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Introduction to Differential Equations
C H A P T E R
1 Introduction to Differential Equations O U T L I N E 1.1 Introduction to Differential Equations: Vocabulary Exercises 1.1 1.2 A Graphical Approach to Solutions: Slope Fields and Direction Fields
Exercises 1.2
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Chapter 1 Summary: Essential Concepts and Formulas
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Chapter 1 Review Exercises
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Introductory Differential Equations introduces and discusses the topics covered in a typical first course in ordinary differential equations. Throughout, we indicate how technologies such as computer algebra systems can be used to enhance the study of differential equations, not only by eliminating some of the computational difficulties that arise in the study of differential equations but also by overcoming some of the visual limitations associated with the solutions of differential equations. The advantages of using technologies like computer algebra systems in the study of differential equations are numerous, but perhaps the most useful is that of being able to produce the graphics associated with solutions of differential equations. This is particularly beneficial in the discussion of applications because many physical situations are modeled with differential equations. For example, in Chapter 5, we see that the motion of a penduIntroductory Differential Equations. DOI: https://doi.org/10.1016/B978-0-12-814948-5.00001-X
lum can be modeled by a differential equation. When we solve the problem of the motion of a pendulum, we use technology to watch the pendulum move. The same is true for studying the differential equations that describe the motion of a mass attached to the end of a spring, as well as many other problems. In having these technologies, the study of differential equations becomes more visual and, consequently, more interesting. Although Chapter 1 is short in length, the vocabulary introduced here is used throughout the text and that subsequent chapters will take advantage of the terminology and techniques discussed here. Nearly all introductions to differential equations begin with German scientist Gottfried Wilhelm Leibniz (1646–1716) and British scientist Isaac Newton (1642–1727), the inventors of calculus. In integral calculus we learn that the area under the graph of a smooth positive function is
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1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
given by a definite integral, but both Leibniz and Newton were more concerned with solving differential equations than finding areas. Many of the methods of solution we discuss in this text are contributed to the great Swiss mathematician Leonhard Euler (1707–1783).
Gottfried Wilhelm Leibniz (1646–1716)
world and indicate how some people use differential equations beyond this course. However, mathematics is also interesting in its own right so mathematical applications are also included throughout the text.
Isaac Newton: We hope this young Isaac Newton was not worried about who discovered calculus first.
Isaac Newton (1642–1727): The controversy as to who (Leibniz or Newton) discovered calculus first became an obsession with Newton during the second half of his life.
Subsequently, many problems, such as determining the motion of a plucked string, not only lead to ordinary and partial differential equations but to other areas of mathematics as well. Differential equations are full of rich and exciting mathematical and applied applications. Consequently, interesting applications are included throughout the text to motivate discussions, make the study of differential equations more interesting and pertinent to the real
Leonhard Euler (1707–1783)
1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS: VOCABULARY We begin our study of differential equations by explaining what a differential equation is. From our experience in calculus, we are famil-
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1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS: VOCABULARY
iar with some differential equations. For example, suppose that the acceleration due to gravity, a(t), (measured in ft/s2 ) of a falling object is a(t) = −32. Then, because a(t) = v (t), where v(t) is the velocity of the object (measured in ft/s), we have v (t) = −32 or dv = −32. dt An equation like this involving a function of a single variable is called an ordinary differential equation (ODE). (If the equation involves partial derivatives, then it is called a partial differential equation (PDE).) In this case, the function to be determined is v = v(t), which depends on the variable t, representing time (measured in seconds). The goal in solving an ODE is to find a function that satisfies the equation. We can solve this ODE through integration: Throughout the text, we will use ODE to refer to an ordinary differential equation and PDE to refer to a partial differential equation.
v(t) =
a(t) dt =
(−32) dt = −32t + C,
where C is an arbitrary constant. This result indicates that v(t) = −32t + C is a solution of the ODE for any choice of the constant C. (We call this a general solution because it involves an arbitrary constant.) In fact, we have found every solution of the ODE because each is expressed as −32t + C. Examples of solutions and the corresponding C values include v(t) = −32t (C = 0), v(t) = −32t + 32 (C = 32), and v(t) = −32t − 8 (C = −8). This shows that there are an infinite number of solutions to the ODE. We can verify that v(t) = −32t + C is a general solution of dv/dt = −32 through substitution: dv d = (−32t + C) = −32. dt dt When we substitute our solution into the left side of the ODE, we obtain −32, the value on
the right side of the ODE. We have verified that v(t) = −32t + C satisfies the ODE for any choice of C. Many times we are given a particular condition that the solution must satisfy. For example, suppose that the object considered earlier has an initial velocity of −64 ft/s. In other words, the velocity at time t = 0 s is represented with the initial condition v(0) = −64. Therefore, we need to find a solution of the ODE that also satisfies the initial condition. We express this initial value problem (IVP) as Throughout the text we will refer to initial value problems as IVPs.
dv = −32, dt
v(0) = −64,
where we solve the initial value problem by first finding a general solution to the ODE and then by applying the initial condition to determine the arbitrary constant. When we substitute t = 0 into the general solution v(t) = −32t + C, we obtain v(0) = −32 · 0 + C = C. We conclude that C = −64 so that v(0) = −64 is satisfied. This means that the solution to the IVP is v(t) = −32t − 64. Notice that unlike the ODE, the IVP has only one solution. Another application of differential equations found in calculus is finding a function when given (a) the slope of the line tangent to the graph of the function at any point (x, y), and (b) a point on the graph. For example, suppose that the slope of the tangent line at any point on the graph of a function y = y(x) is given by dy = 3x 2 − 4x, dx and further that the graph passes through the point (1, 4), which means that y(1) = 4. In this case, the ordinary differential equation is given by dy/dx = 3x 2 − 4x, where we need to find the function y = y(x) that satisfies the initial condi-
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1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
tion y(1) = 4. Therefore, we solve the IVP dy = 3x 2 − 4x, dx
y(1) = 4.
As in the previous example, we find a general solution to dy/dx = 3x 2 − 4x through integration. This yields y = y(x) = x 3 − 2x 2 + C. Now, when we apply the initial condition, we find that y(1) = 13 − 2 · 13 + C = −1 + C = 4 so that C = 5, which means that the solution to the IVP is y = y(x) = x 3 − 2x 2 + 5. Fig. 1.1 shows the graph of the solution together with a portion of the tangent line at the point (1, 4). Notice that the slope of the tangent line at the point (1, 4) is the value of dy/dx evaluated if x = 1: if x = 1, dy/dx = 3 · 12 − 4 · 1 = −1. This observation will be useful in Section 1.2 in helping us better understand the behavior of solutions of differential equations. The previous examples are similar in that they each involve an ordinary differential equation in which the highest order derivative is the first derivative. We call equations of this type first order ordinary differential equations because the order of a differential equation is the order of the highest order derivative appearing in the equation. Example 1.1. Determine the order of each of the following differential equations: (a) dy/dx = x 2 /y 2 cos y, (b) uxx + uyy = 0, (c) (dy/dx)4 = y + x, (d) d 2 x/dt 2 + 2dx/dt + 3x = sin t . Solution: (a) This equation is first order because it includes only one first order derivative, dy/dx. (b) This equation is classified as second order because the highest order derivatives, uxx , representing ∂ 2 u/∂x 2 , and uyy representing ∂ 2 u/∂y 2 , are of order two.
FIGURE 1.1 Solution to dy/dx = 3x 2 − 4x, y(1) = 4 along with a segment of the tangent line at (1, 4).
The equation uxx + uyy = 0 arises in many areas of study, which include fluid flows as well as electrostatic and gravitational potential, is often called Laplace’s equation after Pierre-Simon Laplace (also known as the Marquis de Laplace) (1749–1827) or the potential equation. Consequently, Laplace’s equation is a second order partial differential equation. (e) This is a first order equation because the highest order derivative is the first derivative. Raising that derivative to the fourth power does not affect the order of the equation. The expressions
dy dx
4 and
d 4y dx 4
do not represent the same quantities: (dy/dx)4 represents the derivative of y with respect to x, dy/dx, raised to the fourth power; d 4 y/dx 4 represents the fourth derivative of y with respect to x. (d) The highest order derivative is d 2 x/dt 2 , so the equation is second order.
1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS: VOCABULARY
Pierre-Simon Laplace (1749, Normandy, France1827, Paris, France): Laplace’s name is probably most remembered in reference to the Laplace transform that we will study in Chapter 8.
