Calculus early transcendental functions 3rd edtion 2007 r smith, r minton

www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net CHAPTER Preliminaries 0-2 y 35 30 25 20 15 10 x FIGURE 0.1 The Fibonacci sequence these is the attempt to find patterns to help us better describe the world The other theme is the interplay between graphs and functions By connecting the powerful equation-solving techniques of algebra with the visual images provided by graphs, you will significantly improve your ability to make use of your mathematical skills in solving real-world problems 0.1 POLYNOMIALS AND RATIONAL FUNCTIONS The Real Number System and Inequalities Although mathematics is far more than just a study of numbers, our journey into calculus begins with the real number system While this may seem to be a fairly mundane starting place, we want to give you the opportunity to brush up on those properties that are of particular interest for calculus The most familiar set of numbers is the set of integers, consisting of the whole numbers and their additive inverses: 0, ±1, ±2, ±3, A rational number is any number of the p 27 form q , where p and q are integers and q = For example, 23 , − 73 and 125 are all rational numbers Notice that every integer n is also a rational number, since we can write it as the n quotient of two integers: n = p The irrational numbers are all those real numbers that cannot be written in the form q , where p and q are integers Recall that rational numbers have decimal expansions that either ¯ = 0.125 and = 0.166666¯ are terminate or repeat For instance, 12 = 0.5, 13 = 0.33333, all rational numbers By contrast, irrational numbers have decimal expansions that not repeat or terminate For instance, three familiar irrational numbers and their decimal expansions are √ = 1.41421 35623 , π = 3.14159 26535 and e = 2.71828 18284 We picture the real numbers arranged along the number line displayed in Figure 0.2 (the real line) The set of real numbers is denoted by the symbol R www.elsolucionario.net www.elsolucionario.net 0-3 SECTION 0.1 Ϫ͙2 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1 Polynomials and Rational Functions ͙3 p e FIGURE 0.2 The real line For real numbers a and b, where a < b, we define the closed interval [a, b] to be the set of numbers between a and b, including a and b (the endpoints), that is, FIGURE 0.3 A closed interval a [a, b] = {x ∈ R | a ≤ x ≤ b}, b as illustrated in Figure 0.3, where the solid circles indicate that a and b are included in [a, b] Similarly, the open interval (a, b) is the set of numbers between a and b, but not including the endpoints a and b, that is, b FIGURE 0.4 An open interval (a, b) = {x ∈ R | a < x < b}, as illustrated in Figure 0.4, where the open circles indicate that a and b are not included in (a, b) You should already be very familiar with the following properties of real numbers THEOREM 1.1 If a and b are real numbers and a < b, then (i) (ii) (iii) (iv) For any real number c, a + c < b + c For real numbers c and d, if c < d, then a + c < b + d For any real number c > 0, a · c < b · c For any real number c < 0, a · c > b · c REMARK 1.1 We need the properties given in Theorem 1.1 to solve inequalities Notice that (i) says that you can add the same quantity to both sides of an inequality Part (iii) says that you can multiply both sides of an inequality by a positive number Finally, (iv) says that if you multiply both sides of an inequality by a negative number, the inequality is reversed We illustrate the use of Theorem 1.1 by solving a simple inequality EXAMPLE 1.1 Solving a Linear Inequality Solve the linear inequality 2x + < 13 Solution We can use the properties in Theorem 1.1 to isolate the x First, subtract from both sides to obtain (2x + 5) − < 13 − www.elsolucionario.net a www.elsolucionario.net CHAPTER Preliminaries 0-4 2x < or Finally, divide both sides by (since > 0, the inequality is not reversed) to obtain x < We often write the solution of an inequality in interval notation In this case, we get the interval (−∞, 4) You can deal with more complicated inequalities in the same way EXAMPLE 1.2 Solving a Two-Sided Inequality Solve the two-sided inequality < − 3x ≤ 10 < − 3x and − 3x ≤ 10 Here, we can use the properties in Theorem 1.1 to isolate the x by working on both inequalities simultaneously First, subtract from each term, to get − < (1 − 3x) − ≤ 10 − < −3x ≤ or Now, divide by −3, but be careful Since −3 < 0, the inequalities are reversed We have −3x > ≥ −3 −3 −3 > x ≥ −3 −3 ≤ x < − , − or We usually write this as or in interval notation as [−3, − 53 ) y You will often need to solve inequalities involving fractions We present a typical example in the following Ϫ4 EXAMPLE 1.3 x Ϫ2 Ϫ4 Ϫ8 FIGURE 0.5 y= x −1 x +2 Solving an Inequality Involving a Fraction x −1 ≥ x +2 Solution In Figure 0.5, we show a graph of the function, which appears to indicate that the solution includes all x < −2 and x ≥ Carefully read the inequality and observe that there are only three ways to satisfy this: either both numerator and denominator are positive, both are negative or the numerator is zero To visualize this, we draw number lines for each of the individual terms, indicating where each is positive, negative or zero and use these to draw a third number line indicating the value of the quotient, as shown in the margin In the third number line, we have placed an “ ϫ ” Solve the inequality www.elsolucionario.net Solution First, recognize that this problem requires that we find values of x such that www.elsolucionario.net 0-5 SECTION 0.1 Ϫ 0 ϩ ϩ xϩ2 Ϫ2 ϩ Ϫ ϫ Ϫ2 ϩ For inequalities involving a polynomial of degree or higher, factoring the polynomial and determining where the individual factors are positive and negative, as in example 1.4, will lead to a solution xϪ1 xϩ2 Polynomials and Rational Functions above the −2 to indicate that the quotient is undefined at x = −2 From this last number line, you can see that the quotient is nonnegative whenever x < −2 or x ≥ We write the solution in interval notation as (−∞, −2) ∪ [1, ∞) xϪ1 Ϫ y EXAMPLE 1.4 20 Solving a Quadratic Inequality Solve the quadratic inequality Ϫ10 FIGURE 0.6 ϩ (1.2) xϩ3 Ϫ3 Ϫ (x + 3)(x − 2) > This can happen in only two ways: when both factors are positive or when both factors are negative As in example 1.3, we draw number lines for both of the individual factors, indicating where each is positive, negative or zero and use these to draw a number line representing the product We show these in the margin Notice that the third number line indicates that the product is positive whenever x < −3 or x > We write this in interval notation as (−∞, −3) ∪ (2, ∞) y = x2 + x − Solution In Figure 0.6, we show a graph of the polynomial on the left side of the inequality Since this polynomial factors, (1.1) is equivalent to ϩ No doubt, you will recall the following standard