Second Order And Higher Order Equations Introduction
mywbut.comSecond Order and Higher Order EquationsIntroductionSecond order and higher order equations occur frequently in science and engineering (likependulum problem etc.) and hence has its own importance. It has its own flavour also. Wedevote this section for an elementary introduction.DEFINITION 8.1.1 (Second Order Linear Differential Equation) The equation(8.1.1)is called a SECOND ORDER LINEAR DIFFERENTIAL EQUATION.Hereis an interval contained incontinuous functions defined oncoefficients of Equation (8.1.1) andfunction.and the functionsthe functionsandandare real valuedare called theis called the non-homogeneous term or the forceEquation (8.1.1) is called linear homogeneous ifand non-homogeneous ifRecall that a second order equation is called nonlinear if it is not linear.EXAMPLE 8.1.21. The equationis a second order equation which is nonlinear.2.3.is an example of a linear second order equation.is a non-homogeneous linear second order equation.1
mywbut.com4.is a homogeneous second order linear equation. Thisequation is called EULER EQUATION OF ORDER 2. Hereand are real constants.DEFINITION 8.1.3 A function defined on is called a solution of Equation (8.1.1) iftwice differentiable and satisfies Equation (8.1.1).EXAMPLE 8.1.41.2.andandisare solutions ofare solutions ofWe now state an important theorem whose proof is simple and is omitted.THEOREM 8.1.5 (Superposition Principle) Letandbe two given solutions of(8.1.2)Then for any two real number(8.1.2).the functionis also a solution of EquationIt is to be noted here that Theorem 8.1.5 is not an existence theorem. That is, it does not assertthe existence of a solution of Equation (8.1.2).DEFINITION 8.1.6 (Solution Space) The set of solutions of a differential equation is calledthe solution space.For example, all the solutions of the Equation (8.1.2) form a solution space. Note thatis also a solution of Equation (8.1.2). Therefore, the solution set of a Equation (8.1.2) is nonempty. A moments reflection on Theorem 8.1.5 tells us that the solution space of Equation(8.1.2) forms a real vector space.Remark 8.1.7 The above statements also hold for any homogeneous linear differentialequation. That is, the solution space of a homogeneous linear differential equation is a realvector space.The natural question is to inquire about its dimension. This question will be answered in asequence of results stated below.We first recall the definition of Linear Dependence and Independence.2
mywbut.comDEFINITION 8.1.8 (Linear Dependence and Linear Independence) Letbe an interval inand letbe continuous functions. we say thatare said to be linearlydependent if there are real numbers and (not both zero) such thatThe functionsdependent.are said to be linearly independent ifare not linearin Equation (8.1.2) and thatTo proceed further and to simplify matters, we assume thatthe functionandare continuous onIn other words, we consider a homogeneous linear equation(8.1.3)whereandare real valued continuous functions defined onThe next theorem, given without proof, deals with the existence and uniqueness of solutions ofEquation (8.1.3) with initial conditionsfor someTHEOREM 8.1.9 (Picard's Theorem on Existence and Uniqueness) Consider the Equation(8.1.3) along with the conditions(8.1.4)where andare prescribed real constants. Then Equation (8.1.3), with initial conditionsgiven by Equation (8.1.4) has a unique solution onA word of Caution: NOTE THAT THE COEFFICIENT OF #MATH4130# MATHEND000# INEQUATION ( ) IS MATHEND000# BEFORE WE APPLY THEOREM , WE HAVE TO ENSURE THISCONDITION.An important application of Theorem 8.1.9 is that the equation (8.1.3) has exactlyindependent solutions. In other words, the set of all solutions overof dimensionlinearlyforms a real vector space3
mywbut.comTHEOREM 8.1.10 Let andbe real valued continuous functions onThen Equation(8.1.3) has exactly two linearly independent solutions. Moreover, ifandare two linearlyindependent solutions of Equation (8.1.3), then the solution space is a linear combination ofand.Proof. Letandbe two unique solutions of Equation (8.1.3) with initial conditions(8.1.5)The unique solutionsandexist by virtue of Theorem 8.1.9. We now claim thatare linearly independent. Consider the system of linear equationsand(8.1.6)whereandare unknowns. If we can show that the only solution for the system (8.1.6) is, then the two solutionsUse initial condition onthe result follows.andandwill be linearly independent.to show that the only solution is indeedWe now show that any solution of Equation (8.1.3) is a linear combination ofbe any solution of Equation (8.1.3) and letdefined byBy Definition 8.1.3,So,andand. LetConsider the functionis a solution of Equation (8.1.3). Also note thatand. Henceandare two solution of Equation (8.1.3) with the same initial conditions.Hence by Picard's Theorem on Existence and Uniqueness (see Theorem 8.1.9),orThus, the equation (8.1.3) has two linearly independent solutions. height6pt width 6pt depth 0ptRemark 8.1.114
