Please carefully review your Digital Proof download for formatting, grammar, and design issues that may need to be corrected.We recommend that you review your book three times, with each time focusing on a different aspect.Once you are satisfied with your review, you can approve your proof and move forward to the next step in the publishing process.To print this proof we recommend that you scale the PDF to fit the size of your printer paper. Check the format, including headers, footers, page numbers, spacing, table of contents, and index.Review any images or graphics and captions if applicable.Read the book for grammatical errors and typos.123Digital ProoferA Friendly Introduct...Authored by Mohammed K A Kaabar6.69" x 9.61" (16.99 x 24.41 cm)Black & White on White paper164 pagesISBN-13: 9781506004532ISBN-10: 1506004539 A Friendly Introduction to Differential Equations Copyright 2015 Mohammed K A Kaabar All Rights Reserved2 M. KaabarA Friendly Introduction to Differential Equations Copyright 2015 Mohammed K A Kaabar All Rights Reserved3About the Author MohammedKaabar hasaBachelorofSciencein TheoreticalMathematicsfromWashingtonState University,Pullman,WA.Heis a graduate student in Applied Mathematics at Washington State University, Pullman,WA,andheisamathtutorattheMath LearningCenter(MLC)atWashingtonState University,Pullman. He istheauthorofAFirstCourseinLinearAlgebra Book,and his research interests are applied optimization, numerical analysis, differentialequations,linearalgebra,andreal analysis.He was invitedtoserveasaTechnical ProgramCommittee(TPC)member inmany conferencessuchasICECCS14,ENCINS15,eQeSS 15,SSCC15,ICSoEB15,CCA14,WSMEAP14, EECSI14,JIEEEC13andWCEEENG12. Heisan onlineinstructoroftwofreeonlinecoursesin numericalanalysis:IntroductiontoNumerical AnalysisandAdvancedNumericalAnalysis atUdemy Inc,SanFrancisco,CA. Heisaformermemberof InstituteofElectricalandElectronicsEngineers (IEEE),IEEEAntennasandPropagationSociety, IEEEConsultantsNetwork,IEEESmartGrid Community,IEEETechnicalCommitteeonRFID, IEEELife SciencesCommunity,IEEEGreen ICT Community,IEEECloudComputingCommunity, IEEE Internet of Things Community, IEEE Committee onEarthObservations,IEEEElectricVehicles Community,IEEEElectronDevicesSociety,IEEE Communications Society, and IEEE Computer Society.Healsoreceivedseveraleducationalawardsand certificatesfromaccreditedinstitutions.Formore informationabouttheauthor andhisfreeonline courses,pleasevisithispersonalwebsite: http://www.mohammed-kaabar.net.Copyright 2015 Mohammed K A Kaabar All Rights Reserved4 M. KaabarTable of Contents Introduction6 1 The Laplace Transform9 1.1Introduction to Differential Equations..........9 1.2 Introduction to the Laplace Transforms..14 1.3 Inverse Laplace Transforms..........24 1.4 Initial Value Problems............271.5 Properties of Laplace Transforms.331.6 Systems of Linear Equations.....451.7 Exercises.....49 2SystemsofHomogeneousLinearDifferential Equations (HLDE)51 2.1 HLDE with Constant Coefficients..................512.2 Method of Undetermined Coefficients.602.3 Exercises....653MethodsofFirstandHigherOrdersDifferential Equations67 3.1Variation Method.673.2 Cauchy-Euler Method....743.3 Exercises...76Copyright 2015 Mohammed K A Kaabar All Rights Reserved54 Extended Methods of First and Higher Orders Differential Equations78 4.1 Bernoulli Method......784.2 Separable Method....................................854.3 Exact Method....................................874.4 Reduced to Separable Method.................904.5 Reduction of Order Method..................924.6 Exercises.....955 Applications of Differential Equations 96 5.1 Temperature Application........................965.2 Growth and Decay Application......1005.3 Water Tank Application......104 Appendices 109 A Determinants.......................109B Vector Spaces........116C Homogenous Systems.........135 Answers to Odd-Numbered Exercises157 Index159 Bibliography163 Copyright 2015 Mohammed K A Kaabar All Rights Reserved6 M. KaabarIntroductionInthisbook,Iwrotefivechapters:TheLaplace Transform,SystemsofHomogenousLinearDifferential Equations(HLDE),MethodsofFirstandHigherOrders DifferentialEquations,ExtendedMethodsofFirstand HigherOrdersDifferentialEquations,andApplicationsof DifferentialEquations.Ialsoaddedexercisesattheendof each chapter above to let students practice additional sets of problemsotherthanexamples,andtheycanalsocheck theirsolutionstosomeoftheseexercisesbylookingat Answers to Odd-Numbered Exercises section at the end of this book. This book is a very useful for college students who studied Calculus II, and other students who want to review some concepts of differentialequationsbeforestudying coursessuchas partialdifferentialequations, applied mathematics,andelectriccircuitsII. Accordingtomy experienceas a mathtutor,Ihavenoticedthatsome studentshavedifficultytounderstandsomeconcepts of laplacetransforms becausemostauthorsofdifferential equationsbooksdidnotstartwithlaplacetransformsasa firstchapter,andtheyleftitattheendoftheirbooks. Therefore,Idecidedtostartwithadifferentapproachby choosing laplace transforms to be in the first chapter of this book.Ifyouhaveanycommentsrelatedtothecontentsof thisbook,pleaseemailyourcommentstomohammed.kaabar@email.wsu.edu.I wish to express my gratitude and appreciation to myfather,mymother,andmy onlylovely13-yearold brother who is sick, and I want to spend every dollar in his heathcare.Iwouldalsoliketogiveaspecialthankstoall administrators and professors of mathematics atWSUfor their educational support. In conclusion, I would appreciate toconsiderthisbookasamilestonefordeveloping more mathbooksthatcanserveourmathematicalsociety inthe area of differential equations.Mohammed K A KaabarCopyright 2015 Mohammed K A Kaabar All Rights Reserved7Table of Laplace Transform 1L{(t)] = _c-st(t)Jtu2L{1] = 1sL{tm] =m!sm+1where m is a positive integer (whole number)3L{sinct] =cs2 + c2L{cos ct] =ss2 +c24L{cbt] =1s -b5L{cbt(t)] = F(s) |s - s -b L-1{F(s) |s - s - b ] = cbt(t)6 L{b(t)u(t -b)] = c-bsL{b(t + b)]L-1{c-bsF(s)] = (t - b)u(t - b)7L{u(t - b)] = c-bssL-1_c-bss _ = u(t -b)8 L|(m)(t)| = smF(s) - sm-1(u) -sm-2i(s) - - (m-1)(s)where m is a positive integer9 L{tm(t)](s) = (-1)m dmP(s)dsmwheremisapositive integerL-1]dmP(s)dsm = (-1)mtm(t)where m is a positive integer10L{(t) - b(t)] = F(s) E(s) (t) - b(t) = _()b(t -)Jt011L __()Jt0_ = F(s)sL-1{F(s) E(s)] = (t) - b(t)12 L{o(t)] = 1 L{o(t - b)] = c-bs13 Assume that (t) is periodic with period P, then:L{(t)] =11 -c-Ps _c-st(t)JtP0Table 1.1.1: Laplace TransformCopyright 2015 Mohammed K A Kaabar All Rights Reserved8 M. KaabarThis page intentionally left blankCopyright 2015 Mohammed K A Kaabar All Rights Reserved9Chapter 1 The Laplace Transform Inthischapter,westartwithanintroductionto DifferentialEquations(DEs)includinglinearDEs, nonlinearDEs,independentvariables,dependent variables,andtheorderofDEs.Then,we definethe laplacetransforms,andwegivesomeexamplesof InitialValueProblems(IVPs).Inaddition,wediscuss theinverselaplacetransforms.Wecoverinthe remaining sections an important concept known as the laplacetransformsofderivatives,andwemention somepropertiesoflaplacetransforms.Finally,we learnhowtosolvesystemsoflinearequations(LEs) using Cramers Rule.1.1 Introductions to Differential EquationsIn this section, we are going to discuss how to determine whether the differential equation is linear or nonlinear, and we will find the order of differential equations. At the end of this section, we will show the purpose of differential equations.Copyright 2015 Mohammed K A Kaabar All Rights Reserved10 M. Kaabarlets start with the definition of differential equation and with a simple example about differential equation.Definition 1.1.1 A mathematical equation is called differential equation if it has two types of variables: dependent and independent variables where the dependent variable can be written in terms of independent variable.Example 1.1.1 Given that yi = 1Sx.a) Find y. (Hint: Find the general solution of yi)b) Determine whether yi = 1Sx is a linear differential equation or nonlinear differential equation. Why?c) What is the order of this differential equation?Solution: Part a: To find y, we need to find the general solution of yiby taking the integral of both sides as follows:_yiJy = _1Sx JxSince ]yiJy = y because the integral of derivative function is the original function itself (In general, ]i(x) Jx = (x)), then y = ]1Sx Jx = 152 x2 + c =7.Sx2 +c. Thus, the general solution is the following:y(x) = 7.Sx2 + c where c is constant. This means that yis called dependent variable because it depends on x, and x is called independent variable because it is independent from y.Part b: To determine whether yi = 1Sx is a linear differential equation or nonlinear differential equation,we need to introduce the following definition:Copyright 2015 Mohammed K A Kaabar All Rights Reserved11Definition 1.1.2 The differential equation is called linear if the dependent variable and all its derivatives are to the power 1. Otherwise, the differential equation is nonlinear.According to the above question, we have the following: y(x) = 7.Sx2 + c where c is constant. Since the dependent variable y and all its derivatives are to the power 1, then using definition 1.1.2, this differential equation is linear.Part c: To find the order of this differential equation, we need to introduce the following definition:Definition 1.1.3 The order of differential equation is the highest derivative in the equation (i.e. The order of yiii + Syii + 2yi = 12x2 + 22 is 3).Using definition 1.1.3, the order of yi = 1Sx is 1.Example 1.1.2 Given that ziii +2zii + zi = 2x3 +22.a) Determine whether ziii + 2zii +zi = 2x3 + 22 is a linear differential equation or nonlinear differential equation. Why?b) What is the order of this differential equation?Solution: Part a: Since z is called dependent variable because it depends on x, and x is called independent variable because it is independent from z, then to determine whether ziii + 2zii + zi = 2x3 +22 is a linear differential equation or nonlinear differential equation, we need to use definition 1.1.2 as follows:Since the dependent variable z and all its derivatives are to the power 1, then this differential equation is linear.Part b: To find the order of ziii + 2zii +zi = 2x3 + 22, we use definition 1.1.3 which implies that the order is 3 because the highest derivative is 3.Copyright 2015 Mohammed K A Kaabar All Rights Reserved12 M. KaabarExample 1.1.3 Given that m(4) + (Smii)3 -m = Vx +1.