Example 1.2. (a) Show that y = c1 sin t + c2 cos t satisfies the second order ordinary differential equation (ODE) y + y = 0 where c1 and c2 are arbitrary constants. (b) Find the solution to the IVP y + y = 0, y(0) = 0, y (0) = 1. Solution: (a) Differentiating, we obtain y = dy/dt = c1 cos t − c2 sin t and y = d 2 y/dt 2 = −c2 cos t − c1 sin t . Therefore, substituting y into y + y (the left side of the differential equation) results in d 2y + y = −c1 sin t − c2 cos t + c1 sin t + c2 cos t dt 2 = 0, so the function satisfies the (ordinary differential equation) ODE for any choice of c1 and c2 . We graph the solution for several values of these constants in Fig. 1.2. Because the solution depends on at least one constant, we say that the functions shown in Fig. 1.3 are members of the family of solutions of the ODE. (b) Evaluating the function at t = 0 yields y(0) = c1 sin 0 + c2 cos 0 = c2 . Then, the initial condition, y(0) = 0, indicates that c2 = 0. Similarly, y (0) = c1 cos 0 − c2 sin 0 = c1 , so c1 = 1 so that the second initial condition, y (0) = 1, is satisfied.
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FIGURE 1.2 Graphs of y = c1 cos t + c2 sin t for various values of c1 and c2 .
Therefore, y(t) = sin t is the solution to the initial value problem (IVP). We graph this solution in Fig. 1.3. Notice that the ODE has an infinite number of solutions while the IVP has only one unique solution. Also observe that the number of initial conditions for the initial value problem (IVP) matches the order of the ordinary differential equation (ODE). The next level of classification tells us whether an equation is linear or nonlinear in terms of the dependent variable. For an ordinary differential equation (ODE), assuming that the independent variable is x and the dependent variable is y, an ordinary differential equation (of order n) is called linear if it can be written as d ny d n−1 y d 2y + a (x) + · · · + a (x) n−1 2 dx n dx n−1 dx 2 dy + a1 (x) (1.1) + a0 (x)y = f (x), dx
an (x)
where the functions ak (x), k = 0, 1, . . . , n, and f (x) are given and an (x) is not the zero function. If f (x) is identically the zero function, the linear equation, (1.1), is said to be homogeneous. You should verify that y(x) = 0, which we call the trivial solution, is a solution to every linear homogeneous equation. If f (x) is not identically the zero function, the linear equation, (1.1), is said to be nonhomogeneous.
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1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.3 Graph of y = y(t) = sin t .
In Chapter 4, we will learn that a general solution of the nth order linear equation, (1.1), is a solution that depends on n arbitrary constants and includes all solutions of the equation. If the equation under consideration cannot be written in the form given by Eq. (1.1), the equation is said to be nonlinear. Therefore, some of the properties that lead to classifying an equation as linear or nonlinear are powers of the dependent variable (or one of its derivatives) and functions of the dependent variable. Example 1.3. Determine which of the followdy ing differential equations are linear: (a) = x3, dx d 2u + u = ex , (c) (y − 1) dx + x cos y dy = 0, (b) dx 2 d 3y dy d 2x dy (d) +y = x, (e) + x 2 y = x, (f) 2 + 3 dx dx dx dt sin x = 0. Solution: (a) This equation is linear because the nonlinear term x 3 is the function f (x) of the independent variable in the general formula for a linear differential equation. (b) This equation is also linear. Using u as the name of the dependent variable does not affect the linearity. (c) If y is the dependent variable, solving for dy/dx gives us dy 1−y = . dx x cos y Because the right side of this equation includes a nonlinear function of y, the equation is nonlinear (in y). However, if x is the dependent variable, solving for dx/dy yields dx cos y = x. dy 1−y
This equation is linear in the dependent variable x. (d) The coefficient of the term dy/dx is y instead of an expression involving only the independent variable x. Hence, this equation is nonlinear in the dependent variable y. (e) This equation is linear. The term x 2 is the coefficient function. (f) For this equation, note that x is the dependent variable; t is the independent variable. This equation, known as the pendulum equation because it models the motion of a simple pendulum, is nonlinear because it involves a nonlinear function of the dependent variable x, sin x. If an ordinary differential equation (ODE) has the form dy/dx = f (x), we can use integration to determine y = y(x), although the result may be in terms of integrals that cannot be evaluated using standard techniques of integration. The following examples of differential equations of this type illustrate some of the typical methods of integration that may be encountered. Example 1.4. Solve the following differential dy dy x equations. (a) , = cos x, (b) = √ 2 dx dx x +1 dy dy dy 1 1 (c) , (d) , = 2 = xex , (e) = dx dx dx x + 16 4 − x2 dy and (f) = sin x tan x. dx Solution: In each case, we integrate the indicated function and graph the solution for several values of the constant of integration. Each solution contains a constant of integration so there are infinitely many solutions to each equation. (a) y = cos x dx = sin x + C (see Fig. 1.4A). (b) To eval x uate y = √ dx, we let u = x 2 + 1 so that x2 + 1
1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS: VOCABULARY
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FIGURE 1.4 (A) Graph of y = sin x + C for various values of C. (B) Graph of y = x 2 + 1 + C for various values of C.
1 du = x dx. Then, 2 2x 1 1 1 y= dx = √ √ du 2 2 u x2 + 1 1 u−1/2 du = u1/2 + C = 2 (See Fig. 1.4B.) = x 2 + 1 + C.
du = 2x dx or
1 dx we use a 2 x + 16 trigonometric substitution. Letting x = 4 tan θ , −π/2 < θ < π/2, so that dx = 4 sec2 θ dθ gives us 1 4 sec2 θ dθ y= 16 + (4 tan θ )2 1 4 sec2 θ dθ = 16 + 16 tan2 θ 1 1 4 sec2 θ dθ = 16 sec2 θ x 1 1 1 = dθ = θ + C = tan−1 + C. 4 4 4 4 (c) To integrate y =
First, we use the identity 1 + tan2 θ = sec2 θ and then resubstitute: x/4 = tan θ so θ = tan−1 (x/4).
(d) To evaluate y = xex by hand, we use the Integration by Parts formula with u = x and dv =
ex dx. Then, du = dx and v = ex . This gives x x y = xe dx = xe − ex dx = xex − ex + C. (e) Use partial fractions to evaluate this integral. First, find the partial fraction decomposition of 1/(4 − x 2 ), which is determined by finding constants A and B that satisfy the equation 1 A B = + . (2 − x)(2 + x) 2 − x 2 + x These values are A = B = 1/4. (Why?) Thus, 1 1 1 + dx y= 4 2−x 2+x 1 = [− ln |2 − x| + ln |2 + x|] + C 4 1 2 + x = ln + C. (See Fig. 1.5.) 4 2−x The Integration by Parts formula states that u dv = u v − v du.
Find C if y(1) = 2.
Notice that the solutions of dy/dx = 1/(4−x 2 ) are undefined if x = −2 or x = 2. This is because
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1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
we can find only an equation involving the independent and dependent variables that the solution satisfies. In this case, we say that we have found an implicit solution. We will see that given an arbitrary differential equation, constructing an explicit or implicit solution is nearly always impossible. Consequently, although mathematicians were first concerned with finding analytic (explicit or implicit) solutions to differential equations, they have since (frequently) turned their attention to addressing properties of the solution and finding algorithms to approximate solutions. FIGURE 1.5 Graph of y =
1 2 + x ln +C for various val4 2−x
ues of C.
1/(4 − x 2 ) is undefined at these two values of x. Later, we will discuss in greater detail the relationship between the differential equation and its solutions. sin2 x (f) In this case, y = sin x tan x dx = dx becos x cause tan x = sin x/ cos x. Now use the identity sin2 x + cos2 x = 1 or sin2 x = 1 − cos2 x and divide: sin2 x y= dx cos x 1 − cos2 x y= dx cos x y = (sec x − cos x) dx y = ln | sec x + tan x| − sin x + C. For −π/2 < x < π/2, + C.
sec x dx = ln(sec x + tan x)
In Example 1.4, each solution is given as a function y = y(x) of the independent variable. In these cases, the solution is said to be explicit. In solving some differential equations, however,
Example 1.5. Verify that the equation 2x 2 + y 2 − 2xy + 5x = 0 satisfies the differential equation dy 2y − 4x − 5 = . dx 2y − 2x Solution: We use implicit differentiation to compute y = dy/dx if 2x 2 + y 2 − 2xy + 5x = 0: 4x + 2y
dy dy − 2x − 2y + 5 = 0 dx dx dy (2y − 2x) = 2y − 4x − 5 dx dy 2y − 4x − 5 = . dx 2y − 2x
The equation 2x 2 + y 2 − 2xy + 5x = 0 satisfies the 2y − 4x − 5 dy = . differential equation dx 2y − 2x Although we cannot solve 2x 2 + y 2 − 2xy + 5x = 0 for y as a function of x (see Fig. 1.6), we can determine the corresponding y value(s) for a given value of x. For example, if x = −1, then 2 + y 2 + 2y − 5 = y 2 + 2y − 3 = (y + 3)(y − 1) = 0. Therefore, the points (−1, −3) and (−1, 1) lie on the graph of 2x 2 + y 2 − 2xy + 5x = 0. (See Fig. 1.7.)
