課件說明案例chapter92c

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.If we let , then .Thus, any exponential function of the formis a solution of Equation 1. All

3、owing C to vary through all the real numbers, we get the family of solutions.General Differential EquationsA differential equation is an equation that contains an unknown function and one or more of its derivatives.The order of a differential equation is the order of the highest derivative (of the u

4、nknown function) 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 when y=f(x) and its derivatives are substituted into the equation.For example, we know that the general

5、 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 differentiate the expression for y:The right side of the differential equation es Therefore, for every

6、 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 + C1t + C2,where C1 and C2 are independent arbitrary constants.This Equation (1) has a two-parameter family

7、of solutions. Thus, an nth-order equation has an nth-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-parameter family of solutions, then the r

8、esulting solution is called a particular solution. For example, the function (2) y(t) = -16t2 + 40t +144is a particular solution of Equation (1). Specific values for constants are usually determined by imposing additional conditions, called side conditions or initial conditions. The function given i

9、n (2) is the solution of (1) which satisfies the side conditions y(0) = 144, y(0) = 40.9.3 Separable Equations ; 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 asB

10、ecause f(x) and g(y) are continuous, they have antiderivatives. Thus, we getThis is a one-parameter family of solutions of the equation.Example 1 (a) Solve the differential equation (b) Find the solution of this equation that satisfies the initial condition Solution (a)where C is an arbitrary consta

11、nt.(b) If we put y(0) =2 in the general solution in part (a), we have C= 8/3.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 a one-parameter

12、family of solutions which defines y implicitly as a function of x. To find a solution which 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 isHomogeneous Equations The first-order differential equation is

13、 homogeneous iff it can be put in the form A homogeneous equation can 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. Th

14、us we solve it by solving the transformed equation and then substitute 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

15、 Linear EquationsA first-order differential equation is linear if it 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

16、 general solution of the first-order linear differential equation.Example 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 Th

17、is equation can be written as follows y + (- 2/x)y = 2x + 1.We haveand this is the general solution. Chapter 17 Second-Order Differential Equations 17.1 Second-Order Linear Equations 17.2 Nonhomogeneous Linear Equations 17.3 Applications of SODE 17.4 Series Solutions17.1 Second-Order Linear Equation

18、sA second-order linear differential equation has the form (1)where P, Q, R, and G are continuous. (2) homogeneousIf for some x, Equation (1) 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) =

19、 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) = c1y1(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 lin

20、early dependent, but f(x) = ex and g(x) = xex are linearly independent.Theorem If y1 and y2 are linearly independent solutions of equation (2), and P(x) is never 0, then the general solution is given by y(x) = c1y1(x)+c2y2(x)where c1 and c2 are any constants.If the differential equation has the form

21、 (3)where a(0), b, and c are constants. If y = erx , then y = rerx and y = r2erx .Substitution into the differential equation 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. (characteris

22、tic equation)By the quadratic equation formula, the roots of the characteristic equation areCase 1. If b2 4ac, 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, t

23、he characteristic equation has only one root: r1 = -b/2a.In this case, we can see that are solutions of the differential equation (3).Thus, the general solution ofis Case 3. If a2 4b, the characteristic equation has two complex conjugate roots: Setting we can write r1 = + i, r2 = - i. The functions

24、y1 = excosx, y2 = exsinx are real-valued solutions. Thus, the general solution ofis y = C1 excosx + C2 exsinx.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 equat

25、ion is the quadratic r2 + 2r 15 = 0.There are two real roots: - 5 and 3. The general solution takes 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

26、equations simultaneously, we find that C1 = 1/8, C2 = -1/8.The solution that satisfies the prescribed 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

27、. The general solution can be written y = C1e-2x + C2 xe-2x. Example 3 Find the general solution of 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 Li

28、near Equations(1)(second-order nonhomogeneous linear differential equation with constant coefficients)(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

29、 (1) and yc is the general solution of Equation (2).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 solution of ay + by + cy = G2(x), then y* = y1 + y2 is a solution of (*).The Method of Undetermined Coefficient

30、sThe method 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

31、a particular 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 a

32、nd its derivatives 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 as

33、sume that 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 =

34、 10, -6A - 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

35、of y 5y +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

36、of y = Ae2x. 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 man

37、ner, we can 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.Example 3 Solve y 4y = xex +cos2x.Solution The complementary equation y 4y = 0 has characteristic equation r2 4 = 0.Thus, the gene

38、ral solution of the complementary equation is given by yc = C1e2x + C2e-2xFor the equation y 4y = xex 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+A+B)ex = xexand A = 1/3, B = 2/9, so yp1(x)=( 1/3x 2/9)exFor the equation y 4y = cos2x, we try yp2(x)=Ccos2x+Dsin2xSubstitution gives 4Ccos