up to 120 MWh) in most of power systems worked with coil short circuit in the rotor, some coil shorts were detected just by advanced methods. But after the invention 800 MWh turbo-generators and higher coil short circuits detection became essential due to the large loss during the possible outage in power supply caused by coil short circuit. Therefore, the purpose of this work is to develop the interturn insulation field winding control procedures using the impulse actions. This method is more efficient comparing to the traditional method, which is associated with the time consumption and application complexity in an emergency. 2. Schematic diagram of the diagnostic tests Schematic diagram of the diagnostic tests for the methods associated...
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...REMOVED FROM THE EXAMINATION ROOM Electronic calculators may be used provided that they cannot store text. 1 of 3 P.T.O. 2 of 3 ECON60081 SECTION A Answer ALL Questions. Each question carries equal weight. Question 1. Consider the problem Ax = b where ⎛ ⎞ ⎞ ⎛ ⎛ ⎞ 2 1 3 1 1 A = ⎝ 3 2 5 ⎠ x = ⎝ 2 ⎠ b = ⎝ 3 ⎠ 1 1 2 3 2 Determine the degrees of freedom and the number of redundant equations of this system. Further, determine the solution(s) if solutions exists. [10 marks] Question 2. Give formal definitions for the following: (a) a convex function, (b) a strictly con vex set, (c) a differentiable function. Further, give an example of a concave function that is not differentiable. [10 marks] Question 3. Find the solution of the following differential equation = 1 + 3 − 2 ˙ where (0) = 5. [10 marks] Question 4. Find the general solution of the following second order differential equation + 4 + 10 = ¨ ˙ [10 marks] Question 5. Consider the following system of nonlinear difference equations ½ 1+1 = 31 − 2 2 2+1 = 2 + 1 Find the equilibria and classify them as sink, source or saddle. [10 marks] Continued 3 of 3 ECON60081 SECTION B Answer ALL Questions. Question 6. Assume ≥ 4. For −1 and −1, solve the utility maximization problem (a) max ( ) = 1 ln(1 + ) + 1 ln(1 + ) subject to the constraint 2 + 3 = .[15 marks] 2 4 (b) Let (∗ () ∗...
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...deterministic or probabilistic decision models. As deterministic models, decisions to bring good final outcomes. A deterministic model is “you get what you expect” risk-free model, which determines the outcome. It also depends on the influence of the uncontrollable the factors that determine the outcome of a decision and the information the decision-maker input as a predicting factor (Arsham, 1996). According to Schrodt (2004), deterministic models was widely used in the early 18th century to study physical processes to develop differential equations by many mathematicians. These differential equation allow values of a variable as function of its value at any point in time and as a common form of the deterministic concept. The differential equations apply to a variety of astronomical and mechanical phenomena and have produced crucial scientific literature (Schrodt, 2004). The model is important in astronomical and mechanical phenomena because of the finite equations...
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...Higher Engineering Mathematics In memory of Elizabeth Higher Engineering Mathematics Sixth Edition John Bird, BSc (Hons), CMath, CEng, CSci, FIMA, FIET, MIEE, FIIE, FCollT AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD PARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Newnes is an imprint of Elsevier Newnes is an imprint of Elsevier The Boulevard, Langford Lane, Kidlington, Oxford OX5 1GB, UK 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA First edition 2010 Copyright © 2010, John Bird, Published by Elsevier Ltd. All rights reserved. The right of John Bird to be identified as the author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher. Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone (+44) (0) 1865 843830; fax (+44) (0) 1865 853333; email: permissions@elsevier.com. Alternatively you can submit your request online by visiting the Elsevier web site at http://elsevier.com/locate/permissions, and selecting Obtaining permission to use Elsevier material. Notice No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products...