(Hint: Do not confuse between m(4)and m4because m(4)means the fourth derivative of m, while m4means the fourth power of m).a) Determine whether m(4) +(Smii)3 -m = Vx + 1is a linear differential equation or nonlinear differential equation. Why?b) What is the order of this differential equation?Solution: Part a: Since m is called dependent variable because it depends on x, and x is called independent variable because it is independent from m, then to determine whether m(4) + (Smii)3 -m = Vx + 1 is a linear differential equation or nonlinear differential equation, we need to use definition 1.1.2 as follows: Since the dependent variable m and all its derivatives are not to the power 1, then this differential equation is nonlinear.Part b: To find the order of m(4) +(Smii)3 -m = Vx + 1, we use definition 1.1.3 which implies that the order is 4 because the highest derivative is 4.The following are some useful notations about differential equations:z(m)is the mthderivative of z.zmis the mthpower of z.The following two examples are a summary of this section:Example 1.1.4 Given that2xb(3) + (x +1)b(2) + Sb = x2cx.Determinewhetherit isalineardifferentialequationornonlinear differential equation.Copyright 2015 Mohammed K A Kaabar All Rights Reserved13Solution: To determine whether it is a linear differential equation or nonlinear differential equation, We need to apply what we have learned from the previous examples in the following five steps:Step 1: b is a dependent variable, and x is an independent variable. Step 2: Since b and all its derivatives are to the power 1, then the above differential equation is linear.Step 3: Coefficients of b and all its derivatives are in terms of the independent variable x.Step 4: Assume that C(x) = x2cx. Then, C(x) must be in terms of x.Step 5: Our purpose from the above differentialequation is to find a solution where b can be written in term of x.Thus, the above differential equation is a linear differential equation of order 3.Example 1.1.5 Given that(w2 + 1)b(4) -Swbi = w2 +1. Determine whether it is a lineardifferentialequationornonlineardifferential equation.Solution: To determine whether it is a linear differential equation or nonlinear differential equation, We need to apply what we have learned from the previous examples in the following five steps:Step 1: b is a dependent variable, and w is an independent variable. Step 2: Since b and all its derivatives are to the power 1, then the above differential equation is linear.Step 3: Coefficients of b and all its derivatives are in terms of the independent variable w.Copyright 2015 Mohammed K A Kaabar All Rights Reserved14 M. KaabarStep 4: Assume that C(w) = w2 +1. Then, C(w) must be in terms of w.Step 5: Our purpose from the above differential equation is to find a solution where b can be written in term of w.Thus, the above differential equation is a linear differential equation of order 4.1.2 Introductions to the Laplace TransformsIn this section, we are going to introduce the definition of the laplace transforms in general, and how can we use this definition to find the laplace transform of any function. Then, we will give several examples about the laplace transforms, and we will show how the table 1.1.1 is helpful to find laplace transforms.Definition 1.2.1 the laplace transform, denoted by L, is defined in general as follows: L{(x)] = _(x)c-sxJxuExample 1.2.1 Using definition 1.2.1, find L{1].Solution: To find L{1] using definition 1.2.1, we need to do the following steps:Step 1: We write the general definition of laplacetransform as follows:L{(x)] = _(x)c-sxJxuCopyright 2015 Mohammed K A Kaabar All Rights Reserved15Step 2: Here in this example, (x) = 1 because L{(x)] =L{1].Step 3: L{1] = _(1)c-sxJxuBy the definition of integral, we substitute ] (1)c-sxJx0with limb- ] (1)c-sxJxb0.Step 4: We need to find limb- ] c-sxJxb0as follows:It is easier to find what it is inside the above box [] c-sxJxb0, and after that we can find the limit of ] c-sxJxb0.Thus, ] c-sxJxb0= -1sc-sxx=bx=0 = -1sc-sb + 1sc-s(0) =-1sc-sb + 1s.Step 5: We need find the limit of -1sc-sb +1sas follows:limb-[-1sc-sb +1s = 1s wheie s > u.To check if our answer is right, we need to look at table 1.1.1 at the beginning of this book. According to table 1.1.1 section 2, we found L{1] = 1swhich is the same answer we got. Thus, we can conclude our example with the following fact:Fact 1.2.1 L{ony constont, soy m] = ms.Example 1.2.2 Using definition 1.2.1, find L{c7x].Solution: To find L{c7x] using definition 1.2.1, we need to do the following steps:Step 1: We write the general definition of laplace transform as follows:L{(x)] = _(x)c-sxJxuCopyright 2015 Mohammed K A Kaabar All Rights Reserved16 M. KaabarStep 2: Here in this example, (x) = c7xbecause L{(x)] = L{c7x].Step 3: L{c7x] = _(c7x )c-sxJxuBy the definition of integral, we substitute ] (c7x)c-sxJx0with limb- ] (c7x)c-sxJxb0.Step 4: We need to find limb- ] (c7x)c-sxJxb0as follows:It is easier to find what it is inside the above box [] (c7x)c-sxJxb0, and after that we can find the limit of ] (c7x)c-sxJxb0.Thus, ] (c7x)c-sxJxb0= ] c(7-s)xJxb0=1(7-s)c(7-s)xx=bx=0 =1(7-s)c(7-s)b -1(7-s)c(7-s)(0) =1(7-s)c(7-s)b -1(7-s).Step 5: We need find the limit of 1(7-s)c(7-s)b -1(7-s)as follows:limb-[1(7-s)c(7-s)b -1(7-s) = -1(7-s) =1(s-7) wheie s > 7.To check if our answer is right, we need to look at table 1.1.1 at the beginning of this book. According to table 1.1.1 section 4, we found L{c7x] =1s-7which is the same answer we got.The following examples are two examples about finding the integrals to review some concepts that will help us finding the laplace transforms.Example 1.2.3 Find ]x3c2xJx.Solution: To find ]x3c2xJx, it is easier to use a method known as the table method than using integration by parts. In the table method, we need to create two columns: one for derivatives of x3, and the other one for integrations of c2x. Copyright 2015 Mohammed K A Kaabar All Rights Reserved17Then, we need to keep deriving x3till we get zero, and we stop integrating when the corresponding row is zero. The following table shows the table method to find ]x3c2xJx:Derivatives Part Integration Partx3c2xSx2 12c2x6x 14c2x618c2xu116c2xTable 1.2.1: Table Method for ]x3c2xJxWe always start with positive sign, followed by negative sign, and so on as we can see in the above table 1.2.1.Now, from the above table 1.2.1, we can find ]x3c2xJx as follows: _x3c2xJx = 12x3c2x -14(S)x2c2x + 18(6)xc2x - 116(6)c2x + CThus, ]x3c2xJx = 12x3c2x -34x2c2x +34xc2x -38c2x + C.In conclusion, we can always use the table method to find integrals like ](polynomiol)cuxJx and ](polynomiol)sin(ox)Jx.Example 1.2.4 Find ]Sx2sin(4x)Jx.Solution: To find ]Sx2sin(4x)Jx, it is easier to use the table method than using integration by parts. In the table method, we need to create two columns: one for derivatives of Sx2, and the other one for integrations of sin(4x). Then, we need to keep deriving Sx2till we get zero, and we stop integrating when the corresponding row is zero. The following table shows the table method to find ]Sx2sin(4x)Jx:Copyright 2015 Mohammed K A Kaabar All Rights Reserved18 M. KaabarDerivatives Part Integration PartSx2sin(4x)6x-14cos(4x)6- 116sin(4x)u164cos(4x)Table 1.2.2: Table Method for ]Sx2sin(4x)JxWe always start with positive sign, followed by negative sign, and so on as we can see in the above table 1.2.2.Now, from the above table 1.2.2, we can find ]Sx2sin(4x)Jxas follows: _Sx2sin(4x)Jx= -14(S)x2 cos(4x) -_- 116] 6x sin(4x)+_ 164] 6cos(4x) +CThus, ]Sx2sin(4x)Jx = -34x2 cos(4x) +38x sin(4x) +332cos(4x) +C.Example 1.2.5 Using definition 1.2.1, find L{x2].Solution: To find L{x2] using definition 1.2.1, we need to do the following steps:Step 1: We write the general definition of laplace transform as follows:L{(x)] = _(x)c-sxJxuStep 2: Here in this example, (x) = x2because L{(x)] = L{x2].Step 3: L{x2] = ] (x2 )c-sxJxu Copyright 2015 Mohammed K A Kaabar All Rights Reserved19By the definition of integral, we substitute ] (x2 )c-sxJx0with limb- ] (x2 )c-sxJxb0.Step 4: We need to find limb- ] (x2 )c-sxJxb0as follows:It is easier to find what it is inside the above box [] (x2 )c-sxJxb0, and after that we can find the limit of ] (x2 )c-sxJxb0.Now, we need to find ] (x2 )c-sxJxb0using the table method. In the table method, we need to create two columns: one for derivatives of x2, and the other one for integrations of c-sx. Then, we need to keep deriving x2till we get zero, and we stop integrating when the corresponding row is zero. The following table shows the table method to find ](x2 )c-sxJx:Derivatives Part Integration Partx2c-sx2x-1sc-sx2 1s2 c-sxu- 1s3c-sxTable 1.2.3: Table Method for ](x2 )c-sxJxWe always start with positive sign, followed by negative sign, and so on as we can see in the above table 1.2.3.Now, from the above table 1.2.3, we can find ]Sx2sin(4x)Jxas follows: Thus, ](x2 )c-sxJx = -1sx2c-sx - 2s2xc-sx- 2s3c-sx +C.Now, we need to evaluate the above integral from 0 to b as follows:] (x2 )c-sxJxb0= -1s x2c-sx- 2s2xc-sx- 2sSc-sxx=bx=0 =[-1s b2c-sb- 2s2bc-sb- 2sSc-sb+ 2sS .Copyright 2015 Mohammed K A Kaabar All Rights Reserved20 M. KaabarStep 5: We need find the limit of [-1sb2c-sb -2s2bc-sb- 2s3c-sb + 2s3 as follows:limb-[-1s b2c-sb- 2s2bc-sb- 2sSc-sb+ 2sS = limb-[u + 2sS =2sS wheie s > u.To check if our answer is right, we need to look at table 1.1.1 at the beginning of this book. According to table 1.1.1 section 2 at the right side, we found L{x2] =2!s2+1 =2s3which is the same answer we got.Example 1.2.6 Using table 1.1.1, find L{sin(Sx)].Solution: To find L{sin(Sx)] using table 1.1.1, we need to do the following steps:Step 1: We look at the transform table (table 1.1.1).Step 2: We look at which section in table 1.1.1 contains sin function.Step 3: We write down what we get from table 1.1.1(Section 3 at the left side) as follows: L{sinct] =cs2 + c2Step 4: We change what we got from step 3 to make it look like L{sin(Sx)] as follows:L{sinSx] =5s2+52 =5s2+25.Thus, L{sinSx] =5s2+52 =5s2+25.Example 1.2.7 Using table 1.1.1, find L{cos(-8x)].Solution: To find L{cos(-8x)] using table 1.1.1, we need to do the following steps:Step 1: We look at the transform table (table 1.1.1).Step 2: We look at which section in table 1.1.1 contains cos function.Copyright 2015 Mohammed K A Kaabar All Rights Reserved21Step 3: We write down what we get from table 1.1.1 (Section 3 at the right side) as follows: L{cos ct] =ss2 +c2Step 4: We change what we got from step 3 to make it look like L{cos(-8x)] as follows:L{cos(-8x)] =ss2+(-8)2 =ss2+64.Thus, L{cos(-8x)] =ss2+(-8)2 =ss2+64.We will give some important mathematical results about laplace transforms.Result 1.2.1 Assume that c is a constant, and (x), g(x) are functions. Then, we have the following:(Hint: F(s) anu 0(s) are the laplace transforms of(x) andg(x), respectively).a) L{g(x)] = 0(s) (i.e. L{8] = 8s = 0(s) where g(x) = 8).b) L|c g(x)| = c L{g(x)] = c 0(s).c) L{(x) +g(x)] = F(s) + 0(s).d) L{(x) g(x)] is not necessary equal to F(s) 0(s).Example 1.2.8 Using definition 1.2.1, find L{yi].Solution: To find L{yi] using definition 1.2.1, we need to do the following steps:Step 1: We write the general definition of laplace transform as follows:L{(x)] = _(x)c-sxJxuStep 2: Here in this example, (x) = yibecause L{(x)] = L{yi].Step 3: L{yi] = ] (yi)c-sxJxu By the definition of integral, we substitute ] (y')c-sxJx0with limb- ] (y')c-sxJxb0.Copyright 2015 Mohammed K A Kaabar All Rights Reserved22 M. KaabarStep 4: We need to find limb- ] y'c-sxJxb0as follows:It is easier to find what it is inside the above box [] y'c-sxJxb0, and after that we can find the limit of ] y'c-sxJxb0.To find ] y'c-sxJxb0, we need to use integration by parts as follows:u = c-sxJ: = yiJxJu = -sc-sxJx : = ]J: = ]yiJx = y_uJ: = u: - _:Ju_yic-sxJx = yc-sx - _y(-sc-sx) JxNow, we can find the limit as