1、Chapter 9 Differential Equations 9.1 Modeling with Differential Equations*9.2 Direction Fields and Eulers Method 9.3 Separable Equations*9.4 Exponential Growth and Decay*9.5 The Logistic Equation 9.6 Linear Equations9.1 Modeling with Differential equationsModels of Population Growth Let t =time, P =

2、the number of individuals in the population. Our assumption that the rate of growth of the population is proportional to the population size is written as the equation where k is the proportionality constant.Example Find a function f such that its graph passes through the point（1，2），and that the slo

3、pe of its tangent line at (x,f(x) is 2x.SolutionUsing the fact that f(1)=2xyo112The initial-value problemInitial conditionor side conditionA particular solutionA general solutionGeneral Differential EquationsA differential equation is an equation that contains an unknown function and one or more of

4、its derivatives.The order of a differential equation is the order of the highest derivative (of the unknown func-tion) that appears in the equation. are first-order equations;are equations of the second-order.A function f is called a solution of a differential equation if the equation is satisfied w

5、hen y=f(x) and its derivatives are substituted into the equation.For example, we know that the general solution of the differential equation is given by where C is an arbitrary constant.Example 1 Show that every member of the family of functionsis a solution of the differential equation Solution We

6、differentiate the expression for y:The right side of the differential equation becomes Therefore, for every value of c, the given function is a solution of the differential equation. c is often called parameter. The set of solutions of the second-order differential equationis given by y(t) = -16t2 +

7、 C1t + C2, (2)where C1 and C2 are independent arbitrary constants.This equation (1) has a two-parameter family of solutions. Generally, the term general solution is often used in place of nth-parameter family of solutions. If specific values are assigned to each of the arbitrary constants in an nth-

8、parameter family of solutions, then the resulting solution is called a particular solution. Specific values for constants are usually determined by imposing additional conditions, called side conditions or initial conditions, and the problem of finding a solution of the differential eqution that sat

9、isfies the initial condition is called an initial-value problem. Thus, an nth-order equation has an nth-parameter family of solutions. For example, the function (2) y(t) = -16t2 + 40t +144is a particular solution of Equationwhich satisfies the side conditions y(0) = 144, y(0) = 40.9.3 Separable Equa

10、tions ; Homogeneous EquationsThe first-order equation is separable iff it can be put in the form y = f(x)g(y)where f(x) and g(y) are continuous. To solve such an equation, we begin by writing asbecause f(x) and g(y) are continuous, they have antiderivatives. Thus, we getThis is the general solution

11、of the equation.Example 1 (a) Solve the differential equation (b) Find the solution of this equation that satisfies the initial condition Solution sowhere C is an arbitrary constant.(a) Note that this equation can be written as(b) If we put y(0) =2 in the general solution in part (a), we have C= 8/3

12、.Thus, the solution of the initial-value problem is Example 2 Find a solution of the differential equation which satisfies the side condition y(2) = 1.Solution We show first that the equation is separable:is the general solutions which defines y implicitly as a function of x. To find a solution whic

13、h satisfies the side condition, we set x = 2 and y = 1 in the one-parameter family: 1 + ln1 = 1/222 2 + C and C = 1.Thus, the particular solution isThe function y=0is also a solutionof the given differ-ential equtionExample 2 Find a one-parameter family of solution of the equation SolutionNote that

14、this equation can be written assowhereThe function y=0 is also a solution of the given differ-ential equtionThus the general solutions iswhere C is an arbitrary constant.Homogeneous Equations The first-order differential equation is homogeneous iff it can be put in the form A homogeneous equation ca

15、n be transformed into a separable equation by setting v = y/x.To see this, write xv = y and differentiate with respect to x: v + xv = y.Substituting y and y into equation, gives v + xv = f(v), and The above equation is separable. Thus we solve it by solving the transformed equation and then substitu

16、te y/x back in for v.Example 1 Find a one-parameter family of solution of the equation xyy = 3x2 + y2.Solution Note that this equation can be written asNow, set vx = y. Then, v + xv = y and Substituting y/x back in for v, we have9.6 Linear EquationsA first-order differential equation is linear if it

17、 can be written in the form y + p(x)y = q(x)where p(x) and q(x) are continuous functions on a given interval.To solve the equation y + p(x)y = q(x), multiply both sides by the integrating factor and integrate both sides.Thus, is the general solution of the first-order linear differential equation.Ex

