Advanced Engineering Mathematics, 7 th Edition Peter V. O’Neil © 2012 Cengage Learning Engineering. All Rights Reserved. CHAPTER 2 Linear Second-Order.

© 2012 Cengage Learning Engineering. All Rights Reserved.2 A second-order differential equation is one containing a second derivative but no higher derivative. The theory of second-order differential equations is vast, and we will focus on linear second-order equations, which have many important uses. CHAPTER 2 : Page 49

© 2012 Cengage Learning Engineering. All Rights Reserved.3 2.1 The Linear Second-Order Equation This section lays the foundations for writing solutions of the second-order linear differential equation. Generally, this equation is CHAPTER 2 : Page 49

© 2012 Cengage Learning Engineering. All Rights Reserved.4 2.1 The Linear Second-Order Equation Notice that this equation “loses” its second derivative at any point where P(x) is zero, presenting technical difficulties in writing solutions. We will therefore begin by restricting the equation to intervals (perhaps the entire real line) on which P(x)  0. On such an interval, we can divide the differential equation by P(x) and confine our attention to the important case (2.1) We will refer to this as the second-order linear differential equation. CHAPTER 2 : Page 49

© 2012 Cengage Learning Engineering. All Rights Reserved.5 2.1 The Linear Second-Order Equation Often, we assume that p and q (coefficient functions) are continuous (at least on the interval where we seek solutions). The function f is called a forcing function for the differential equation, and in some applications, it can have finitely many jump discontinuities. CHAPTER 2 : Page 49

© 2012 Cengage Learning Engineering. All Rights Reserved.6 2.1 The Linear Second-Order Equation To get some feeling for what we are dealing with, consider a simple example y  − 12x = 0. Since y  = 12x, we can integrate once to obtain and then once again to get with c and k as arbitrary constants. CHAPTER 2 : Page 49-50

© 2012 Cengage Learning Engineering. All Rights Reserved.7 2.1 The Linear Second-Order Equation It seems natural that the solution of a second-order differential equation, which involves two integrations, should contain two arbitrary constants. For any choices of c and k, we can graph the corresponding solution, obtaining integral curves. Figure 2.1 shows integral curves for several choices of c and k. CHAPTER 2 : Page 50 Usually we said those two arbitrary constants are independent. Here “independent” means a different set of constants will results in different solution functions.

© 2012 Cengage Learning Engineering. All Rights Reserved.9 2.1 The Linear Second-Order Equation Unlike the first-order case, there may be many integral curves through a given point in the plane. In this example, if we specify that y(0) = 3, then we must choose k = 3, leaving c still arbitrary. These solutions through (0, 3) are y(x) = 2x 3 +cx + 3. Some of these curves are shown in Figure 2.2. CHAPTER 2 : Page 50

© 2012 Cengage Learning Engineering. All Rights Reserved.11 2.1 The Linear Second-Order Equation We single out exactly one of these curves if we specify its slope at (0, 3). For example, if we specify that y(0) = −1, then y(0) = c = −1, so y(x) = 2x 3 −x + 3. This is the only solution passing through (0, 3) with slope −1. CHAPTER 2 : Page 51

© 2012 Cengage Learning Engineering. All Rights Reserved.12 2.1 The Linear Second-Order Equation To sum up, in this example, we obtain a unique solution by specifying a point that the graph must pass through, together with the slope this solution must have at this point. Thus, by specifying two conditions for a point it is possible to get a unique solution. Such kind of conditions are referred to initial conditions. CHAPTER 2 : Page 51

© 2012 Cengage Learning Engineering. All Rights Reserved.13 2.1 The Linear Second-Order Equation This leads us to define the initial value problem for the linear second-order differential equation as the problem in which x 0, A, and B are given. We will state, without proof, an existence theorem for this initial value problem. CHAPTER 2 : Page 51

© 2012 Cengage Learning Engineering. All Rights Reserved.14 THEOREM 2.1 Existence of Solutions Let p, q, and f be continuous on an open interval I. Then the initial value problem has a unique solution on I. CHAPTER 2 : Page 51

© 2012 Cengage Learning Engineering. All Rights Reserved.15 2.1 The Linear Second-Order Equation We now have an idea of the kind of problem we will be solving and of some conditions under which we are guaranteed a solution. Now we want to develop a strategy to follow to solve linear equations and initial value problems. This strategy will be in two steps, beginning with the case that f (x) is identically zero. CHAPTER 2 : Page 51

© 2012 Cengage Learning Engineering. All Rights Reserved.17 The Structure of Solutions If y 1 and y 2 are solutions and c 1 and c 2 are numbers, we call c 1 y 1 + c 2 y 2 a linear combination of y 1 and y 2. It is an important property of the homogeneous linear equation (2.2) that a linear combination of solutions is again a solution. CHAPTER 2 : Page 51

© 2012 Cengage Learning Engineering. All Rights Reserved.19 THEOREM 2.2 － Proof Let y 1 and y 2 be solutions, and let c 1 and c 2 be numbers. Substitute c 1 y 1 + c 2 y 2 into the differential equation: because of the assumption that y 1 and y 2 are both solutions. CHAPTER 2 : Page 51

© 2012 Cengage Learning Engineering. All Rights Reserved.20 2.1 The Linear Second-Order Equation The point to taking linear combinations c 1 y 1 +c 2 y 2 is to generate new solutions from y 1 and y 2. However, if y 2 is itself a constant multiple of y 1, say y 2 = ky 1, then the linear combination is just a constant multiple of y 1, so y 2 does not contribute any new information. This leads to the following definition. CHAPTER 2 : Page 52

© 2012 Cengage Learning Engineering. All Rights Reserved.21 2.1 The Linear Second-Order Equation CHAPTER 2 : Page 52 Two functions are linearly independent on an open interval I (which can be the entire real line) if neither function is a constant multiple of the other for all x in the interval. If one function is a constant multiple of the other on the entire interval, then these functions are called linearly dependent. The definition of “linear independency” is the same as that you will learn in Linear Algebra.  Anyone cannot be represented by the others. Now, it is simple to define for only two functions.

© 2012 Cengage Learning Engineering. All Rights Reserved.22 EXAMPLE 2.1 cos(x) and sin(x) are solutions of y  + y = 0 over the entire real line. These solutions are linearly independent, because there is no number k such that cos(x) = k sin(x) or sin(x) = k cos(x) for all x. Because these solutions are linearly independent, linear combinations c 1 cos(x) + c 2 sin(x) give us new solutions, not just constant multiples of one of the known solutions. CHAPTER 2 : Page 52

© 2012 Cengage Learning Engineering. All Rights Reserved.23 2.1 The Linear Second-Order Equation There is a simple test to determine whether two solutions of equation (2.2) are linearly independent or dependent on an open interval I. Define the Wronskian W(y 1, y 2 ) of two solutions y 1 and y 2 to be the 2 × 2 determinant Often we denote this Wronskian as just W(x). CHAPTER 2 : Page 52 Why need Wronskian? It is simple to test linear independence for two functions. But for more functions or power series functions, it may not be easy to check.