definition xϪ2 ϩ Ϫ Ϫ3 ϩ (x ϩ 3)(x Ϫ 2) DEFINITION 1.1 The absolute value of a real number x is |x| = x, −x, if x ≥ if x < Make certain that you read Definition 1.1 correctly If x is negative, then −x is positive This says that |x| ≥ for all real numbers x For instance, using the definition, |− | = −(−4) = 4, notice that for any real numbers a and b, |a · b| = |a| · |b| NOTES However, For any two real numbers a and b, |a − b| gives the distance between a and b (See Figure 0.7.) |a + b| = |a| + |b|, in general (To verify this, simply take a = and b = −2 and compute both quantities.) However, it is always true that |a + b| ≤ |a| + |b| ͉a Ϫ b͉ a b FIGURE 0.7 The distance between a and b This is referred to as the triangle inequality The interpretation of |a − b| as the distance between a and b (see the note in the margin) is particularly useful for solving inequalities involving absolute values Wherever possible, we suggest that you use this interpretation to read what the inequality means, rather than merely following a procedure to produce a solution www.elsolucionario.net x Ϫ6 Ϫ4 Ϫ2 Ϫ (1.1) x + x − > 10 www.elsolucionario.net CHAPTER Preliminaries EXAMPLE 1.5 0-6 Solving an Inequality Containing an Absolute Value Solve the inequality |x − 2| < 5 Ϫ ϭ Ϫ3 2ϩ5ϭ7 FIGURE 0.8 |x − 2| < (1.3) Solution Before you start trying to solve this, take a few moments to read what it says Since |x − 2| gives the distance from x to 2, (1.3) says that the distance from x to must be less than So, find all numbers x whose distance from is less than We indicate the set of all numbers within a distance of in Figure 0.8 You can now read the solution directly from the figure: −3 < x < or in interval notation: (−3, 7) Many inequalities involving absolute values can be solved simply by reading the inequality correctly, as in example 1.6 Solving an Inequality with a Sum Inside an Absolute Value Solve the inequality |x + 4| ≤ 7 Ϫ4 Ϫ ϭ Ϫ11 Ϫ4 (1.4) Solution To use our distance interpretation, we must first rewrite (1.4) as Ϫ4 ϩ ϭ |x − (−4)| ≤ This now says that the distance from x to −4 is less than or equal to We illustrate the solution in Figure 0.9, from which it follows that −11 ≤ x ≤ or [−11, 3] FIGURE 0.9 | x + |≤ Recall that for any real number r > 0, |x| < r is equivalent to the following inequality not involving absolute values: −r < x < r In example 1.7, we use this to revisit the inequality from example 1.5 EXAMPLE 1.7 An Alternative Method for Solving Inequalities Solve the inequality |x − 2| < y Solution This is equivalent to the two-sided inequality (x2, y2) y2 −5 < x − < Adding to each term, we get the solution Distance ͉y2 Ϫ y1͉ −3 < x < 7, or in interval notation (−3, 7), as before y1 (x1, y1) ͉x2 Ϫ x1͉ x1 x2 FIGURE 0.10 Distance x Recall that the distance between two points (x1 , y1 ) and (x2 , y2 ) is a simple consequence of the Pythagorean Theorem and is given by d{(x1 , y1 ), (x2 , y2 )} = We illustrate this in Figure 0.10 (x2 − x1 )2 + (y2 − y1 )2 www.elsolucionario.net EXAMPLE 1.6 www.elsolucionario.net 0-7 SECTION 0.1 EXAMPLE 1.8 Polynomials and Rational Functions Using the Distance Formula Find the distance between the points (1, 2) and (3, 4) Solution The distance between (1, 2) and (3, 4) is d{(1, 2), (3, 4)} = (3 − 1)2 + (4 − 2)2 = √ 4+4= √ Year 1960 1970 1980 1990 x The federal government conducts a nationwide census every 10 years to determine the population Population data for the last several decades are shown in the accompanying table One difficulty with analyzing these data is that the numbers are so large This problem is remedied by transforming the data We can simplify the year data by defining x to be the number of years since 1960 Then, 1960 corresponds to x = 0, 1970 corresponds to x = 10 and so on The population data can be simplified by rounding the numbers to the nearest million The transformed data are shown in the accompanying table and a scatter plot of these data points is shown in Figure 0.11 Most people would say that the points in Figure 0.11 appear to form a straight line (Use a ruler and see if you agree.) To determine whether the points are, in fact, on the same line (such points are called colinear), we might consider the population growth in each of the indicated decades From 1960 to 1970, the growth was 24 million (That is, to move from the first point to the second, you increase x by 10 and increase y by 24.) Likewise, from 1970 to 1980, the growth was 24 million However, from 1980 to 1990, the growth was only 22 million Since the rate of growth is not constant, the data points not fall on a line Notice that to stay on the same line, y would have had to increase by 24 again The preceding argument involves the familiar concept of slope U.S Population 179,323,175 203,302,031 226,542,203 248,709,873 y 179 10 203 20 227 30 249 Transformed data y 250 200 DEFINITION 1.2 For x1 = x2 , the slope of the straight line through the points (x1 , y1 ) and (x2 , y2 ) is the number y2 − y1 m= (1.5) x2 − x1 150 100 50 x 10 20 30 When x1 = x2 , the line through (x1 , y1 ) and (x2 , y2 ) is vertical and the slope is undefined FIGURE 0.11 Population data y , We often describe slope as “the change in y divided by the change in x,” written x Rise or more simply as (see Figure 0.12a on the following page) Run The slope of a straight line is the same no matter which two points on the line you select Referring to Figure 0.12b (where the line has positive slope), notice that for any four points A, B, D and E on the line, the two right triangles ABC and DEF are similar Recall that for similar triangles, the ratios of corresponding sides must be the same In this case, this says that y y = x x www.elsolucionario.net Equations of Lines www.elsolucionario.net CHAPTER Preliminaries 0-8 y y B (x2, y2) y2 ⌬y A ⌬y ϭ y2 Ϫ y1 ϭ Rise y1 ⌬x E C ⌬yЈ (x1, y1) D ⌬x ϭ x2 Ϫ x1 x1 F ⌬xЈ ϭ Run x x2 x FIGURE 0.12a FIGURE 0.12b Slope Similar triangles and slope and so, the slope is the same no matter which two points on the line are selected Furthermore, a line is the only curve with constant slope Notice that a line is horizontal if and only if its slope is zero EXAMPLE 1.9 Finding the Slope of a Line Find the slope of the line through the points (4, 3) and (2, 5) Solution From (1.5), we get m= EXAMPLE 1.10 y2 − y1 5−3 = = = −1 x2 − x1 2−4 −2 Using Slope to Determine if Points Are Colinear Use slope to determine whether the points (1, 2), (3, 10) and (4, 14) are colinear Solution First, notice that the slope of the line joining (1, 2) and (3, 10) is m1 = y2 − y1 10 − = = = x2 − x1 3−1 Similarly, the slope through the line joining (3, 10) and (4, 14) is m2 = y2 − y1 14 − 10 = = x2 − x1 4−3 Since the slopes are the same, the points must be colinear Recall that if you know the slope and a point through which the line must pass, you have enough information to graph the line The easiest way to graph