follows:limb-_y'c-sxJxb0= limb-yc-sx - limb-_y(-sc-sx)buJxlimb-_y'c-sxJxb0= limb-yc-sx|x = bx = u+ _y(sc-sx)uJxBecause ] y(sc-sx)0Jx = limb-] y(-sc-sx)b0Jx.limb-_y'c-sxJxb0= limb-(yc-sb - yc-s(u)) + _y(sc-sx)uJxlimb-_y'c-sxJxb0= limb-(y(b)c-sb- y(u)c-s(u)) + s _yc-sxuJxSince we have]yc-sx0Jx, then using result 1.2.1 (] yc-sx0Jx = (s)), we obtain the following:limb-_y'c-sxJxb0= limb-(y(b)c-sb - y(u)) + s _yc-sxuJxCopyright 2015 Mohammed K A Kaabar All Rights Reserved23_y'c-sxJxb0= limb-(y(b)c-sb - y(u)) + s(s)_y'c-sxJxb0= (y(b)c-s() - y(u)) + s(s)_y'c-sxJxb0= (u - y(u)) + s(s)_y'c-sxJxb0= s(s) - y(u)Thus, L{yi] = s(s) -y(u).We conclude this example with the following results:Result 1.2.2 Assume that (x) is a function, and F(s) is the laplace transform of (x). Then, we have the following:a) L{i(x)] = L|(1)(x)| = sF(s) -(u).b) L{ii(x)] = L|(2)(x)| = s2F(s) - s(u) -i(u).c) L{iii(x)] = L|(3)(x)| = s3F(s) - s2(u) - si(u) -ii(u).d) L|(4)(x)| = s4F(s) -s3(u) - s2i(u) -sii(u) -iii(u).For more information about this result, it is recommended to look at section 8 in table 1.1.1. If you look at it, you will find the following:L|(m)(t)| = smF(s) - sm-1(u) -sm-2i(s) - - (m-1)(s).Result 1.2.3 Assume that c is a constant, and (x) is a function where F(s) is the laplace transform of (x). Then, we have the following: L{c(x)] = cL{(x)] = cF(s).Copyright 2015 Mohammed K A Kaabar All Rights Reserved24 M. Kaabar1.3 Inverse Laplace TransformsIn this section, we will discuss how to find the inverse laplace transforms of different types of mathematical functions, and we will use table 1.1.1 to refer to the laplace transforms.Definition 1.3.1 The inverse laplace transform, denoted by L-1, is defined as a reverse laplace transform, and to find the inverse laplace transform, we need to think about which function has a laplace transform that equals to the function in the inverse laplace transform. For example, suppose that (x) is a function where F(s) is the laplace transform of (x). Then, the inverse laplace transform is L-1{F(s)] = (x). (i.e. L-1]1s we need to think which function has a laplace transform that equals to 1s , in this case the answer is 1).Example 1.3.1 Find L-1]34s .Solution: First of all, L-1]34s = L-1]S4[1s.Using definition 1.3.1 and table 1.1.1, the answer is 34. Example 1.3.2 Find L-1]ss2+2.Solution: Using definition 1.3.1 and the right side of section 3 in table 1.1.1, we find the following: L{cos ct] =ss2 +c2 ss2+2is an equivalent toss2+(V2)2Copyright 2015 Mohammed K A Kaabar All Rights Reserved25Since L|cos V2t| =ss2+(V2)2, then this means that L-1]ss2+2 = cos V2t. Example 1.3.3 Find L-1] -3s2+9.Solution: Using definition 1.3.1 and the left side of section 3 in table 1.1.1, we find the following: L{sinct] =cs2 + c2 -3s2+9is an equivalent to -3s2+(-3)2Since L{sin(-S)t] = L{sin(-St)] =-3s2+(-3)2, then this means that L-1] -3s2+9 = sin(-St). Example 1.3.4 Find L-1] 1s+8.Solution: Using definition 1.3.1 and section 4 in table 1.1.1, we find the following: L{cbt] =1s-b. 1s+8is an equivalent to 1s-(-8)Since L{c-8t] =1s+8, then this means that L-1] 1s+8 =c-8t. Result 1.3.1 Assume that c is a constant, and (x) is a function where F(s) anu 0(s) are the laplace transforms of (x), anu g(x), iespectively. a) L-1{cF(s)] = cL-1{F(s)] = c(x).b) L-1{F(s) +0(s)] = L-1{F(s)] + L-1{0(s)] = (x) +g(x).Example 1.3.5 Find L-1]5s2+4.Solution: Using result 1.3.1, L-1]5s2+4 = SL-1]1s2+4. Now, by using the left side of section 3 in table 1.1.1, we need to make it look like cs2+c2because L{sinct] =cs2+c2. Copyright 2015 Mohammed K A Kaabar All Rights Reserved26 M. KaabarTherefore, we do the following: 52L-1](2)1s2+4 = 52L-1]2s2+4, and 2s2+4is an equivalent to 2s2+(2)2. Since 52L{sin(2)t] = L ]52sin(2t) = 52[2s2+(2)2 =5s2+4, then by using definition 1.3.1, this means that L-1]5s2+4 =52sin(2t). Example 1.3.6 Find L-1]72s-3.Solution: Using result 1.3.1, L-1]72s-3 = 7L-1]12s-3. Now, by using section 4 in table 1.1.1, we need to make it look like 1s-bbecause L{cbt] =1s-b. Therefore, we do the following: 7L-1]12s-3 = 7L-1_12[s-32_ =72L-1_1[s-32_.Since 72L ]c32t = L ]72c32t = 72_1[s-32_ =72s-3, then by using definition 1.3.1, this means that L-1]72s-3 = 72c32t. Example 1.3.7 Find L-1]1+3ss2+9.Solution: L-1]1+3ss2+9 = L-1]1s2+9 +3ss2+9. Now, by usingsection 3 in table 1.1.1, we need to make it look like cs2+c2and ss2+c2because L{sinct] =cs2+c2 anu L{cos ct] =ss2+c2. Therefore, we do the following:L-1]1s2+9 +3ss2+9 = 13L-1]1(3)s2+9 +SL-1]ss2+9.Since 13L{sin(St)] +SL{cos(St)] = [1s2+9 +3ss2+9 , then by using definition 1.3.1, this means that L-1]1+3ss2+9 =L-1]1s2+9 +L-1] 3ss2+9 = 13sin(St) +Scos(St). Example 1.3.8 Find L-1]6s4.Copyright 2015 Mohammed K A Kaabar All Rights Reserved27Solution: Using definition 1.3.1 and the right side of section 2 in table 1.1.1, we find the following: L{tm] =m!sm+1where m is a positive integer. 6s4is an equivalent to 3!s3+1Since L{t3] =3!s3+1 = 6s4, then this means that L-1]6s4 = t3. 1.4 Initial Value ProblemsInthissection,wewillintroducethemaintheoremof differential equations known as Initial Value Problems (IVP), andwewilluseit withwhatwehavelearned from sections 1.2 and 1.3 to find the largest interval on the x-axis.Definition1.4.1Givenon(x)y(n) + on-1(x)y(n-1) + +o0(x)y(x) = K(x).Assumethaton(m) = u forevery m e I whereI issomeinterval,and on(x), on-1(x), ., o0(x), K(x) arecontinuousonI.Suppose thaty(w), yi(w), ., y(n-1)(w) forsomew e I. Then,thesolutiontothedifferentialequationsis unique which means that there exists exactly oney(x)intermsofx,andthistypeofmathematicalproblems is called Initial Value Problems (IVP).Example 1.4.1 Find the largest interval on the x -oxisso that x-3x+2y(3)(x) +2y(2)(x) + Vx +1yi(x) = Sx +7 has a solution. Given yi(S) = 1u, y(S) = 2, y(2)(S) = -S.Copyright 2015 Mohammed K A Kaabar All Rights Reserved28 M. KaabarSolution: Finding largest interval on the x -oxismeans that we need to find the domain for the solution of the above differential equation in other words we need to find for what values of x the solution of the above differential equation holds. Therefore, we do the following:x-3x+2y(3)(x) +2y(2)(x) +Vx + 1yi(x) = Sx + 7Using definition 1.4.1, we also suppose the following:o3(x) = x - Sx + 2o2(x) = 2o1(x) = Vx + 1K(x) = Sx + 7Now,weneedtodeterminetheintervalofeach coefficient above as follows:o3(x) = x-3x+2hasasolution whichiscontinuouseverywhere (+) except x = -2 anu x = S.o2(x) = 2 hasasolution whichiscontinuouseverywhere (+).o1(x) = Vx + 1 hasasolutionwhichiscontinuouson the interval |-1, ).K(x) = Sx + 7 hasasolutionwhichiscontinuous everywhere (+).Thus, the largest interval on the x - oxis is (S, ).Example 1.4.2 Find the largest interval on the x -oxisso that (x2 +2x - S)y(2)(x) +1x+3y(x) = 1u has a solution. Given yi(2) = 1u, y(2) = 4.Copyright 2015 Mohammed K A Kaabar All Rights Reserved29Solution: Finding largest interval on the x -oxismeans that we need to find the domain for the solution of the above differential equation in other words we need to find for what values of x the solution of the above differential equation holds. Therefore, we do the following:(x2 +2x - S)y(2)(x) +1x+3y(x) = 1uUsing definition 1.4.1, we also suppose the following:o2(x) = (x2 +2x - S)o1(x) = uo0(x) =1x + SK(x) = 1uNow,weneedtodeterminetheintervalofeach coefficient above as follows:o2(x) = (x2 +2x - S) = (x - 1)(x + S) hasasolution whichiscontinuouseverywhere(+)exceptx = 1 and x = -S.o0(x) =1x+3hasasolutionwhichiscontinuous everywhere (+) except x = -S.K(x) = 1u hasasolutionwhichiscontinuous everywhere (+).Thus, the largest interval on the x - oxis is (1, ).Example 1.4.3 Solve the following Initial Value Problem (IVP): yi(x) + Sy(x) = u. Given y(u) = S.Solution: yi(x) + Sy(x) = u is a linear differential equation of order 1. First, we need to find the domain for the solution of the above differential equation in Copyright 2015 Mohammed K A Kaabar All Rights Reserved30 M. Kaabarother words we need to find for what values of x the solution of the above differential equation holds. Therefore, we do the following:1yi(x) +Sy(x) = uUsing definition 1.4.1, we also suppose the following:o1(x) = 1o0(x) = SK(x) = uThe domain of solution is (-, ).Now,tofindthesolutionoftheabovedifferential equation,weneedtotakethelaplacetransformfor both sides as followsL{yi(x)] +L{Sy(x)] = L{u]L{yi(x)] +SL{y(x)] = u because (L{u] = u).(s(s) -y(u)) +S(s) = u from result 1.2.2.s(s) - y(u) +S(s) = u(s)(s +S) = y(u)Wesubstitutey(u) = S becauseitisgiveninthe question itself.(s)(s +S) = S(s) =3(s+3)Tofind asolution,weneedtofindtheinverselaplace transform as follows:L-1{(s)] = L-1]S(s+S) = SL-1]1(s+S) andweusetable 1.1.1 section 4.Copyright 2015 Mohammed K A Kaabar All Rights Reserved31y(x) = Sc-3x(It iswrittenintermsofx insteadoftbecause we need it in terms of x).Then,wewillfindyi(x) byfindingthederivativeof what we got above (y(x) = Sc-3x) as follows:yi(x) = (Sc-3x)i = (S)(-S)c-3x = -9c-3x.Finally,tocheckoursolutionifitisright,we substitutewhatwegotfromy(x) = Sc-3xandyi(x) =-9c-3xin yi(x) +Sy(x) = u as follows:-9c-3x +S(Sc-3x) = -9c-3x +(9c-3x) = uThus, our solution is correct which is y(x) = Sc-3xand yi(x) = -9c-3x.Example 1.4.4 Solve the following Initial Value Problem (IVP): y(2)(x) + Sy(x) = u. Given y(u) = u, and yi(u) = 1.Solution: y(2)(x) +Sy(x) = u is a linear differential equation of order 2. First, we need to find the domain for the solution of the above differential equation in other words we need to find for what values of x the solution of the above differential equation holds. Therefore, we do the following:1y(2)(x) +Sy(x) = uUsing definition 1.4.1, we also suppose the following:o2(x) = 1o1(x) = uo0(x) = SK(x) = uThe domain of solution is (-, ).Copyright 2015 Mohammed K A Kaabar All Rights Reserved32 M. KaabarNow,tofindthesolutionoftheabovedifferential equation,weneedtotakethelaplacetransformfor both sides as followsL{y(2)(x)] + L{Sy(x)] = L{u]L{y(2)(x)] + SL{y(x)] = u because (L{u] = u).(s2(s) -sy(u) -yi(u)) + S(s) = u from result 1.2.2.s2(s) - sy(u) -yi(u) +S(s) = u(s)(s2 +S) = sy(u) + yi(u)Wesubstitutey(u) = u,andyi(u) = 1 becauseitis given in the question itself.(s)(s2 +S) = (s)(u) +1(s)(s2 +S) = u +1(s)(s2 +S) = 1(s) =1(s2 +S)Tofind asolution,weneedtofindtheinverselaplace transform as follows:L-1{(s)] = L-1_1(s2+S)_ = 1VSL-1_1_s2+(VS)2]_ andweuse table 1.1.1 at the left side of section 3.y(x) =1V3sin(VS x) (It is written in terms of x instead of t because we need it in terms of x).Then,wewillfindyi(x) byfindingthederivativeof what we got above [y(x) = 1V3sin(VS x) as follows:Copyright 2015 Mohammed K A Kaabar All Rights Reserved33yi(x) = _ 1VSsin(VSx)]i= 1VS(VS) cos(VS x) = cos(VS x)Now,wewillfindy(2)(x) byfindingthederivativeof what we got above as follows:y(2)(x) = (cos(VS x))i= -VS sin(VSx)Finally,tocheckoursolutionifitisright,we substitute what we got from y(x) and y(2)(x) in y(2)(x) + Sy(x) = u as follows:-VS sin(VSx) + VS sin(VS x) = uThus, our solution is correct which is y(x) =1V3sin(VS x) and y(2)(x) = -VS sin(VSx).1.5 Properties of Laplace TransformsIn this section, we discuss several properties of laplace transforms such as shifting, unit step function, periodic function, and convolution. We start with some examples of shifting property.Example 1.5.1 Find L{c3xx3].Solution: By using shifting property at the left side of section 5 in table 1.1.1, we obtain:L{cbx(x)] = F(s) |s - s -bLet b = S, and (x) = xS.F(s) = L{x3].Copyright 2015 Mohammed K A Kaabar All Rights Reserved34 M. KaabarHence, L{c3xx3] = L{x3] |s - s -S =3!