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...Exercise 1 – Find the first derivative and the second derivative of the following functions Answer: Applying constant function and power function rule (A) Y = 3 + 10X + 5X2 dY/dX = 0 + 1.10.X1-1 +2.5.X2-1 dY/dX = 10 + 10X d2Y/ dX2 = 0 + 1.10.X1-1 d2Y/ dX2 = 10 (B) Y = 2X (4 + X3 ) Y = 8X + 2X4 dY/dX = 1.8.X1-1 + 4.2.X4-1 dY/dX = 8 + 8X3 d2Y/ dX2 = 0 + 3.8.X3-1 d2Y/ dX2 = 24X2 (C) Y = 3 /X2 Y = 3X-2 dY/dX = -2.3.X-2-1 dY/dX = -6X-3 dY/dX = -6/X3 d2Y/ dX2 = -3.-6X-3-1 d2Y/ dX2 = 18X-4 d2Y/ dX2 = 18/X4 (D) Y = 18T – 2T2 dY/dT = 1.18.T1-1 – 2.2.T2-1 dY/dT = 18 – 4T d2Y/ dT2 = 0 – 1.4.T1-1 d2Y/ dT2 = - 4 Exercise 2 - Find the partial Derivative of Y with respect to X Answer (A) Y = 10 + 3Z + 2X ∂Y/∂X = 0 + 0 + 1.2.X1-1 ∂Y/∂X = 2 (B) Y= 18Z + X2 + Z.X ∂Y/∂X = 0 + 2.1.X2-1 + Z ∂Y/∂X = 2X + Z Application - The nursing home industry is growing rapidly because the aging of American population. According to the study of an economist, the average cost per patient day of a nursing home can be approximated by C = A – 0.16B + 0.002B2 Where, B is the nursing home’s number of patient days per year ( in thousands) and A is the number that depends on the location and other factors but not on B. Based on the information , how big must a nursing home be ( in terms of patient – days) to minimize the cost per patient day ? Answer – C= ƒ (A, B) Where C is Avg. Cost per patient day A is variable depends on location and other factors ...
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...Beams in this chapter, we describe methods for determining the equation of the deflection curve of beams and finding deflection and slope at specific points along the axis of the beam 9.2 Differential Equations of the Deflection Curve consider a cantilever beam with a concentrated load acting upward at the free end the deflection in the y v is the displacement direction of the axis the angle of rotation (also called slope) is the angle between the x axis and the tangent to the deflection curve point m1 is located at distance x point m2 is located at distance x + dx slope at slope at m1 m2 is is +d denote O' the center of curvature and the radius of curvature, then d = ds is and the curvature 1 = 1 C = d C ds the sign convention is pictured in figure slope of the deflection curve dv C dx for = tan or ds j dx d C dx d C = dx = cos j 1 and d 2v CC dx2 = dv tan-1 C dx tan dv C dx j , then small = 1 C = 1 C = = if the materials of the beam is linear elastic = 1 C = M C EI [chapter 5] then the differential equation of the deflection curve is obtained d C dx d2v = CC dx2 = M C EI and v dV CC dx d 4v CC dx4 = -q q -C EI it can be integrated to find ∵ dM CC dx d 3v CC dx3 = V then V = C EI = 2 sign conventions for M, V and q are shown the above equations can be written in a simple form EIv" = M EIv"' = V EIv"" = -q this equations are valid only when Hooke's law applies and when the slope and the...
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...Syllabus Cambridge International A Level Further Mathematics Syllabus code 9231 For examination in June and November 2013 Contents Cambridge A Level Further Mathematics Syllabus code 9231 1. Introduction ..................................................................................... 2 1.1 1.2 1.3 1.4 Why choose Cambridge? Why choose Cambridge International A Level Further Mathematics? Cambridge Advanced International Certificate of Education (AICE) How can I find out more? 2. Assessment at a glance .................................................................. 5 3. Syllabus aims and objectives ........................................................... 7 4. Curriculum content .......................................................................... 8 4.1 Paper 1 4.2 Paper 2 5. Mathematical notation................................................................... 17 6. Resource list .................................................................................. 22 7 Additional information.................................................................... 26 . 7 .1 7 .2 7 .3 7 .4 7 .5 7 .6 Guided learning hours Recommended prior learning Progression Component codes Grading and reporting Resources Cambridge A Level Further Mathematics 9231. Examination in June and November 2013. © UCLES 2010 1. Introduction 1.1 Why choose Cambridge? University of Cambridge International Examinations (CIE) is the world’s largest provider of international...
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