© 2012 Cengage Learning Engineering. All Rights Reserved.24 THEOREM 2.3 Properties of the Wronskian Suppose y 1 and y 2 are solutions of equation (2.2) on an open interval I. 1.Either W(x)=0 for all x in I, or W(x)  0 for all x in I. 2.y 1 and y 2 are linearly independent on I if and only if W(x)  0 on I. The proof will be given later. CHAPTER 2 : Page 52

© 2012 Cengage Learning Engineering. All Rights Reserved.25 2.1 The Linear Second-Order Equation Conclusion (2) is called the Wronskian test for linear independence. Two solutions are linearly independent on I exactly when their Wronskian is nonzero on I. In view of conclusion (1), we need only check the Wronskian at a single point of I, since the Wronskian must be either identically zero on the entire interval or nonzero on all of I. It cannot vanish for some x and be nonzero for others in I. CHAPTER 2 : Page 52

© 2012 Cengage Learning Engineering. All Rights Reserved.26 THEOREM 2.3 Properties of the Wronskian Note that the above conclusions are for y 1 and y 2 are solutions of equation (2.2) on an open interval I. Suppose y 1 and y 2 are two functions. 1.There is no entirely zero or nonzero property. 2.y 1 and y 2 are linearly independent on I if and only if W(x)  0 for some x on I. CHAPTER 2 : Page 52

© 2012 Cengage Learning Engineering. All Rights Reserved.27 EXAMPLE 2.2 Check by substitution that y 1 (x) = e 2x and y 2 (x) = xe 2x are solutions of y  − 4y + 4y = 0 for all x. The Wronskian is and this is never zero, so y 1 and y 2 are linearly independent solutions. CHAPTER 2 : Page 52

© 2012 Cengage Learning Engineering. All Rights Reserved.28 2.1 The Linear Second-Order Equation In many cases, it is obvious whether two functions are linearly independent or dependent. However, the Wronskian test is important, as we will see shortly (for example, in Section 2.4.1 and in the proof of Theorem 2.4). CHAPTER 2 : Page 53

© 2012 Cengage Learning Engineering. All Rights Reserved.29 2.1 The Linear Second-Order Equation We are now ready to determine what is needed to find all solutions of the homogeneous linear equation y  + p(x)y + q(x)y = 0. We claim that, if we can find two linearly independent solutions, then every solution must be a linear combination of these two solutions. CHAPTER 2 : Page 53

© 2012 Cengage Learning Engineering. All Rights Reserved.30 THEOREM 2.4 Let y 1 and y 2 be linearly independent solutions of y  + py + qy = 0 on an open interval I. Then every solution on I is a linear combination of y 1 and y 2. CHAPTER 2 : Page 53

© 2012 Cengage Learning Engineering. All Rights Reserved.31 THEOREM 2.4 CHAPTER 2 : Page 53 This provides a strategy for finding all solutions of the homogeneous linear equation on an open interval. 1. Find two linearly independent solutions y 1 and y 2. 2. The linear combination y = c 1 y 1 + c 2 y 2 contains all possible solutions.

© 2012 Cengage Learning Engineering. All Rights Reserved.32 THEOREM 2.4 CHAPTER 2 : Page 53 For this reason, we call two linearly independent solutions y 1 and y 2 a fundamental set of solutions on I, and we call c 1 y 1 + c 2 y 2 the general solution of the differential equation on I. Once we have the general solution c 1 y 1 + c 2 y 2 of the differential equation, we can find the unique solution of an initial value problem by using the initial conditions to determine c 1 and c 2.

© 2012 Cengage Learning Engineering. All Rights Reserved. Let φ be any solution of on I. We want to show that there are numbers c 1 and c 2 such that (not to show is a solution, which is Theorem 2.2.) Choose any x 0 in I. Let and. Then φ is the unique solution of the initial value problem 33 THEOREM 2.4 － Proof CHAPTER 2 : Page 53

© 2012 Cengage Learning Engineering. All Rights Reserved.34 THEOREM 2.4 － Proof Now consider the two algebraic equations in the two unknowns c 1 and c 2 : Because y 1 and y 2 are linearly independent, their Wronskian W(x) is nonzero. These two algebraic equations therefore yield CHAPTER 2 : Page 53

© 2012 Cengage Learning Engineering. All Rights Reserved.35 THEOREM 2.4 － Proof CHAPTER 2 : Page 53 With these choices of c 1 and c 2, it is easy to prove is a solution of φ is also a solution of the initial value problem. Since this problem has the unique solution, then we can claim, as we wanted to show.

© 2012 Cengage Learning Engineering. All Rights Reserved.36 EXAMPLE 2.3 e x and e 2x are solutions of y  − 3y + 2y = 0. Therefore, every solution has the form This is the general solution of y  − 3y + 2y = 0. CHAPTER 2 : Page 54

© 2012 Cengage Learning Engineering. All Rights Reserved.37 EXAMPLE 2.3 If we want to satisfy the initial conditions y(0) = −2, y(0) = 3, choose the constants c 1 and c 2 so that y(0) = c 1 + c 2 = −2 y'(0) = c 1 + 2c 2 = 3. Then c 1 = −7 and c 2 = 5, so the unique solution of the initial value problem is y(x) = −7e x + 5e 2x. CHAPTER 2 : Page 54

© 2012 Cengage Learning Engineering. All Rights Reserved.38 The Nonhomogeneous Case We now want to know what the general solution of equation (2.1) looks like when f (x) is nonzero at least for some x. In this case, the differential equation is nonhomogeneous. The main difference between the homogeneous and nonhomogeneous cases is that, for the nonhomogeneous equation, sums and constant multiples of solutions need not be solutions. CHAPTER 2 : Page 54

© 2012 Cengage Learning Engineering. All Rights Reserved.39 EXAMPLE 2.4 We can check by substitution that sin(2x) +2x and cos(2x) + 2x are solutions of the nonhomogeneous equation y  + 4y = 8x. However, if we substitute the sum of these solutions, sin(2x) + cos(2x) + 4x, into the differential equation, we find that this sum is not a solution. Furthermore, if we multiply one of these solutions by 2, taking, say, 2 sin(2x) + 4x, we find that this is not a solution either. CHAPTER 2 : Page 54

© 2012 Cengage Learning Engineering. All Rights Reserved.40 2.1 The Linear Second-Order Equation However, given any two solutions Y 1 and Y 2 of equation (2.1), we find that their difference Y 1 − Y 2 is a solution, not of the nonhomogeneous equation, but of the associated homogeneous equation (2.2). To see this, substitute Y 1 − Y 2 into equation (2.2): CHAPTER 2 : Page 54

© 2012 Cengage Learning Engineering. All Rights Reserved.41 2.1 The Linear Second-Order Equation But the general solution of the associated homogeneous equation (2.2) has the form c 1 y 1 + c 2 y 2, where y 1 and y 2 are linearly independent solutions of the homogeneous equation (2.2). Since Y 1 − Y 2 is a solution of this homogeneous equation, then for some numbers c 1 and c 2 : which means that CHAPTER 2 : Page 54

© 2012 Cengage Learning Engineering. All Rights Reserved.42 2.1 The Linear Second-Order Equation But Y 1 and Y 2 are any solutions of equation (2.1). This means that, given any one solution Y 2 of the nonhomogeneous equation, any other solution has the form c 1 y 1 + c 2 y 2 + Y 2 for some constants c 1 and c 2. We will summarize this discussion as a general conclusion, in which we will use Y p (for particular solution) instead of Y 2 of the discussion. CHAPTER 2 : Page 54

© 2012 Cengage Learning Engineering. All Rights Reserved.43 THEOREM 2.5 Let Y p be any solution of the nonhomogeneous equation (2.1). Let y 1 and y 2 be linearly independent solutions of equation (2.2). Then the expression c 1 y 1 + c 2 y 2 + Y p contains every solution of equation (2.1). CHAPTER 2 : Page 55

© 2012 Cengage Learning Engineering. All Rights Reserved.44 THEOREM 2.5 For this reason, we call c 1 y 1 + c 2 y 2 + Y p the general solution of equation (2.1). Theorem 2.5 suggests a strategy for finding all solutions of the nonhomogeneous equation (2.1). 1.Find two linearly independent solutions y 1 and y 2 of the associated homogeneous equation y  + p(x)y + q(x)y = 0. CHAPTER 2 : Page 55

© 2012 Cengage Learning Engineering. All Rights Reserved.45 THEOREM 2.5 2. Find any particular solution Y p of the nonhomogeneous equation y  + p(x)y + q(x)y = f (x). 3. The general solution of y  + p(x)y + q(x)y = f (x) is y(x) = c 1 y 1 (x) + c 2 y 2 (x) + Y p (x) in which c 1 and c 2 can be any real numbers. CHAPTER 2 : Page 55

© 2012 Cengage Learning Engineering. All Rights Reserved.46 THEOREM 2.5 If there are initial conditions, use these to find the constants c 1 and c 2 to solve the initial value problem. CHAPTER 2 : Page 55