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1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS: VOCABULARY
Example 1.6 (Duffing’s Equation). Duffing’s equation is the second order nonlinear equation d 2x dx +k − x + x 3 = cos ωt, dt dt 2
(1.2)
where k, , and ω are positive constants. Sources: See texts like Jordan and Smith’s Nonlinear Ordinary Differential Equations, [16].
Write Duffing’s equation as a system of first order equations
FIGURE 1.6 Graph of 2x 2 + y 2 − 2xy + 5x = 0.
Solution: Let y = x . Then, y = x and substituting into Duffing’s equation gives us x + kx − x + x 3 = cos ωt
Find an equation of the line tangent to the graph of 2x 2 + y 2 − 2xy + 5x = 0 at the points (−1, −3) and (−1, 1).
In the same manner that we consider systems of equations in algebra, we can also consider systems of differential equations. For example, if x and y represent functions of t , we will learn in Chapter 6 to solve the system of linear equations
dx/dt = ax + by , dy/dt = cx + dy where a, b, c, and d represent constants and differentiation is with respect to t . We will see that systems of differential equations arise naturally in many physical situations that are modeled with more than one equation and involve more than one dependent variable. In addition, we will see that it is often useful to write a differential equation of order greater than one as a system of first order equations, especially when the original equation is nonlinear.
y + ky − x + x 3 = cos ωt y = x − x 3 − ky + cos ωt. Thus, Duffing’s equation is equivalent to the nonlinear system
x = y y = x − x 3 − ky + cos ωt Note that a system of differential equations can consist of more than two equations. For example, the basic equations that describe the competition between two organisms, with population densities x1 and x2 , respectively, in a chemostat are Sources: See Smith and Waltman’s The Theory of the Chemostat, [27], for a detailed discussion of chemostat models.
⎧ m1 S m2 S ⎪ x1 − x2 S = 1 − S − ⎪ ⎪ ⎪ a + S a 1 2+S ⎪ ⎨ m1 S x1 = x1 −1 a ⎪ 1+S ⎪ ⎪ ⎪ m2 S ⎪ ⎩x2 = x2 −1 a2 + S
,
(1.3)
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1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.7 Notice that near the points (−1, −3) and (−1, 1) the implicit solution looks like a function. In fact, when we zoom in near the points (−1, −3) and (−1, 1), we see what appears to be the graph of a function.
where denotes differentiation with respect to t ; S = S(t), x1 = x1 (t), and x2 = x2 (t). For Eqs. (1.3), we remark that S denotes the concentration of the nutrient available to the competitors with population densities x1 and x2 . We investigate chemostat models in more detail in Chapter 7.
7. 8. 9. 10.
EXERCISES 1.1 Determine if each of the following equations is an ordinary differential equation or a partial differential equation. If the equation is an ordinary differential equation, then determine (a) the order of the ordinary differential equation and (b) if the equation is linear or nonlinear. 1. 2. 3. 4. 5.
d 2y
dy + − 2y = x 3 dx 2 dx dy y + y 4 = sin x dx 2 ∂ 2y 2 ∂ y , c > 0 constant = c ∂t 2 ∂x 2 y − 2y + 5y + y = ex ( = d/dx; y = y(x)) 2 dy +y =0 dx
d 2y dy +t + 2y = 0 2 dt dt 2 2 1 ∂ z ∂ z ∂ 2z = + , (z = z(t, x, y)) c2 ∂t 2 ∂x 2 ∂y 2 u ux + ut = 0, (u = u(t, x)) 2 2 d y + 2y = 2x x dx 2 d 2x + 2 sin x = sin 2t (x = x(t)) dt 2 ut + u ux = σ uxx , σ constant
6. t 2
11.
12. (2x − 1) dx − dy = 0 13. (2t − y) dt − dy = 0 ∂u ∂u 14. = u (u = u(x, y)) ∂x ∂y 15. (2x − y) dx − y dy = 0 16. Write each of the following second order equations as a system of first order equations. d 2 x dx (a) − − 6x = 0 dt dt 2 d 2x dx (b) 4 2 + 4 + 37x = 0 dt dt d 2x (c) L 2 + g sin x = 0, L, g positive condt stants, x = x(t) dx d 2x − μ(1 − x 2 ) + x = 0, μ > 0 con(d) 2 dt dt stant
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EXERCISES 1.1
d 2x dx +(b −t) −ax = 0, a, b constants dt dt 2 In Exercises 17–28, verify that each of the given functions is a solution to the corresponding differential equation. (A, B, and C represent constants.) (e) t
17. dy/dx + 2y = 0, y(x) = e−2x , y(x) = 5e−2x 2 18. dy/dx + xy = 0, y(x) = e−x /2 19. dy/dx + y = sin x, y(x) = e−x − 12 cos x + 1 2 sin x d 2 y dy 20. − − 12y = 0, y(t) = e4t , y(t) = e−3t dt dt 2 21. y + 9y = 0, y = A + Be−9t ( denotes d/dt) 22. x + 3x − 10x = 0, x(t) = Ae2t + Be−5t 23. x + x = t cos t − cos t, x = A cos t + B sin t + 1 2 1 1 4 t sin t − 2 t sin t + 4 t cos t 24. y − 12y + 40y = 0, y = e6x cos 2x, y = e6x sin 2x ( denotes d/dx) 25. y − 4y = 0, y = A + Be2x + Ce−2x 26. y − 2y = 0, y = A + Bt + Ce2t 27. x 2 y − 12x y + 42y = 0, y = Ax 6 + Bx 7 28. t 2 y + 3t y + 5y = 0, y = t −1 (A cos(2 ln t) + B sin(2 ln t)) In Exercises 29–33, verify that the given equation satisfies the differential equation. Use the equation to determine y for the given value of x (or t). 29. dy/dx = −x/y, x 2 + y 2 = 16, x = 0 30. 3y(t 2 + y)dt + t (t 2 + 6y)dy = 0, t 3 y + 3ty 2 = 8, t = 2 31. dy/dx = −2y/x − 3, x 3 + x 2 y = 100, x = 1 32. y cos t dt + (2y + sin t) dy = 0, y 2 + y sin t = 1, t =0 33. (y/x + cos y) dx + (ln x − x sin y) dy = 0, y ln x + x cos y = 0, x = 1 In Exercises 34–43, use integration to find a solution to the differential equation. 34. 35. 36. 37. 38. 39.
dy/dx = (x 2 − 1)(x 3 − 3x)3 2 dy/dx = x sin √x 2 dy/dx = x/ x − 16 dy/dx = 1/(x ln x) dy/dx = x ln x dy/dx = xe−x
40. 41. 42. 43. 44. 45. 46.