18、ample 1 Solve the differential equationSolution An integrating factor is Multiplying both sides of the differential equation by we get Integrating both sides, we haveExample 2 Find the general solution of xy 2y = 2x2 + x.Solution This equation can be written as follows y + (- 2/x)y = 2x + 1.We havea

19、nd this is the general solution. Example 3 Find the general solution of Solution This equation can be written as followsWe haveand this is the general solution. Example 2 Find a one-parameter family of solution of the equation SolutionC is an arbitrary constant.Chapter 17 Second-Order Differential E

20、quations 17.1 Second-Order Linear Equations 17.2 Nonhomogeneous Linear Equations 17.3 Applications of SODE 17.4 Series Solutions17.1 Second-Order Linear EquationsA second-order linear differential equation has the form (1)where P, Q, R, and G are continuous. (2) homogeneousIf for some x, Equation (1

21、) is nonhomogeneous.Theorem If y1(x) and y2(x) are both solutions of the linear homogeneous equation (2) and c1 and c2 are any constants, then the functiony(x) = c1y1(x)+c2y2(x)is also a solution of Equation (2).Proof Since y1 and y2 are solutions of Equation (2), we have Therefore, Thus, y(x) = c1y

22、1(x)+c2y2(x) is a solution of Equation (2).linearly independent solutions y1 and y2 : y1 /y2 constantFor instance, the functions f(x) = x2 and g(x) = 5x2 are linearly dependent, but f(x) = ex and g(x) = xex are linearly independent.Theorem If y1 and y2 are linearly independent solutions of equation

23、(2), and P(x) is never 0, then the general solution of equation (2) is given by y(x) = c1y1(x)+c2y2(x)where c1 and c2 are any constants.If the differential equation has the form (3)where a(0), b, and c are constants. If y = erx , then y = rerx and y = r2erx .Substitution into the differential equati

24、on gives ar2erx + brerx + cerx = 0 and, since erx 0, ar2 + br + c = 0. This show that the function y = erx satisfies the differential equation iff ar2 + br + c = 0. (characteristic equation)kriktristik By the quadratic equation formula, the roots of the characteristic equation areCase 1. If b2 4ac,

25、the characteristic equation has two distinct real roots:Both are linearly independent solutions of the differential equation (3). Therefore, the general solution ofis Case 2. If b2 = 4ac, the characteristic equation has only one root: r1 = -b/2a.In this case, we can see that are solutions of the dif

26、ferential equation (3).Thus, the general solution ofis Case 3. If b2 4ac, the characteristic equation has two complex conjugate roots: Setting we can write r1 = + i, r2 = - i. The functions y1 = excosx, y2 = exsinx are real-valued solutions. Thus, the general solution ofis y = ex (C1 cosx + C2 sinx)

27、.kndu,geit Example 1 Find the general solution of the equation y + 2y 15y = 0.Then find the particular solution that satisfies the side conditions y(0) = 0, y(0) = - 1.Solution The characteristic equation is the quadratic r2 + 2r 15 = 0.There are two real roots: - 5 and 3. The general solution takes

28、 the form y = C1e-5x +C2e3x. Differentiating the general solution, we have y = -5C1e-5x + 3C2e3x.The conditions y(0) = 0, y(0) = - 1 are satisfied iff C1 + C2 = 0 and -5C1 + 3C2 = -1.Solving these two equations simultaneously, we find that C1 = 1/8, C2 = -1/8.The solution that satisfies the prescrib

29、ed side conditions is the functionExample 2 Find the general solution of the equation y + 4y + 4y = 0.Solution The characteristic equation is the quadratic r2 + 4r + 4 = 0.The number 2 is the only root. The general solution can be written y = C1e-2x + C2 xe-2x. Example 3 Find the general solution of

30、 the equation y + y + 3y = 0.Solution The characteristic equation is r2 + r + 3 =0.The quadratic formula shows that there are two complex roots:The general solution takes the form17.1 Nonhomogeneous Linear Equations(1)(second-order nonhomogeneous linear differential equation with constant coefficien