(s)3+1 |s - s -S =3!(s)4 |s - s -S.Now, we need to substitute s with s -S as follows:S!(s - S)4 =6(s - S)4Thus, L{c3xx3] =3!(s-3)4 =6(s-3)4.Example 1.5.2 Find L{c-2xsin(4x)].Solution: By using shifting property at the left side of section 5 in table 1.1.1, we obtain:L{cbx(x)] = F(s) |s - s -bLet b = -2, and (x) = sin(4x).F(s) = L{sin(4x)].Hence, L{c-2xsin(4x)] = L{sin(4x)] |s - s + 2 =4s2 +16 |s - s + 2Now, we need to substitute s with s +2 as follows:4(s + 2)2 +16Thus, L{c-2xsin(4x)] =4(s+2)2+16.Example 1.5.3 Find L-1]s(s-2)2+4.Solution: Since we have a shift such as s -2, we need to do the following:L-1_s(s -2)2 +4_ = L-1_ s -2 +2(s -2)2 + 4_= L-1_s - 2(s -2)2 +4 +2(s -2)2 +4_Copyright 2015 Mohammed K A Kaabar All Rights Reserved35L-1_s(s - 2)2 + 4_ = L-1_s - 2(s -2)2 + 4_ + L-1_2(s - 2)2 + 4_By using shifting property at the right side of section 5 in table 1.1.1, we obtain:L-1{F(s) |s - s -b ] = cbx(x)Let b = 2, F1(s) = L-1]s-2(s-2)2+4, and F2(s) = L-1]2(s-2)2+4.L-1]s(s-2)2+4 = L-1{F1(s) |s - s - 2 ] + L-1{F2(s) |s - s - 2 ]Thus, L-1]s(s-2)2+4 = c2x cos(2x) +c2x sin(2x).Example 1.5.4 Find L-1]s(s+2)3.Solution: Since we have a shift such as s +2, we need to do the following:L-1_s(s +2)3 _ = L-1_s +2 -2(s + 2)3 _ = L-1_ s + 2(s +2)3 -2(s +2)3_L-1_s(s + 2)S _ = L-1_ s + 2(s + 2)S _ +L-1_-2(s + 2)S_L-1_s(s + 2)S _ = L-1_1(s + 2)2 _ +L-1_-2(s + 2)S_By using shifting property at the right side of section 5 in table 1.1.1, we obtain:L-1{F(s) |s - s -b ] = cbx(x)Let b = -2, F1(s) = L-1]1(s+2)2, and F2(s) = L-1]-2(s+2)3.L-1]s(s+2)3 = L-1{F1(s) |s - s +2 ] + L-1{F2(s) |s - s +2 ]Thus, L-1]s(s+2)3 = c-2xx -c-2xx2.Example 1.5.5 Find L-1]1s2-4.Copyright 2015 Mohammed K A Kaabar All Rights Reserved36 M. KaabarSolution: L-1]1s2-4 = L-1]1(s-2)(s+2). Since the numerator has a polynomial of degree 0 (x0 = 1), and the denominator a polynomial of degree 2, then this means the degree of numerator is less than the degree of denominator. Thus, in this case, we need to use the partial fraction as follows:1(s - 2)(s + 2) =o(s - 2) +b(s + 2)It is easier to use a method known as cover method than using the traditional method that takes long time to finish it. In the cover method, we cover the original, say (s -2), and substitute s = 2 in 1(s-2)(s+2)to find the value of o. Then, we cover the original, say (s + 2), and substitute s = -2 in 1(s-2)(s+2)to find the value of b.Thus, o = 14and b = -14. This implies that 1(s - 2)(s + 2) =14(s -2) +-14(s + 2)Now, we need to do the following:L-1_1s2 -4 _ = L-1_14(s - 2) +-14(s +2) _L-1_1s2 - 4_ = L-1_14(s - 2) _ +L-1_-14(s + 2)_L-1_1s2 - 4_ = 14L-1_1(s - 2)_ - 14L-1_1(s + 2)_Copyright 2015 Mohammed K A Kaabar All Rights Reserved37Thus, L-1]1s2-4 = 14c2x - 14c-2x.Now, we will introduce a new property from table 1.1.1 in the following two examples.Example 1.5.6 Find L{xcx].Solution: By using the left side of section 9 in table 1.1.1, we obtain:L{tm(t)](s) = (-1)mJmF(s)Jsm= (-1)mF(m)(s)where L{(x)] = F(s), and (x) = cx.Hence, L{xcx] = (-1)1F(1)(s) = -F(1)(s)Now, we need to find F(1)(s) as follows: This means that we first need to find the laplace transform of (x), and then we need to find the first derivative of the result from the laplace transform.F(1)(s) = Fi(s) = (L{(x)])i = (L{cx])i = [ 1s-1i= -1(s-1)2Thus, L{xcx] = (-1)1F(1)(s) = -F(1)(s) = -[-1(s-1)2 =1(s-1)2.Example 1.5.7 Find L{x2sin(x)].Solution: By using the left side of section 9 in table 1.1.1, we obtain:L{tm(t)](s) = (-1)mJmF(s)Jsm= (-1)mF(m)(s)where L{(x)] = F(s), and (x) = sin(x).Hence, L{x2sin(x)] = (-1)2F(2)(s) = F(2)(s)Copyright 2015 Mohammed K A Kaabar All Rights Reserved38 M. KaabarNow, we need to find F(2)(s) as follows: This means that we first need to find the laplace transform of (x), and then we need to find the second derivative of the result from the laplace transform.F(2)(s) = Fii(s) = (L{(x)])ii = (L{sin(x)])ii = [1s2+1ii=[-2s(s2+1)2i= -2(s2+1)2+8s2(s2+1)(s2+1)4Thus, L{x2sin(x)] = -2(s2+1)2+8s2(s2+1)(s2+1)4.Example 1.5.8 Solve the following Initial Value Problem (IVP): y(2)(x) + Sy(1)(x) +6y(x) = 1. Given y(u) = yi(u) = u.Solution: y(2)(x) +Sy(1)(x) +6y(x) = 1 is a linear differential equation of order 2. First, we need to find the domain for the solution of the above differential equation in other words we need to find for what values of x the solution of the above differential equation holds. Therefore, we do the following:1y(2)(x) +Sy(1)(x) + 6y(x) = 1Using definition 1.4.1, we also suppose the following:o2(x) = 1o1(x) = So0(x) = 6K(x) = 1The domain of solution is (-, ).Now,tofindthesolutionoftheabovedifferential equation,weneedtotakethelaplacetransformfor both sides as followsCopyright 2015 Mohammed K A Kaabar All Rights Reserved39L{y(2)(x)] + L{Sy(1)(x)] +L{6y(x)] = L{1]L{y(2)(x)] + SL{y(1)(x)] +6L{y(x)] = 1sbecause [L{1] = 1s.(s2(s) -sy(u) -yi(u)) + S(s(s) - y(u)) +6(s) = 1sfrom result 1.2.2. We substitute y(u) = u, and yi(u) = ubecause it is given in the question itself.(s2(s) -u - u) + S(s(s) - u) + 6(s) = 1ss2(s) + Ss(s) + 6(s) = 1s(s)(s2 +Ss + 6) = 1s(s) =1s(s2 + Ss +6) =1s(s +S)(s +2)Tofind asolution,weneedtofindtheinverselaplace transform as follows:y(x) = L-1{(s)] = L-1]1s(s+S)(s+2)Since the numerator has a polynomial of degree 0 (x0 = 1), and the denominator a polynomial of degree 3, then this means the degree of numerator is less than the degree of denominator. Thus, in this case, we need to use the partial fraction as follows:1s(s + S)(s +2) = os +b(s + S) +c(s + 2)Now, we use the cover method. In the cover method, we cover the original, say s, and substitute s = u in 1s(s+3)(s+2)to find the value of o. We cover the original, Copyright 2015 Mohammed K A Kaabar All Rights Reserved40 M. Kaabarsay (s +S), and substitute s = -S in 1s(s+3)(s+2)to find the value of b. Then, we cover the original, say (s + 2), and substitute s = -2 in 1s(s+3)(s+2)to find the value of c. Thus, = 16, b = 13and c = -12. This implies that 1s(s + S)(s +2) = 16s +1S(s + S) +-12(s + 2)Now, we need to do the following:L-1_1s(s +S)(s +2)_ = L-1_16s +1S(s + S) +-12(s +2)_L-1_1s(s + S)(s + 2)_ = 16L-1_1s_ +1SL-1_1(s + S)_-12L-1_1(s + 2)_Thus, y(x) = L-1]1s(s+3)(s+2) = 16+ 13c-3x -12c-2x.Definition 1.5.1 Given o > u. Unit Step Function is defined as follows: u(x - o) = _u i u x < o1 i o x < Result 1.5.1 Given o u. Then, we obtain:a) u(x - u) = u(x) = 1 for every u x .b) u(x - ) = u for every u x .Example 1.5.9 Find u(x -7)|x = 8.Solution: Since x = 8 is between o = 7 and , then by using definition 1.5.1, we obtain:u(x - 7)|x = 8 = u(8 - 7) = 1.Copyright 2015 Mohammed K A Kaabar All Rights Reserved41Example 1.5.10 Find u(x - S)|x = 1.Solution: Since x = 1 is between 0 and = S , then by using definition 1.5.1, we obtain:u(x - S)|x = 1 = u(1 - 7) = u.Example 1.5.11 Find L{u(x -S)].Solution: By using the left side of section 7 in table 1.1.1, we obtain:L{u(t - b)] = c-bssThus, L{u(x -S)] = c-3ss.Example 1.5.12 Find L-1]c-2ss .Solution: By using the right side of section 7 in table 1.1.1, we obtain:L-1_c-bss _ = u(t -b)Thus, L-1]c-2ss = u(x -2) = _u i u x < 21 i 2 x < Example 1.5.13 Given (x) = _S i 1 x < 4-x i 4 x < 1u(x + 1) i 1u x < Rewrite (x) in terms of unit Stcp Function.Solution: To re-write (x) in terms of unit Stcp Function, we do the following:(x) = S(u(x - 1) - u(x - 4))+(-x)(u(x - 4) - u(x - 1u))+(x + 1)(u(x -1u) - u).Copyright 2015 Mohammed K A Kaabar All Rights Reserved42 M. KaabarNow, we need to check our unit step functions as follows: We choose x = 9.(9) = S(u(9 - 1) - u(9 -4))+(-9)(u(9 - 4) - u(9 -1u))+(9 +1)(u(9 -1u) - u)(9) = S(1 -1) +(-9)(1 - u) + (1u)(u -u)(9) = S(u) + (-9)(1) +(1u)(u)(9) = u + (-9)(1) + u = -9.Thus, our unit step functions are correct.Example 1.5.14 Find L{xu(x - 2)].Solution: By using the upper side of section 6 in table 1.1.1, we obtain:L{b(x)u(x -b)] = c-bsL{b(x +b)]where b = 2, and b(x) = x.Hence, L{xu(x - 2)] = c-2sL{b(x + 2)] = c-2sL{x + 2] =c-2s [1s2 +2s.Definition 1.5.2 Convolution, denoted by -, is defined as follows: (x) - b(x) = _()b(x - )Jx0where (x) and b(x) are functions. (Note: do not confuse between multiplication and convolution).Result 1.5.2 L{(x) - b(x)] = L{b(x) - (x)] =F(s) E(s) where (x) and b(x) are functions. (The proof of this result left as an exercise 16 in section 1.7). Copyright 2015 Mohammed K A Kaabar All Rights Reserved43Result 1.5.3 L{(x) b(x)] = L{b(x)] L{(x)].Example 1.5.15 Use definition 1.5.2 to find L|] sin()Jx0|.Solution: By using definition 1.5.2and section 10 in table 1.1.1, we obtain:L __sin()Jx0_ = L{1 - sin(x)] = L{1] L{sin(x)] = 1s 1(s2 + 1)=1s(s2 +1)Thus, L|] sin()Jx0| =1s(s2+1).Definition 1.5.3 (x) is a periodic function on |u, ) if (x) has a period P such that (b) = (b -P) for every b P.Example 1.5.16 Given (x) is periodic on |u, ) such that the first period of (x) is given by the following piece-wise continuous function:_ S i u x < 2-2 i 2 x < 8a) Find the 8thperiod of this function.b) Suppose P = 8. Find (1u).c) Suppose P = 8. Find (Su).Solution: Part a: By using section 13 in table 1.1.1, we obtain:L{(x)] =11 - c-Ps _c-st(x)JxP0Since we need find the 8thperiod, then this means that P = 8, and we can apply what we got above as follows: Copyright 2015 Mohammed K A Kaabar All Rights Reserved44 M. KaabarL{(x)] =11 -c-8s_c-st(x)Jx80=11 -c-8s_(x)c-stJx80Using the given first period function, we obtain:L{(x)] =11 -c-8s_(x)c-stJx80=11 -c-8s _S_c-stJx -2_c-stJx8220_=11 -c-8s_-Ss c-st_x = 2x = u+ 2s c-st_x = 8x = 2_=11 -c-8s _-Ss (c-2t -1) +2s (c-8t -c-2t)_Part b: By using definition 1.5.3, we obtain:(1u) = (1u - 8) = (2) = -2 from the given first period function.Part c: By using definition 1.5.3, we obtain:(Su) = (Su - 8) = (22).(22) = (22 - 8) = (14).(14) = (14 - 8) = (6) = -2 from the given first period function.Definition 1.5.4 Suppose that i > u is fixed, and o < ] < i is chosen arbitrary. Then, we obtain:o](t -i) =`11u i u t < (i - ])12]i (i - ]) t < (i + ])u t (i +])o](t -i) is called Dirac Delta Function.Result 1.5.4 o(t -i) = lim-0+ o](t - i).Copyright 2015 Mohammed K A Kaabar All Rights Reserved45Example 1.5.17 Find L{o(t +12)].Solution: By using the right side of section 12 in table 1.1.1, we obtain: L{o(t -b)] = c-bsThus, L{o(t +12)] = c12s.1.6 Systems of Linear EquationsMost of the materials of this section are taken fromsection 1.8 in my published book titled A First Course in Linear Algebra: Study Guide for the Undergraduate Linear Algebra Course, First Edition1, because it is very important to give a review from linear algebra about Cramers rule, and how some concepts of linear algebra can be used to solve some problems in differential equations. In this section, we discuss how to use what we have learned from previous sectionssuch as initial value problems (IVP), and how to use Cramers rule to solve systems of linear equations.Definition 1.6.1 Given n n system of linear equations. Let WX = A be the matrix form of the given system:Wlllllx1x2x3.xn11111=lllllo1o2o3.on11111Copyright 2015 Mohammed K A Kaabar All Rights Reserved46 M. KaabarThe system has a unique solution if and only if uet(W) = u. Cramers Rule tells us how to find x1, x2 , ., xnas follows:Let W= _1 S 41 2 17 4 S_ Then, the solutions for the system of linear equations are:x1 = uet _o1S 4. 