−2(x + 5) dy = dx (x + 2)(x − 4) x − x2 dy = 2 dx (x + √ 1)(x + 1) 2 dy/dx = x − 16/x dy/dx = (4 − x 2 )3/2 dy/dx = 1/(x 2 − 16) dy/dx = cos x cot x dy/dx = sin3 x tan x
In Exercises 45–54, use the indicated conditions with the indicated solution to determine the solution to the given problem. 47. dy/dx + 2y = 0, y(0) = 2, y(x) = Ae−2x 48. dy/dt + y = sin t, y(0) = −1, y(t) = Ae−t − 1 1 2 cos t + 2 sin t 49. y − y − 12y = 0, y(0) = 1, y (0) = −1, y = Ae4x + Be−3x 50. y + 9y = 0, y(0) = 2, y (0) = −1, y = A + Be−9x 51. y − 2y = 0, y(0) = 0, y (0) = 1, y (0) = 3, y = A + Bx + Ce2x 52. y − 4y = 0, y(0) = 1, y (0) = −1, y (0) = 0, y = A + Be2x + Ce−2x 53. t 2 y − 12t y + 42y = 0, y(1) = 0, y (1) = −1, y(t) = At 6 + Bt 7 54. x 2 y + 3xy + 5y = 0, y(1) = 0, y (1) = 1, y(x) = x −1 (A cos(2 ln x) + B sin(2 ln x)) In Exercises 55–58, solve the initial value problem. 55. 56. 57. 58. 59.
dy/dx = 4x 3 − x + 2, y(0) = 1 dy/dt = sin 2t − cos y(0) = 0 2t, dy/dx = x −2 cos x −1 , y(2/π) = 1 dy/dx = (ln x)/x, y(1) = 0 The velocity of a falling object with mass m that is subjected to air resistance proportional the instantaneous velocity v of the object is found by solving the initial value
m dv/dt = mg − cv problem , where c > v(0) = v0 0 is the proportionality constant. (a) Given that a general solution to m dv/dt = mg − cv is v(t) = mg/c +
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1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
Ke−ct/m , find the solution of this initial value problem. (b) Determine limt→∞ v(t). 60. The number of cells in a bacteria colony after t hours is determined by solving the ini
dP /dt = kP . tial value problem P (0) = P0 (a) Given that a general solution of dP /dt = kP is P (t) = Cekt , use the initial condition to find C. (b) Find the value of k so that the population doubles in 8 hours. 61. In 1840, the Belgian mathematicianbiologist Pierre F. Verhulst (1804–1849) developed the logistic equation, dP /dt = rP − aP 2 , where r and a are positive constants, to predict the population P (t) in certain countries. (a) Given that a general solution to this r equation is P (t) = , find the a + Ce−rt solution that satisfies P (0) = P0 , where P0 > 0 is constant. (b) Determine limt→∞ P (t).
63.
64. 65.
66.
67. 68. 69.
Pierre Francois Verhulst (1804–1849)
62. The differential equation dS/dt + 3S/(t + 100) = 0, where S(t) is the number of pounds of salt in a particular tank at time t, is used to approximate the amount of salt in the tank containing a salt-water mixture in
70.
71.
which pure water is allowed to flow into the tank while the mixture is allowed to flow out of the tank. If S(t) = 15,000,000/(t + 100)3 , show that S satisfies dS/dt + 3S/(t + 100) = 0. What is the initial amount of salt in the tank? As t → ∞, what happens to the amount of salt in the tank? The displacement (measured from x = 0) of a mass attached to the end of a spring at time t is given by x(t) = 3 cos 4t + 94 sin 4t. Show that x satisfies the ordinary differential equation x + 16x = 0. What is the initial position of the mass? What is the initial velocity of the mass? Show that u(x, y) = ln x 2 + y 2 satisfies Laplace’s equation, uxx + uyy = 0. The temperature in a thin rod of length 2π after t minutes at a position x between 0 and 2π is given by u(x, t) = 3 − e−16kt cos 4x. Show that u satisfies ut = k uxx . What is the initial temperature (t = 0) at x = π ? What happens to the temperature at each point in the wire as t → ∞? The displacement u of a string of length 1 at time t position x, where x is measured from x = 0 is given by u(x, t) = sin πx cos t. Show that u satisfies π 2 utt = uxx . What is the value of u at the endpoints x = 0 and x = 1 for all values of t? Find the value(s) of m so that y = x m is a solution of x 2 y − 2xy + 2y = 0. Find the value(s) of k so that y = ekt is a solution of y − 3y − 18y = 0. d 2x dy Use the fact that e y = e2x + 2e2x y dx dx dy + 2e2x y = ex . and integration to solve e2x dx dy d x + ex y and Use the fact that (e y) = ex dx dx dy integration to solve ex + ex y = xex . dx The time-independent Schrödinger equation is given by −
h2 d 2 ψ(x) + U (x)ψ(x) = Eψ(x). 2m dx 2
13
EXERCISES 1.1
If U (x) = 0, find conditions on E so that ψ(x) = A sin(nπx/L) is a solution of the time-independent Schrödinger equation.
equation, and (b) graph the solution on the indicated interval(s). Throughout Introductory Differential Equations, we use graphs of solutions of differential equations. In some cases, we are able to predict what the graph of a solution should look like. If the graph of our proposed solution does not appear as predicted we know that either we made a mistake in constructing our proposed solution or our conjecture about the general shape of the graph of the solution is wrong. In other cases, we will find that it is easier to examine the graph of a solution than it is to examine the solution (if we are able to construct one in the first place).
The great physicist Erwin Schrödinger (1887–1961) received a Nobel prize for his work in 1933.
72. A singular solution of a differential equation is a solution that cannot be derived from the general solution of the differential equation. Use implicit differentiation to show that −1/x + 2/x 2 + 1/y − 1/y 2 = C is a general (implicit) solution of the differential equation dy/dx = (x − 4)y 3 /[x 3 (y − 2)]. Is y = 0 a solution of this differential equation? Is y = 0 a singular solution? 73. Show that x + x 2 /y = C is a general (implicit) solution of the differential equation dy/dx = (y 2 + 2xy)/x 2 . Is y = 0 a solution of this differential equation? Is y = 0 a singular solution? 74. The current I (t) in an L-R circuit, which contains a resistor, an inductor, and a voltage source, satisfies the differential equation RI + LdI /dt = E(t), where R and L are constants representing the resistance and the inductance and E(t) is the voltage source. Is this equation linear or nonlinear? Determine the order of the equation. In Exercises 75–77, (a) verify that the indicated function is a solution of the given differential
75. xy + y = cos x, y = (sin x)/x; [−2π, 0) ∪ (0, 2π] 76. 16y + 24y + 153y = 0, y = e−3t/4 cos 3t; [0, 3π/2] 77. x 3 y + x 2 y + xy − 40y = 0, y = x −1 sin(3 ln x); (0, π] 78. (a) Verify that ⎧ √ √ √ ⎨x = e−t 100 3 sin 3t + 20 cos 3t 3 √ √ √ ⎩y = e−t − 40 3 sin 3t + 20 cos 3t 3 is a solution
of the system of differential dx/dt = 4y equations . dy/dt = −x − 2y (b) Graph x(t),
y(t), and the parametric x = x(t) equations for 0 ≤ t ≤ 2π . y = y(t) 79. (a) Show that (x 2 + y 2 )2 = 5xy is an implicit solution of [4x(x 2 + y 2 ) − 5y] dx + [4y(x 2 + y 2 ) − 5x] dy = 0. (b) Graph (x 2 + y 2 )2 = 5xy on the rectangle [−2, 2] × [−2, 2]. (c) Approximate all points on the graph of (x 2 + y 2 )2 = 5xy with x-coordinate 1.
14
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
(d) Approximate all points on the graph of (x 2 + y 2 )2 = 5xy with y-coordinate −0.319. Often the calculus and algebra encountered in solving differential equations can be quite complicated. Computer algebra systems are capable of performing the integration and algebraic simplification encountered when solving many differential equations. Having access to these tools can be a great advantage: a large number of problems can be solved quickly, we make conjectures as to the general form of a solution to different forms of differential equations, and these tools allow us to check and verify our work.
80. Solve dy/dx = sin4 x, y(0) = 0 and graph the resulting solution on the interval [0, 4π ]. 81. A general solution of y (4) + 25 2 y − 5y + 629 −x/2 (c cos 3x + 1 16 y = 0 is given by y = e x/2 c2 sin 3x) + e (c3 cos 2x + c4 sin 2x), where c1 , c2 , c3 , and c4 are constants. Solve the initial value problem
629 y (4) + 25 2 y − 5y + 16 y = 0 y(0) = 0, y (0) = 1, y (0) = −1, y (0) = 1 and graph the resulting solution. 82. A general solution of the system
dx/dt = 4y dy/dt = −4x
x = −c1 cos 4t + c2 sin 4t , where c1 and y = c2 cos 4t + c1 sin 4t c2 are⎧constants. Solve the initial value prob⎪ ⎨dx/dt = 4y lem dy/dt = −4x and then graph ⎪ ⎩ x(0) = 4, y(0) = 0 x(t), y(t), and the parametric equations
x = x(t) . y = y(t) is
83. A general solution of the system
dx/dt = −5x + 4y dy/dt = 2x + 2y
x = −c1 e3t − 4c2 e−6t , where c1 and c2 y = 2c1 e3t + c2 e−6t are constants. Solve the initial value prob⎧ ⎪ x = −5x + 4y ⎨ lem y = 2x + 2y and then graph ⎪ ⎩ x(0) = 4, y(0) = 0 x(t), y(t), and the parametric equations
x = x(t) . y = y(t)
is
1.2 A GRAPHICAL APPROACH TO SOLUTIONS: SLOPE FIELDS AND DIRECTION FIELDS • Systems of Ordinary Differential Equations and Direction Fields • Relationship Between Systems of First Order and Higher Order Equations Suppose that we are asked to solve the ordi2 nary differential equation dy/dx = e−x . In this case, we do not attempt to solve this equation through integration as we did in the previous section because of the presence of the function 2 f (x) = e−x on the right side of the ordinary differential equation. (See Exercise 21 at the end of this section.) Instead, we can gain insight into the behavior of solutions of this ordinary differential equation through a graphical approach by considering the slope of the tangent line to solutions of the ordinary differential equation. Recall from Section 1.1 that the differential equation gives the slope of the tangent line to solutions of the ordinary differential equation at the given point (x, y) in the xy-plane. Therefore, if we wish to determine the slope of the tangent
15
1.2 A GRAPHICAL APPROACH TO SOLUTIONS
2
FIGURE 1.8 (A) Several line segments in the slope field for dy/dx = e−x . (B) One view of the slope field for the equation. (C) A different view of the slope field for the equation.