31、ts)(2)is called the complementary equation.Theorem The general solution of the nonhomogeneous differential equation (1) can be written as y(x) = yp(x)+yc(x)where yp is a particular solution of Equation (1) and yc is the general solution of Equation (2).The Method of Undetermined CoefficientsThe meth

32、od of undetermined coefficients works for equations (1) ay + by + cy = G(x), where the function G(x) has a special form. A Particular Solution of ay + by + cy = G(x)Note: If yp satisfies the equation ay + by + cy = 0, try xyp; if xyp also satisfies the reduced equation, then x2yp will give a particu

33、lar solution. Example 1 Find a particular solution of each of the following differential equations: (a) y + 2y +5y = 10e-2x. (b) y + 2y +y = 10cos3x + 6sin3x.Solution (a) We assume that the equation has a solution of the form yp = Ae-2x. Then (yp)= -2Ae-2x and (yp)= 4Ae-2x.Substituting yp and its de

34、rivatives into the equation, we have 4Ae-2x + 2(-2Ae-2x) + 5Ae-2x = 10e-2x 5Ae-2x = 10e-2x.Therefore, 5A = 10 and A = 2.Thus, a particular solution of the equation (a) is yp = 2e-2x.We can verify that the general solution of the equation (a) is y = C1e-xcos2x + C2e-xsin2x + 2e-2x. (b) We assume that

35、 the equation has a solution of the form yp = Acos3x + Bsin3x. Then (yp)= -3Asin3x + 3Bcos3x and (yp)= -9Acos3x -9Bsin3x.Substituting yp and its derivatives into the equation, we have (-8A + 6B)cos3x + (-6A -8B)sin3x = 10cos3x + 6sin3xThis equation will be satisfied for all x iff -8A + 6B = 10, -6A

36、- 8B = 6.The solution of this pair of equations is A = -29/25, B =3/25.Thus, a particular solution of the equation (b) is yp = -29/25cos3x +3/25sin3x.We can verify that the general solution of the equation (b) is y = C1e-x + C2xe-x -29/25cos3x +3/25sin3x Example 2 Find the general solution of y 5y +

37、6y = 4e2x.Solution The complementary equation y 5y +6y = 0 has characteristic equation r2 5r + 6 = (r 2)(r 3) = 0Thus, the general solution of the complementary equation is given by yc = C1e2x + C2e3xNoting that y = Ae2x satisfies the equation y 5y +6y = 0, we should try yp = Axe2x instead of y = Ae

38、2x. The derivatives of yp are (yp) = Ae2x + 2Axe2x, (yp) = 4Ae2x + 4Axe2x. Substituting into the differential equation, we have 4Ae2x + 4Axe2x 5(Ae2x + 2Axe2x) + 6Axe2x = 4e2x -Ae2x = 4e2x and A = -4.Thus, yp = C1e2x + C2e3x 4xe2x is the general solution of y 5y +6y = 4e2x. In a similar manner, we c

39、an verify that the form of a particular solution of y 4y +4y = e2x is yp = Ax2e2x since e2x and xe2x are thesolutions of the complementary equation y 4y +4y = 0.THE SUPERPOSITION PRINCIPLE Given the equation (*) ay + by + cy = G1(x) + G2(x). If y1 is a solution of ay + by + cy = G1(x) and y2 is a so

40、lution of ay + by + cy = G2(x), then y* = y1 + y2 is a solution of (*).Example 3 Solve y 4y = xex+cos2x.Solution The complementary equation y 4y = 0 has characteristic equation r2 4 = 0.Thus, the general solution of the complementary equation is given by yc = C1e2x + C2e-2xFor the equation y 4y = xe

41、x we try yp1(x)=(Ax+B)exThen (yp1) = (Ax+A+B)ex, (yp1) = (Ax+2A+B)ex, so substitution in the equation gives (Ax+2A+B)ex_ 4(Ax+B)ex= xex Or -3Ax+2A-3B=xThus -3A=1and 2A-3B=0,so A=-1/3,B=-2/9,and yp1(x)=-(1/3 x+2/9)exFor the equation y 4y = cos2x we try yp2(x)=Ccos2x+Dsin2xsubstitution in the equation