mywbut.com1. Observe that the solution space of Equation (8.1.3) forms a real vector space ofdimension2. The solutionsandcorresponding to the initial conditionsare called a FUNDAMENTAL SYSTEM of solutions for Equation (8.1.3).3. Note that the fundamental system for Equation (8.1.3) is not unique.Consider anon-singular matrixwithfundamental system for the differential Equation 8.1.3 andof the matrixThat is, ifLetbe aThen the rowsalso form a fundamental system for Equation 8.1.3.is a fundamental system for Equation 8.1.3 thenis also a fundamental system wheneverEXAMPLE 8.1.12is a fundamental system forNote thatEXERCISE 8.1.13is also a fundamental system. Here the matrix is1. State whether the following equations are SECOND-ORDER LINEAR or SECOND-ORDERNON-LINEAR equaitons.1.2.3.4.2. By showing thatandare solutions of5
mywbut.comconclude thatandare also solutions ofform a fundamental set of solutions?3. Given thatbasis.Doforms a basis for the solution space ofandfind anotherMore on Second Order EquationsIn this section, we wish to study some more properties of second order equations which havenice applications. They also have natural generalisations to higher order equations.DEFINITION 8.2.1 (General Solution) Letforandbe a fundamental system of solutions(8.2.1)The general solutionwhereandof Equation (8.2.1) is defined byare arbitrary real constants. Note thatis also a solution of Equation (8.2.1).In other words, the general solution of Equation (8.2.1) is aparameters being-parameter family of solutions, theand6
mywbut.comSecond Order equations with ConstantCoefficientsDEFINITION 8.3.1 Letandbe constant real numbers. An equation(8.3.1)is called a SECOND ORDER HOMOGENEOUS LINEAR EQUATION WITH CONSTANTCOEFFICIENTS.Let us assume thatto be a solution of Equation (8.3.1) (wherebe determined). To simplify the matter, we denoteis a constant, and is toandIt is easy to note thatis a solution of Equation (8.3.1) if and only ifNow, it is clear that(8.3.2)Equation (8.3.2) is called the CHARACTERISTIC EQUATION of Equation (8.3.1). Equation (8.3.2)is a quadratic equation and admits 2 roots (repeated roots being counted twice).Case 1: LetThenandbe real roots of Equation (8.3.2) withare two solutions of Equation (8.3.1) and moreover they are linearlyindependent (sinceEquation (8.3.1).Case 2: LetThen). That is,forms a fundamental system of solutions ofbe a repeated root ofNow,7
mywbut.comButand therefore,Hence,andare two linearly independent solutions of Equation (8.3.1). In this case,we have a fundamental system of solutions of Equation (8.3.1).Case 3: LetSo,be a complex root of Equation (8.3.2).is also a root of Equation (8.3.2). Before we proceed, we note:LEMMA 8.3.2 Letbe a solution of Equation (8.3.1), where and are realvalued functions. Then and are solutions of Equation (8.3.1). In other words, the real partand the imaginary part of a complex valued solution (of a real variable ODE Equation (8.3.1))are themselves solution of Equation (8.3.1).Proof. exercise. height6pt width 6pt depth 0ptLetbe a complex root ofThenis a complex solution of Equation (8.3.1). By Lemma 8.3.2,are solutions of Equation (8.3.1). It is easy to note thatas good as sayingEquation (8.3.1).EXERCISE 8.3.3andandare linearly independent. It isforms a fundamental system of solutions of1. Find the general solution of the follwoing equations.1.2.3.4.where2. Solve the following IVP's.is a real constant.1.2.3.8
mywbut.com4.3. Find two linearly independent solutionsandof the following equations.1.2.3.4.4. Show that the IVPAlso, in each case, findhas a unique solution for any real number5. Consider the problem(8.3.3)6.7. Show that it has a solution if and only ifshow that ifCompare this with Exercise 4. Also,then there are infinitely many solutions to (8.3.3).9