2
2
FIGURE 1.9 (A) Slope field for dy/dx = e−x . (B) Using the slope field to sketch the solution to dy/dx = e−x , y(−2) = −1. (C) A different view of using the slope field to sketch the solution to dy/dx = e
line to the solutions to the ordinary differential equation that passes through (0, 1) at this point, we substitute (0, 1) into the right side of 2 dy/dx = e−x . Because the right side only depends on x, we have slope dy 2 = e−(0) = 1. dx In fact, the slope of the line tangent to solutions at√all points of the form (0, y) is 1. At the point ( ln 2, 4), we find that the slope is √ dy 1 2 −1 = e−( ln 2) = e− ln 2 = eln 2 = . dx 2
Again, we obtain the same slope at√all points √ ( ln 2, y) ≈ (0.832555, y) and (− ln 2, y) ≈
−x 2
, y(−2) = −1.
(−0.832555, y). In Fig. 1.8, we draw several short line segments using points of the form (0, y) for y = 0, ±1, ±2, ±3. Notice that we use a triangle with base length 1 and height length 1 to assist in sketching the tangent lines with slope 1 at these points. We use a triangle with base length 2 and height length 1 to help us draw the lines of slope 1/2 √ that are tangent to solutions at the points (− ln 2, y) for y = 0, ±1, ±2, ±3. By drawing a set of short line segments representing the tangent lines to solutions of the ordinary differential equation at numerous points in the xy-plane, we construct the slope field of the ordinary differential equa2 tion. We show the slope field for dy/dx = e−x on the square [−2, 2] × [−2, 2] in Fig. 1.9A. (Note that because constructing a slope field is time
16
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.10 (A) Graphs of x(t) = sin t and y(t) = cos t for 0 ≤ t ≤ 2π . (B) Graph of the parametric equations x(t) = sin t and y(t) = cos t for 0 ≤ t ≤ 2π .
consuming, we usually let a computer algebra system do the work for us.) Observe that at each point along the y-axis, the slope is 1 as we predicted earlier. Notice also that the slope appears to be zero for larger values of |x|. This is because for large values of x (in absolute value), 2 dy/dx ≈ 0 because limx→±∞ e−x = 0. Therefore, we expect solutions to “flatten out” as x increases. We can use the slope field to investigate the solution to an initial value problem such as dy 2 = e−x , dx
y(−2) = −1
by starting at the point (−2, −1) and tracing the solution by following the tangent slopes. This solution is sketched in Fig. 1.9B. Thus, although we were not able to determine explicit formulas for either the general solution of the ODE or the solution to the IVP, we were able to determine some properties of the solutions by using the slope field of the differential equation.
Systems of Ordinary Differential Equations and Direction Fields We can also consider systems of differential equations. In Chapter 6, we will learn how to solve systems of first order ordinary differential equations of the form
dx/dt = ax + by dy/dt = cx + dy
,
where a, b, c, and d are given constants. In the case of this system, we solve for x = x(t) and y = y(t). For example, if we consider the system
dx/dt = y , dy/dt = −x we can verify that the parametric equations
x(t) = sin t y(t) = cos t satisfy the system because d dx = (sin t) = cos t = y and dt dt dy d = (cos t) = − sin t = −x. dt dt We can graph each function separately as we do in Fig. 1.10A. (The graph of x(t) is the solid curve; that of y(t) is dashed.) Another option is to graph them as a pair of parametric equations as in Fig. 1.10B. As we recall, the graph of this pair of parametric equations is a circle of radius 1 centered at the origin because x 2 = sin2 t and y 2 = cos2 t , so that x 2 + y 2 = sin2 t + cos2 t = 1. However, we must indicate the orientation of the curve (the direction of increasing parameter value t ). For this pair of parametric equations, we find that at t = 0, x(0) = sin 0 = 0 and y(0) = cos 0 = 1. Therefore, the point (0, 1) corresponds to t = 0. Similarly, the point (1, 0) corresponds to t = π/2 because x(π/2) = sin π/2 = 1 and y(π/2) = cos π/2 = 0. This means that
1.2 A GRAPHICAL APPROACH TO SOLUTIONS
the solution moves from (0, 1) to (1, 0) as t increases. To determine if the orientation is clockwise or counter-clockwise, we test a t -value between 0 and π/2. Choosing t = π/4, we find the √ 2 and y(π/4) = cos π/4 = x(π/4) = sin π/4 = 1/ √ 1/ 2 so the orientation is clockwise. The parametric equations {x(t) = sin t, y(t) = cos t} satisfy the initial value problem
dx/dt = y, x(0) = 0 dy/dt = −x, y(0) = 1 because they satisfy the system of differential equations as well as the two initial conditions. Notice that in the case of an initial value problem involving a system of differential equations, an initial condition is given for each of the variables x and y that depend on t. In the parametric plot, the solution to this initial value problem passes through the point (x(0), y(0)) = (0, 1). Another way to view a system of two ordinary differential equations is through the use of a direction field, which is similar to a slope field. For example, for the first order system
dx/dt = ax + by , dy/dt = cx + dy we first write it as a first order equation with dy/dt cx + dy dy = = . dx dx/dt ax + by Observe that the procedure described here can be used for any system of the form
dx/dt = f (x, y) . dy/dt = g(x, y)
Then, we can consider the slope field associated with this differential equation. For example, if we refer back to the system
dx/dt = y , dy/dt = −x
17
we obtain the first order equation dy/dx = −x/y, so we can determine the slope of tangent lines to solutions at points √ xy-plane. For √ in the example, at the point (1/ 2, 1/ 2), the solution to this system √ that passes through this point has −1/ 2 slope √ = −1. In a similar manner, we can 1/ 2 find the slope at other points in the plane. However, as we mentioned in our earlier discussion, we must indicate the orientation when we graph parametric equations, so we consider the vector dx/dt, dy/dt = dx/dt i + dy/dt j with components from the system of differential equations. In the case of this system, we consider √ dx/dt, dy/dt = y, −x . At the √ point √ (1/ 2, √ 1/ 2), we obtain the vector 1/ 2, −1/√ 2 . This √ 2, 1/ 2) means that the solution√ through (1/ √ has tangent vector 1/ 2, −1/ 2 . The direction field up of tangent vectors such √ √ is made as 1/ 2, −1/ 2 to solutions at points in the plane, so it is similar to the slope field for dy/dy = −x/y shown in Fig. 1.11A, except that vectors are used to indicate the orientation of solutions. In Fig. 1.11B, we show the direction field for this system. The vectors in the direction field indicate that solutions to this system are circles in the xy-plane that are directed clockwise. We graph several solutions along with the direction field in Fig. 1.11C. A collection of solutions in the xy-plane is called the phase portrait of the system. Notice that at points in the first quadrant, where x > 0 and y > 0, dx/dt = y > 0 and dy/dt = −x < 0. This means that x(t) increases and y(t) decreases along solutions in the first quadrant. In the second quadrant, where x < 0 and y > 0, dx/dt = y > 0 and dy/dt = −x > 0. Therefore, x(t) and y(t) increase along solutions in the second quadrant. We can perform a similar analysis for points in the other two quadrants. Generally, we will use standard mathematical notation throughout Introductory Differential Equations so i = 1, 0 and j = 0, 1 .