42、 gives -8Ccos2x-8Dsin2x=cos2xThus C=-1/8,D=0and yp2(x)=-1/8cos2xBy the superposition principle ,the general solutionIs y= C1e2x + C2e-2x )-(1/3 x+2/9)ex -1/8cos2xThe Method of Variation of ParametersThe complementary equationits general solution is given by Where and are linearly independent solutio

43、nsletis a particular solution of the nonhomogeneous linear differential equation ay + by + cy = G(x),Find a particular solution of the differential equation:ExampleSolutionThe auxiliary eqution isWith ,so the solution ofisUsing Variation of Parametersletis a particular solution of the nonhomogeneous

44、 linear differential equation thenits solution isIntegrating these equations, we haveSo a particular solution of the differential equation isThree type higher order differential equationsIntegrating the equation, we havesois the general solution of the equation.Let ,thenSubstituting and into the equ

45、ation, we haveExampleSolveSolutionLet ,thenSubstituting and into the equation, we haveSubstituting back in for p, we haveis the general solution of the equation.Let ,thenSubstituting and into the equation, we haveExampleSolveSolutionLet ,thenSubstituting and into the equation, we haveSubstituting ba

46、ck in for p, we haveis the general solution of the equation.ExercisesDefinition: The chang is called the difference of at and is denoted by whereDifference EquationsExample Find andSolution By the definition of the difference,we haveProperties of DifferenceExample If ,Find Solution the second-order

47、differencethe three-order differenceOtherwise Example If , Find and Solutionwe haveorDefinitionA difference equation is an equation that contains an unknown function and its differences.General Difference EquationsTwo forms can be transformed into each other. The order of a difference equation is th

48、e maximum difference between the subscripts of unknown functi-on. Definitionnth-order difference equationSecond-order difference equationA function f is called a solution of a difference equation if the equation is satisfied when y=f(x) is substituted into the equation.Definitionan nth-order equatio

49、n has an nth-parameter family of solutions and it is also called general solution. If specific values are assigned to each of the arbitrary constants in an nth-parameter family of solutions, then the resulting solution is called a particular solution. Specific values for constants are usually determ

50、ined by imposing additional conditions, called side conditions or initial conditions, and the problem of finding a solution of the differential eqution that satisfies the initial condition iscalled an initial-value problem. An initial-value problem A nth-order linear difference equation with constan

51、t coefficients has the form (1)where are constant and (2)If for some x, Equation (1) is nonhomogeneous.nth-order homogeneous linear difference equation with constant coefficientsTheorem If are solutions of the linear homogeneous equation (2) and are any constants, then the functionis also a solution

52、 of equation (2).Theorem If are linearly independent solutions of equation (2), then the general solution of equation (2) is given bywhere are any constants.Theorem The general solution of the non-homogeneous differential equation (1) can be written as where is a particular solution of equation (1)

53、and is the general solution of equation (2).First-order difference equation with constant coefficientswhere are constants andFirst-order homogeneous linear difference equation with constant coefficientsIts general solution iswhere is any constants. A Particular Solution ofTry when ,Try whenTry whenN

54、ote: If satisfies the equation try if also satisfies the reduced equation, then will give a particular solution.Find a particular solution of the difference equation:ExampleSolution We assume that the equation has a solution of the form Substituting into the equation, we have and Thus, a particular

55、solution of the equation isFind a particular solution of the difference equation:ExampleSolution We assume that the equation has a solution of the form Substituting into the equation, we have and Thus, a particular solution of the equation isExample Find the general solution of the equation then fin

56、d the particular solution that satisfies the side conditionSolution The general solution of takes the form where is any constants.We assume that the equation has a solution of the form Substituting into the equation, we have and Thus, a particular solution of the equation isTherefore, the general so

57、lution of the difference equation iswhere is any constants.The conditions y(0) = 1 is satisfied iff Thus,the solution that satisfies the side conditions is the functionFind a particular solution of the difference equation:ExampleSolution We assume that the equation has a solution of the form Substit

58、uting into the equation, we have it is This equation will be satisfied for all x iffthus Therefore, a particular solution of the equation isSecond-order difference equation with constant coefficientswhere are constants andSecond-order homogeneous linear difference equation with constant coefficientsThe general solution of the difference equation We assume that the equation has a solution of the formSubstitution into the difference equation givesSince soThis show that the function satisfies the differential equation iff . (characteristic equation)By the quadratic equation formula, the roots