a line is to plot two points and then draw the line through them In this case, you need only to find a second point www.elsolucionario.net www.elsolucionario.net 0-9 SECTION 0.1 Polynomials and Rational Functions y EXAMPLE 1.11 Graphing a Line If a line passes through the point (2, 1) with slope 23 , find a second point on the line and then graph the line y2 − y1 Solution Since slope is given by m = , we take m = 23 , y1 = and x1 = 2, x2 − x1 to obtain 2 y2 − = x2 − x You are free to choose the x-coordinate of the second point For instance, to find the point at x2 = 5, substitute this in and solve From Ϫ1 y2 − y2 − = = , 5−2 FIGURE 0.13a Graph of straight line m= ⌬y ϭ ⌬x ϭ 3 x Ϫ1 FIGURE 0.13b = y x The slope of 23 says that if we move three units to the right, we must move two units up to stay on the line, as illustrated in Figure 0.13b In example 1.11, the choice of x = was entirely arbitrary; you can choose any x-value you want to find a second point Further, since x can be any real number, you can leave x as a variable and write out an equation satisfied by any point (x, y) on the line In the general case of the line through the point (x0 , y0 ) with slope m, we have from (1.5) that y − y0 m= (1.6) x − x0 Multiplying both sides of (1.6) by (x − x0 ), we get Using slope to find a second point y − y0 = m(x − x0 ) or POINT-SLOPE FORM OF A LINE y = m(x − x0 ) + y0 y Equation (1.7) is called the point-slope form of the line EXAMPLE 1.12 Finding the Equation of a Line Given Two Points Find an equation of the line through the points (3, 1) and (4, −1), and graph the line −1 − −2 = = −2 Using (1.7) with slope 4−3 m = −2, x-coordinate x0 = and y-coordinate y0 = 1, we get the equation of the line: Solution From (1.5), the slope is m = x Ϫ1 (1.7) y = −2(x − 3) + FIGURE 0.14 y = −2(x − 3) + (1.8) To graph the line, plot the points (3, 1) and (4, −1), and you can easily draw the line seen in Figure 0.14 www.elsolucionario.net we get = y2 − or y2 = A second point is then (5, 3) The graph of the line is shown in Figure 0.13a An alternative method for finding a second point is to use the slope y www.elsolucionario.net 10 CHAPTER Preliminaries 0-10 In example 1.12, you may be tempted to simplify the expression for y given in (1.8) As it turns out, the point-slope form of the equation is often the most convenient to work with So, we will typically not ask you to rewrite this expression in other forms At times, a form of the equation called the slope-intercept form is more convenient This has the form y = mx + b, where m is the slope and b is the y-intercept (i.e., the place where the graph crosses the y-axis) In example 1.12, you simply multiply out (1.8) to get y = −2x + + or y = −2x + As you can see from Figure 0.14, the graph crosses the y-axis at y = Theorem 1.2 presents a familiar result on parallel and perpendicular lines Two (nonvertical) lines are parallel if they have the same slope Further, any two vertical lines are parallel Two (nonvertical) lines of slope m and m are perpendicular whenever the product of their slopes is −1 (i.e., m · m = −1) Also, any vertical line and any horizontal line are perpendicular y Since we can read the slope from the equation of a line, it’s a simple matter to determine when two lines are parallel or perpendicular We illustrate this in examples 1.13 and 1.14 20 10 Ϫ4 x Ϫ2 EXAMPLE 1.13 Find an equation of the line parallel to y = 3x − and through the point (−1, 3) Ϫ10 Solution It’s easy to read the slope of the line from the equation: m = The equation of the parallel line is then Ϫ20 y = 3[x − (−1)] + FIGURE 0.15 or simply y = 3(x + 1) + We show a graph of both lines in Figure 0.15 Parallel lines y EXAMPLE 1.14 Finding the Equation of a Perpendicular Line Find an equation of the line perpendicular to y = −2x + and intersecting the line at the point (1, 2) x Ϫ2 Finding the Equation of a Parallel Line Ϫ2 Ϫ4 FIGURE 0.16 Perpendicular lines Solution The slope of y = −2x + is −2 The slope of the perpendicular line is then −1/(−2) = 12 Since the line must pass through the point (1, 2), the equation of the perpendicular line is (x − 1) + 2 We show a graph of the two lines in Figure 0.16 y= We now return to this subsection’s introductory example and use the equation of a line to estimate the population in the year 2000 www.elsolucionario.net THEOREM 1.2 GTBL001-15 P2: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 15-29 SECTION 15.4 glucose/100 ml blood two hours after the injection and 78 mg glucose/100 ml blood three hours after the injection 40 Show that the data in exercise 39 are inconsistent with the case < ω < α 41 Consider an RLC-circuit with capacitance C and charge Q(t) at time t The energy in the circuit at time t is given by [Q(t)]2 u(t) = Show that the charge in a general RLC-circuit 2C has the form Q(t) = e−(R/L)t/2 |Q cos ωt + c2 sin ωt|, R − 4L/C The relative where Q = Q(0) and ω = 2L energy loss from time t = to time t = 2π is given by ω u(2π/ω) − u(0) Uloss = and the inductance quality factor u(0) 2π is defined by Using a Taylor polynomial approximation Uloss of e x , show that the inductance quality factor is approximately L ω R Power Series Solutions of Differential Equations 1249 √ where k = 2m E/h ¯ Then set up the equations (−a) = and (a) = Both equations are true if c1 = c2 = 0, but in this case the solution would be (x) = To find nontrivial solutions (that is, nonzero solutions), find all values of k such that cos ka = or sin ka = Then, solve for the energy E in terms of a, m and h ¯ These are the only allowable energy levels for the particle Finally, determine what happens to the energy levels as a increases without bound Imagine a hole drilled through the center of the Earth What would happen to a ball dropped in the hole? Galileo conjectured that the ball would undergo simple harmonic motion, which is the periodic motion of an undamped spring or pendulum This solution requires no friction and a nonrotating Earth Gm m The force due to gravity of two objects r units apart is , r2 where G is the universal gravitation constant and m and m are the masses of the objects Let R be the radius of the Earth and y the displacement from the center of the Earth y R EXPLORATORY EXERCISES In quantum mechanics, the possible locations of a particle are described by its wave function (x) The wave function satisfies Schrăodingers wave equation h (x) + V (x) (x) = E (x) 2m Here, h¯ is Planck’s constant, m is mass, V (x) is the potential function for external forces and E is the particle’s energy In the case of a bound particle with an infinite square well of width 2a, the potential function is V (x) = for −a ≤ x ≤ a We will show that the particle’s energy is quantized by solving the boundary value problem consisting of the differential h¯ equation (x) + v(x) (x) = E (x) plus the boundary 2m conditions (−a) = and (a) = The theory of boundary