18
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.11 (A) Slope field for dy/dx = −x/y. (B) Direction field for dx/dt = y, dy/dt = −x. (C) Direction field for dx/dt = y, dy/dt = −x and several solution curves.
Relationship Between Systems of First Order and Higher Order Equations
dx/dt = y . If dy/dt = −x we differentiate the equation dx/dt = y with respect to t , we obtain d 2 x/dt 2 = dy/dt. Therefore, if we equate dy/dt = −x and dy/dt = d 2 x/dt 2 , we have d 2 x/dt 2 = −x or d 2 x/dt 2 + x = 0. We say that the system of two first order ODEs is equivalent to the second order ODE d 2 x/dt 2 + x = 0. Often, however, we begin with a second order ODE of the form d 2 x/dt 2 + b dx/dt + c x = f (t) and would like to write the equation as a system of first order ODEs. We do this by letting dx/dt = y and by differentiating this equation with respect to t to obtain d 2 x/dt 2 = dy/dt. Solving the second order ODE for d 2 x/dt 2 , we find d 2 x/dt 2 = −b dx/dt − c x + f (t) so that dy/dt = −b dx/dt − c x + f (t). Replacing dx/dt in this equation with y, we obtain the following equivalent system of first order ODEs:
dx/dt = y . dy/dt = −b y − c x + f (t)
gree greater than two into a system of first order ODEs.
Again, consider the system
By writing the second order ODE as a system of first order ODEs, we investigate the behavior of the solution of the second order ODE by observing the behavior in the direction field and phase portrait of the corresponding system. A similar procedure is used to transform an ODE of de-
EXERCISES 1.2 In Exercises 1–4, use the slope field to determine if the indicated path is that of a solution to the differential equation.
1.
dy = −y/x dx
2.
dy = x2 + y2 dx
19
EXERCISES 1.2
dy 3. = x/y dx
7.
dy =x −y dx
dy = x2 − y 4. dx
8.
dy = y2 − x2 dx
In Exercises 5–8, use the slope field to sketch the solutions of the differential equation that pass through the given points.
5.
dy = x/y dx
6.
dy = x2 − y dx
9. Graph the slope field for dy/dt = sin y. Determine limt→∞ y(t) if y(0) = y0 where (a) y0 = −3π/2; (b) y0 = −π/2; (c) y0 = π/2; (d) y(0) = 3π/2. 10. Graph the slope field for dy/dx = sin x. Does limx→∞ y(x) exist for any initial condition y(0) = y0 ? Solve the ODE and find limx→∞ y(x). Does this match the graphical result? 11. Graph the slope field for dy/dx = e−x . Does limx→∞ y(x) exist for any initial condition y(0) = y0 ? Solve the ODE and find limx→∞ y(x). Does this match the graphical result? 12. Graph the slope field for dy/dx = 1/(x 2 + 1). Does limx→∞ y(x) exist for any initial condition y(0) = y0 ? Solve the ODE and find limx→∞ y(x). Does this match the graphical result? In Exercises 13–14, use the direction field of the given system to sketch the graph of the solution that satisfies the indicated initial conditions. Determine limt→∞ x(t) and limt→∞ y(t) in each case (if they exist).
20
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.12 Figure for Exercise 23.
dx/dt = y ; (a) x(0) = 0, y(0) = dy/dt = x 2; (b) x(0) = −2, y(0) = 0; (c) x(0) = −2,
13. System:
y(0) = 2;
14. System: y(0) = 5;
dx/dt = 3x − 2y dy/dt = 4x − y
;
x(0) = 0,
d 2 x/dt 2 + 4x = 0 d 2 x/dt 2 − 5 dx/dt = 0 d 2 x/dt 2 + 4 dx/dt + 13x = 0 x − 6x + 7x = 0 ( = d/dt; x = x(t)) x + 16x = sin t x + 4x + 13x = e−t (a) Use a computer algebra system to solve 2 dy/dx = e−x . (b) Graph the solution to 2 the IVP dy/dx = e−x , y(0) = a for a = −2, −1, 0, 1 2. (c) Graph the slope field of the differential equation together with the solutions in (b). (d) Do the solutions appear to match the results described at the beginning of the section? 22. (a) Use a computer algebra system to graph the slope field for dy/dx = sin(2x − y). (b) Use the slope fie to graph the solution y(x) that satisfies the initial condition y(0) = 5. Does limx→∞ y(x) appear to exist? 23. Consider the systems
15. 16. 17. 18. 19. 20. 21.
In Exercises 15–20, write the second order equation as a system of first order equations.
dx/dt = −y dy/dt = x
and
dx/dt = −y dy/dt = −x
.
(a) For any given initial conditions, in a brief paragraph explain why you think that the solutions of the systems are similar or different.
EXERCISES 1.2
21
FIGURE 1.13 Figure for Exercise 24.
(b) Fig. 1.12 shows the direction field associated with each system. Use the direction field to help you graph the solutions that satisfy these initial conditions (i) x(0) = 0.5, y(0) = 0; (ii) x(0) = −0.25, y(0) = 0; (iii) x(0) = 0, y(0) = 0.75; and (iv) x(0) = 0, y(0) = −0.5. (c) How do your graphs affect your conjecture in (a)? 24. Consider the systems
dx/dt = x/2 dy/dt = y
and
dx/dt = −x/2 dy/dt = −y
.
(a) For any given initial conditions, in a brief paragraph explain why you think that the solutions of the systems are similar or different. (b) Fig. 1.13 shows the direction field associated with each system. Use the direction field to help you graph the solutions that satisfy these initial condition (i) x(0) = 0.5, y(0) = 0.25; (ii) x(0) = −0.25, y(0) = −0.5; (iii) x(0) = −0.5, y(0) = 0.75; and (iv) x(0) = 0.75, y(0) = −0.5. (c) How do your graphs affect your conjecture in (a)?
25. (Competing Species) The system of equa
dx/dt = x(a − b1 x − b2 y) where a, tions dy/dt = y(c − d1 x − d2 y) b1 , b2 , c, d1 , and d2 represent positive constants can be used to model the size of the population of two species, represented by x(t) and y(t), competing for a common food supply. (a) Fig. 1.14A shows the direction field for the system if a = 1, b1 = 2, b2 = 1, c = 1, d1 = 0.75, and d2 = 2. (i) Use the direction field to graph various solutions if both x(0) and y(0) are positive. (ii) Use the direction field and your graphs to approximate limt→∞ x(t) and limt→∞ y(t). (b) Fig. 1.14B shows the direction field for the system if a = 1, b1 = 1, b2 = 1, c = 0.67, d1 = 0.75, and d2 = 1. (i) Use the direction field to graph various solutions if both x(0) and y(0) are positive. (ii) Use the direction field and your graphs to determine the fate of the species with population y(t). What happens to the species with population x(t)?
22
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.14 (A) Figure for Exercise 25 (a). (B) Figure for Exercise 25 (b).
CHAPTER 1 SUMMARY: ESSENTIAL CONCEPTS AND FORMULAS Differential Equation (DE) An equation that contains the derivative or differentials of one or more dependent variables with respect to one or more dependent variables. Ordinary Differential Equation (ODE) If a differential equation contains only ordinary derivatives (of one or more dependent variables) with respect to a single independent variable, the equation is called an ordinary differential equation. Partial Differential Equation (PDE) A differential equation that contains the partial derivatives or differentials of one or more independent variables with respect to more than one independent variable is called a partial differential equation. Linear Ordinary Differential Equation A linear ODE is an equation that can be written in the form an (x)y (n) + an−1 (x)y (n−1) + · · · + a2 (x)y + a1 (x)y + a0 y = f (x), where y = y(x). Order of an Equation The order of the highest order derivative in a differential equation is called the order of the equation.
Solution A solution of a differential equation on a given interval is a function that is continuous on the interval and has all the necessary derivatives that are present in the differential equation such that when substituted into the equation yields an identity for all values on the interval. Explicit Solution A solution given as a function of the independent variable. Implicit Solution A solution given as a relation such as f (x, y) = 0, f (t, x) = 0, or f (t, y) = 0. Trivial Solution y = 0 is always a solution of the nth order linear homogeneous equation. Slope Field A collection of line segments that indicate the slope of the tangent line to the solution(s) of a differential equation. Phase Portrait A collection
of graphs of soludx/dt = f (x, y) tions to the system in dy/dt = g(x, y) the xy-plane. Direction Field A collection of vectors that indicate the slope and direction of the tangent line to the solutions of a system of differential equations.
CHAPTER 1 REVIEW EXERCISES
23
CHAPTER 1 REVIEW EXERCISES
lution to determine the solution(s) to the given problem.