© 2012 Cengage Learning Engineering. All Rights Reserved.47 EXAMPLE 2.5 We will find the general solution of y  + 4y = 8x. It is routine to verify that sin(2x) and cos(2x) are linearly independent solutions of y  + 4y = 0. Observe also that Y p (x) = 2x is a particular solution of the nonhomogeneous equation. CHAPTER 2 : Page 55

© 2012 Cengage Learning Engineering. All Rights Reserved.48 EXAMPLE 2.5 Therefore, the general solution of y  + 4y =8x is y = c 1 sin(2x) + c 2 cos(2x) + 2x. This expression contains every solution of the given nonhomogeneous equation by choosing different values of the constants c 1 and c 2. CHAPTER 2 : Page 55

© 2012 Cengage Learning Engineering. All Rights Reserved.51 2.1 The Linear Second-Order Equation We now have strategies for solving equations (2.1) and (2.2) and the initial value problem. We must be able to find two linearly independent solutions of the homogeneous equation and any one particular solution of the nonhomogeneous equation. We now will develop important cases in which we can carry out these steps. CHAPTER 2 : Page 55

© 2012 Cengage Learning Engineering. All Rights Reserved.53 2.2 Reduction of Order Given y'' + p(x)y' + q(x)y = 0, we want two independent solutions. Reduction of order is a technique for finding a second solution, if we can somehow produce a first solution. CHAPTER 2 : Page 56

© 2012 Cengage Learning Engineering. All Rights Reserved.54 2.2 Reduction of Order Suppose we know a solution y 1, which is not identically zero. We will look for a second solution of the form y 2 (x) = u(x)y 1 (x). Compute y' 2 = u'y 1 + uy' 1, y'' 2 = u''y 1 +2u'y' 1 + uy'' 1. In order for y 2 to be a solution we need u''y 1 +2u'y' 1 + uy'' 1 + p[u'y 1 + uy' 1 ] + quy 1 = 0. CHAPTER 2 : Page 57

© 2012 Cengage Learning Engineering. All Rights Reserved.55 2.2 Reduction of Order Rearrange terms to write this equation as u''y 1 + u'[2y' 1 + py 1 ] + u[y'' 1 + py' 1 + qy 1 ] = 0. The coefficient of u is zero because y 1 is a solution. Thus we need to choose u so that u''y 1 + u' [2y' 1 + py 1 ] = 0. On any interval in which y 1 (x)  0, we can write CHAPTER 2 : Page 57

© 2012 Cengage Learning Engineering. All Rights Reserved.56 2.2 Reduction of Order To help focus on the problem of determining u, denote a known function because y 1 (x) and p(x) are known. Then u'' + g(x)u' = 0. CHAPTER 2 : Page 57

© 2012 Cengage Learning Engineering. All Rights Reserved.57 2.2 Reduction of Order Let v = u' to get v' + g(x)v = 0. This is a linear first-order differential equation for v, with general solution v(x)= Ce  g(x)dx. Since we need only one second solution y 2, we will take C = 1, so v(x)= e  g(x)dx. CHAPTER 2 : Page 57

© 2012 Cengage Learning Engineering. All Rights Reserved.58 2.2 Reduction of Order Finally, since v = u', If we can perform these integrations and obtain u(x), then y 2 = uy 1 is a second solution of y'' + py' + qy = 0. Further, CHAPTER 2 : Page 57

© 2012 Cengage Learning Engineering. All Rights Reserved.59 2.2 Reduction of Order Since v(x) is an exponential function, v(x)  0. And the preceding derivation was carried out on an interval in which y 1 (x)  0. Thus W(x)  0 and y 1 and y 2 form a fundamental set of solutions on this interval. The general solution of y'' + py' + qy = 0 is c 1 y 1 + c 2 y 2. We do not recommend memorizing formulas for g, v and then u. Given one solution y 1, substitute y 2 = uy 1 into the differential equation and, after the cancellations that occur because y 1 is one solution, solve the resulting equation for u(x). CHAPTER 2 : Page 57

© 2012 Cengage Learning Engineering. All Rights Reserved.60 EXAMPLE 2.6 Suppose we are given that y 1 (x) = e  2x is one solution of y'' + 4y' + 4y = 0. To find a second solution, let y 2 (x) = u(x)e  2x. Then Substitute these into the differential equation to get CHAPTER 2 : Page 58

© 2012 Cengage Learning Engineering. All Rights Reserved.61 EXAMPLE 2.6 Some cancellations occur because e  2x is one solution, leaving u''e  2x = 0, or u'' = 0. Two integrations yield u(x) = cx + d. Since we only need one second solution y 2, we only need one u, so we will choose c = 1 and d = 0. CHAPTER 2 : Page 58

© 2012 Cengage Learning Engineering. All Rights Reserved.62 EXAMPLE 2.6 This gives u(x) = x and y 2 (x) = xe  2x. Now for all x. Therefore, y 1 and y 2 form a fundamental set of solutions for all x, and the general solution of y'' + 4y' + 4y = 0 is y(x) = c 1 e  2x + c 2 xe  2x. CHAPTER 2 : Page 58

© 2012 Cengage Learning Engineering. All Rights Reserved.63 EXAMPLE 2.7 Suppose we want the general solution of y''  (3 / x)y' + (4 / x 2 )y = 0 for x >0, and somehow we find one solution y 1 (x) = x 2. Put y 2 (x) = x 2 u(x) and compute y’ 2 = 2xu + x 2 u’ and y'' 2 = 2u + 4xu' + x 2 u''. Substitute into the differential equation to get CHAPTER 2 : Page 58

© 2012 Cengage Learning Engineering. All Rights Reserved.64 EXAMPLE 2.7 Then x 2 u'' + xu' = 0. Since the interval of interest is x >0, we can write this as xu'' + u' = 0. With v = u', this is xv' + v = (xv)' = 0, so xv = c. We will choose c = 1. Then CHAPTER 2 : Page 58

© 2012 Cengage Learning Engineering. All Rights Reserved.65 EXAMPLE 2.7 so u = ln(x) + d, and we choose d = 0 because we need only one suitable u. Then y 2 (x) = x 2 ln(x) is a second solution. Further, for x >0, CHAPTER 2 : Page 59

© 2012 Cengage Learning Engineering. All Rights Reserved.66 EXAMPLE 2.7 Then x 2 and x 2 ln(x) form a fundamental set of solutions for x >0. The general solution is for x >0 is y(x) = c 1 x 2 + c 2 x 2 ln(x). CHAPTER 2 : Page 59

© 2012 Cengage Learning Engineering. All Rights Reserved.68 2.3 The Constant Coefficient Case We have outlined strategies for solving second-order linear homogeneous and nonhomogeneous differential equations. In both cases, we must begin with two linearly independent solutions of a homogeneous equation. This can be a difficult problem. However, when the coefficients are constants, we can write solutions fairly easily. CHAPTER 2 : Page 60

© 2012 Cengage Learning Engineering. All Rights Reserved.69 2.3 The Constant Coefficient Case Consider the constant-coefficient linear homogeneous equation y  +ay +by = 0 (2.3) in which a and b are constants (numbers). A method suggests itself if we read the differential equation like a sentence. CHAPTER 2 : Page 60

© 2012 Cengage Learning Engineering. All Rights Reserved.70 2.3 The Constant Coefficient Case We want a function y such that the second derivative, plus a constant multiple of the first derivative, plus a constant multiple of the function itself is equal to zero for all x. This behavior suggests an exponential function e λx, because derivatives of e λx are constant multiples of e λx. We therefore try to find λ so that e λx is a solution. CHAPTER 2 : Page 60

© 2012 Cengage Learning Engineering. All Rights Reserved.71 2.3 The Constant Coefficient Case Substitute e λx into equation (2.3) to get Since e λx is never zero, the exponential factor cancels, and we are left with a quadratic equation for λ: λ 2 + aλ + b = 0. (2.4) CHAPTER 2 : Page 60

© 2012 Cengage Learning Engineering. All Rights Reserved.72 2.3 The Constant Coefficient Case The quadratic equation (2.4) is the characteristic equation of the differential equation (2.3). Notice that the characteristic equation can be read directly from the coefficients of the differential equation, and we need not substitute e λx each time. The characteristic equation has roots leading to the following three cases. CHAPTER 2 : Page 60