2 1on4 S_uet(W)x2 = uet _1 o141 . 17 onS_uet(W)x3 = uet _1 S o11 2 .7 4 on_uet(W)Example 1.6.1 Solve the following system of linear equations using Cramers Rule:_2x1 + 7x2 = 1S-1ux1 +Sx2 = -4 Solution: First of all, we write 2 2 system in the form WX = A according to definition 1.6.1.j 2 7-1u S[ jx1x2[ = j1S-4[Since Win this form is j 2 7-1u S[, thenuet(W) = (2 S) - (7 (-1u)) = 6 - (-7u) = 76 = u.The solutions for this system of linear equations are:Copyright 2015 Mohammed K A Kaabar All Rights Reserved47x1 = uet j1S 7-4 S[uet(W)= uet j1S 7-4 S[76= 6776x2 = uet j 2 1S-1u -4[uet(W)= uet j 2 1S-1u -4[76= 12276Thus, the solutions arex1 = 6776 anu x2 = 12276 .Example 1.6.2 Solve for n(t) and m(t):_nii(t) -4m(t) = u ......Equotion 1n(t) +2mi(t) = Sc2t .... . .Equotion 2 Given that n(u) = 1, ni(u) = 2, anu m(u) = 1.Solution: First, we need to take the laplace transform of both sides for each of the above two equations.For Equotion 1: We take the laplace transform of both sides:L{n''(t)] +L{-4m(t)] = L{u]L{n''(t)] -4L{m(t)] = L{u](s2N(s) - sn(u) -ni(u)) -4H(s) = uNow,wesubstitutewhatisgiveninthisquestionto obtain the following:(s2N(s) - s - 2) - 4H(s) = uThus, s2N(s) -4H(s) = s +2.For Equotion 2: We take the laplace transform of both sides:L{n(t)] +L{2m'(t)] = L{Sc2t]L{n(t)] +2L{m'(t)] = SL{c2t]N(s) +2(sH(s) -m(u)) =Ss - 2Copyright 2015 Mohammed K A Kaabar All Rights Reserved48 M. KaabarNow,wesubstitutewhatisgiveninthisquestionto obtain the following:N(s) +2(sH(s) - 1) =Ss - 2N(s) +2sH(s) =Ss -2+2 = 2s + 1s - 2Thus, N(s) + 2sH(s) = 2s+1s-2. From what we got from Equation 1 and Equation 2, we need to find N(s) and H(s) as follows:_s2N(s) -4H(s) = s +2N(s) + 2sH(s) = 2s +1s - 2Now, we use Cramers rule as follows:N(s) = uet _ s + 2 -42s + 1s - 22s_uet js2-41 2s[= 2s2 + 4s1+8s + 4s - 22s3 +4= (s - 2)(2s2 +4s) +8s + 4(s -2)(2s3 + 4)= 2s3 + 4s2 - 4s2 -8s + 8s + 4(s -2)(2s3 + 4)=1s -2Hence, n(t) = L-1{N(s)] = L-1] 1s-2 = c2t.Further, we can use one of the given equations to find m(t) as follows: nii(t) - 4m(t) = uWe need to find the second derivative of n(t).ni(t) = 2c2t.nii(t) = 4c2t.Now, we can find m(t) as follows:Copyright 2015 Mohammed K A Kaabar All Rights Reserved49nii(t) - 4m(t) = u - m(t) = nii(t)4= 4c2t4= c2t.Thus, m(t) = c2t.1.7 Exercises 1.Find L-1]10(s-4)4.2.Find L-1]s+5(s-1)2+16.3.Find L-1] s+5(s+3)4.4.Find L-1]53s-10.5.Find L-1]2s2-6s+13.6.Find L-1]5s2-7s-8.7.Solve the following Initial Value Problem (IVP): 2yi(x) +6y(x) = u. Given y(u) = -4.8.Solve the following Initial Value Problem (IVP): yii(x) + 4y(x) = u. Given y(u) = 2, and yi(u) = u.9.Find L-1]4(s-1)2(s+3).10. Find L{u[x -n2 sin(x)].11. Find L{u(x -2)cSx].12. Given (x) = _ 4 i u x < Sc2xi S x < Rewrite (x) in terms of unit Stcp Function.13. Find L-1]sc-4xs2+4.Copyright 2015 Mohammed K A Kaabar All Rights Reserved50 M. Kaabar14. Solve the following Initial Value Problem (IVP): yii(x)+Syi(x) - 4y(x) = (u). Given (x) = ]1 i u x < Su 0tbcrwiscy(u) = yi(u) = u.15. Solve the following Initial Value Problem (IVP): yii(x)+7yi(x) - 8y(x) = (x) where (x) = _ S i u x < S-2 i S x < y(u) = yi(u) = u.16. Prove result 1.5.2.17. Solve the following Initial Value Problem (IVP): y(3)(x)+yi(x) = u(x -S). Given y(u) = yi(u) = yii(u) =u.18. Find L|] cos(2)c(Sx+2)Jx0|.19. Find L-1]2s(s2+4)2.20. Solve for r(t) and k(t):_ri(t) -2k(t) = u ...... .. Equotion 1rii(t) - 2ki(t) = u ...... .Equotion 2 Given that k(u) = 1, ri(u) = 2, anu r(u) = u.21. Solve for w(t) and b(t):_w(t) - _b()Jtu= 1 ...... Equotion 1wii(t) +bi(t) = 4 .. ..... .Equotion 2 Given that w(u) = 1, wi(u) = b(u) = u.22. Find (t) such that (t) = c-3s +] ()Jt0.(Hint: Use laplace transform to solve this problem)Copyright 2015 Mohammed K A Kaabar All Rights Reserved51Chapter 2 Systems of Homogeneous Linear Differential Equations (HLDE) In this chapter, we start introducing the homogeneous lineardifferentialequations(HLDE)withconstant coefficients.Inaddition,wediscusshowtofind the generalsolutionofHLDE.Attheendofthischapter, weintroduceanewmethodcalledUndetermined Coefficient Method.2.1 HLDE with Constant CoefficientsInthissection,wediscusshowtofindthegeneral solutionofthehomogeneouslineardifferential equations (HLDE) with constant coefficients.To give an introduction about HLDE, it is important to start with the definition of homogeneous system.*Definition 2.1.1 Homogeneous System is defined as a m n system of linear equations that has all zero constants. (i.e. the following is an example of Copyright 2015 Mohammed K A Kaabar All Rights Reserved52 M. Kaabarhomogeneous system): _2o +b - c + J = uSo + Sb + Sc + 4J = u-b +c -J = u*Definition 2.1.1 is taken from section 3.1 in my published book titled A First Course in Linear Algebra: Study Guide for the Undergraduate Linear Algebra Course, First Edition1.Example 2.1.1 Describe the following differential equation: yii(x)+Syi(x) = u.Solution: Since the above differential equation has a zero constant, then according to definition 2.1.1, it is a homogeneous differential equation. In addition, it is linear because the dependent variable y and all its derivatives are to the power 1. For the order of this homogeneous differential equation, since the highest derivative is 2, then the order is 2. Thus, yii(x)+Syi(x) = u is a homogeneous linear differential equation of order 2.Example 2.1.2 Describe the following differential equation: Sy(3)(x)-2yi(x) + 7y(x) = u.Solution: Since the above differential equation has a zero constant, then according to definition 2.1.1, it is a homogeneous differential equation. In addition, it is linear because the dependent variable y and all its derivatives are to the power 1. For the order of this homogeneous differential equation, since the highest derivative is 3, then the order is 3. Copyright 2015 Mohammed K A Kaabar All Rights Reserved53Thus, Sy(3)(x)-2yi(x) +7y(x) = u is a homogeneous linear differential equation of order 3.Example 2.1.3 Describe the following differential equation: Sy(2)(x)-2yi(x) = 12.Solution: Since the above differential equation has a nonzero constant, then according to definition 2.1.1, it is a non-homogeneous differential equation. In addition, it is linear because the dependent variable yand all its derivatives are to the power 1. For the order of this non-homogeneous differential equation, since the highest derivative is 2, then the order is 2. Thus, Sy(2)(x)-2yi(x) = 12 is a non-homogeneous linear differential equation of order 2.Example 2.1.4 Find the general solution the following Initial Value Problem (IVP): 2yi(x) + 4y(x) = u. Given: yi(u) = 1. (Hint: Use the concepts of section 1.4)Solution: 2yi(x) + 4y(x) = u is a homogeneous linear differential equation of order 1. First, we need to find the domain for the solution of the above differential equation in other words we need to find for what values of x the solution of the above differential equation holds. Therefore, we do the following:2yi(x) +4y(x) = uUsing definition 1.4.1, we also suppose the following:o1(x) = 2o0(x) = 4K(x) = uThe domain of solution is (-, ).Copyright 2015 Mohammed K A Kaabar All Rights Reserved54 M. KaabarNow,tofindthesolutionoftheabovedifferential equation,weneedtotakethelaplacetransformfor both sides as followsL{2yi(x)] +L{4y(x)] = L{u]2L{yi(x)] + 4L{y(x)] = u because (L{u] = u).2(s(s) -yi(u)) + 4(s) = u from result 1.2.2.2s(s) -2yi(u) + 4(s) = u(s)(2s +4) = 2yi(u)(s)(2s +4) = 2(1)(s) =22s +4 =22(s +2) =1s + 2Tofind asolution,weneedtofindtheinverselaplace transform as follows:y(x) = L-1{(s)] = L-1] 1s+2 = c-2x.Thus, the general solution y(x) = cc-2x, for some constant c. Here, c = 1.Result 2.1.1 Assume that m1y(n)(x) + m2y(x) = u is ahomogeneous linear differential equation of order nwith constant coefficients m1and m2. Then, a) m1y(n)(x) +m2y(x) = u must have exactly nindependent solutions, say 1(x), 2(x), ., n(x).b) Every solution of m1y(n)(x) + m2y(x) = u is of the form: c11(x) + c22(x) ++ cnn(x), for some constants c1, c2, ., cn.Copyright 2015 Mohammed K A Kaabar All Rights Reserved55Result 2.1.2 Assume that y1 = ck1x, y2 = ck2x, , yn = cknxare independent if and only if k1, k2, ., knare distinct real numbers.Example 2.1.5 Given yii(x) - 4y(x) = u,y(u) = 4, yi(u) = 1u. Find the general solution for y(x). (Hint: Use results 2.1.1 and 2.1.2)Solution: yii(x) - 4y(x) = u is a homogeneous linear differential equation of order 2. In this example, we will use a different approach from example 2.1.4 (laplace transform approach) to solve it. Since yii(x) - 4y(x) = u is HLDE with constant coefficients, then we will do the following: Let y(x) = ckx, we need to find k.First of all, we will find the first and second derivatives as follows:yi(x) = kckxyii(x) = k2ckxNow, we substitute y(x) = ckxand yi(x) = kckxinyii(x) - 4y(x) = u as follows:k2ckx -4ckx = uckx(k2 - 4) = uckx(k - 2)(k + 2) = uThus, k = 2 or k = -2. Then, we use our values to substitute k in our assumption which is y(x) = ckx:at k = 2, y1(x) = c2xCopyright 2015 Mohammed K A Kaabar All Rights Reserved56 M. Kaabarat k = -2, y2(x) = c-2xHence, using result 2.1.1, the general solution for y(x)is: yhomo(x) = c1c2x +c2c-2x, for some c1, c2 e +. (Note: bomo denotes to homogeneous). Now, we need to find the values of c1 anu c2as follows:at x = u, yhomo(u) = c1c2(0) +c2c-2(0)yhomo(u) = c1c0 + c2c0yhomo(u) = c1 + c2Since y(u) = 4, then c1 + c2 = 4 ........... (1)yhomoi(x) = 2c1c2x -2c2c-2xat x = u, yhomoi(u) = 2c1c2(0) - 2c2c-2(0)yhomoi(u) = 2c1c0 -2c2c0yhomoi(u) = 2c1 -2c2Since yi(u) = 1u, then 2c1 - 2c2 = 1u........ (2)From (1) and (2),c1 = 4.S andc2 = -u.S.Thus, the general solution is:yhomo(x) = 4.Sc2x - u.Sc-2x.Example 2.1.6 Given y(3)(x) - Sy(2)(x) + 6yi(x) = u.Find the general solution for y(x). (Hint: Use results 2.1.1 and 2.1.2, and in this example, no need to find the values of c1, c2, anu c3)Solution: y(3)(x) -Sy(2)(x) +6yi(x) = u is a homogeneous linear differential equation of order 3. Since y(3)(x) -Sy(2)(x) + 6yi(x) = u is HLDE with Copyright 2015 Mohammed K A Kaabar All Rights Reserved57constant coefficients, then we will do the following: Let y(x) = ckx, we need to find k.First of all, we will find the first, second, and thirdderivatives as follows:yi(x) = kckxyii(x) = k2ckxy(3)(x) = k3ckxNow, we substitute yi(x) = kckx, yii(x) = k2ckx, and y(3)(x) = k3ckxin y(3)(x) -Sy(2)(x) +6yi(x) = u as follows:k3ckx -Sk2ckx + 6kckx = uckx(k3 -Sk2 +6k) = uckx(k(k2 - Sk +6)) = uckx(k(k -2)(k -S)) = uThus, k = u, k = 2 or k = S. Then, we use our values to substitute k in our assumption which is y(x) = ckx:at k = u, y1(x) = c(0)x = 1at k = 2, y2(x) = c2xat k = S, y3(x) = c3xThus, using result 2.1.1, the general solution for y(x)is: yhomo(x) = c1 +c2c2x +c3c3x, for some c1, c2, c3 e +. (Note: bomo denotes to homogeneous). Example 2.1.7 Given y(5)(x) - y(4)(x) -2y(3)(x) = u.Copyright 2015 Mohammed K A Kaabar All Rights Reserved58 M. KaabarFind the general solution for y(x). (Hint: Use results 2.1.1 and 2.1.2, and in this example, no need to find the values of c1, c2, c3, c4 anu c5)Solution: y(5)(x) -y(4)(x) - 2y(3)(x) = u is a homogeneous linear differential equation of order 5. Since y(5)(x) -y(4)(x) -2y(3)(x) = u is HLDE with constant coefficients, then we will do the following: Let y(x) = ckx, we need to find k.First of all, we will find the first, second, third, fourth, and fifth derivatives as follows:yi(x) = kckxyii(x) = k2ckxy(3)(x) = k3ckxy(4)(x) = k4ckxy(5)(x) = k5ckxNow, we substitute y(5)(x) = k5ckx, y(4)(x) = k4ckx, and y(3)(x) = k3ckxin y(5)(x) -y(4)(x) - 2y(3)(x) = uas follows:k5ckx -k4ckx -2k3ckx = uckx(k5 -k4 -2k3) = uckx(k3(k2 - k -2)) = uckx(k3(k -2)(k +1)) = uThus, k = u, k = u, k = u, k = 2 or k = -1. Then, we use our values to substitute k in our assumption which is y(x) = ckx:Copyright 2015 Mohammed K A Kaabar All Rights Reserved59at k = u, y1(x) = c(0)x = 1at k = u, y2(x) = c(0)x = 1 xat k = u, y3(x) = c(0)x = 1 x2because k3 = Spon{1, x, x2] = u (Note: Spon{1, x, x2] = umeans o 1 + b x +c x2 = u)In other words Spon{1, x, x2] is the set of all linear combinations of 1, x, and x2.at k = 2, y4(x) = c2xat k = -1, y5(x) = c-xThus, using result 2.1.1, the general solution for y(x)is: yhomo(x) = c1 +c2x +c3x2+c4c2x + c5c-x, for some c1, c2, c3, c4, c5 e +. (Note: bomo denotes to homogeneous). Example 2.1.8 Given y1(x) = c3x, y2(x) = c-3x, anuy3(x) = cx . Are y1(x), y3(x), anu y3(x) independent?Solution: We cannot write y3(x) as a linear combination of y1(x) and y2(x) as follows:cx = (FixcJ Numbcr) c3x + (FixcJ Numbcr) c-3xThus, y1(x), y3(x), anu y3(x) are independent.Example 2.1.9 Given y1(x) = c(x+3), y2(x) = c3, anuy3(x) = cx . Are y1(x), y3(x), anu y3(x) independent?Solution: We can write y1(x) as a linear combination of y2(x) and y3(x) as follows:c(x+3) = cx c3. Thus, y1(x), y3(x), anu y3(x) are dependent (not independent).Copyright 2015 Mohammed K A Kaabar All Rights Reserved60 M. Kaabar2.2 Method of Undetermined CoefficientsInthissection,wediscusshowtousewhatwehave learnedfromsection2.1tocombineitwithwhatwe will learn from section 2.2 in order to find the general solutionusingamethodknownasundetermined coefficientsmethod.Inthismethod,wewillfinda generalsolutionconsistingofhomogeneoussolution and particular solution together.Wegivethefollowingexamplestointroducethe undetermined coefficient method.Example 2.2.1 Given yi(x) +Sy(x) = x, y(u) = 1. Find the general solution for y(x). (Hint: No need to find the value of c1)Solution: Since yi(x) + Sy(x) = x does not have a constant coefficient, then we need to use the undetermined coefficients method as follows:Step 1: We need to find the homogeneous solution by letting yi(x) +Sy(x) equal to zero as follows:yi(x) +Sy(x) = u. Now, it is a homogeneous linear differential equation of order 1. Since yi(x) +Sy(x) = u is a HLDE with constant coefficients, then we will do the following: Copyright 2015 Mohammed K A Kaabar All Rights Reserved61Let y(x) = ckx, we need to find k.First of all, we will find the first, second, third, fourth, and fifth derivatives as follows:yi(x) = kckxNow, we substitute yi(x) = kckxin yi(x) +Sy(x) = u as follows:kckx + Sckx = uckx(k + S) = uThus, k = -S. Then, we use our value to substitute k in our assumption which is y(x) = ckx:at k = -S, y1(x) = c(-3)x = c-3xThus, using result 2.1.1, the general solution for y(x)is: yhomo(x) = c1c-3x, for some c1 e +. (Note: bomodenotes to homogeneous). Step 2: We need to find the particular solution as follows: Since yi(x) +Sy(x) equals x, then the particular solution should be in the following form:yputcuIu(x) = o +bx becausex isapolynomialofthe firstdegree,andthegeneralformforfirstdegree polynomial is o + bx.Now, we need to find o and b as follows:yputcuIui(x) = bWe substitute yputcuIui(x) = b in yi(x) +Sy(x) = x.b +S(o +bx) = xb + So +Sbx = xCopyright 2015 Mohammed K A Kaabar All Rights Reserved62 M. Kaabarat x = u, we obtain: b +So + (Sb)(u) = ub + So +u = ub +So = u .....................(1)at x = 1, we obtain: b +So + (Sb)(1) = 1b + So +Sb = 14b + So = 1 ........... ..........(2)From (1) and (2), we get: o = -19and b = 13.Thus, yputcuIu(x) = -19 + 13x.Step 3: We need to find the general solution as follows:ygcncuI(x) = yhomo(x) + yputcuIu(x)Thus, ygcncuI(x) = c1c-3x +[-19 + 13x.Example 2.2.2 Given yi(x) +Sy(x) = c-3x, y(u) = 1.Find the general solution for y(x). (Hint: No need to find the value of c1)Solution: In this example, we will have the same homogeneous solution as we did in example 2.2.1 but the only difference is the particular solution. We will repeat some steps in case you did not read example 2.2.1. Since yi(x) +Sy(x) = c-3xdoes not have a constant coefficient, then we need to use the undetermined coefficients method as follows:Step 1: We need to find the homogeneous solution by letting yi(x) +Sy(x) equal to zero as follows:yi(x) +Sy(x) = u. Now, it is a homogeneous linear differential equation of order 1. Copyright 2015 Mohammed K A Kaabar All Rights Reserved63Since yi(x) +Sy(x) = u is a HLDE with constant coefficients, then we will do the following: Let y(x) = ckx, we need to find k.First of all, we will find the first, second, third, fourth, and fifth derivatives as follows:yi(x) = kckxNow, we substitute yi(x) = kckxin yi(x) +Sy(x) = u as follows:kckx + Sckx = uckx(k + S) = uThus, k = -S. Then, we use our value to substitute k in our assumption which is y(x) = ckx:at k = -S, y1(x) = c(-3)x = c-3xThus, using result 2.1.1, the general solution for y(x)is: yhomo(x) = c1c-3x, for some c1 e +. (Note: bomodenotes to homogeneous). Step 2: We need to find the particular solution as follows: Since yi(x) +Sy(x) equals c-3x, then the particular solution should be in the following form:yputcuIu(x) = (oc-3x)xNow, we need to find o as follows:yputcuIui(x) = (oc-3x) -S(oxc-3x)WesubstituteyputcuIui(x) = (oc-3x) - S(oxc-3x) in yi(x) +Sy(x) = c-3x.|(oc-3x) -S(oxc-3x)] +S(oc-3x)x = c-3xCopyright 2015 Mohammed K A Kaabar All Rights Reserved64 M. Kaabaroc-3x = c-3xo = 1Thus, yputcuIu(x) = (1c-3x)x = xc-3x.Step 3: We need to find the general solution as follows:ygcncuI(x) = yhomo(x) + yputcuIu(x)Thus, ygcncuI(x) = c1c-3x +xc-3x.Result 2.2.1 Suppose that you have a linear differential equationwiththeleastderivative,saym,andthis differentialequation equals toapolynomialofdegree w. Then, we obtain the following:yputcuIu = |Polynomiol o cgrcc w]xm.Result 2.2.2 Suppose that you have a linear differential equation,thenthegeneralsolutionisalwayswritten as: ygcncuI(x) = yhomogcncous(x) +yputcuIu(x).Example 2.2.3 Given y(4)(x) - 7y(3)(x) = x2. Describe yputcuIu(x) but do not find it.Solution: To describe yputcuIu(x), we do the following:By using result 2.2.1, we obtain the following: yputcuIu = |o +bx +cx2]x3.Example 2.2.4 Given y(4)(x) - 7y(3)(x) = S. Describe yputcuIu(x) but do not find it.Solution: To describe yputcuIu(x), we do the following:By using result 2.2.1, we obtain the following: yputcuIu = |o]x3 = ox3.Copyright 2015 Mohammed K A Kaabar All Rights Reserved65Example 2.2.5 Given y(2)(x) - Sy(x) = x2cx. Describe yputcuIu(x) but do not find it.Solution: To describe yputcuIu(x), we do the following:Inthisexample,welookatx2,andwewriteitas: o + bx +cx2, and then we multiply it by cx.yputcuIu = |o +bx +cx2]cx.Example 2.2.6 Given y(2)(x) - Sy(x) = sin(Sx)cx.Describe yputcuIu(x) but do not find it.Solution: To describe yputcuIu(x), we do the following:In this example, we look at sin(Sx), and we write it as: (o sin(Sx) +bsin(Sx)), and then we multiply it by cx.yputcuIu = |o sin(Sx) +bsin(Sx)]cx.2.3 Exercises 1.Given y(2)(x) +2yi(x) +y(x) = u. Find the general solution for y(x). (Hint: Use results 2.1.1 and 2.1.2, and in this exercise, no need to find the values of c1, anu c2)2.Given y(3)(x) -y(2)(x) = Sx. Find the general solution for y(x). (Hint: No need to find the value of c1, c2, anu c3) 3.Given y(4)(x) -y(3)(x) = Sx2. Find the general solution for y(x). (Hint: No need to find the value of c1, c2, c3, anu c4)Copyright 2015 Mohammed K A Kaabar All Rights Reserved66 M. Kaabar4.Given y(3)(x) -y(2)(x) = cx. Find the general solution for y(x). (Hint: No need to find the value of c1, c2, anu c3)5.Given y(3)(x) -y(2)(x) = sin(2x). Find the general solution for y(x). (Hint: No need to find the value of c1, c2, anu c3)6.Given y(2)(x) -Sy(x) = Sxsin(Sx). Describe yputcuIu(x) but do not find it.7.Given y(2)(x) -Sy(x) = cx2. Describe yputcuIu(x)but do not find it.Copyright 2015 Mohammed K A Kaabar All Rights Reserved67Chapter 3 Methods of First and Higher Orders Differential Equations Inthischapter,we introducetwonewmethodscalled VariationMethodandCauchy-EulerMethod inorder tosolvefirstandhigherordersdifferentialequations.Inaddition,wegiveseveralexamplesaboutthese methods,andthedifferencebetweenthemandthe previous methods in chapter 2.3.1 Variation MethodInthissection,we discusshowtofindtheparticular solution using Variation Method. For the homogeneous solution,itwillbesimilartowhatwelearnedin chapter 2.Definition 3.1.1 Given o2(x)y(2) + o1(x)yi = K(x) is a linear differential equation of order 2. Assume that y1(x) and y2(x) are independent solution to the homogeneous solution. Then, the particular solution using Variation Method is written as:Copyright 2015 Mohammed K A Kaabar All Rights Reserved68 M. KaabaryputcuIu(x) = b1(x)y1(x) + b2(x)y2(x). To find b1(x)and b2(x), we need to solve the following two equations:b1i(x)y1(x) + b2i(x)y2(x) = ub1i(x)y1i(x) +b2i(x)y2i(x) = K(x)o2(x)Definition 3.1.2 Given o3(x)y(3) + +o1(x)yi = K(x) is a linear differential equation of order 3. Assume that y1(x), y2(x) and y3(x) are independent solution to the homogeneous solution. Then, the particular solution using Variation Method is written as:yputcuIu(x) = b1(x)y1(x) + b2(x)y2(x) + b3(x)y3(x). To find b1(x), b2(x) and b3(x), we need to solve the following three equations:b1i(x)y1(x) + b2i(x)y2(x) + b3i(x)y3(x) = ub1i(x)y1i(x) +b2i(x)y2i(x) + b3i(x)y3i(x) = ub1(2)(x)y1(2)(x) + b2(2)(x)y2(2)(x) + b3(2)(x)y3(2)(x) = K(x)o3(x)Example 3.1.1 Given y(2) + Syi = 1x . Find the general solution for y(x). (Hint: No need to find the values of c1and c2)Solution: Since y(2) +Syi = 1xdoes not have a constant coefficient, then we need to use the variation method as follows: Step 1: We need to find the homogeneous solution by letting y(2) + Syiequal to zero as follows:y(2) + Syi = u. Now, it is a homogeneous linear differential equation of order 2. Copyright 2015 Mohammed K A Kaabar All Rights Reserved69Since y(2) + Syi = u is a HLDE with constant coefficients, then we will do the following: Let y(x) = ckx, we need to find k.First of all, we will find the first and second derivatives as follows:yi(x) = kckxyii(x) = k2ckxNow, we substitute yi(x) = kckxand yii(x) = k2ckxin y(2) + Syi = u as follows:k2ckx + Skckx = uckx(k2 +Sk) = uckx(k(k +S)) = uThus, k = u and k = -S. Then, we use our values to substitute k in our assumption which is y(x) = ckx:at k = u, y1(x) = c(0)x = c0 = 1at k = -S, y2(x) = c(-3)x = c-3xNotice that y1(x) and y2(x) are independent.Thus, using result 2.1.1, the general homogenoussolution for y(x) is: yhomo(x) = c1 + c2c-3x, for some c1anu c2 e +. (Note: bomo denotes to homogeneous). Step 2: We need to find the particular solution using definition 3.1.1 as follows: Since y(2) + Syiequals 1x, then the particular solution should be in the following form: yputcuIu(x) = b1(x)y1(x) + b2(x)y2(x). To find b1(x) and b2(x), we need to solve the following two equations:Copyright 2015 Mohammed K A Kaabar All Rights Reserved70 M. Kaabarb1i(x)y1(x) + b2i(x)y2(x) = ub1i(x)y1i(x) +b2i(x)y2i(x) = K(x)o2(x)y1(x) = 1 ------y1i(x) = uy2(x) = c-3x------y2i(x) = -Sc-3xNow, we substitute what we got above in the particular solution form as follows:b1i(x)(1) +b2i(x)(c-3x) = ub1i(x)(u) +b2i(x)(-Sc-3x) = 1x1b1i(x)(1) +b2i(x)(c-3x) = u ....... . ....