value problems is different from that of the initial value problems in this chapter, which typically have unique solutions In fact, in this exercise we specifically want more than one solution Start with the differential equation and show that for V (x) = 0; the general solution is (x) = c1 cos kx + c2 sin kx, For a ball at position y with |y| ≤ R, the ball is attracted to the center of the Earth as if the Earth were a single particle located at the origin with mass ρv, where ρ is the density of the Earth and v is the volume of the sphere of radius |y| (This assumes a constant density and a spherical Earth.) If M is the mass of the Earth, show that if you neglect damping, the position of the ball GM GM satisfies the equation y + y = Use g = to simR R plify this Find the motion of the ball Does the period depend on the starting position? Compare the motions of balls dropped simultaneously from the Earth’s surface and halfway to the center of the Earth Explore the motion of a ball thrown from the surface of the Earth at y = R with initial velocity −R/100 15.4 POWER SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS So far in this chapter, we have seen how to solve only those second-order equations with constant coefficients, such as y − 6y + 9y = What if the coefficients aren’t constant? For instance, suppose you wanted to solve the equation y + 2x y + 2y = www.elsolucionario.net P1: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 15-29 SECTION 15.4 glucose/100 ml blood two hours after the injection and 78 mg glucose/100 ml blood three hours after the injection 40 Show that the data in exercise 39 are inconsistent with the case < ω < α 41 Consider an RLC-circuit with capacitance C and charge Q(t) at time t The energy in the circuit at time t is given by [Q(t)]2 u(t) = Show that the charge in a general RLC-circuit 2C has the form Q(t) = e−(R/L)t/2 |Q cos ωt + c2 sin ωt|, R − 4L/C The relative where Q = Q(0) and ω = 2L energy loss from time t = to time t = 2π is given by ω u(2π/ω) − u(0) Uloss = and the inductance quality factor u(0) 2π is defined by Using a Taylor polynomial approximation Uloss of e x , show that the inductance quality factor is approximately L ω R Power Series Solutions of Differential Equations 1249 √ where k = 2m E/h ¯ Then set up the equations (−a) = and (a) = Both equations are true if c1 = c2 = 0, but in this case the solution would be (x) = To find nontrivial solutions (that is, nonzero solutions), find all values of k such that cos ka = or sin ka = Then, solve for the energy E in terms of a, m and h ¯ These are the only allowable energy levels for the particle Finally, determine what happens to the energy levels as a increases without bound Imagine a hole drilled through the center of the Earth What would happen to a ball dropped in the hole? Galileo conjectured that the ball would undergo simple harmonic motion, which is the periodic motion of an undamped spring or pendulum This solution requires no friction and a nonrotating Earth Gm m The force due to gravity of two objects r units apart is , r2 where G is the universal gravitation constant and m and m are the masses of the objects Let R be the radius of the Earth and y the displacement from the center of the Earth y R EXPLORATORY EXERCISES In quantum mechanics, the possible locations of a particle are described by its wave function (x) The wave function satisfies Schrăodingers wave equation h¯ (x) + V (x) (x) = E (x) 2m Here, h¯ is Planck’s constant, m is mass, V (x) is the potential function for external forces and E is the particle’s energy In the case of a bound particle with an infinite square well of width 2a, the potential function is V (x) = for −a ≤ x ≤ a We will show that the particle’s energy is quantized by solving the boundary value problem consisting of the differential h¯ equation (x) + v(x) (x) = E (x) plus the boundary 2m conditions (−a) = and (a) = The theory of boundary value problems is different from that of the initial value problems in this chapter, which typically have unique solutions In fact, in this exercise we specifically want more than one solution Start with the differential equation and show that for V (x) = 0; the general solution is (x) = c1 cos kx + c2 sin kx, For a ball at position y with |y| ≤ R, the ball is attracted to the center of the Earth as if the Earth were a single particle located at the origin with mass ρv, where ρ is the density of the Earth and v is the volume of the sphere of radius |y| (This assumes a constant density and a spherical Earth.) If M is the mass of the Earth, show that if you neglect damping, the position of the ball GM GM satisfies the equation y + y = Use g = to simR R plify this Find the motion of the ball Does the period depend on the starting position? Compare the motions of balls dropped simultaneously from the Earth’s surface and halfway to the center of the Earth Explore the motion of a ball thrown from the surface of the Earth at y = R with initial velocity −R/100 15.4 POWER SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS So far in this chapter, we have seen how to solve only those second-order equations with constant coefficients, such as y − 6y + 9y = What if the coefficients aren’t constant? For instance, suppose you wanted to solve the equation y + 2x y + 2y = www.elsolucionario.net P1: OSO/OVY P1: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 1250 CHAPTER 15 Second-Order Differential Equations 15-30 We leave it as an exercise to show that substituting y = er x in this case does not lead to a solution However, in many cases such as this, we can find a solution by assuming that the solution can be written as a power series, such as y= ∞ an x n n=0 The idea is to substitute this series into the differential equation and then use the resulting equation to determine the coefficients, a0 , a1 , a2 , an Before we see how to this in general, we illustrate this for a simple equation whose solution is already known, to demonstrate that we arrive at the same solution using either method EXAMPLE 4.1 Power Series Solution of a Differential Equation Use a power series to determine the general solution of y + y = Solution First, observe that this equation has constant coefficients and its general solution is y = c1 sin x + c2 cos x, where c1 and c2 are constants We now look for a solution of the equation in the form of the power series y = a0 + a1 x + a2 x + a3 x + · · · = ∞ an x n n=0 To substitute this into the equation, we first need to obtain representations for y and y Assuming that the power series is convergent and has a positive radius of convergence, recall that we can differentiate term-by-term to obtain the derivatives y = a1 + 2a2 x + 3a3 x + · · · = ∞ nan x n−1 n=1 y = 2a2 + 6a3 x + · · · = and ∞ n(n − 1)an x n−2 n=2 Substituting these power series into the differential equation, we get 0=y +y= REMARK 