© 2012 Cengage Learning Engineering. All Rights Reserved.73 Case 1: Real, Distinct Roots This occurs when a 2 − 4b > 0. The distinct roots are and are linearly independent solutions, and in this case, the general solution of equation (2.3) is CHAPTER 2 : Page 60

© 2012 Cengage Learning Engineering. All Rights Reserved.74 EXAMPLE 2.8 From the differential equation we immediately read the characteristic equation as having real, distinct roots 3 and −2. The general solution is CHAPTER 2 : Page 60

© 2012 Cengage Learning Engineering. All Rights Reserved.75 Case 2: Repeated Roots This occurs when a 2 − 4b = 0 and the root of the characteristic equation is λ= −a/2. One solution of the differential equation is e −ax/2. We need a second, linearly independent solution. We will invoke a method called reduction of order, which will produce a second solution if we already have one solution. Attempt a second solution y(x) = u(x)e −ax/2. CHAPTER 2 : Page 61

© 2012 Cengage Learning Engineering. All Rights Reserved.78 Case 2: Repeated Roots Since b − a 2 /4 = 0 in this case and e −ax/2 never vanishes, this equation reduces to u  = 0. This has solutions u(x) = cx + d with c and d as arbitrary constants. Therefore, any function y = (cx + d)e −ax/2 is also a solution of equation (2.3) in this case. CHAPTER 2 : Page 61

© 2012 Cengage Learning Engineering. All Rights Reserved.79 Case 2: Repeated Roots Since we need only one solution that is linearly independent from e −ax/2, choose c = 1 and d = 0 to get the second solution xe −ax/2. The general solution in this repeated roots case is This is often written as y = e −ax/2 (c 1 + c 2 x). CHAPTER 2 : Page 61

© 2012 Cengage Learning Engineering. All Rights Reserved.80 Case 2: Repeated Roots It is not necessary to repeat this derivation every time we encounter the repeated root case. Simply write one solution e −ax/2, and a second, linearly independent solution is xe −ax/2. CHAPTER 2 : Page 61

© 2012 Cengage Learning Engineering. All Rights Reserved.81 EXAMPLE 2.9 We will solve y  + 8y + 16y = 0. The characteristic equation is with repeated root λ= −4. The general solution is CHAPTER 2 : Page 61

© 2012 Cengage Learning Engineering. All Rights Reserved.82 Case 3: Complex Roots The characteristic equation has complex roots when a 2 − 4b <0. Because the characteristic equation has real coefficients, the roots appear as complex conjugates α +iβ and α −iβ in which α can be zero but β is nonzero. CHAPTER 2 : Page 61

© 2012 Cengage Learning Engineering. All Rights Reserved.83 Case 3: Complex Roots Now the general solution is or (2.5) This is correct, but it is sometimes convenient to have a solution that does not involve complex numbers. CHAPTER 2 : Page 61-62

© 2012 Cengage Learning Engineering. All Rights Reserved.84 Case 3: Complex Roots We can find such a solution using an observation made by the eighteenth century Swiss mathematician Leonhard Euler, who showed that, for any real number β, Problem 22 suggests a derivation of Euler’s formula. By replacing x with −x, we also have CHAPTER 2 : Page 62

© 2012 Cengage Learning Engineering. All Rights Reserved.86 Case 3: Complex Roots If we choose c 1 = c 2 = 1/2, we obtain the particular solution e αx cos(βx). And if we choose c 1 = 1/2i = −c 2, we obtain the particular solution e αx sin(βx). Since these solutions are linearly independent, we can write the general solution in this complex root case as (2.6) in which c 1 and c 2 are arbitrary constants. We may also write this general solution as (2.7) CHAPTER 2 : Page 62

© 2012 Cengage Learning Engineering. All Rights Reserved.87 Case 3: Complex Roots Either of equations (2.6) or (2.7) is the preferred way of writing the general solution in Case 3, although equation (2.5) also is correct. We do not repeat this derivation each time we encounter Case 3. Simply write the general solution (2.6) or (2.7), with α ±iβ the roots of the characteristic equation. CHAPTER 2 : Page 62

© 2012 Cengage Learning Engineering. All Rights Reserved.88 EXAMPLE 2.10 Solve y  + 2y + 3y =0. The characteristic equation is with complex conjugate roots. With α = −1 and, the general solution is CHAPTER 2 : Page 53

© 2012 Cengage Learning Engineering. All Rights Reserved.89 EXAMPLE 2.11 Solve y  + 36y = 0. The characteristic equation is with complex roots λ= ±6i. Now α = 0 and β = 6, so the general solution is CHAPTER 2 : Page 53

© 2012 Cengage Learning Engineering. All Rights Reserved.90 2.3 The Constant Coefficient Case We are now able to solve the constant coefficient homogeneous equation in all cases. Here is a summary. Let λ 1 and λ 2 be the roots of the characteristic equation CHAPTER 2 : Page 62-63

© 2012 Cengage Learning Engineering. All Rights Reserved.91 2.3 The Constant Coefficient Case Then: 1.1. If λ 1 and λ 2 are real and distinct, 2.If λ 1 =λ 2, 3.If the roots are complex α ± iβ, CHAPTER 2 : Page 63

© 2012 Cengage Learning Engineering. All Rights Reserved.93 2.4 The Nonhomogeneous Equation From Theorem 2.4, the keys to solving the nonhomogeneous linear equation (2.1) are to find two linearly independent solutions of the associated homogeneous equation and a particular solution Y p for the nonhomogeneous equation. We can perform the first task at least when the coefficients are constant. We will now focus on finding Y p, considering two methods for doing this. CHAPTER 2 : Page 64

© 2012 Cengage Learning Engineering. All Rights Reserved.94 2.4.1 Variation of Parameters Suppose we know two linearly independent solutions y 1 and y 2 of the associated homogeneous equation. One strategy for finding Y p is called the method of variation of parameters. Look for functions u 1 and u 2 so that CHAPTER 2 : Page 64

© 2012 Cengage Learning Engineering. All Rights Reserved.95 2.4.1 Variation of Parameters To see how to choose u 1 and u 2, substitute Y p into the differential equation. We must compute two derivatives. First, Simplify this derivative by imposing the condition that (2.8) CHAPTER 2 : Page 64 Why we can have this extra condition?

© 2012 Cengage Learning Engineering. All Rights Reserved.96 2.4.1 Variation of Parameters We have two known u 1 and u 2. But, when substitute them into the differential equation y  + p(x)y + q(x)y = f (x), we only have one equation. It is impossible to find exact solutions. There are lots of solution for u 1 and u 2 and we cannot find them. If we can introduce one extra condition, we may be able to find solutions easily. Any conditions are fine. But can work well. CHAPTER 2 : Page 64

© 2012 Cengage Learning Engineering. All Rights Reserved.98 2.4.1 Variation of Parameters Rearrange terms to write The two terms in square brackets are zero, because y 1 and y 2 are solutions of y  + p(x)y + q(x)y = 0. The last equation therefore reduces to (2.9) CHAPTER 2 : Page 64

© 2012 Cengage Learning Engineering. All Rights Reserved.99 2.4.1 Variation of Parameters Equations (2.8) and (2.9) can be solved for u ' 1 and u ' 2 to get (2.10) where W(x) is the Wronskian of y 1 (x) and y 2 (x). CHAPTER 2 : Page 64-65

© 2012 Cengage Learning Engineering. All Rights Reserved.100 2.4.1 Variation of Parameters We know that W(x)  0, because y 1 and y 2 are assumed to be linearly independent solutions of the associated homogeneous equation. Integrate equations (2.10) to obtain (2.11) CHAPTER 2 : Page 65 Be sure that this formula is obtained from the use of y  + p(x)y + q(x)y = f (x) Note that the leading coefficient is 1 (to check what f (x) is).