(1)b2i(x)(-Sc-3x) = 1x.......... . ...... (2)By solving (1) and (2), b2i(x) = - 13xc3xandb1i(x) = 13xc3x(c-3x) = 13x.Sinceitisimpossibletointegrateb2i(x) = - 13xc3xto find b2(x), then it is enough to write as:b2(x) = ] - 13tc3tx0Jt.Since it is possible to integrate b1i(x) = 13xto find b1(x), then we do the following:b1(x) = ] 13xJx = 13(ln|x|), x > u.Thus, we write the particular solutionas follows:yputcuIu(x) = 1S(ln|x|) +c-3x __- 1Stc3tx0Jt_Step 3: We need to find the general solution as follows:Copyright 2015 Mohammed K A Kaabar All Rights Reserved71ygcncuI(x) = yhomo(x) + yputcuIu(x)Thus, ygcncuI(x) = (c1 + c2c-3x) + 13(ln|x|) +c-3x [] - 13tc3tx0Jt, for some c1anu c2 e +.Example 3.1.2 Given y(2) + 6yi +8y = c-4x . Find the general solution for y(x). (Hint: No need to find the values of c1and c2)Solution: Since y(2) +6yi + 8y = c-4xdoes not have a constant coefficient, then we need to use the variation method as follows: Step 1: We need to find the homogeneous solution by letting y(2) + 6yi +8y equal to zero as follows:y(2) + 6yi +8y = u. Now, it is a homogeneous linear differential equation of order 2. Since y(2) + 6yi +8y = u is a HLDE with constant coefficients, then we will do the following: Let y(x) = ckx, we need to find k.First of all, we will find the first and second derivatives as follows:yi(x) = kckxyii(x) = k2ckxNow, we substitutey(x) = ckx, yi(x) = kckxand yii(x) = k2ckxin y(2) +6yi +8y = u as follows:k2ckx + 6kckx + 8ckx = uckx(k2 +6k + 8) = uckx((k + 2)(k + 4)) = uCopyright 2015 Mohammed K A Kaabar All Rights Reserved72 M. KaabarThus, k = -2 and k = -4. Then, we use our values to substitute k in our assumption which is y(x) = ckx:at k = -2, y1(x) = c(-2)x = c-2xat k = -4, y2(x) = c(-4)x = c-4xNotice that y1(x) and y2(x) are independent.Thus, using result 2.1.1, the general homogenous solution for y(x) is: yhomo(x) = c1c-2x +c2c-4x, for some c1anu c2 e +. (Note: bomo denotes to homogeneous). Step 2: We need to find the particular solution using definition 3.1.1 as follows: Since y(2) + 6yi +8y equals c-4x, then the particular solution should be in thefollowing form: yputcuIu(x) = b1(x)y1(x) + b2(x)y2(x). To find b1(x) and b2(x), we need to solve the following two equations:b1i(x)y1(x) + b2i(x)y2(x) = ub1i(x)y1i(x) +b2i(x)y2i(x) = K(x)o2(x)y1(x) = c-2x------y1i(x) = -2c-2xy2(x) = c-4x------y2i(x) = -4c-4xNow, we substitute what we got above in the particular solution form as follows:b1i(x)(c-2x) +b2i(x)(c-4x) = ub1i(x)(-2c-2x) +b2i(x)(-4c-4x) = c-4x1b1i(x)(c-2x) + b2i(x)(c-4x) = u ....... . .... . .(1)b1i(x)(-2c-2x) +b2i(x)(-4c-4x) = c-4x........ (2)Copyright 2015 Mohammed K A Kaabar All Rights Reserved73Bysolving(1) and (2), andusingCramersrule,we obtain:b1i(x) = x1 =dct_0 c-4xc-4x-4c-4x_dctj c-2xc-4x-2c-2x-4c-4x[=-c-8x-4c-6x +2c-6x = -c-8x-2c-6x= 12c-2xBy substituting b1i(x) in (1) to find b2i(x) as follows:_12c-2x] (c-2x) +b2i(x)(c-4x) = ub2i(x) = -12Sinceitispossibletointegrateb1i(x) = 12c-2xtofind b1(x), then we do the following:b1(x) = ]12c-2xJx = -14c-2x.Sinceitispossibletointegrateb2i(x) = -12tofind b2(x), then we do the following:b2(x) = ]-12Jx = -12x.Thus, we write the particular solution as follows:yputcuIu(x) = -14c-2x(c-2x) -12x(c-4x)Step 3: We need to find the general solution as follows:ygcncuI(x) = yhomo(x) + yputcuIu(x)Thus, ygcncuI(x) = (c1c-2x +c2c-4x) + [-14c-2x(c-2x) -12x(c-4x), for some c1anu c2 e +.Copyright 2015 Mohammed K A Kaabar All Rights Reserved74 M. Kaabar3.2 Cauchy-Euler MethodIn this section, we will show how to use Cauchy-Euler Methodtofindthegeneralsolutionfordifferential equations that do not have constant coefficients.To introduce this method, we start with some examples as follows:Example 3.2.1 Given xy(2) -yi + 1xy = u. Find the general solution for y(x). (Hint: No need to find the values of c1and c2)Solution: Since xy(2) - yi + 1xy = u does not have constant coefficients, then we need to use the Cauchy-Euler method by letting y = xk, and after substitution all terms must be of the same degree as follows:First of all, we will find the first and second derivatives as follows:yi = kxk-1yii = k(k - 1)xk-2Now, we substitute y = xk, yi = kxk-1andyii = k(k -1)xk-2in xy(2) -yi + 1xy = u as follows:xk(k -1)xk-2 -kxk-1 +1xxk = uk(k -1)xk-1 -kxk-1 +xk-1 = uxk-1(k(k - 1) - k +1) = uxk-1(k2 -k - k +1) = uCopyright 2015 Mohammed K A Kaabar All Rights Reserved75xk-1(k2 - 2k + 1) = uxk-1((k -1)(k - 1)) = uThus, k = 1 and k = 1. Then, we use our values to substitute k in our assumption which is y = xk:at k = 1, y1 = x1 = xat k = 1, y2 = x1 = x ln(x)In the above case, we multiplied x by ln(x) because we had a repeating for x, and in Cauchy-Euler Method, we should multiply any repeating by natural logarithm.Thus, the general solution for y(x) is:y(x) = c1x + c2xln(x), for some c1anu c2 e +.Example 3.2.2 Given x3y(2) - x2yi +xy = u. Find the general solution for y(x). (Hint: No need to find the values of c1and c2)Solution: Since x3y(2) +x2yi + xy = u does not have constant coefficients, then we need to use the Cauchy-Euler method by letting y = xk, and after substitution all terms must be of the same degree as follows:First of all, we will find the first, second and thirdderivatives as follows:yi = kxk-1yii = k(k - 1)xk-2yiii = k(k -1)(k - 2)xk-3Now, we substitute y = xk, yi = kxk-1, and yii = k(k -1)xk-2in x3y(2) + x2yi +xy = u as follows:Copyright 2015 Mohammed K A Kaabar All Rights Reserved76 M. Kaabarx3(k(k - 1)xk-2) + x2(kxk-1) + x(xk) = u(k(k - 1)xk+1) +(kxk+1) + (xk+1) = uxk+1(k2 -k + k +1) = uxk+1(k2 +1) = uThus, k = _V1 = _i = u _ (1)(i).Then, we use our values to substitute k in our assumption which is y = xk:Since we have two parts (real and imaginary), then by using the Cauchy-Euler Method, we need to write our solution as follows:y1 = x(cuI put)cos(Imoginory port ln(x)= x(0)cos(1 ln(x)) = cos(ln(x))y2 = x(cuI put)sin(Imoginory port ln(x)= x(0)sin(1 ln(x)) = sin(ln(x))Thus, the general solution for y(x) is:y(x) = c1cos(ln(x)) +c2sin(ln(x)), for somec1anu c2 e +.3.3 Exercises 1.Given y(2) +yi +4y = u. Find the general solution for y(x). (Hint: No need to find the values of c1and c2)2.Given y(2) +Syi + 7y = u. Find the general solution for y(x). (Hint: No need to find the values of c1and c2)3.Given x3y(3) +xyi = u. Find the general solution for y(x). (Hint: No need to find the values of c1, c2and c3)Copyright 2015 Mohammed K A Kaabar All Rights Reserved774.Given x3y(3) -2xyi = u. Find the general solution for y(x). (Hint: No need to find the values of c1, c2and c3)5.Given x2y(2) +yi = 2x2. Is it possible to find the general solution for y(x) using Cauchy-Euler Method? Why? Copyright 2015 Mohammed K A Kaabar All Rights Reserved78 M. KaabarChapter 4 Extended Methods of First and Higher Orders Differential Equations In this chapter, we discuss some new methods such as BernulliMethod,SeparableMethod,ExactMethod, ReducedtoSeparableMethodandReductionofOrder Method.Weusethesemethodstosolvefirstand higherorderslinearandnon-lineardifferential equations.Inaddition,wegiveexamplesaboutthese methods,andthedifferencesbetweenthemandthe previous methods in chapter 2 and chapter 3.4.1 Bernoulli MethodIn this section, we start with two examples about using integralfactortosolvefirstorderlineardifferential equations. Then, we introduce Bernulli Method to solve someexamplesoffirstordernon-lineardifferential equations.Definition 4.1.1 Given o1(x)yi +o2(x)y = K(x),wheie o1(x) = u is a linear differential equation of order 1. Dividing both sides by o1(x), we obtain:Copyright 2015 Mohammed K A Kaabar All Rights Reserved79o1(x)o1(x)yi + o2(x)o1(x)y = K(x)o1(x)yi +o2(x)o1(x)y = K(x)o1(x)Assume that g(x) = u2(x)u1(x)and F(x) = K(x)u1(x), then:yi +g(x)y = F(x) ........ (1)Thus, the solution using Integral Factor Method is written in the following steps:Step 1: Multiply both sides of (1) by lettingI = c]g(x)dx:yic] g(x)dx +g(x)yc]g(x)dx = F(x)c]g(x)dxStep 2: yic] g(x)dx +g(x)yc]g(x)dx = |y c]g(x)dx]i.(2)Step 3: |y c]g(x)dx]i= F(x)c]g(x)dx ........ . (S)Step 4: Integrate both sides of (S), we obtain:_|y c]g(x)dx]iJx = _(F(x)c]g(x)dx) Jxy c]g(x)dx = _(F(x)c]g(x)dx) JxStep 5: By solving for y, and substituting I = c] g(x)dxwe obtain:y I = _(F(x))(I) Jxy = ](F(x))(I) JxIy = ]I F(x) JxIThus, the final solution is: y = ]I F(x) JxIExample 4.1.1 Given x2yi - 2xy = 4x3. Find the general solution for y(x). (Hint: Use integral factor method and no need to find the value of c)Copyright 2015 Mohammed K A Kaabar All Rights Reserved80 M. KaabarSolution: Since x2yi -2xy = 4x3does not have constant coefficients, and it is a first order non-linear differential equation, then by using definition 4.1.1, we need to use the integral factor method by letting I = c]g(x)dx, where g(x) = u2(x)u1(x) = -2xx2 = -2xand F(x) = K(x)u1(x) = 4x3x2 = 4xHence, I = c]g(x)dx = c]-2xdx= c-2In(x) = cIn(x-2) = 1x2The general solution is written as follows:y = ]I F(x) JxIy = ] 1x2 4x Jx1x2y = ]4xJx1x2y = 4 ln(x) +c1x2y = 4x2ln(x) +cx2Thus, the general solution is: y = 4x2ln(x) + cx2for some c e +.Example 4.1.2 Given (x +1)yi + y = S. Find the general solution for y(x). (Hint: Use integral factor method and no need to find the value of c)Copyright 2015 Mohammed K A Kaabar All Rights Reserved81Solution: Since (x + 1)yi +y = S does not have constant coefficients, and it is a first order non-linear differential equation, then by using definition 4.1.1, we need to use the integral factor method by letting I = c]g(x)dx, where g(x) = u2(x)u1(x) =1(x+1)andF(x) = K(x)o1(x) =S(x + 1)Hence, I = c]g(x)dx = c]1(x+1)dx= cIn(x+1) = (x + 1)The general solution is written as follows:y = ]I F(x) JxIy = ](x + 1) S(x +1)Jx(x +1)y = ]S Jx(x +1)y = Sx + c(x +1)y =Sx(x +1)+c(x + 1)Thus, the general solution is: y =5x(x+1) +c(x+1)for some c e +.Definition 4.1.2 Given yi +g(x)y = (x)ynwhere n e +and n = u and n = 1 is a non-linear differential equation of order 1. Thus, the solution using Bernoulli Method is written in the following steps:Step 1: Change it to first order linear differential equation by letting w = y1-n.Copyright 2015 Mohammed K A Kaabar All Rights Reserved82 M. KaabarStep 2: Find the derivative of both sides for w = y1-nas follows:JwJx = (1 -n)y1-n-1 JyJxJwJx = (1 -n)y-n JyJx............... .. (1)Step 3: Solve (1) for ddxas follows:JyJx =11 -nyn JwJx .................