4.1 Notice that when we change ∞ n(n − 1)an x n−2 to n=2 ∞ (n + 2)(n + 1)an+2 x n , the n=0 index in the sequence increases by (for example, an becomes an+2 ), while the initial value of the index decreases by ∞ n(n − 1)an x n−2 + n=2 ∞ an x n (4.1) n=0 The immediate objective here is to combine the two series in (4.1) into one power series Since the powers in the one series are of the form x n−2 and in the other series are of the form x n , we will first need to rewrite one of the two series Notice that we have y = ∞ n(n − 1)an x n−2 = 2a2 + · 2a3 x + · 3a4 x + · · · n=2 = ∞ n=0 (n + 2)(n + 1)an+2 x n www.elsolucionario.net GTBL001-15 P2: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 15-31 SECTION 15.4 Power Series Solutions of Differential Equations 1251 Substituting this into equation (4.1) gives us 0=y +y= ∞ (n + 2)(n + 1)an+2 x n + n=0 = ∞ ∞ an x n n=0 [(n + 2)(n + 1)an+2 + an ]x n (4.2) n=0 Read equation (4.2) carefully; it says that the power series on the right converges to the constant function f (x) = In view of this, all of the coefficients must be zero That is, = (n + 2)(n + 1)an+2 + an , for n = 0, 1, 2, We solve this for the coefficient with the largest index, to obtain an+2 = −an , (n + 2)(n + 1) (4.3) for n = 0, 1, 2, Equation (4.3) is called the recurrence relation, which we use to determine all of the coefficients of the series solution The general idea is to write out (4.3) for a number of specific values of n and then try to recognize a pattern that the coefficients follow From (4.3), we have for the even-indexed coefficients that −a0 −1 = a0 , 2·1 2! −a2 a4 = = a0 = a0 , 4·3 4·3·2·1 4! −1 −a4 a6 = = a0 , 6·5 6! −a6 = a0 a8 = 8·7 8! and so on (Try to write down a10 by recognizing the pattern, without referring to the recurrence relation.) Since we can write each even-indexed coefficient as a2n , for some n, we can now write down a simple formula that works for any of these coefficients We have (−1)n a0 , a2n = (2n)! a2 = for n = 0, 1, 2, Similarly, using (4.3), we have that the odd-indexed coefficients are −1 −a1 = a1 , 3·2 3! −a3 a5 = = a1 , 5·4 5! −1 −a5 a7 = = a1 , 7·6 7! −a7 a9 = = a1 9·8 9! and so on Since we can write each odd-indexed coefficient as a2n+1 (or alternatively as a2n−1 ), for some n, note that we have the following simple formula for the odd-indexed coefficients: (−1)n a2n+1 = a1 , (2n + 1)! a3 = www.elsolucionario.net P1: OSO/OVY P1: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 1252 CHAPTER 15 Second-Order Differential Equations 15-32 for n = 0, 1, 2, Since we have now written every coefficient in terms of either a0 or a1 , we can rewrite the solution by separating the a0 terms from the a1 terms We have y= ∞ an x n = a0 + a1 x + a2 x + a3 x + · · · n=0 = a0 − = a0 ∞ n=0 1 x + x + · · · + a1 x − x + x + · · · 2! 4! 3! 5! ∞ (−1)n 2n (−1)n 2n+1 x + a1 x (2n)! (2n + 1)! n=0 y1 (x) y2 (x) = a0 y1 (x) + a1 y2 (x), (4.4) where y1 (x) and y2 (x) are two solutions of the differential equation (assuming the series converge) At this point, you should be able to easily check that both of the indicated power series converge absolutely for all x, by using the Ratio Test Beyond this, you might also recognize that the series solutions y1 (x) and y2 (x) that we obtained are, in fact, the Maclaurin series expansions of cos x and sin x, respectively In light of this, (4.4) is an equivalent solution to that found by using the methods of section 15.1 The method used to solve the differential equation in example 4.1 is certainly far more complicated than the methods we used in section 15.1 for solving the same equation However, this new method can be used to solve a wider range of differential equations than those solvable using our earlier methods We now return to the equation mentioned in the introduction to this section EXAMPLE 4.2 Solving a Differential Equation with Variable Coefficients Find the general solution of the differential equation y + 2x y + 2y = Solution First, observe that since the coefficient of y is not constant, we have little choice but to look for a series solution of the equation As in example 4.1, we begin by assuming that we may write the solution as a power series, y= ∞ an x n n=0 As before, we have y = ∞ nan x n−1 n=1 ∞ y = and n(n − 1)an x n−2 n=2 Substituting these three power series into the equation, we get = y + 2x y + 2y = ∞ n(n − 1)an x n−2 + 2x n=2 = ∞ n=2 n(n − 1)an x n−2 + ∞ n=1 ∞ n=1 2nan x n + ∞ n=0 2an x n , nan x n−1 + ∞ an x n n=0 (4.5) www.elsolucionario.net GTBL001-15 P2: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 15-33 SECTION 15.4 Power Series Solutions of Differential Equations 1253 where in the middle term, we moved the x into the series and combined powers of x In order to combine the three series, we must only rewrite the first series so that its general term is a multiple of x n , instead of x n−2 As we did in example 4.1, we write ∞ n(n − 1)an x n−2 = n=2 ∞ (n + 2)(n + 1)an+2 x n n=0 and so, from (4.5), we have 0= ∞ n(n − 1)an x n−2 + n=2 = ∞ ∞ n=1 ∞ ∞ 2an x n n=0 (n + 2)(n + 1)an+2 x n + n=0 = 2nan x n + ∞ ∞ 2nan x n + n=0 2an x n n=0 [(n + 2)(n + 1)an+2 + 2nan + 2an ]x n n=0 = ∞ [(n + 2)(n + 1)an+2 + 2(n + 1)an ]x n (4.6) n=0 ∞ 2nan x n = To get this, we used the fact that n=1 ∞ 2nan x n (Notice that the first term in n=0 the series on the right is zero!) Reading equation (4.6) carefully, note that we again have a power series converging to the zero function, from which it follows that all of the coefficients must be zero: = (n + 2)(n + 1)an+2 + 2(n + 1)an , REMARK 4.2 Always solve for the coefficient with the largest index for n = 0, 1, 2, Again solving for the coefficient with the largest index, we get the recurrence relation 2(n + 1)an an+2 = − (n + 2)(n + 1) 2an an+2 = − or n+2 Much like we saw in example 4.1, the recurrence relation tells us that all of the even-indexed coefficients are related to a0 and all of the odd-indexed coefficients are related to a1 In order to try to recognize the pattern, we write out a number of terms, using the recurrence relation We have a2 = − a0 = −a0 , 2 a4 = − a2 = a0 , 2 a6 = − a4 = − a0 , 3! a8 = − a6 = a0 4! and so on At this point, you should recognize the pattern for these coefficients (If not, write out a few more terms.) Note that we can write the even-indexed coefficients as a2n = (−1)n a0 , n! www.elsolucionario.net P1: OSO/OVY P1: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net CHAPTER 15 Second-Order Differential Equations 15-34 for n = 0, 1, 2, Be sure to match this formula against those coefficients calculated above to see that they match Continuing with the odd-indexed coefficients, we have from the recurrence relation that a3 = − a1 , 22 a5 = − a3 = a1 , 5·3 23 a7 = − a5 = − a1 , 7·5·3 24 a1 a9 = − a7 = 9·7·5·3 and so on While you might recognize the pattern here, it’s hard to write down this pattern succinctly Observe that the products in the denominators are not quite factorials Rather, each is the product of the first so many odd numbers The solution to this is to write this as a factorial, but then cancel out all of the even integers in the product In particular, note that 2·4 = 9·7·5·3 2·3 2·2 2·1 9! = 24 · 4! , 9! so that a9 becomes 24 24 · 24 · 4! 