Schaum's Easy Outlines Of Differential Equations
SCHAUM’S Easy OUTLINESDIFFERENTIALEQUATIONS
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SCHAUM’S Easy OUTLINESDIFFERENTIALEQUATIONSBased on Schaum’sO u t l i n e o f T h e o r y a n d P ro b l e m s o fD i f f e re n t i a l E q u a t i o n s , S e c o n d E d i t i o nb y R i c h a r d B r o n s o n , Ph.D.Abridgement EditorE r i n J . B r e d e n s t e i n e r , Ph.D.SCHAUM’S OUTLINE SERIESM c G R AW - H I L LNew York Chicago San Francisco Lisbon London MadridMexico City Milan New Delhi San JuanSeoul Singapore Sydney Toronto
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For more information about this title, click here.ContentsChapter 1Chapter 2Chapter 3Chapter 4Chapter 5Chapter 6Chapter 7Chapter 8Chapter 9Chapter 10Chapter 11Basic Concepts and ClassifyingDifferential EquationsSolutions of First-OrderDifferential EquationsApplications of First-OrderDifferential EquationsLinear Differential Equations:Theory of SolutionsSolutions of Linear HomogeneousDifferential Equations withConstant CoefficientsSolutions of LinearNonhomogeneous Equationsand Initial-Value ProblemsApplications of Second-OrderLinear Differential EquationsLaplace Transforms and InverseLaplace TransformsSolutions by Laplace TransformsMatrices and the MatrixExponentialSolutions of Linear DifferentialEquations with ConstantCoefficients by Matrix MethodsvCopyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.18202933394755656978
vi DIFFERENTIAL EQUATIONSChapter 12Chapter 13Chapter 14Chapter 15AppendixIndexPower Series SolutionsGamma and Bessel FunctionsNumerical MethodsBoundary-Value Problemsand Fourier SeriesLaplace Transforms8598104115124133
Chapter 1Basic Conceptsand ClassifyingDifferentialEquationsIn This Chapter: Differential EquationsNotationSolutionsInitial-Value and Boundary-ValueProblemsStandard and Differential FormsLinear EquationsBernoulli EquationsHomogeneous EquationsSeparable EquationsExact Equations1Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
2 DIFFERENTIAL EQUATIONSDifferential EquationsA differential equation is an equation involving an unknown function andits derivatives.Example 1.1: The following are differential equations involving the unknown function y.dy 5x 3dxey42d2y dy 1 2 dx dx 2d3yd2y (sinx) 5 xy 0dx 3dx 2(1.1)(1.2)(1.3)372 d2y dy 3 dy 2 3 y dx y dx 5 x dx (1.4) 2 y 2 y 02 4 t x 2(1.5)A differential equation is an ordinary differential equation if the unknownfunction depends on only one independent variable. If the unknown function depends on two or more independent variables, the differential equation is a partial differential equation. In this book we will be concernedsolely with ordinary differential equations.Example 1.2: Equations 1.1 through 1.4 are examples of ordinary differential equations, since the unknown function y depends solely on the variable x. Equation 1.5 is a partial differential equation, since y depends onboth the independent variables t and x.