mywbut.comNon Homogeneous EquationsThroughout this section, denotes an interval inwe assume thatandare realvalued continuous function defined onNow, we focus the attention to the study of nonhomogeneous equation of the form(8.4.1)We assume that the functionsandare known/given. The non-zero function(8.4.1) is also called the non-homogeneous term or the forcing function. The equationin(8.4.2)is called the homogeneous equation corresponding to (8.4.1).Consider the set of all twice differentiable functions defined onthis set byWe define an operatoronThen (8.4.1) and (8.4.2) can be rewritten in the (compact) form(8.4.3)(8.4.4)The ensuing result relates the solutions of (8.4.1) and (8.4.2).THEOREM 8.4.11. Letand(8.4.2).2. Letbe two solutions of (8.4.1) onbe any solution of (8.4.1) onand letThenis a solution ofbe any solution of (8.4.2). Thenis a solution of (8.4.1) on10
mywbut.comProof. Observe thatWe therefore haveThe linearity of(8.4.2).is a linear transformation on the set of twice differentiable function onimplies thator equivalently,is a solution ofFor the proof of second part, note thatimplies thatThus,is a solution of (8.4.1). height6pt width 6pt depth 0ptThe above result leads us to the following definition.DEFINITION 8.4.2 (General Solution) A general solution of (8.4.1) on(8.4.1) of the formwhere(8.4.2) andis a solution ofis a general solution of the corresponding homogeneous equationis any solution of (8.4.1) (preferably containing no arbitrary constants).We now prove that the solution of (8.4.1) with initial conditions is unique.THEOREM 8.4.3 (Uniqueness) Suppose thatIVPLetandbe two solutions of the(8.4.5)ThenProof. Letfor allThensatisfiesBy the uniqueness theorem 8.1.9, we haveheight6pt width 6pt depth 0ptonOr in other words,on11
mywbut.comRemark 8.4.4 The above results tell us that to solve (i.e., to find the general solution of (8.4.1))or the IVP (8.4.5), we need to find the general solution of the homogeneous equation (8.4.2) anda particular solutionof (8.4.1). To repeat, the two steps needed to solve (8.4.1), are:1. compute the general solution of (8.4.2), and2. compute a particular solution of (8.4.1).Then add the two solutions.Stephas been dealt in the previous sections. The remainder of the section is devoted to stepi.e., we elaborate some methods for computing a particular solutionEXERCISE 8.4.5of (8.4.1).1. Find the general solution of the following equations:1.(You may note here thatis a particular solution.)2.(First show that2. Solve the following IVPs:1.is a particular solution.)(It is given thatis a particularsolution.)2.(First guess a particular solution usingthe idea given in Exercise 8.4.5.1b )3. Letwhereandandbe two continuous functions. Let's be particular solutions ofare continuous functions. Show thatis a particular solutionof12
mywbut.comVariation of ParametersIn the previous section, calculation of particular integrals/solutions for some special cases havebeen studied. Recall that the homogeneous part of the equation had constant coefficients. In thissection, we deal with a useful technique of finding a particular solution when the coefficients ofthe homogeneous part are continuous functions and the forcing functionhomogeneous term) is piecewise continuous. Supposesolutions ofand(or the non-are two linearly independent(8.5.1)onwhereandare arbitrary continuous functions defined onis a solution of (8.5.1) for any constantsso thatandWe now vary"andThen we know thatto functions of(8.5.2)is a solution of the equation(8.5.3)where is a piecewise continuous function defined ontheorem.The details are given in the followingTHEOREM 8.5.1 (Method of Variation of Parameters) Letfunctions defined onand letandbe a piecewise continuous function ontwo linearly independent solutions of (8.5.1) ongiven bybe continuousLetThen a particular solutionandbeof (8.5.3) is(8.5.4)13
mywbut.comwhereis the Wronskian ofandthe indefinite integrals of the respective arguments.)Proof. Letand(Note that the integrals in (8.5.4) arebe continuously differentiable functions (to be determined) such that(8.5.5)is a particular solution of (8.5.3). Differentiation of (8.5.5) leads to(8.5.6)We chooseandso that(8.5.7)Substituting (8.5.7) in (8.5.6), we have(8.5.8)SinceAsis a particular solution of (8.5.3), substitution of (8.5.5) and (8.5.8) in (8.5.3), we getandare solutions of the homogeneous equation (8.5.1), we obtain the condition(8.5.9)We now determine and from (8.5.7) and (8.5.9). By using the Cramer's rule for a linearsystem of equations, we get(8.5.10)(note thatandfor anyare linearly independent solutions of (8.5.1) and hence the Wronskian,). Integration of (8.5.10) give us14