© 2012 Cengage Learning Engineering. All Rights Reserved.101 2.4.1 Variation of Parameters Once we have u 1 and u 2, we have a particular solution Y p = u 1 y 1 + u 2 y 2, and the general solution of y  + p(x)y + q(x)y = f (x) is CHAPTER 2 : Page 65

© 2012 Cengage Learning Engineering. All Rights Reserved.102 EXAMPLE 2.12 Find the general solution of for −π/4 < x <π/4. The characteristic equation of y  +4y=0 is λ 2 +4=0 with complex roots λ= ±2i. Two linearly independent solutions of the associated homogeneous equation y  +4y = 0 are CHAPTER 2 : Page 65

© 2012 Cengage Learning Engineering. All Rights Reserved.107 2.4.2 Undetermined Coefficients We will discuss a second method for finding a particular solution of the nonhomogeneous equation, which, however, applies only to the constant coefficient case y  + ay + by = f (x). The idea behind the method of undetermined coefficients is that sometimes we can guess a general form for Y p (x) from the appearance of f (x). CHAPTER 2 : Page 66

© 2012 Cengage Learning Engineering. All Rights Reserved.108 EXAMPLE 2.13 We will find the general solution of y  − 4y = 8x 2 − 2x. It is routine to find the general solution c 1 e 2x + c 2 e −2x of the associated homogeneous equation. We need a particular solution Y p (x) of the nonhomogeneous equation. CHAPTER 2 : Page 66

© 2012 Cengage Learning Engineering. All Rights Reserved.109 EXAMPLE 2.13 Because f (x) = 8x 2 − 2x is a polynomial and derivatives of polynomials are polynomials, it is reasonable to think that there might be a polynomial solution. Furthermore, no such solution can include a power of x higher than 2. If Y p (x) had an x 3 term, this term would be retained by the −4y term of y  −4y, and 8x 2 − 2x has no such term. CHAPTER 2 : Page 66

© 2012 Cengage Learning Engineering. All Rights Reserved.110 EXAMPLE 2.13 This reasoning suggests that we try a particular solution Y p (x) = Ax 2 + Bx + C. Compute y(x) = 2Ax + B and y  (x) = 2A. Substitute these into the differential equation to get Write this as CHAPTER 2 : Page 66

© 2012 Cengage Learning Engineering. All Rights Reserved.111 EXAMPLE 2.13 This second-degree polynomial must be zero for all x if Y p is to be a solution. But a second-degree polynomial has only two roots, unless all of its coefficients are zero. Therefore, −4A − 8 = 0, −4B + 2 = 0, and 2A − 4C = 0. CHAPTER 2 : Page 66

© 2012 Cengage Learning Engineering. All Rights Reserved.112 EXAMPLE 2.13 Solve these to get A = −2, B = 1/2, and C = −1. This gives us the particular solution The general solution is CHAPTER 2 : Page 66

© 2012 Cengage Learning Engineering. All Rights Reserved.113 EXAMPLE 2.14 Find the general solution of y  + 2y − 3y = 4e 2x. The general solution of y  + 2y − 3y = 0 is c 1 e −3x +c 2 e x. CHAPTER 2 : Page 66

© 2012 Cengage Learning Engineering. All Rights Reserved.114 EXAMPLE 2.14 Now look for a particular solution. Because derivatives of e 2x are constant multiples of e 2x, we suspect that a constant multiple of e 2x might serve. Try Y p (x)= Ae 2x. Substitute this into the differential equation to get CHAPTER 2 : Page 67

© 2012 Cengage Learning Engineering. All Rights Reserved.116 EXAMPLE 2.15 Find the general solution of y  − 5y + 6y = −3sin(2x). The general solution of y  − 5y + 6y = 0 is c 1 e 3x +c 2 e 2x. CHAPTER 2 : Page 67

© 2012 Cengage Learning Engineering. All Rights Reserved.117 EXAMPLE 2.15 We need a particular solution Y p of the nonhomogeneous equation. Derivatives of sin(2x) are constant multiples of sin(2x) or cos(2x). Derivatives of cos(2x) are also constant multiples of sin(2x) or cos(2x). This suggests that we try a particular solution Y p (x) = A cos(2x) + B sin(2x). Notice that we include both sin(2x) and cos(2x) in this first attempt, even though f (x) just has a sin(2x) term. CHAPTER 2 : Page 67

© 2012 Cengage Learning Engineering. All Rights Reserved.119 EXAMPLE 2.15 Rearrange terms to write [2B + 10A + 3]sin(2x) = [−2A + 10B]cos(2x). But sin(2x) and cos(2x) are not constant multiples of each other unless these constants are zero. Therefore, 2B +10A + 3 = 0 = −2A + 10B. CHAPTER 2 : Page 67

© 2012 Cengage Learning Engineering. All Rights Reserved.121 2.4.2 Undetermined Coefficients The method of undetermined coefficients has a trap built into it. Consider the following. CHAPTER 2 : Page 67

© 2012 Cengage Learning Engineering. All Rights Reserved.122 EXAMPLE 2.16 Find a particular solution of y  + 2y − 3y = 8e x. Reasoning as before, try Y p (x)= Ae x. Substitute this into the differential equation to obtain Ae x + 2Ae x − 3Ae x = 8e x. But then 8e x = 0, which is a contradiction. CHAPTER 2 : Page 67

© 2012 Cengage Learning Engineering. All Rights Reserved.123 2.4.2 Undetermined Coefficients The problem here is that e x is also a solution of the associated homogeneous equation, so the left side will vanish when Ae x is substituted into y  + 2y − 3y = 8e x. Whenever a term of a proposed Y p (x) is a solution of the associated homogeneous equation, multiply this proposed solution by x. If this results in another solution of the associated homogeneous equation, multiply it by x again. CHAPTER 2 : Page 68

© 2012 Cengage Learning Engineering. All Rights Reserved.124 EXAMPLE 2.17 Revisit Example 2.16. Since f (x)=8e x, our first impulse was to try Y p (x)= Ae x. But this is a solution of the associated homogeneous equation, so multiply by x and try Y p (x)= Axe x. Now CHAPTER 2 : Page 68

© 2012 Cengage Learning Engineering. All Rights Reserved.125 EXAMPLE 2.17 Substitute these into the differential equation to get This reduces to 4Ae x = 8e x, so A = 2, yielding the particular solution Y p (x) = 2xe x. The general solution is CHAPTER 2 : Page 68

© 2012 Cengage Learning Engineering. All Rights Reserved.126 EXAMPLE 2.18 Solve y  − 6y + 9y = 5e 3x. The associated homogeneous equation has the characteristic equation (λ − 3) 2 = 0 with repeated roots λ = 3. The general solution of this associated homogeneous equation is c 1 e 3x +c 2 xe 3x. CHAPTER 2 : Page 68

© 2012 Cengage Learning Engineering. All Rights Reserved.127 EXAMPLE 2.18 For a particular solution, we might first try Y p (x) = Ae 3x, but this is a solution of the homogeneous equation. Multiply by x and try Y p (x) = Axe 3x. This is also a solution of the homogeneous equation, so multiply by x again and try Y p (x) = Ax 2 e 3x. If this is substituted into the differential equation, we obtain A = 5/2, so a particular solution is Y p (x)=5x 2 e 3x /2. The general solution is CHAPTER 2 : Page 68

© 2012 Cengage Learning Engineering. All Rights Reserved.128 2.4.2 Undetermined Coefficients The method of undetermined coefficients is limited by our ability to “guess” a particular solution from the form of f (x), and unlike variation of parameters, requires that the coefficients of y and y be constant. Here is a summary of the method. Suppose we want to find the general solution of CHAPTER 2 : Page 68

© 2012 Cengage Learning Engineering. All Rights Reserved.129 2.4.2 Undetermined Coefficients Step 1. –Write the general solution of the associated homogeneous equation with y 1 and y 2 linearly independent. We can always do this in the constant coefficient case. CHAPTER 2 : Page 68

© 2012 Cengage Learning Engineering. All Rights Reserved.130 2.4.2 Undetermined Coefficients Step 2. –We need a particular solution Y p of the nonhomogeneous equation. This may require several steps. Make an initial attempt of a general form of a particular solution using f (x) and perhaps Table 2.1 as a guide. If this is not possible, this method cannot be used. If we can solve for the constants so that this first guess works, then we have Y p. CHAPTER 2 : Page 68

© 2012 Cengage Learning Engineering. All Rights Reserved.131 2.4.2 Undetermined Coefficients Step 3. –If any term of the first attempt is a solution of the associated homogeneous equation, multiply by x. If any term of this revised attempt is a solution of the homogeneous equation, multiply by x again. Substitute this final general form of a particular solution into the differential equation and solve for the constants to obtain Y p. CHAPTER 2 : Page 69

© 2012 Cengage Learning Engineering. All Rights Reserved.134 2.4.2 Undetermined Coefficients CHAPTER 2 : Page 69 Table 2.1 provides a list of functions for a first try at Y p (x) for various functions f (x) that might appear in the differential equation. In this list, P(x) is a given polynomial of degree n, Q(x) and R(x) are polynomials of degree n with undetermined coefficients for which we must solve, and c and β are constants. If there is a homogeneous solution in f (x), we need to multiply Q(x) (or R(x)) with x , where  is the smallest nonnegative integer that makes x  Q(x) does not have any homogeneous solution.