CHAPTER 1: Basic Concepts and Classification3Note!The order of a differential equation is the order ofthe highest derivative appearing in the equation.Example 1.3: Equation 1.1 is a first-order differential equation; 1.2, 1.4,and 1.5 are second-order differential equations. (Note in 1.4 that the order of the highest derivative appearing in the equation is two.) Equation1.3 is a third-order differential equation.NotationThe expressions y ′, y ′′, y ′′′, y ( 4 ) ,., y ( n ) are often used to represent, respectively, the first, second, third, fourth, . . ., nth derivatives of y with respect to the independent variable under consideration. Thus, y ′′ represents d 2 y / dx 2 if the independent variable is x, but represents d 2 y / dp 2if the independent variable is p. Observe that parenthesis are used in y(n)to distinguish it from the nth power, yn. If the independent variable istime, usually denoted by t, primes are often replaced by dots. Thus,y , y, and y represent, dy / dt, d 2 y / dt 2 , and d 3 y / dt 3 , respectively.SolutionsA solution of a differential equation in the unknown function y and theindependent variable x on the interval is a function y(x) that satisfiesthe differential equation identically for all x in .Example 1.4: Is y( x ) c1 sin 2 x c2 cos 2 x, where c1 and c2 are arbitrary constants, a solution of y ′′ 4 y 0 ?Differentiating y, we find y 2c1cos2x 2c2 sin2x and y 4c1 sin 2 x 4c2 cos 2 x . Hence,
4 DIFFERENTIAL EQUATIONSy ′′ 4 y ( 4c1 sin 2 x 4c2 cos 2 x ) 4(c1 sin 2 x c2 cos 2 x ) ( 4c1 4c1 )sin 2 x ( 4c2 4c2 )cos 2 x 0Thus, y c1 sin 2 x c2 cos 2 x satisfies the differential equation for allvalues of x and is a solution on the interval ( , ) .Example 1.5: Determine whether y x 2 1 is a solution of ( y ′) 4 y 2 1.Note that the left side of the differential equation must be nonnegative forevery real function y(x) and any x, since it is the sum of terms raised tothe second and fourth powers, while the right side of the equation is negative. Since no function y(x) will satisfy this equation, the given differential equation has no solutions.We see that some differential equations have infinitely many solutions (Example 1.4), whereas other differential equations have no solutions (Example 1.5). It is also possible that a differential equation has exactly one solution. Consider ( y ′) 4 y 2 0 , which for reasons identicalto those given in Example 1.5 has only one solution y 0 .You Need to KnowA particular solution of a differential equation is anyone solution. The general solution of a differentialequation is the set of all solutions.Example 1.6: The general solution to the differential equation in Example 1.4 can be shown to be (see Chapters Four and Five) y c1 sin 2 x c2 cos 2 x. That is, every particular solution of the differentialequation has this general form. A few particular solutions are: (a) y 5 sin 2 x 3 cos 2 x (choose c1 5 and c2 3 ), (b) y sin 2 x (choosec1 1 and c2 0 ), and (c) y 0 (choose c1 c2 0 ).
CHAPTER 1: Basic Concepts and Classification5The general solution of a differential equation cannot always be expressed by a single formula. As an example consider the differential equation y ′ y 2 0 , which has two particular solutions y 1 / x and y 0 .Initial-Value and Boundary-Value ProblemsA differential equation along with subsidiaryconditions on the unknown function and its derivatives, all given at the same value of the independent variable, constitutes an initial-valueproblem. The subsidiary conditions are initialconditions. If the subsidiary conditions are given at more than one value of the independentvariable, the problem is a boundary-value problem and the conditions are boundary conditions.Example 1.7: The problem y ′′ 2 y ′ e x ; y( p ) 1, y ′( p ) 2 is an initialvalue problem, because the two subsidiary conditions are both given atx p . The problem y ′′ 2 y ′ e x ; y(0) 1, y(1) 1 is a boundary-valueproblem, because the two subsidiary conditions are given at x 0 andx 1.A solution to an initial-value or boundary-value problem is a function y(x) that both solves the differential equation and satisfies all givensubsidiary conditions.Standard and Differential FormsStandard form for a first-order differential equation in the unknown function y(x) isy ′ f ( x, y)(1.6)where the derivative y ′ appears only on the left side of 1.6. Many, butnot all, first-order differential equations can be written in standard formby algebraically solving for y ′ and then setting f(x,y) equal to the rightside of the resulting equation.