(2)Step 4: Since we assumed that w = y1-n, theny = w[ 11-n, and hence yn = w[ n1-n.Step 5: Substitute what we got above inddx + g(x)y = (x)ynas follows:11 -nyn JwJx + g(x)w[ 11-n= (x)w[ n1-n... .... . (S)Step 6: Divide (S) by 11-nynas follows:JwJx + g(x)w[ 11-n11 -nyn= (x)w[ n1-n11 -nyn JwJx + g(x)(1 -n)w[ 11-ny-n = (x)(1 - n)w[ n1-ny-nStep 7: After substitution, we obtain:JwJx + g(x)(1 -n)yy-n = (x)(1 - n)yny-n JwJx + g(x)(1 -n)y1-n = (x)(1 -n)yn-n JwJx + g(x)(1 -n)y1-n = (x)(1 -n)y0 JwJx + g(x)(1 -n)y1-n = (x)(1 -n)Step 8: We substitute w = y1-nin the above equation as follows:JwJx + g(x)(1 -n)w = (x)(1 -n) Copyright 2015 Mohammed K A Kaabar All Rights Reserved83Thus, the final solution is:JwJx + (1 - n)g(x)w = (1 - n)(x) In the following example, we will show how to useBernoulli Method, and we will explore the relationship between Bernoulli Method and Integral Factor Method.Example 4.1.3 Given xyi + Sx2y = (6x2)y3. Find the general solution for y(x). (Hint: Use Bernoulli method and no need to find the value of c)Solution: Since xyi +Sx2y = (6x2)y3does not have constant coefficients, and it is a first order non-linear differential equation, then by using definition 4.1.2, we need to do the following by letting w = y1-n, where in this example n = S, and g(x) = Sx2and (x) = 6x2.Since we assumed that w = y1-3 = y-2, theny = w[ 11-n= w[ 11-3= w-12 =1Vw, and ddx = -12y3 dwdxWe substitute what we got above in xyi +Sx2y =(6x2)y3as follows:x _-12y3 JwJx] +Sx2_ 1Vw] = (6x2) _1wVw] ..... .(1)Now, we divide (1) by -12xy3as follows:JwJx + Sx2_ 1Vw]-12xy3= (6x2)y3-12xy3JwJx +(-2)Sx _ 1Vw] y-3 = (-2)6x .......... (2)Copyright 2015 Mohammed K A Kaabar All Rights Reserved84 M. KaabarThen, we substitute y =1Vwin (2) as follows:JwJx +(-2)Sx(y)y-3 = (-2)6xJwJx + (-2)Sxy1-3 = (-2)6xJwJx +(-2)Sxy-2 = (-2)6x .............(S)Now, we substitute w = y-2in (S) as follows:JwJx +(-2)Sxw = (-2)6xJwJx -6xw = -12x ............... . ... (4)Then, we solve (4) for w(x) as follows:To solve (4), we need to use the integral factor method:Hence, I = c]g(x)dx = c]-6x dx = c-6x22= c-3x2The general solution for w(x) is written as follows:w = ]I F(x) JxIw = ]I (-12x) JxIw = ]c-3x2 (-12x) Jxc-3x2w = 2 ]c-3x2 (-6x) Jxc-3x2w = 2c-3x2+cc-3x2w = 2c-3x2c-3x2 +cc-3x2Copyright 2015 Mohammed K A Kaabar All Rights Reserved85w = 2 +cc-3x2w = 2 + cc3x2The general solution for w(x) is: w(x) = 2 + cc3x2.Thus, the general solution for y(x) is:y(x) =1w(x) =12 +cc3x2for some c e +.4.2 Separable MethodInthissection,we willsolvesomedifferential equations using a method known as Separable Method. Thismethodiscalledseparablebecauseweseparate two different terms from each other.Definition 4.2.1 The standard form of Separable Methodis written as follows:(All in tcrms o x)Jx - (All in tcrms o y)Jy = uNote: it does not matter whether it is the above form or in the following form:(All in tcrms o y)Jy - (All in tcrms o x)Jx = uExample 4.2.1 Solve the following differential equation: ddx =3(x+3)Solution: By using definition 4.2.1, we need to rewrite the above equation in a way that each term is separated from the other term as follows:Copyright 2015 Mohammed K A Kaabar All Rights Reserved86 M. KaabarJyJx =y3(x +S) =1(x +S)1y3............... (1)Now, we need to do a cross multiplication for (1) as follows:1y3Jy =1(x + S)Jx1y3Jy -1(x +S)Jx = u ............ ....(2)Then, we integrate both sides of (2) as follows:__ 1y3Jy -1(x + S)Jx] = _u__ 1y3]Jy - __1(x +S)] Jx = c_(y-3)Jy -__1(x +S)] Jx = c-12y-2 - ln(|(x +S)|) = cThus, the general solution is : -12y-2 - ln(|(x +S)|) = cExample 4.2.2 Solve the following differential equation: ddx = c3+2xSolution: By using definition 4.2.1, we need to rewrite the above equation in a way that each term is separated from the other term as follows:Copyright 2015 Mohammed K A Kaabar All Rights Reserved87JyJx = c3+2x = c3 c2x = c2xc-3 ............ (1)Now, we need to do a cross multiplication for (1) as follows:c-3Jy = c2xJxc-3Jy -c2xJx = u .............. ...(2)Then, we integrate both sides of (2) as follows:_(c-3Jy -c2xJx) = _u_(c-3)Jy -_(c2x) Jx = c-1Sc-3 - 12c2x = cThus, the general solution is : -1Sc-3 - 12c2x = c4.3 Exact MethodInthissection,wewillsolvesomedifferential equations usingamethodknown asExactMethod.In otherwords,thismethodiscalledtheAnti-Implicit Derivative Method.Definition 4.3.1 The standard form of Exact Methodis written as follows:JyJx = -FxF .................... ..(1)Copyright 2015 Mohammed K A Kaabar All Rights Reserved88 M. KaabarThen, we solve(1) to find F(x, y), and our general solution will be as follows: F(x, y) = c for some constant c e +. In other words, the standard form for exact first order differential equation is: FJy +FxJx = u, and it is considered exact if Fx = Fx.Note: Fx(x, y) is defined as the first derivative with respect to x and considering y as a constant, while F(x, y) is defined as the first derivative with respect to y and considering x as a constant.Example 4.3.1 Given x2 +y2 -4 = u. Find ddx.Solution: By using definition 4.3.1, we first find Fx(x, y)by finding the first derivative with respect to x and considering y as a constant as follows: Fx(x, y) = 2x.Then, we find F(x, y) by finding the first derivative with respect to y and considering x as a constant as follows: F(x, y) = 2y.Thus, ddx = -PxPj = -2x2 = -x.Example 4.3.2 Given y3cx +Sxy2 - x3 +xy -1S = u.Find ddx.Solution: By using definition 4.3.1, we first find Fx(x, y)by finding the first derivative with respect to x and considering y as a constant as follows:Fx(x, y) = y3cx +Sy2 -Sx2 +y.Then, we find F(x, y) by finding the first derivative with respect to y and considering x as a constant as follows: F(x, y) = Sy2cx +6xy + x.Thus, ddx = -PxPj = -(3cx+32-3x2+)(32cx+6x+x).Copyright 2015 Mohammed K A Kaabar All Rights Reserved89Example 4.3.3 Solve the following differential equation: (-Sx +y)Jy -(Sx +Sy)Jx = uSolution: First of all, we need to check for the exact method as follows: We rewrite the above differential equation according to definition 4.3.1:(-Sx +y)Jy +-(Sx + Sy)Jx = u(-Sx +y)Jy +(-Sx - Sy)Jx = u ........ ... (1)Thus, from the above differential equation, we obtain:Fx(x, y) = (-Sx -Sy) and F(x, y) = (-Sx +y)Now, we need to check for the exact method by findingFx(x, y) = Fx(x, y) as follows:We first find Fx(x, y) by finding the first derivative of Fx(x, y) with respect to y and considering x as a constant as follows: Fx(x, y) = -SThen, we find Fx(x, y) by finding the first derivative of F(x, y) with respect to x and considering y as a constant as follows: Fx(x, y) = -SSince Fx(x, y) = Fx(x, y) = -S, then we can use the exact method.Now, we choose either Fx(x, y) = (-Sx -Sy) or F(x, y) = (-Sx + y), and then we integrate. We will choose F(x, y) = (-Sx + y) and we will integrate it as follows:_F(x, y)Jy = _(-Sx + y)Jy = -Sxy +12y2 + (x) .(2)F(x, y) = -Sxy + 12y2 +(x)We need to find (x) as follows:Copyright 2015 Mohammed K A Kaabar All Rights Reserved90 M. KaabarSince we selected F(x, y) = (-Sx + y) previously for integration, then we need to find Fx(x, y) for (2) as follows:Fx(x, y) = -Sy +i(x) .......... . ...... (S)Now, we substitute Fx(x, y) = (-Sx - Sy) in (S) as follows:(-Sx -Sy) = -Sy +i(x)i(x) = -Sx -Sy +Sy = -Sx ............ (4)Then, we integrate both sides of (4) as follows:_i(x)Jx = _-SxJx(x) = _-SxJx = -S2x2Thus, the general solution of the exact method is : F(x, y) = c-Sxy +12y2 +-S2x2 = c4.4 Reduced to Separable MethodInthissection,wewillsolvesomedifferential equationsusingamethodknownasReducedto Separable Method. Definition 4.4.1 The standard form of Reduced to Separable Method is written as follows:ddx = (ox +by + c) where o, b = u.Example 4.4.1 Solve the following differential equation: ddx =sIn(5x+)cos(5x+)-2sIn(5x+)- S.Copyright 2015 Mohammed K A Kaabar All Rights Reserved91Solution: By using definition 4.4.1, we first let u = Sx +y, and then, we need to find the first derivative of both sides of u = Sx +y.JuJx = S +JyJx....................(1)Now, we solve (1) for ddxas follows:JyJx = JuJx -S ....................(2)Then, we substitute u = Sx + y and (2) inddx =sIn(5x+)cos(5x+)-2sIn(5x+)-S as follows:JuJx - S =sin(u)cos(u) - 2sin(u)- SJuJx =sin(u)cos(u) -2sin(u)...............(S)Now, we can use the separable method to solve (S) as follows:By using definition 4.2.1, we need to rewrite (S) in a way that each term is separated from the other term as follows:JuJx =sin(u)cos(u) -2sin(u) =1cos(u) - 2sin(u)sin(u)..... .. (4)Now, we need to do a cross multiplication for (4) as follows:cos(u) - 2sin(u)sin(u)Ju = 1Jxcos(u) - 2sin(u)sin(u)Ju -1Jx = u ...........(S)Then, we integrate both sides of (S) as follows:__cos(u) -2sin(u)sin(u)Ju - 1Jx_ = _uCopyright 2015 Mohammed K A Kaabar All Rights Reserved92 M. Kaabar__cos(u) -2sin(u)sin(u)_Ju -_(1) Jx = c__cos(u)sin(u) -2 sin(u)sin(u) _Ju - x = c__cos(u)sin(u) - 2_Ju -x = cln(|sin(u)|) - 2u -x = c ............ . ... (6)Now, we substitute u = Sx + y in (6) as follows:ln(|sin(Sx + y)|) - 2(Sx +y) -x = cThus, the general solution is : ln(|sin(Sx + y)|) - 2(Sx +y) -x = c4.5 Reduction of Order MethodInthissection,wewillsolvedifferentialequations using a method called Reduction of Order Method. Definition 4.5.1 Reduction of Order Method is valid method only for second order differential equations, and one solution to the homogenous part must be given. For example, given (x + 1)y(2)(x) - yi(x) = u,and y1(x) = 1. To find y2(x), the differential equation must be written in the standard form (Coefficient of y(2)must be 1) as follows:Copyright 2015 Mohammed K A Kaabar All Rights Reserved93(x + 1)(x + 1)y(2)(x) - yi(x)(x + 1) =u(x + 1)1y(2)(x) - yi(x)(x + 1) = uy(2)(x) +-1(x + 1)yi(x) = u .............(1)Now, let H(x) =-1(x+1), and substitute it in (1) as follows:y(2)(x) + H(x)yi(x) = uHence, y2(x) is written as follows:y2(x) = y1(x) _c]-M(x)dxy12(x)JxIn our example, y2(x) = 1 ]c]1(x+1)dx(1)2Jx =1 ]cIn(x+1)1Jx = ]cIn(x+1)Jx = ](x +1) Jx = 12x2 +x.Thus, the homogenous solution is written as follows:yhomogcnous(x) = c1 +c2[12x2 + x, for some c1, c2 e +. Example 4.5.1 Given the following differential equation: xy(2)(x) +(x + 1)yi(x) - (2x + 1)y = xc7x, and y1(x) = cxis a solution to the associated homogenous part. Find yhomogcnous(x)? (Hint: Find first y2(x), and then write yhomogcnous(x))Solution: By using definition 4.5.1, To find y2(x), the differential equation must be equal to zero and must Copyright 2015 Mohammed K A Kaabar All Rights Reserved94 M. Kaabaralso be written in the standard form (Coefficient of y(2)must be 1) as follows:We divide both sides ofxy(2)(x) + (x +1)yi(x) - (2x +1)y = u by x as follows:xxy(2)(x) + (x +1)xyi(x) -2x + 1xy = ux1y(2)(x) + (x +1)xyi(x) -2x + 1xy = u ..... ...(1)Now, let (x) = (x+1)x= [1 + 1x , and substitute it in (1)as follows:y(2)(x) + H(x)yi(x) -2x + 1xy = uHence, y2(x) is written as follows:y2(x) = y1(x) _c]-M(x)dxy12(x)JxIn example 4.5.1, y2(x) = cx ]c][1+1xdx(cx)2Jx =cx _c-x cIn[1xc2xJx = cx _c-3xxJxSince it is impossible to integrate cx ]c-3xxJx, then it is enough to write it as: cx ]c-3ttx0Jt.Therefore, y2(x) = cx ]c-3ttx0Jt.Thus, the homogenous solution is written as follows:yhomogcnous(x) = c1cx + c2[cx ]c-3ttx0Jt, for some c1, c2 e +. Copyright 2015 Mohammed K A Kaabar All Rights Reserved954.6 Exercises1.Given (x + 1)yi +xy = (x+1)42. Find the general solution for y(x). (Hint: Use Bernoulli method and no need to find the value of