© 2012 Cengage Learning Engineering. All Rights Reserved.135 2.4.3 The Principle of Superposition CHAPTER 2 : Page 69 Suppose we want to find a particular solution of It is routine to check that, if Y j is a solution of then Y 1 + Y 2 + ··· + Y N is a particular solution of the original differential equation.

© 2012 Cengage Learning Engineering. All Rights Reserved.136 EXAMPLE 2.19 CHAPTER 2 : Page 69 Find a particular solution of y  + 4y = x + 2e −2x. To find a particular solution, consider two problems: –Problem 1: y  + 4y = x –Problem 2: y  + 4y = 2e −2x

© 2012 Cengage Learning Engineering. All Rights Reserved.137 EXAMPLE 2.19 CHAPTER 2 : Page 69-70 Using undetermined coefficients, we find a particular solution of Problem 1 and a particular solution of Problem 2. A particular solution of the given differential equation is Using this, the general solution is

© 2012 Cengage Learning Engineering. All Rights Reserved.139 2.5 Spring Motion CHAPTER 2 : Page 70 A spring suspended vertically and allowed to come to rest has a natural length L. An object (bob) of mass m is attached at the lower end, pulling the spring d units past its natural length. The bob comes to rest in its equilibrium position and is then displaced vertically a distance y 0 units (Figure 2.3) and released from rest or with some initial velocity. We want to construct a model allowing us to analyze the motion of the bob.

© 2012 Cengage Learning Engineering. All Rights Reserved.140 2.5 Spring Motion CHAPTER 2 : Page 70 Let y(t) be the displacement of the bob from the equilibrium position at time t, and take this equilibrium position to be y = 0. Down is chosen as the positive direction. Now consider the forces acting on the bob. Gravity pulls it downward with a force of magnitude mg. By Hooke’s law, the spring exerts a force ky on the object. k is the spring constant, which is a number quantifying the “stiffness" of the spring.

© 2012 Cengage Learning Engineering. All Rights Reserved.141 2.5 Spring Motion CHAPTER 2 : Page 70 At the equilibrium position, the force of the spring is −kd, which is negative because it acts upward. If the object is pulled downward a distance y from this position, an additional force −ky is exerted on it. The total force due to the spring is therefore −kd − ky. The total force due to gravity and the spring is mg −kd −ky. Since at the equilibrium point this force is zero, then mg = kd. The net force acting on the object due to gravity and the spring is therefore just −ky.

© 2012 Cengage Learning Engineering. All Rights Reserved.142 2.5 Spring Motion CHAPTER 2 : Page 70-71 There are forces tending to retard or damp out the motion. These include air resistance or perhaps viscosity of a medium in which the object is suspended. A standard assumption (verified by observation) is that the retarding forces have magnitude proportional to the velocity y. Then for some constant c called the damping constant, the retarding forces equal cy. The total force acting on the bob due to gravity, damping, and the spring itself is −ky − cy.

© 2012 Cengage Learning Engineering. All Rights Reserved.143 2.5 Spring Motion CHAPTER 2 : Page 71 Finally, there may be a driving force f (t) acting on the bob. In this case, the total external force is

© 2012 Cengage Learning Engineering. All Rights Reserved.146 2.5 Spring Motion CHAPTER 2 : Page 71 This is the spring equation. Solutions give the displacement of the bob as a function of time and enable us to analyze the motion under various conditions.

© 2012 Cengage Learning Engineering. All Rights Reserved.147 2.5.1 Unforced Motion CHAPTER 2 : Page 71 The motion is unforced if f (t) = 0. Now the spring equation is homogeneous, and the characteristic equation has roots As we might expect, the solution for the displacement, and hence the motion of the bob, depends on the mass, the amount of damping, and the stiffness of the spring.

© 2012 Cengage Learning Engineering. All Rights Reserved.150 Case 1: c 2 − 4km > 0 CHAPTER 2 : Page 72 Since m and k are positive, c 2 − 4km < c 2, so < c and λ 1 < 0. Therefore, regardless of initial conditions. In the case that c 2 − 4km > 0, the motion simply decays to zero as time increases. This case is called overdamping.

© 2012 Cengage Learning Engineering. All Rights Reserved.151 EXAMPLE 2.20 Overdamping CHAPTER 2 : Page 72 Let c = 6, k = 5, and m = 1. Now the general solution is y(t) = c 1 e −t + c 2 e −5t. Suppose the bob was initially drawn upward 4 feet from equilibrium and released downward with a speed of 2 feet per second. Then y(0) = −4 and y(0) = 2, and we obtain

© 2012 Cengage Learning Engineering. All Rights Reserved.152 EXAMPLE 2.20 Overdamping CHAPTER 2 : Page 72 Figure 2.4 is a graph of this solution. Keep in mind here that down is the positive direction. Since −9 + e −4t 0, then y(t) < 0, and the bob always remains above the equilibrium point. Its velocity y(t) = e −t (9 − 5e −4t )/2 decreases to zero as t increases, so the bob moves downward toward equilibrium with decreasing velocity, approaching arbitrarily close to but never reaching this position and never coming completely to rest.

© 2012 Cengage Learning Engineering. All Rights Reserved.154 Case 2: c 2 − 4km = 0 CHAPTER 2 : Page 73 In this case, the general solution of the spring equation is This case is called critical damping. While y(t)→0 as t→∞, as with overdamping, now the bob can pass through the critical point, as the following example shows.

© 2012 Cengage Learning Engineering. All Rights Reserved.155 EXAMPLE 2.21 Critical Damping CHAPTER 2 : Page 73 Let c =2 and k =m =1. Then y(t) = (c 1 +c 2 t)e −t. Suppose the bob is initially pulled up four feet above the equilibrium position and then pushed downward with a speed of 5 feet per second. Then y(0) = −4, and y(0) = 5. So

© 2012 Cengage Learning Engineering. All Rights Reserved.156 EXAMPLE 2.21 Critical Damping CHAPTER 2 : Page 73 Since y(4) = 0, the bob reaches the equilibrium four seconds after it was released and then passes through it. In fact, y(t) reaches its maximum when t = 5 seconds, and this maximum value is y(5) = e −5, which is about 0.007 units below the equilibrium point. The velocity y(t) = (5−t)e −t is negative for t >5, so the bob’s velocity decreases after the five second point.

© 2012 Cengage Learning Engineering. All Rights Reserved.157 EXAMPLE 2.21 Critical Damping CHAPTER 2 : Page 73 Since y(t)→0 as t→∞, the bob moves with decreasing velocity back toward the equilibrium point as time increases. Figure 2.5 is a graph of this displacement function for 2 ≤ t ≤ 8. In general, when critical damping occurs, the bob either passes through the equilibrium point exactly once, as in Example 2.21, or never reaches it at all, depending on the initial conditions.