6 DIFFERENTIAL EQUATIONSThe right side of 1.6 can always be written as a quotient of two other functions M(x,y) and N(x,y). Then 1.6 becomes dy / dx M ( x, y) / N ( x, y), which is equivalent to the differential formM ( x, y)dx N ( x, y)dy 0(1.7)Linear EquationsConsider a differential equation in standard form 1.6. If f(x,y) can be written as f ( x, y) p( x ) y q( x ) (that is, as a function of x times y, plus another function of x), the differential equation is linear. First-order lineardifferential equations can always be expressed asy ′ p( x ) y q( x )(1.8)Linear equations are solved in Chapter Two.Bernoulli EquationsA Bernoulli differential equation is an equation of the formy ′ p( x ) y q( x ) y n(1.9)where n denotes a real number. When n 1 or n 0, a Bernoulli equationreduces to a linear equation. Bernoulli equations are solved in ChapterTwo.Homogeneous EquationsA differential equation in standard form (1.6) is homogeneous iff (tx, ty) f ( x, y)(1.10)for every real number t. Homogeneous equations are solved in ChapterTwo.
CHAPTER 1: Basic Concepts and Classification7Note!In the general framework of differential equations,the word “homogeneous” has an entirely differentmeaning (see Chapter Four). Only in the contextof first-order differential equations does “homogeneous” have the meaning defined above.Separable EquationsConsider a differential equation in differential form (1.7). If M(x,y) A(x)(a function only of x) and N(x,y) B(y) (a function only of y), the differential equation is separable, or has its variables separated. Separableequations are solved in Chapter Two.Exact EquationsA differential equation in differential form (1.7) is exact if M ( x, y) N ( x, y) y x(1.11)Exact equations are solved in Chapter Two (where a more precise definition of exactness is given).
Chapter 2Solutions ofFirst-OrderDifferentialEquationsIn This Chapter: Separable EquationsHomogeneous EquationsExact EquationsLinear EquationsBernoulli EquationsSolved ProblemsSeparable EquationsGeneral SolutionThe solution to the first-order separable differential equation (see Chapter One).A( x )dx B( y)dy 0(2.1)8Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
CHAPTER 2: Solutions of First-Order Differential Equations9is A( x )dx B( y)dy c(2.2)where c represents an arbitrary constant.(See Problem 2.1)The integrals obtained in Equation 2.2 may be, for all practical purposes, impossible to evaluate. In such case, numerical techniques (seeChapter 14) are used to obtain an approximate solution. Even if the indicated integrations in 2.2 can be performed, it may not be algebraicallypossible to solve for y explicitly in terms of x. In that case, the solution isleft in implicit form.Solutions to the Initial-Value ProblemThe solution to the initial-value problemA( x )dx B( y)dy 0;y ( x 0 ) y0(2.3)can be obtained, as usual, by first using Equation 2.2 to solve the differential equation and then applying the initial condition directly to evaluate c.Alternatively, the solution to Equation 2.3 can be obtained fromx x0yA( s)ds B(t )dt 0(2.4)y0where s and t are variables of integration.Homogeneous EquationsThe homogeneous differential equationdy f ( x, y)dx(2.5)having the property f(tx, ty) f(x, y) (see Chapter One) can be transformed into a separable equation by making the substitution
10 DIFFERENTIAL EQUATIONSy xv(2.6)along with its corresponding derivativedydv v xdxdx(2.7)The resulting equation in the variables v and x is solved as a separabledifferential equation; the required solution to Equation 2.5 is obtained byback substitution.Alternatively, the solution to 2.5 can be obtained by rewriting the differential equation asdx1 dy f ( x, y)(2.8)x yu(2.9)dxdu u ydydy(2.10)and then substitutingand the corresponding derivativeinto Equation 2.8. After simplifying, the resulting differential equationwill be one with variables (this time, u and y) separable.Ordinarily, it is immaterial which method of solution is used. Occasionally, however, one of the substitutions 2.6 or 2.9 is definitely superior to the other one. In such cases, the better substitution is usually apparent from the form of the differential equation itself.(See Problem 2.2)Exact EquationsDefining PropertiesA differential equationM(x, y)dx N(x, y)dy 0(2.11)