mywbut.com(8.5.11)( without loss of generality, we set the values of integration constants to zero). Equations (8.5.11)and (8.5.5) yield the desired results. Thus the proof is complete. height6pt width 6pt depth 0ptBefore, we move onto some examples, the following comments are useful.Remark 8.5.21. The integrals in (8.5.11) exist, becauseandare continuous functions anda piecewise continuous function. Sometimes, it is useful to write (8.5.11) in the formwhereandis a fixed point ingiven by (8.5.4) assumes the formIn such a case, the particular solutionisas(8.5.12)for a fixed pointand for any2. Again, we stress here that, andare assumed to be continuous. They need not beconstants. Also, is a piecewise continuous function on3. A word of caution. While using (8.5.4), one has to keep in mind that the coefficient ofin (8.5.3) isEXAMPLE 8.5.31. Find the general solution of15
mywbut.comSolution: The general solution of the corresponding homogeneous equationis given byHere, the solutionsandandare linearly independent overTherefore, a particular solution,by Theorem 8.5.1, is(8.5.13)So, the required general solution iswhereis given by (8.5.13).2. Find a particular solution ofSolution: Verify that the given equation is16
mywbut.comand two linearly independent solutions of the corresponding homogeneous part areandHereBy Theorem 8.5.1, a particular solutionis given byThe readers should note that the methods of Section 8.7 are not applicable as the givenequation is not an equation with constant coefficients.EXERCISE 8.5.41. Find a particular solution for the following problems:1.2.wherefor all3.4.2. Use the method of variation of parameters to find the general solution of1.for all2.for all3. Solve the following IVPs:1.wherewith17
mywbut.com2.for allwithandHigher Order Equations with ConstantCoefficientsThis section is devoted to an introductory study of higher order linear equations with constantcoefficients. This is an extension of the study oforder linear equations with constantcoefficients (see, Section 8.3).The standard form of a linearbyorder differential equation with constant coefficients is given(8.6.1)whereis a linear differential operator of orderwith constant coefficients,constants (called the coefficients of the linear equation) and the functionbeing realis a piecewise18
mywbut.comcontinuous function defined on the intervalderivative ofIfWe will be using the notationfor thethen (8.6.1) which reduces to(8.6.2)is called a homogeneous linear equation, otherwise (8.6.1) is called a non-homogeneous linearequation. The functionis also known as the non-homogeneous term or a forcing term.DEFINITION 8.6.1 A functiondifferentiable andRemark 8.6.21. Ifanddefined onis called a solution of (8.6.1) ifandtimesalong with its derivatives satisfy (8.6.1).are any two solutions of (8.6.1), thenHence, if is a solution of (8.6.2) andsolution of (8.6.1).2. Letisis also a solution of (8.6.2).is a solution of (8.6.1), thenis abe two solutions of (8.6.2). Then for any constants (need not be real)is also a solution of (8.6.2). The solutionis called the superposition ofand3. Note thatis a solution of (8.6.2). This, along with the super-position principle,ensures that the set of solutions of (8.6.2) forms a vector space overThis vector spaceis called the SOLUTION SPACE or space of solutions of (8.6.2).As in Section 8.3, we first take up the study of (8.6.2). It is easy to note (as in Section 8.3) thatfor a constantwhere,(8.6.3)19
mywbut.comDEFINITION 8.6.3 (Characteristic Equation) The equationin (8.6.3), is called the CHARACTERISTIC EQUATION of (8.6.2).whereis definedNote thatis of polynomial of degree with real coefficients. Thus, it has zeros(counting with multiplicities). Also, in case of complex roots, they will occur in conjugate pairs.In view of this, we have the following theorem. The proof of the theorem is omitted.THEOREM 8.6.4(8.6.3)1. Ifare distinct roots ofare the2. Ifis a solution of (8.6.2) on any intervalif and only ifthenlinearly independent solutions of (8.6.2).is a repeated root oftimes, thenof multiplicityi.e.,is a zero of (8.6.3) repeatedare linearly independent solutions of (8.6.2), corresponding to the root3. Ifis a root ofis a complex root ofofthen so is the complex conjugateThen the corresponding linearly independent solutions of (8.6.2) areThese are complex valued functions ofnote thatHowever, using super-position principle, we20