© 2012 Cengage Learning Engineering. All Rights Reserved.159 Case 3: c 2 − 4km < 0 CHAPTER 2 : Page 73-74 Here the spring constant and mass of the bob are sufficiently large that c 2 < 4km and the damping is less dominant. This is called underdamping. The general underdamped solution has the form in which

© 2012 Cengage Learning Engineering. All Rights Reserved.160 Case 3: c 2 − 4km < 0 CHAPTER 2 : Page 74 Since c and m are positive, y(t)→0 as t→∞, as in the other two cases. This is not surprising in the absence of an external driving force. However, with underdamping, the motion is oscillatory because of the sine and cosine terms in the displacement function. The motion is not periodic however because of the exponential factor e −ct/2m, which causes the amplitudes of the oscillations to decay as time increases.

© 2012 Cengage Learning Engineering. All Rights Reserved.161 EXAMPLE 2.22 Underdamping CHAPTER 2 : Page 74 Let c = k = 2 and m = 1. The general solution is Suppose the bob is driven downward from a point three feet above equilibrium with an initial speed of two feet per second. Then y(0) = −3, and y(0) = 2. The solution is

© 2012 Cengage Learning Engineering. All Rights Reserved.162 EXAMPLE 2.22 Underdamping CHAPTER 2 : Page 74 The behavior of this solution is visualized more easily if we write it in phase angle form. Choose C and δ so that

© 2012 Cengage Learning Engineering. All Rights Reserved.165 EXAMPLE 2.22 Underdamping CHAPTER 2 : Page 75 Then, and the solution can be written in phase angle form as The graph is a cosine curve with decaying amplitude squeezed between graphs of and. Figure 2.6 shows y(t) and these two exponential functions as reference curves.

© 2012 Cengage Learning Engineering. All Rights Reserved.167 EXAMPLE 2.22 Underdamping CHAPTER 2 : Page 75 The bob passes back and forth through the equilibrium point as t increases. Specifically, it passes through the equilibrium point exactly when y(t) = 0, which occurs at times for n = 0, 1, 2, ···.

© 2012 Cengage Learning Engineering. All Rights Reserved.169 2.5.2 Forced Motion CHAPTER 2 : Page 75 Different driving forces will result in different motion. We will analyze the case of a periodic driving force f (t)= A cos(ωt). Now the spring equation (2.12) is (2.13)

© 2012 Cengage Learning Engineering. All Rights Reserved.170 2.5.2 Forced Motion CHAPTER 2 : Page 75 We have solved the associated homogeneous equation in all cases on c, k, and m. For the general solution of equation (2.13), we need only a particular solution. Application of the method of undetermined coefficients yields the particular solution

© 2012 Cengage Learning Engineering. All Rights Reserved.171 2.5.2 Forced Motion CHAPTER 2 : Page 75 It is customary to denote to write We will analyze some specific cases to get some insight into the motion with this forcing function.

© 2012 Cengage Learning Engineering. All Rights Reserved.172 Case 1: Overdamped Forced Motion CHAPTER 2 : Page 75-76 Overdamping occurs when c 2 − 4km > 0. Suppose c = 6, k = 5, m = 1 and If the bob is released from rest from the equilibrium position, then y(t) satisfies the initial value problem The solution is

© 2012 Cengage Learning Engineering. All Rights Reserved.173 Case 1: Overdamped Forced Motion CHAPTER 2 : Page 76 A graph of this solution is shown in Figure 2.7. As time increases, the exponential terms decay to zero, and the displacement behaves increasingly like oscillating up and down through the equilibrium point with approximate period. Contrast this with the overdamped motion without the forcing function in which the bob began above the equilibrium point and moved with decreasing speed down toward it but never reached it.

© 2012 Cengage Learning Engineering. All Rights Reserved.175 Case 2: Critically Damped Forced Motion CHAPTER 2 : Page 76 Let c = 2, m = k = 1, ω = 1, and A = 2. Assume that the bob is released from rest from the equilibrium point. Now the initial value problem is with the solution

© 2012 Cengage Learning Engineering. All Rights Reserved.176 Case 2: Critically Damped Forced Motion CHAPTER 2 : Page 76 Figure 2.8 is a graph of this solution, which is a case of critically damped forced motion. As t increases, the term with the exponential factor decays (although not as fast as in the overdamping case where there is no factor of t). Nevertheless, after sufficient time, the motion settles into nearly (but not exactly because −te −t is never zero for t > 0) a sinusoidal motion back and forth through the equilibrium point.

© 2012 Cengage Learning Engineering. All Rights Reserved.178 Case 3: Underdamped Forced Motion CHAPTER 2 : Page 76 Let c = k = 2, m = 1,, and, so c 2 − 4km < 0. Suppose the bob is released from rest at the equilibrium position. The initial value problem is with the solution

© 2012 Cengage Learning Engineering. All Rights Reserved.179 Case 3: Underdamped Forced Motion CHAPTER 2 : Page 77 This is underdamped forced motion. Unlike overdamping and critical damping, the exponential term e −t has a trigonometric factor sin(t). Figure 2.9 is a graph of this solution. As time increases, the term becomes less influential and the motion settles nearly into an oscillation back and forth through the equilibrium point with a period of nearly

© 2012 Cengage Learning Engineering. All Rights Reserved.181 2.5.3 Resonance CHAPTER 2 : Page 77 In the absence of damping, an important phenomenon called resonance can occur. Suppose c = 0, but there is still a driving force A cos(ωt). Now the spring equation (2.12) is

© 2012 Cengage Learning Engineering. All Rights Reserved.182 2.5.3 Resonance CHAPTER 2 : Page 77 From the particular solution Y p found in Section 2.5.2, with c = 0, we find that this spring equation has general solution in which

© 2012 Cengage Learning Engineering. All Rights Reserved.183 2.5.3 Resonance CHAPTER 2 : Page 77 This number is called the natural frequency of the spring system, and it is a function of the stiffness of the spring and the mass of the bob. ω is the input frequency and is contained in the driving force. This general solution assumes that the natural and input frequencies are different. Of course, the closer we choose the natural and input frequencies, the larger the amplitude of the cos(ωt) term in the solution.

© 2012 Cengage Learning Engineering. All Rights Reserved.184 2.5.3 Resonance CHAPTER 2 : Page 77 Resonance occurs when the natural and input frequencies are the same. Now the differential equation is (2.14)

© 2012 Cengage Learning Engineering. All Rights Reserved.185 2.5.3 Resonance CHAPTER 2 : Page 77 The solution derived for the case when ω  ω 0 does not apply to equation (2.14). To find the general solution in the present case, first find the general solution of the associated homogeneous equation This has the general solution

© 2012 Cengage Learning Engineering. All Rights Reserved.186 2.5.3 Resonance CHAPTER 2 : Page 78 Now we need a particular solution of equation (2.14). To use the method of undetermined coefficients, we might try a function of the form a cos(ω 0 t) + b sin(ω 0 t). However, these are solutions of the associated homogeneous equation, so instead we try

© 2012 Cengage Learning Engineering. All Rights Reserved.188 2.5.3 Resonance CHAPTER 2 : Page 78 The general solution is This solution differs from the case ω  ω 0 in the factor of t in the particular solution. Because of this, solutions increase in amplitude as t increases. This phenomenon is called resonance.

© 2012 Cengage Learning Engineering. All Rights Reserved.189 2.5.3 Resonance CHAPTER 2 : Page 78 As an example, suppose c 1 = c 2 =ω 0 = 1 and A/2m = 1. Now the solution is Figure 2.10 displays the increasing amplitude of the oscillations with time.

© 2012 Cengage Learning Engineering. All Rights Reserved.191 2.5.3 Resonance CHAPTER 2 : Page 78 While there is always some damping in the real world, if the damping constant is close to zero when compared to other factors and if the natural and input frequencies are (nearly) equal, then oscillations can build up to a sufficiently large amplitude to cause resonance-like behavior. This caused the collapse of the Broughton Bridge near Manchester, England, in 1831 when a column of soldiers marching across maintained a cadence (input frequency) that happened to closely match the natural frequency of the material of the bridge.