22·4 · 4! a1 = a1 = a1 9·7·5·3 9! 9! More generally, we now have a9 = 0.75 a2n+1 = 0.5 0.25 Ϫ5 Ϫ2.5 2.5 y = y1 (x) = e−x for n = 0, 1, Now that we have expressions for all of the coefficients, we can write the solution of the differential equation as y= FIGURE 15.16a ∞ = a0 ∞ (−1) 2n (−1)n 22n n! 2n+1 x + a1 x n! (2n + 1)! n=0 n y1 (x) y2 (x) = a0 y1 (x) + a1 y2 (x), 0.4 0.2 Ϫ3 Ϫ2 Ϫ1 Ϫ0.2 (a2n x 2n + a2n+1 x 2n+1 ) n=0 ∞ n=0 0.6 ∞ an x n = n=0 y (−1)n 22n n! a1 , (2n + 1)! x Ϫ0.4 Ϫ0.6 FIGURE 15.16b 10-term approximation to y = y2 (x) where y1 and y2 are two power series solutions of the differential equation We leave it as an exercise to use the Ratio Test to show that both of these series converge absolutely for all x You might recognize y1 (x) as the Maclaurin series expansion for e−x , but in practice recognizing series solutions as power series of familiar functions is rather unlikely To give you an idea of the behavior of these functions, we draw a graph of y1 (x) in Figure 15.16a and of y2 (x) in Figure 15.16b We obtained the graph of y2 (x) by plotting the partial sums of the series From examples 4.1 and 4.2, you might get the idea that if you look for a series solution, you can always recognize the pattern of the coefficients and write the pattern down www.elsolucionario.net 1254 GTBL001-15 P2: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 15-35 SECTION 15.4 Power Series Solutions of Differential Equations 1255 succinctly Unfortunately, the pattern is most often difficult to see and even more difficult to write down compactly Still, series solutions are a valuable means of solving a differential equation In the worst case, you can always compute a number of the coefficients of the series from the recurrence relation and then use the first so many terms of the series as an approximation to the actual solution In example 4.3, we illustrate the more common case where the coefficients are a bit more challenging to find EXAMPLE 4.3 A Series Solution Where the Coefficients Are Harder to Find Use a power series to find the general solution of Airy’s equation y − x y = Solution As before, we assume that we may write the solution as a power series ∞ y= an x n n=0 ∞ y = Again, we have nan x n−1 n=1 y = and ∞ n(n − 1)an x n−2 n=2 Subsituting these power series into the equation, we get = y − xy = ∞ n(n − 1)an x n−2 − x n=2 ∞ = ∞ an x n n=0 n(n − 1)an x n−2 − n=2 ∞ an x n+1 n=0 In order to combine the two preceding series, we must rewrite one or both series so that they both have the same power of x For simplicity, we rewrite the first series only We have 0= ∞ n(n − 1)an x n−2 − n=2 = ∞ ∞ an x n+1 n=0 (n + 3)(n + 2)an+3 x n+1 − n=−1 an x n+1 n=0 = (2)(1)a2 + ∞ ∞ ∞ (n + 3)(n + 2)an+3 x n+1 − n=0 = 2a2 + ∞ an x n+1 n=0 [(n + 3)(n + 2)an+3 − an ]x n+1 , n=0 where we wrote out the first term of the first series and then combined the two series, once both had an index that started with n = Again, this is a power series expansion of the zero function and so, all of the coefficients must be zero That is, = 2a2 (4.7) www.elsolucionario.net P1: OSO/OVY P1: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 1256 CHAPTER 15 Second-Order Differential Equations and 15-36 = (n + 3)(n + 2)an+3 − an , (4.8) for n = 0, 1, 2, Equation (4.7) says that a2 = and (4.8) gives us the recurrence relation an+3 = an , (4.9) (n + 3)(n + 2) for n = 0, 1, 2, Notice that here, instead of having all of the even-indexed coefficients related to a0 and all of the odd-indexed coefficients related to a1 , (4.9) tells us that every third coefficient is related In particular, notice that since a2 = 0, (4.9) now says that a2 = 0, 5·4 a8 = a5 = 8·7 and so on So, every third coefficient starting with a2 is zero But, how we concisely write down something like this? Think about the notation a2n and a2n+1 that we have used previously You can view a2n as a representation of every second coefficient starting with a0 Likewise, a2n+1 represents every second coefficient starting with a1 In the present case, if we want to write down every third coefficient starting with a2 , we write a3n+2 We can now observe that a5 = a3n+2 = 0, for n = 0, 1, 2, Continuing on with the remaining coefficients, we have from (4.9) that a3 = a0 , 3·2 1 a3 = a0 , a6 = 6·5 6·5·3·2 1 a9 = a6 = a0 9·8 9·8·6·5·3·2 and so on Hopefully, you see the pattern that’s developing for these coefficients The trouble here is that it’s not as easy to write down this pattern as it was in the first two examples Notice that the denominator in the expression for a9 is almost 9!, but with every third factor in the product deleted Since we don’t have a way of succinctly writing this down, we write the coefficients by indicating the pattern, as follows: a3n = (3n − 2)(3n − 5) · · · · · a0 , (3n)! where this is not intended as a literal formula, as explicit substitution of n = or n = would result in negative values Rather, this is an indication of the general pattern Similarly, the recurrence relation gives us a1 , 4·3 1 a7 = a4 = a1 , 7·6 7·6·4·3 1 a10 = a7 = a1 10 · 10 · · · · · a4 = www.elsolucionario.net GTBL001-15 P2: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 15-37 SECTION 15.4 Power Series Solutions of Differential Equations 1257 and so on More generally, we can establish the pattern: a3n+1 = (3n − 1)(3n − 4) · · · · · a1 , (3n + 1)! where again, this is not intended as a literal formula Now that we have found all of the coefficients, we can write down the solution, by separately writing out every third term of the series, as follows: y= ∞ an x n = n=0 = a0 ∞ a3n x 3n + a3n+1 x 3n+1 + a3n+2 x 3n+2 n=0 ∞ n=0 ∞ (3n − 2)(3n − 5) · · · · · 3n (3n − 1)(3n − 4) · · · · · 3n+1 x + a1 x (3n)! (3n + 1)! n=0 y1 (x) y2 (x) = a0 y1 (x) + a1 y2 (x) We leave it as an exercise to use the Ratio Test to show that the power series defining y1 and y2 are absolutely convergent for all x You may have noticed that in all three of our examples, we assumed that there was a solution of the form y= ∞ an x n = a0 + a1 x + a2 x + · · · , n=0 only to arrive at the general solution y = a0 y1 (x) + a1 y2 (x), where y1 and y2 were power series solutions of the equation This is in fact not coincidental One can show that (at least for certain equations) this is always the case One clue as to why this might be so lies in the following Suppose that we want to solve the initial value problem consisting of a secondorder differential equation and the initial conditions y(0) = A and y (0) = B Taking y(x) = ∞ an x n gives us n=0 y (x) = ∞ nan x n−1 = a1 + 2a2 x + 3a3 x + · · · n=0 So, imposing the initial conditions, we have A = y(0) = a0 + a1 (0) + a2 (0)2 + · · · = a0 and B = y (0) = a1 + 2a2 (0) + 3a3 (0)2 + · · · = a1 So, irrespective of the particular equation we’re solving, we always have y(0) = a0 and y (0) = a1 You might ask what you’d if the initial conditions were imposed at some point other than at x = 0, say at x = x0 In this case, we look for a power series solution of the form y= ∞ an (x − x0 )n n=0 It’s easy to show that in this case, we still have y(x0 ) = a0 and y (x0 ) = a1 www.elsolucionario.net P1: OSO/OVY P1: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 1258 CHAPTER 15 Second-Order Differential Equations 15-38 In the exercises, we explore finding series solutions about a variety of different points BEYOND FORMULAS This section connects two important threads of calculus: solutions of differential equations and infinite series In Chapter 8, we expressed known functions like sin x and cos x as power series Here, we just extend that idea to unknown solutions of differential equations For second-order homogeneous equations, keep in mind that the general solution has the format c1 y1 (x) + c2 y2 (x) You should think about the problems in this section as following this strategy: write the solution as a power series, substitute into the differential equation and find relationships between the coefficients of the power series, remembering that two of the coefficients will be left as arbitrary constants EXERCISES 15.4 WRITING EXERCISES After substituting a power series representation into a differential equation, the next step is always to rewrite one or more of the series, so that all series have the same exponent (Typically, we want x n ) Explain why this is an important step For example, what would we be unable to if the exponents were not the same? The recurrence relation is typically solved for the coefficient with the largest index Explain why this is an important step Explain why you can’t solve equations with nonconstant coefficients, such as y + 2x y + 2y = 0, by looking for a solution in the form y = er x The differential equations solved in this section are actually of a special type, where we find power series solutions centered at what is called an ordinary point For the equation x y + y + 2y = 0, the point x = is not an ordinary point Discuss what goes wrong here if you look for a power series ∞ an x n solution of the form In exercises 1–8, find the recurrence relation and general power ∞ ∞ y= an (x − 1)n n=0 10 Find a series solution of y + y + (x − 2)y = in the form ∞ an (x − 2)n n=0 11 Find a series solution of Airy’s equation y − x y = in the ∞ form an (x − 1)n [Hint: First rewrite the equation in the n=0 form y − (x − 1)y − y = 0.] 12 Find a series solution of Airy’s equation y − x y = in the ∞ form an (x − 2)n n=0 13 Solve the initial value problem y + 2x y + 4y = 0, y(0) = 5, y (0) = −7 (See exercise 1.) 14 Solve the initial value problem y + 4x y + 8y = 0, y(0) = 2, y (0) = π (See exercise 2.) 15 Solve the initial value problem y + (1 − x)y − y = 0, y(1) = −3, y (1) = 12 (See exercise 9.) 16 Solve the initial value problem y + y + (x − 2)y = 0, y(2) = 1, y (2) = −1 (See exercise 10.) n=0 series solution of the form Find a series solution of y + (1 − x)y − y = in the form an x n n‫؍‬0 y + 2x y + 4y = y + 4x y + 8y = y − x y − y = y − x y − 2y = y − x y = y + 2x y = y − x y = y + x y − 2y = 17 Determine the radius of convergence of the power series solutions about x0 = of y − x y − y = (See exercise 3.) 18 Determine the radius of convergence of the power series solutions about x0 = of y − x y − 2y = (See exercise 4.) 19 Determine the radius of convergence of the power series solutions about x0 = of y + (1 − x)y − y = (See exercise 9.) 20 Determine the radius of convergence of the power series solutions about x0 = of y − x y = (See exercise 11.) www.elsolucionario.net GTBL001-15 P2: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 15-39 CHAPTER 15 21 Find a series solution of the form y = ∞ an x n to the equation n=0 x y + x y + x y = (Bessel’s equation of order 0) 22 Find a series solution of the form y = ∞ an x n to the equation n=0 x y + x y + (x − 1)y = (Bessel’s equation of order 1) 2 23 Determine the radius of convergence of the series solution found in example 4.3 24 Determine the radius of convergence of the series solution found in problem 12 Review Exercises 1259 28 Use the technique of exercise 25 to find the fifth-degree Taylor polynomial for the solution of the initial value problem y + y − (e x )y = 0, y(0) = 2, y (0) = 29 Use the technique of exercise 25 to find the fifth-degree Taylor polynomial for the solution of the initial value problem y + x y + (sin x)y = 0, y(π) = 0, y (π ) = 30 Use the technique of exercise 25 to find the fifth-degree Taylor polynomial for the solution of the initial value problem y + (cos x)y + x y = 0, y( π2 ) = 3, y π2 = 25 For the initial value problem y + 2x y − x y = 0, y(0) = 2, y (0) = −5, substitute in x = and show that y (0) = Then take y = −2x y + x y and show that y = −2x y + (x − 2)y + y Conclude that y (0) = 12 Then compute y (4) (x) and find y (4) (0) Finally, compute y (5) (x) and find y (5) (0) Write out the fifth-degree Taylor polynox2 mial for the solution, P5 (x) = y(0) + y (0)x + y (0) + x3 x4 x5 y (0) + y (4) (0) + y (5) (0) 3! 4! 5! 26 Use the technique of exercise 25 to find the fifth-degree Taylor polynomial for the solution of the initial value problem y + x y − (cos x)y = 0, y(0) = 3, y (0) = The equation y − 2xy + 2ky = for some integer k ≥ is known as Hermite’s equation Following our procedure for finding series solutions in powers of x, show that, in fact, one of the series solutions is simply a polynomial of degree k For this polynomial solution, choose the arbitrary constant such that the leading term of the polynomial is 2k x k The polynomial is called the Hermite polynomial Hk (x) Find the Hermite polynomials H0 (x), H1 (x), , H5 (x) 27 Use the technique of exercise 25 to find the fifth-degree Taylor polynomial for the solution of the initial value problem y + e x y − (sin x)y = 0, y(0) = −2, y (0) = The Chebyshev polynomials are polynomial solutions of the equation (1 − x )y − x y + k y = 0, for some integer k ≥ Find polynomial solutions for k = 0, 1, and EXPLORATORY EXERCISES Review Exercises WRITING EXERCISES The following list includes terms that are defined and theorems that are stated in this chapter For each term or theorem, (1) give a precise definition or statement, (2) state in general terms what it means and (3) describe the types of problems with which it is associated Nonhomogeneous equation Second-order differential equation Method of undetermined coefficients Damping Resonance Recurrence relation The current in an electrical circuit satisfies the same differential equation as