mywbut.comare also solutions of (8.6.2). Thus, in the case ofbeing a complex root ofwe have the linearly independent solutionsEXAMPLE 8.6.51. Find the solution space of the differential equationSolution: Its characteristic equation isBy inspection, the roots ofsolutions areareSo, the linearly independentand the solution space is2. Find the solution space of the differential equation21
mywbut.comSolution: Its characteristic equation isBy inspection, the roots ofsolutions areareSo, the linearly independentand the solution space is3. Find the solution space of the differential equationSolution: Its characteristic equation isBy inspection, the roots ofaresolutions areSo, the linearly independentand the solution space isFrom the above discussion, it is clear that the linear homogeneous equation (8.6.2), admitslinearly independent solutions since the algebraic equation(counting with multiplicity).has exactlyrootsDEFINITION 8.6.6 (General Solution) Letsolution of (8.6.2). Thenbe any set oflinearly independentis called a general solution of (8.6.2), whereEXAMPLE 8.6.7are arbitrary real constants.22
mywbut.com1. Find the general solution ofSolution: Note that 0 is the repeated root of the characteristic equationgeneral solution isSo, the2. Find the general solution ofSolution: Note that the roots of the characteristic equationareSo, the general solution isEXERCISE 8.6.81. Find the general solution of the following differential equations:1.2.3.2. Find a linear differential equation with constant coefficients and of orderthe following solutions:1.2.which admitsandand3.and3. Solve the following IVPs:1.2.4. Euler Cauchy Equations:Letbe given constants. The equation23
mywbut.com(8.6.4)5.6. is called the homogeneous Euler-Cauchy Equation (or just Euler's Equation) of degree(8.6.4) is also called the standard form of the Euler equation. We define7.8. Then substituting9.10. So,we getis a solution of (8.6.4), if and only if(8.6.5)11.12. Essentially, for finding the solutions of (8.6.4), we need to find the roots of (8.6.5), whichis a polynomial inWith the above understanding, solve the following homogeneousEuler equations:1.2.3.For an alternative method of solving (8.6.4), see the next exercise.13. Consider the Euler equation (8.6.4) withLetandandLetor equivalentlyThen1. show thator equivalently2. using mathematical induction, show that3. with the new (independent) variable , the Euler equation (8.6.4) reduces to anequation with constant coefficients. So, the questions in the above part can besolved by the method just explained.24
mywbut.comWe turn our attention toward the non-homogeneous equation (8.6.1). If(8.6.1) and ifis any solution ofis the general solution of the corresponding homogeneous equation (8.6.2), thenis a solution of (8.6.1). The solutionthe GENERAL SOLUTION of (8.6.1).involvesarbitrary constants. Such a solution is calledSolving an equation of the form (8.6.1) usually means to find a general solution of (8.6.1). Thesolutionis called a PARTICULAR SOLUTION which may not involve any arbitrary constants.Solving (8.6.1) essentially involves two steps (as we had seen in detail in Section 8.3).Step 1: a) Calculation of the homogeneous solutionandb) Calculation of the particular solutionIn the ensuing discussion, we describe the method of undetermined coefficients to determineNote that a particular solution is not unique. In fact, ifis a solution of (8.6.1) andis anysolution of (8.6.2), thenis also a solution of (8.6.1). The undetermined coefficientsmethod is applicable for equations (8.6.1).25
mywbut.comMethod of Undetermined CoefficientsIn the previous section, we have seen than a general solution of(8.7.6)can be written in the formwhereis a general solution ofandthis, in this section, we shall attempt to obtainis a particular solution of (8.7.6). In view offor (8.7.6) using the method of undeterminedcoefficients in the following particular cases of1.2.3.Case I.We first assume thatis not a root of the characteristic equation, i.e.,Note thatTherefore, let us assume that a particular solution is of the formwhereSinceThus,an unknown, is an undetermined coefficient. Thuswe can chooseto obtainis a particular solution of26
mywbut.comModification Rule: Ifis a root of the characteristic equation, i.e.,(i.e.,andand obtain the value ofEXAMPLE 8.7.1by substitutingwith multiplicity) then we take,of the formin1. Find a particular solution ofSolution: HerewithNote thatThis on substitution givesSo, we chooseandAlso, the characteristic polynomial,is not a root ofThus, we assumewhich gives a particular solution as2. Find a particular solution ofSolution: The characteristic polynomial isClearly,andandhas multiplicityThus, we assumeSubstituting it in the given equation,we have27