In Exercises 1–5, determine (a) if the equation is an ordinary differential or partial differential equation; (b) the order of the differential equation; and (c) if the equation is linear or nonlinear.
20. dy/dx + 2y = x 2 , y(0) = 1, y = 14 − 12 x + 1 2 −2x 2 x + Ae 21. y + 4y = t , y(0) = 1, y(π/4) = π/16, y = t/4 + A cos 2t + B sin 2t 22. x 2 y + 5xy + 4y = 0, y(1) = 1, y (1) = 0, y = Ax −2 + Bx −2 ln x
dy/dt = y a ux + ut = 0, u = u(x, t), a constant d 2 y/dx 2 + 2 dy/dx + y = 0 m x + kx = sin t, m and k positive constants; x = x(t) ∂φ ∂ 2 φ ∂ 2 φ = 2 5. ∂x ∂x 2 ∂y
1. 2. 3. 4.
In Exercises 6–13, verify that the given function is a solution of the corresponding differential equation. (A and B denote constants.) dy/dx + y cos x = 0, y = e− sin x dy/dx − y = sin x, y = (ex − cos x − sin x)/2 y + 4y − 5y = 0, y = e−5x , y = ex y − 6y + 45y = 0, y = e3x (cos 6x − sin 6x) xy − xy − 16y = 0, y = Ax 5 + Bx −3 x 2 y + 3xy + 2y = 0, y = x −1 (cos(ln x) − sin(ln x)) 12. d 2 y/dx 2 + 2 dy/dx + 2y = x, y = (x − 1)/2 13. y − 7y + 12y = 2, y = Ae3x + Be4x + 1/6 6. 7. 8. 9. 10. 11.
In Exercises 14 and 15, verify that the given implicit function satisfies the differential equation. 14. (2x − 3y) dx + (2y − 3x)dy = 0, x 2 − 3xy + y2 = 1 15. (y cos(xy) + sin x) dx + x cos(xy) dy = 0, sin(xy) − cos x = 0 In Exercises 16–19, find a solution of the differential equation. 16. dy/dx = xe−x 17. dy/dx = x 2 sin x 2x 2 − x + 1 dy = 18. 2 dx (x − 1)(x √ + 1) 2 19. dy/dx = x / x 2 − 1 2
Exercises 20–22, use the indicated initial or boundary conditions with the given general so-
Exercises 23 and 24 solve the initial-value problem. Graph the solution on an appropriate interval. 23. dy/dx = cos2 x sin x, y(0) = 0 24. dy/dx = 13 (4x − 9)(x − 3)−2/3 , y(0) = 0 25. The temperature on the surface of a steel ball at time t is given by u(t) = 70e−kt + 30 (in o F) where k is a positive constant. Show that u satisfies the first order equation du/dt = −k(u − 30). What is the initial temperature (t = 0) on the surface of the ball? What happens to the temperature as t → ∞? 26. The displacement (measured from x = 0) of a mass attached to the end of a spring at √ 1 −t time t is given by x(t) = 4 e (cos 35t + √ √9 sin 35t). Show that x satisfies the ordi35 nary differential equation x +2x +36x = 0. What is the initial displacement of the mass? What is the initial velocity of the mass? 27. For a particular wire of length 1 foot, the temperature at time t hours at a position of x feet from the end (x = 0) of the wire is es2 2 timated by u(x, t) = e−π kt sin πx − e−4π kt × sin 2πx. Show that u satisfies the equation ut = k uxx . What is the initial temperature (t = 0) at x = 1? What happens to the temperature at each point in the wire as t → ∞? 28. The height u of a long string at time t and position x where x is measured from the middle of the string (x = 0) is given by u(x, t) = sin x cos 2t . Show that u satisfies the wave equation utt = 4uxx . What is the initial height (t = 0) at x = 0?
24
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.15 (A) Figure for Exercise 30. (B) Figure for Exercise 31.
29. Show that u(x, y) = tan−1 (y/x) satisfies Laplace’s equation uxx + uyy = 0. 30. The slope field of y = 2(y − y 2 ) is shown in Fig. 1.15A. Sketch the graph of the solution that satisfies the indicated initial condition. Also, determine if limx→∞ y(x) exists. (a) y(0) = 0.5; (b) y(0) = 1.5; (c) y(0) = −0.5. 31. The direction field for {dx/dt = x, dy/dt = 2y} is shown in Fig. 1.15B. Is it pos-
sible to select initial conditions so that limt→∞ x(t) = 0 and limt→∞ y(t) = 0? (Briefly explain.) (−1)k × 32. The function J1 (x) = ∞ k=0 k!(k + 1)!22k+1 x 2k+1 is called the Bessel function of order 1. Verify that J1 (x) is a solution of Bessel’s equation of order 1, x 2 y + xy + 2 x − 1 y = 0.
1 Introduction to Differential Equations O U T L I N E 1.1 Introduction to Differential Equations: Vocabulary Exercises 1.1 1.2 A Graphical Approach to Solutions: Slope Fields and Direction Fields
Exercises 1.2
18
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Chapter 1 Summary: Essential Concepts and Formulas
22
14
Chapter 1 Review Exercises
23
Introductory Differential Equations introduces and discusses the topics covered in a typical first course in ordinary differential equations. Throughout, we indicate how technologies such as computer algebra systems can be used to enhance the study of differential equations, not only by eliminating some of the computational difficulties that arise in the study of differential equations but also by overcoming some of the visual limitations associated with the solutions of differential equations. The advantages of using technologies like computer algebra systems in the study of differential equations are numerous, but perhaps the most useful is that of being able to produce the graphics associated with solutions of differential equations. This is particularly beneficial in the discussion of applications because many physical situations are modeled with differential equations. For example, in Chapter 5, we see that the motion of a penduIntroductory Differential Equations. DOI: https://doi.org/10.1016/B978-0-12-814948-5.00001-X
lum can be modeled by a differential equation. When we solve the problem of the motion of a pendulum, we use technology to watch the pendulum move. The same is true for studying the differential equations that describe the motion of a mass attached to the end of a spring, as well as many other problems. In having these technologies, the study of differential equations becomes more visual and, consequently, more interesting. Although Chapter 1 is short in length, the vocabulary introduced here is used throughout the text and that subsequent chapters will take advantage of the terminology and techniques discussed here. Nearly all introductions to differential equations begin with German scientist Gottfried Wilhelm Leibniz (1646–1716) and British scientist Isaac Newton (1642–1727), the inventors of calculus. In integral calculus we learn that the area under the graph of a smooth positive function is
1
Copyright © 2018 Elsevier Inc. All rights reserved.
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1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
given by a definite integral, but both Leibniz and Newton were more concerned with solving differential equations than finding areas. Many of the methods of solution we discuss in this text are contributed to the great Swiss mathematician Leonhard Euler (1707–1783).
Gottfried Wilhelm Leibniz (1646–1716)
world and indicate how some people use differential equations beyond this course. However, mathematics is also interesting in its own right so mathematical applications are also included throughout the text.
Isaac Newton: We hope this young Isaac Newton was not worried about who discovered calculus first.
Isaac Newton (1642–1727): The controversy as to who (Leibniz or Newton) discovered calculus first became an obsession with Newton during the second half of his life.