CHAPTER 2: Solutions of First-Order Differential Equations11is exact if there exists a function g(x, y) such thatdg(x, y) M(x, y)dx N(x, y)dy(2.12)Note!Test for exactness: If M(x,y) and N(x,y) are continuous functions and have continuous first partialderivatives on some rectangle of the xy-plane, thenEquation 2.11 is exact if and only if M (x , y ) N(x, y) y x(2.13)Method of SolutionTo solve Equation 2.11, assuming that it is exact, first solve the equations g( x, y) M ( x, y) x(2.14) g( x, y) N ( x, y) y(2.15)for g(x, y). The solution to 2.11 is then given implicitly byg(x, y) c(2.16)where c represents an arbitrary constant.Equation 2.16 is immediate from Equations 2.11 and 2.12. If 2.12 issubstituted into 2.11, we obtain dg(x, y(x)) 0. Integrating this equation(note that we can write 0 as 0 dx), we have dg( x, y( x )) 0 dx , which,in turn, implies 2.16.Integrating FactorsIn general, Equation 2.11 is not exact. Occasionally, it is possible to transform 2.11 into an exact differential equation by a judicious multiplication. A function I(x, y) is an integrating factor for 2.11 if the equation
12 DIFFERENTIAL EQUATIONSI(x, y)[M(x, y)dx N(x, y)dy] 0(2.17)is exact. A solution to 2.11 is obtained by solving the exact differentialequation defined by 2.17. Some of the more common integrating factorsare displayed in Table 2.1 and the conditions that follow:If1 M N g( x ), a function of x alone, thenN y x I ( x, y) e Ifg ( x ) dx(2.18)1 M N h( y), a function of y alone, thenM y x h ( y ) dyI ( x, y) e (2.19)If M yf(xy) and N xg(xy), thenI ( x, y) 1xM yN(2.20)In general, integrating factors are difficult to uncover. If a differentialequation does not have one of the forms given above, then a search foran integrating factor likely will not be successful, and other methods ofsolution are recommended.(See Problems 2.3–2.6)Linear EquationsMethod of SolutionA first-order linear differential equation has the form (see Chapter One)y ′ p( x ) y q( x )(2.21)An integrating factor for Equation 2.21 isI( x) e p ( x ) dx(2.22)
CHAPTER 2: Solutions of First-Order Differential Equations13Table 2.1which depends only on x and is independent of y. When both sides of 2.21are multiplied by I(x), the resulting equationI ( x ) y ′ p( x ) I ( x ) y I ( x )q( x )(2.23)is exact. This equation can be solved by the method described previously. A simpler procedure is to rewrite 2.23 asd ( yI ) Iq( x )dx
14 DIFFERENTIAL EQUATIONSintegrate both sides of this last equation with respect to x, and then solvethe resulting equation for y. The general solution for Equation 2.21 isy I ( x )q( x )dx cI( x)where c is the constant of integration.(See Problem 2.7)Bernoulli EquationsA Bernoulli differential equation has the formy ′ p( x ) y q( x ) y n(2.24)where n is a real number. The substitutionz y1 n(2.25)transforms 2.24 into a linear differential equation in the unknown function z(x).(See Problem 2.8)Solved ProblemsSolved Problem 2.1 Solvedy x 2 2 .dxyThis equation may be rewritten in the differential form( x 2 2)dx ydy 0which is separable with A(x) x2 2 and B(y) y. Its solution is (x2 2)dx ydy cor1 31x 2 x y2 c32
CHAPTER 2: Solutions of First-Order Differential Equations15Solving for y, we obtain the solution in implicit form asy2 2 3x 4x k3with k 2c. Solving for y explicitly, we obtain th
Note! The orderof a differential equation is the order of the highest derivative appearing in the equation. Example 1.3:Equation 1.1 is a first-order differential equation; 1.2, 1.4, and 1.5 are second-order differential equations. (Note in 1.4 that the or-der of the highest derivative appearing in the equation is two.)
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