mywbut.comSolving forwe getand thus a particular solution is3. Find a particular solution ofSolution: The characteristic polynomial iswe getandThus, usingand hence a particular solution is4. SolveEXERCISE 8.7.2 Find a particular solution for the following differential equations:1.2.3.Case II.We first assume thatHere, we assume thatis not a root of the characteristic equation, i.e.,is of the formand then comparing the coefficients ofthe values of andand(why!) inobtain28
mywbut.comModification Rule: Ifmultiplicityis a root of the characteristic equation, i.e.,withthen we assume a particular solution asand then comparing the coefficients inEXAMPLE 8.7.3obtain the values ofand1. Find a particular solution ofSolution: Here,andThuswhich is not a root of thecharacteristic equationNote that the roots ofThus, let us assumeThis gives usComparing the coefficients ofandOn solving forandandwe getareon both sides, we getSo, a particular solutionis2. Find a particular solution of29
mywbut.comSolution: Here,andThuswhich is a root with multiplicityof the characteristic equationSo, letSubstituting this in the given equation and comparingthe coefficients ofandon both sides, we getandThus, aparticular solution isEXERCISE 8.7.4 Find a particular solution for the following differential equations:1.2.3.Case III.SupposeThen we assume thatand then compare the coefficient ofModification Rule: Ifmultiplicityinto obtain the values ofis a root of the characteristic equation, i.e.,forwiththen we assume a particular solution asand then compare the coefficient ofinto obtain the values offorEXAMPLE 8.7.5 Find a particular solution of30
mywbut.comSolution: Aswe assumewhich on substitution in the given differential equation givesComparing the coefficients of different powers ofandThus, a particular solution isFinally, note that ifofand solving, we getis a particular solution ofandis a particular solutionthen a particular solution ofis given byIn view of this, one can use method of undetermined coefficients for the cases, wherelinear combination of the functions described above.EXAMPLE 8.7.6 Find a particular soltution ofis aSolution: We can divide the problem into two problems:1.2.For the first problem, a particular solution (Example 8.7.3.2) isFor the second problem, one can check thatis a particular solution.31
mywbut.comThus, a particular solution of the given problem isEXERCISE 8.7.7 Find a particular solution for the following differential equations:1.2.3.4.5.6.32
Recall that a second order equation is called nonlinear if it is not linear.) is called linear homogeneous if and non-homogeneous if . E. XAMPLE . 8.1.2 . 1. The equation . is a second order equation which is nonlinear. 2. is an example of a linear second order equation. 3. is a non-homogeneous linear second order equation. mywbut.com 1
third-order Σ ADC is similar to that of a first-order CT Σ ADC. Its robustness against RC product variation is higher than that of a third-order purely CT Σ ADC. Fig. 11 shows the simulated SQNR of the proposed third-order Σ ADC and the proposed second-order NS SAR quantizer. At the OSR of 20, the second-order NS SAR
Lecture 15: Ordinary Differential Equations: Second Order 1. Key points Simutaneous 1st order ODEs and linear stability analysis. 2nd order linear ODEs (homogeneous and inhomogeneous. Maple DEplot Eigenvectors 2. General Remarks Second order ODEs are much harder to solve than first order ODEs. First of all, a second order
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Higher order linear differential equations with arbitrary order and variable coefficients are reduced in this work. The method is based on the decomposition of their coefficients and the approach reduces the order until second order equation is produced. The method to find closed-form solutions to the second order equation is then developed.
Second-Order Linear Equations - Spring System Intro - 1 Second-Order Linear Equations - Spring System So far we have seen examples of rst-order DEs, or equations with rst derivatives of some unknown function. From here on in the course, we will study di erential equations with second or higher derivatives. One classic source of di erential .
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