the displacement function for a mass attached to a spring The particular solution of a nonhomogeneous equation mu + ku + cu = F has the same form as the forcing function F Resonance cannot occur if there is damping A recurrence relation can always be solved to find the solution of a differential equation TRUE OR FALSE State whether each statement is true or false and briefly explain why If the statement is false, try to “fix it” by modifying the given statement to a new statement that is true The form of the solution of ay + by + cy = depends on the value of b2 − 4ac In exercises 1–6, find the general solution of the differential equation y + y − 12y = y + 4y + 4y = y + y + 3y = y + 3y − 8y = www.elsolucionario.net P1: OSO/OVY P1: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 1260 CHAPTER 15 Second-Order Differential Equations 15-40 Review Exercises y − y − 6y = e3t + t + In exercises 21 and 22, find the recurrence relation and a general y − 4y = 2e + 16 cos 2t power series solution of the form ∞ 2t an (x − 1)n n‫؍‬0 In exercises 7–10, solve the initial value problem y + 2y − 8y = 0, y(0) = 5, y (0) = −2 y + 2y + 5y = 0, y(0) = 2, y (0) = y + 4y = cos t, y(0) = 1, y (0) = 10 y − 4y = 2e2t + 16 cos 2t 11 A spring is stretched inches by a 4-pound weight The weight is then pulled down an additional inches and released Neglect damping Find an equation for the position of the weight at any time t and graph the position function 12 In exercise 11, if an external force of cos ωt pounds is applied to the weight, find the value of ω that would produce resonance If instead ω = 10, find and graph the position of the weight 13 A series circuit has an inductor of 0.2 henry, a resistor of 160 ohms and a capacitor of 10−2 farad The initial charge on the capacitor is 10−4 coulomb and there is no initial current Find the charge on the capacitor and the current at any time t 14 In exercise 13, if the resistor is removed and an impressed voltage of sin ωt volts is applied, find the value of ω that produces resonance In this case, what would happen to the circuit? In exercises 15 and 16, determine the form of a particular solution 15 u + 2u + 5u = 2e−t sin 2t + 4t − cos 2t 16 u + 2u − 3u = (3t + 1)et − e−3t cos 2t 17 A spring is stretched inches by a 4-pound weight The weight is then pulled down an additional inches and set in motion with a downward velocity of ft/s A damping force equal to 0.4u slows the motion of the spring An external force of magnitude sin 2t pounds is applied Completely set up the initial value problem and then find the steady-state motion of the spring 18 A spring is stretched inches by an 8-pound weight The weight is then pushed up inches and set in motion with an upward velocity of ft/s A damping force equal to 0.2u slows the motion of the spring An external force of magnitude cos 3t pounds is applied Completely set up the initial value problem and then find the steady-state motion of the spring In exercises 19 and 20, find the recurrence relation and a general ∞ power series solution of the form an x n n‫؍‬0 19 y − 2x y − 4y = 20 y + (x − 1)y = 21 y − 2x y − 4y = 22 y + (x − 1)y = In exercises 23 and 24, solve the initial value problem 23 y − 2x y − 4y = 0, y(0) = 4, y (0) = 24 y − 2x y − 4y = 0, y(1) = 2, y (1) = EXPLORATORY EXERCISES A pendulum that is free to rotate through 360 degrees has two equilibrium points One is hanging straight down and the other is pointing straight up The θ = π equilibrium is unstable and is classified as a saddle point This means that for most but not all initial conditions, solutions that start near θ = π will get farther away Explain why with initial conditions θ(0) = π and θ (0) = 0, the solution is exactly θ(t) = π However, explain why initial conditions θ(0) = 3.1 and θ (0) = would have a solution that gets farther from θ = π For the model θ (t) + Lg θ(t) = 0, show that if v = π Lg , then the initial conditions θ(0) = and θ (0) = v produce a solution that reaches the state θ = π and θ = Physically, explain why the pendulum would remain at θ = π and then explain why the solution of our model does not get “stuck” at θ = π Explain why for any starting angle θ, there exist two initial angular velocities that will balance the pendulum at θ = π The undamped pendulum model θ (t) + Lg sin θ(t) = is equivalent to the system of equations (with y1 = θ and y2 = θ ) y1 = y2 , g y2 = − sin y1 L Use a CAS to sketch the phase portrait of this system of equations near the equilibrium point (π, 0) Explain why the phase portrait shows an unstable equilibrium point with a small set of initial conditions that lead to the equilibrium point In exploratory exercise of section 15.3, we investigate the motion of a ball dropped in a hole drilled through a nonrotating Earth Here, we investigate the motion taking into account the Earth’s rotation (See Andrew Simoson’s article in the June 2004 Mathematics Magazine.) We describe the motion in polar coordinates with respect to a fixed plane through the equator Define unit vectors ur = cos θ, sin θ and www.elsolucionario.net GTBL001-15 P2: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net 15-41 CHAPTER 15 Review Exercises 1261 Review Exercises uθ = − sin θ, cos θ If the ball has position vector r ur , show that its acceleration is given by velocity from the rotation of the Earth For the gravitational force f (r ) = −c2 r , show that a solution is [r θ (t) + 2r (t)θ (t)]uθ + {r (t) − r (t)[θ (t)]2 }ur Since gravity acts in the radial direction only, r θ (t) + 2r (t)θ (t) = Show that this implies that r (t)θ (t) = k for some constant k (This is the law of conservation of angular momentum.) If the acceleration due to gravity is f (r )ur for some function f, show that k2 = f (r ) r3 Initial conditions are r (0) = R, r (0) = 0, θ(0) = and 2π θ (0) = Here, Q is the period of one revolution of the Q Earth and we assume that the ball inherits the initial angular r (t) − r (t) = or r (θ) = R cos2 ct + k2 c2 R sin2 ct 1 R2 cos2 θ + c2 R k2 sin2 θ Show that this converts to x2 y2 + =1 R (k/Rc)2 and describe the path of the ball www.elsolucionario.net P1: OSO/OVY GTBL001-15 P2: OSO/OVY QC: OSO/OVY GTBL001-Smith-v16.cls T1: OSO November 24, 2005 17:29 www.elsolucionario.net www.elsolucionario.net P1: OSO/OVY 1262 ... familiar is through experience, by working example after example 60 Ϫ2 EXPLORATORY EXERCISES GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS y Ϫ4 row must she win to raise the reported winning percentage... other criminals Rossmo served 21 years with the Vancouver Police Department His mentors were Professors Paul and Patricia Brantingham of Simon Fraser University The Brantinghams developed Crime... example 2.2 was produced by a process of trial and error with thoughtful corrections You are unlikely to get a perfect picture on the first try However, you can enlarge the graphing window (i.e.,