© 2012 Cengage Learning Engineering. All Rights Reserved.192 2.5.3 Resonance CHAPTER 2 : Page 78-79 More recently the Tacoma Narrows Bridge in the state of Washington experienced increasing oscillations driven by high winds, causing the concrete roadbed to oscillate in sensational fashion until it collapsed into Puget Sound. This occurred on November 7, 1940. At one point, one side of the roadbed was about twenty-eight feet above the other as it thrashed about. Unlike the Broughton Bridge, local news crews were on hand to film this, and motion pictures of the collapse are available in many engineering and science schools.

© 2012 Cengage Learning Engineering. All Rights Reserved.193 2.5.4 Beats CHAPTER 2 : Page 79 In the absence of damping, an oscillatory driving force can also cause a phenomenon called beats. Suppose ω  ω 0, and consider

© 2012 Cengage Learning Engineering. All Rights Reserved.194 2.5.4 Beats CHAPTER 2 : Page 79 Assume that the object is released from rest from the equilibrium position, so y(0) = y(0) = 0. The solution is The behavior of this solution reveals itself more clearly if we write it as

© 2012 Cengage Learning Engineering. All Rights Reserved.195 2.5.4 Beats CHAPTER 2 : Page 79 This formulation exhibits a periodic variation of amplitude in the solution, depending on the relative sizes of ω 0 + ω and ω 0 − ω. This periodic variation is called a beat. As an example, suppose ω 0 +ω =5 and ω 0 −ω =1, and the constants are chosen so that

© 2012 Cengage Learning Engineering. All Rights Reserved.198 2.5.5 Analogy with an Electrical Circuit CHAPTER 2 : Page 79-80 In an RLC circuit with electromotive force E(t), the differential equation for the current is Since i = q, this is a second-order differential equation for the charge:

© 2012 Cengage Learning Engineering. All Rights Reserved.199 2.5.5 Analogy with an Electrical Circuit CHAPTER 2 : Page 80 Assuming that the resistance, inductance, and capacitance are constant, this equation is exactly analogous to the spring equation with a driving force, which has the form

© 2012 Cengage Learning Engineering. All Rights Reserved.200 2.5.5 Analogy with an Electrical Circuit CHAPTER 2 : Page 80 This means that solutions of the spring equation immediately translate into solutions of the circuit equation with the following identifications: –Displacement function y(t) ⇐⇒ charge q(t) –Velocity y(t) ⇐⇒ current i(t) –Driving force f (t) ⇐⇒ electromotive force E(t) –Mass m ⇐⇒ inductance L –Damping constant c ⇐⇒ resistance R –Spring modulus k ⇐⇒ reciprocal 1/C of the capacitance.

© 2012 Cengage Learning Engineering. All Rights Reserved.201 2.6 Euler’s Differential Equation CHAPTER 2 : Page 81 If A and B are constants, the second-order differential equation x 2 y  + Axy + By = 0 (2.15) is called Euler’s equation. Euler’s equation is defined on the half-lines x > 0 and x 0, and a simple adjustment will yield solutions on x < 0.

© 2012 Cengage Learning Engineering. All Rights Reserved.202 2.6 Euler’s Differential Equation A change of variables will convert Euler’s equation to a constant coefficient linear second-order homogeneous equation, which we can always solve. Let or, equivalently, t = ln(x). If we substitute x = e t into y(x), we obtain a function of t as CHAPTER 2 : Page 81

© 2012 Cengage Learning Engineering. All Rights Reserved.203 2.6 Euler’s Differential Equation To convert Euler’s equation to an equation in t, we need to convert derivatives of y(x) to derivatives of Y(t). First, by the chain rule, we have CHAPTER 2 : Page 81

© 2012 Cengage Learning Engineering. All Rights Reserved.205 2.6 Euler’s Differential Equation Therefore, Substitute these into Euler’s equation to obtain the transformed differential equation or (2.16) This is a constant coefficient equation which we know how to solve. CHAPTER 2 : Page 82

© 2012 Cengage Learning Engineering. All Rights Reserved.206 2.6 Euler’s Differential Equation We need not go through this derivation whenever we encounter an Euler equation. The coefficients of equation (2.16) can be read directly from those of the Euler equation. Solve this transformed equation for Y (t), then replace t = ln(x) to obtain the solution y(x) of the Euler equation. CHAPTER 2 : Page 82

© 2012 Cengage Learning Engineering. All Rights Reserved.207 2.6 Euler’s Differential Equation In doing this, it is useful to recall that, for any number r and for x > 0, Furthermore, for any positive quantity k. Thus, for example, CHAPTER 2 : Page 82

© 2012 Cengage Learning Engineering. All Rights Reserved.209 EXAMPLE 2.23 This constant coefficient linear homogeneous equation has general solution Replace t = ln(x) to obtain the general solution of the Euler equation: for x > 0. CHAPTER 2 : Page 82

© 2012 Cengage Learning Engineering. All Rights Reserved.210 EXAMPLE 2.24 Consider the Euler equation x 2 y  − 5xy + 9y = 0. The transformed equation is with the general solution Y(t) = c 1 e 3t +c 2 te 3t. The Euler equation has the general solution for x > 0. CHAPTER 2 : Page 82

© 2012 Cengage Learning Engineering. All Rights Reserved.213 2.6 Euler’s Differential Equation Another way of finding solutions for Euler’s DE is to check the solution of the form x r for x 2 y  + Axy + By = 0 and to find r.  r 2 + (A-1)r + Br = 0. As usual, we solve an initial value problem by finding the general solution of the differential equation and then using the initial conditions to determine the constants. CHAPTER 2 : Page 83

© 2012 Cengage Learning Engineering. All Rights Reserved.217 2.6 Euler’s Differential Equation Note when you apply variation parameter for Euler’ DE, the formula be sure that the equation is because in the transformation, the original form is y  + p(x)y + q(x)y = f (x). CHAPTER 2 : Page 82

© 2012 Cengage Learning Engineering. All Rights Reserved.218 2.6 Euler’s Differential Equation But when you apply underdetermined coefficient for Euler’ DE, to guess the possible particular forms, the equation must be considered in a constant coefficient form, then the equation must be written as and then we can check the form of f (x) to find possible particular solution forms. CHAPTER 2 : Page 82

© 2012 Cengage Learning Engineering. All Rights Reserved.220 Higher Order Differential Equations Basically, those methodologies used for second order DSs can also be applied to higher order DEs. Concept definitions for higher order DEs: --Linear dependency/independency: y 1, y 2, …, y n are said to be linearly independent if it is impossible to write any one of those functions as a linear combination of the others. Or c 1 y 1 +c 2 y 2 + …+c n y n =0 only if c 1 =c 2 = …=c n =0 CHAPTER 2 : Page 83

© 2012 Cengage Learning Engineering. All Rights Reserved.221 Higher Order Differential Equations --General solution: y 1, y 2, …, y n are linearly independent solutions of an n-th order homogeneous linear DE, then the general solution is a linear combination of those functions. -- The constant coefficient linear DE can also be solved in a similar way (finding roots for the associated characteristic equation) and so does Euler’s equations. CHAPTER 2 : Page 83

© 2012 Cengage Learning Engineering. All Rights Reserved.222 Higher Order Differential Equations --For nonhomogeneous DEs, the variation parameter and undetermined coefficient approaches can also be employed to find a particular solution. --For variation parameter, for an n-th order DE, there are n unknown functions, but only have one equation. Thus, we need to introduce n-1 extra equations. The similar idea can be used. CHAPTER 2 : Page 83

Advanced Engineering Mathematics, 7 th Edition Peter V. O’Neil © 2012 Cengage Learning Engineering. All Rights Reserved. CHAPTER 2 Linear Second-Order.

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Advanced Engineering Mathematics, 7 th Edition Peter V. O’Neil © 2012 Cengage Learning Engineering. All Rights Reserved. CHAPTER 2 Linear Second-Order.

Similar presentations

Presentation on theme: "Advanced Engineering Mathematics, 7 th Edition Peter V. O’Neil © 2012 Cengage Learning Engineering. All Rights Reserved. CHAPTER 2 Linear Second-Order."— Presentation transcript:

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About project

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