Subsequently, many problems, such as determining the motion of a plucked string, not only lead to ordinary and partial differential equations but to other areas of mathematics as well. Differential equations are full of rich and exciting mathematical and applied applications. Consequently, interesting applications are included throughout the text to motivate discussions, make the study of differential equations more interesting and pertinent to the real
Leonhard Euler (1707–1783)
1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS: VOCABULARY We begin our study of differential equations by explaining what a differential equation is. From our experience in calculus, we are famil-
3
1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS: VOCABULARY
iar with some differential equations. For example, suppose that the acceleration due to gravity, a(t), (measured in ft/s2 ) of a falling object is a(t) = −32. Then, because a(t) = v (t), where v(t) is the velocity of the object (measured in ft/s), we have v (t) = −32 or dv = −32. dt An equation like this involving a function of a single variable is called an ordinary differential equation (ODE). (If the equation involves partial derivatives, then it is called a partial differential equation (PDE).) In this case, the function to be determined is v = v(t), which depends on the variable t, representing time (measured in seconds). The goal in solving an ODE is to find a function that satisfies the equation. We can solve this ODE through integration: Throughout the text, we will use ODE to refer to an ordinary differential equation and PDE to refer to a partial differential equation.
v(t) =
a(t) dt =
(−32) dt = −32t + C,
where C is an arbitrary constant. This result indicates that v(t) = −32t + C is a solution of the ODE for any choice of the constant C. (We call this a general solution because it involves an arbitrary constant.) In fact, we have found every solution of the ODE because each is expressed as −32t + C. Examples of solutions and the corresponding C values include v(t) = −32t (C = 0), v(t) = −32t + 32 (C = 32), and v(t) = −32t − 8 (C = −8). This shows that there are an infinite number of solutions to the ODE. We can verify that v(t) = −32t + C is a general solution of dv/dt = −32 through substitution: dv d = (−32t + C) = −32. dt dt When we substitute our solution into the left side of the ODE, we obtain −32, the value on
the right side of the ODE. We have verified that v(t) = −32t + C satisfies the ODE for any choice of C. Many times we are given a particular condition that the solution must satisfy. For example, suppose that the object considered earlier has an initial velocity of −64 ft/s. In other words, the velocity at time t = 0 s is represented with the initial condition v(0) = −64. Therefore, we need to find a solution of the ODE that also satisfies the initial condition. We express this initial value problem (IVP) as Throughout the text we will refer to initial value problems as IVPs.
dv = −32, dt
v(0) = −64,
where we solve the initial value problem by first finding a general solution to the ODE and then by applying the initial condition to determine the arbitrary constant. When we substitute t = 0 into the general solution v(t) = −32t + C, we obtain v(0) = −32 · 0 + C = C. We conclude that C = −64 so that v(0) = −64 is satisfied. This means that the solution to the IVP is v(t) = −32t − 64. Notice that unlike the ODE, the IVP has only one solution. Another application of differential equations found in calculus is finding a function when given (a) the slope of the line tangent to the graph of the function at any point (x, y), and (b) a point on the graph. For example, suppose that the slope of the tangent line at any point on the graph of a function y = y(x) is given by dy = 3x 2 − 4x, dx and further that the graph passes through the point (1, 4), which means that y(1) = 4. In this case, the ordinary differential equation is given by dy/dx = 3x 2 − 4x, where we need to find the function y = y(x) that satisfies the initial condi-
4
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
tion y(1) = 4. Therefore, we solve the IVP dy = 3x 2 − 4x, dx
y(1) = 4.
As in the previous example, we find a general solution to dy/dx = 3x 2 − 4x through integration. This yields y = y(x) = x 3 − 2x 2 + C. Now, when we apply the initial condition, we find that y(1) = 13 − 2 · 13 + C = −1 + C = 4 so that C = 5, which means that the solution to the IVP is y = y(x) = x 3 − 2x 2 + 5. Fig. 1.1 shows the graph of the solution together with a portion of the tangent line at the point (1, 4). Notice that the slope of the tangent line at the point (1, 4) is the value of dy/dx evaluated if x = 1: if x = 1, dy/dx = 3 · 12 − 4 · 1 = −1. This observation will be useful in Section 1.2 in helping us better understand the behavior of solutions of differential equations. The previous examples are similar in that they each involve an ordinary differential equation in which the highest order derivative is the first derivative. We call equations of this type first order ordinary differential equations because the order of a differential equation is the order of the highest order derivative appearing in the equation. Example 1.1. Determine the order of each of the following differential equations: (a) dy/dx = x 2 /y 2 cos y, (b) uxx + uyy = 0, (c) (dy/dx)4 = y + x, (d) d 2 x/dt 2 + 2dx/dt + 3x = sin t . Solution: (a) This equation is first order because it includes only one first order derivative, dy/dx. (b) This equation is classified as second order because the highest order derivatives, uxx , representing ∂ 2 u/∂x 2 , and uyy representing ∂ 2 u/∂y 2 , are of order two.
FIGURE 1.1 Solution to dy/dx = 3x 2 − 4x, y(1) = 4 along with a segment of the tangent line at (1, 4).
The equation uxx + uyy = 0 arises in many areas of study, which include fluid flows as well as electrostatic and gravitational potential, is often called Laplace’s equation after Pierre-Simon Laplace (also known as the Marquis de Laplace) (1749–1827) or the potential equation. Consequently, Laplace’s equation is a second order partial differential equation. (e) This is a first order equation because the highest order derivative is the first derivative. Raising that derivative to the fourth power does not affect the order of the equation. The expressions
dy dx
4 and
d 4y dx 4
do not represent the same quantities: (dy/dx)4 represents the derivative of y with respect to x, dy/dx, raised to the fourth power; d 4 y/dx 4 represents the fourth derivative of y with respect to x. (d) The highest order derivative is d 2 x/dt 2 , so the equation is second order.
1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS: VOCABULARY
Pierre-Simon Laplace (1749, Normandy, France1827, Paris, France): Laplace’s name is probably most remembered in reference to the Laplace transform that we will study in Chapter 8.
Example 1.2. (a) Show that y = c1 sin t + c2 cos t satisfies the second order ordinary differential equation (ODE) y + y = 0 where c1 and c2 are arbitrary constants. (b) Find the solution to the IVP y + y = 0, y(0) = 0, y (0) = 1. Solution: (a) Differentiating, we obtain y = dy/dt = c1 cos t − c2 sin t and y = d 2 y/dt 2 = −c2 cos t − c1 sin t . Therefore, substituting y into y + y (the left side of the differential equation) results in d 2y + y = −c1 sin t − c2 cos t + c1 sin t + c2 cos t dt 2 = 0, so the function satisfies the (ordinary differential equation) ODE for any choice of c1 and c2 . We graph the solution for several values of these constants in Fig. 1.2. Because the solution depends on at least one constant, we say that the functions shown in Fig. 1.3 are members of the family of solutions of the ODE. (b) Evaluating the function at t = 0 yields y(0) = c1 sin 0 + c2 cos 0 = c2 . Then, the initial condition, y(0) = 0, indicates that c2 = 0. Similarly, y (0) = c1 cos 0 − c2 sin 0 = c1 , so c1 = 1 so that the second initial condition, y (0) = 1, is satisfied.
5
FIGURE 1.2 Graphs of y = c1 cos t + c2 sin t for various values of c1 and c2 .
Therefore, y(t) = sin t is the solution to the initial value problem (IVP). We graph this solution in Fig. 1.3. Notice that the ODE has an infinite number of solutions while the IVP has only one unique solution. Also observe that the number of initial conditions for the initial value problem (IVP) matches the order of the ordinary differential equation (ODE). The next level of classification tells us whether an equation is linear or nonlinear in terms of the dependent variable. For an ordinary differential equation (ODE), assuming that the independent variable is x and the dependent variable is y, an ordinary differential equation (of order n) is called linear if it can be written as d ny d n−1 y d 2y + a (x) + · · · + a (x) n−1 2 dx n dx n−1 dx 2 dy + a1 (x) (1.1) + a0 (x)y = f (x), dx
an (x)
where the functions ak (x), k = 0, 1, . . . , n, and f (x) are given and an (x) is not the zero function. If f (x) is identically the zero function, the linear equation, (1.1), is said to be homogeneous. You should verify that y(x) = 0, which we call the trivial solution, is a solution to every linear homogeneous equation. If f (x) is not identically the zero function, the linear equation, (1.1), is said to be nonhomogeneous.
6
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.3 Graph of y = y(t) = sin t .
In Chapter 4, we will learn that a general solution of the nth order linear equation, (1.1), is a solution that depends on n arbitrary constants and includes all solutions of the equation. If the equation under consideration cannot be written in the form given by Eq. (1.1), the equation is said to be nonlinear. Therefore, some of the properties that lead to classifying an equation as linear or nonlinear are powers of the dependent variable (or one of its derivatives) and functions of the dependent variable. Example 1.3. Determine which of the followdy ing differential equations are linear: (a) = x3, dx d 2u + u = ex , (c) (y − 1) dx + x cos y dy = 0, (b) dx 2 d 3y dy d 2x dy (d) +y = x, (e) + x 2 y = x, (f) 2 + 3 dx dx dx dt sin x = 0. Solution: (a) This equation is linear because the nonlinear term x 3 is the function f (x) of the independent variable in the general formula for a linear differential equation. (b) This equation is also linear. Using u as the name of the dependent variable does not affect the linearity. (c) If y is the dependent variable, solving for dy/dx gives us dy 1−y = . dx x cos y Because the right side of this equation includes a nonlinear function of y, the equation is nonlinear (in y). However, if x is the dependent variable, solving for dx/dy yields dx cos y = x. dy 1−y
This equation is linear in the dependent variable x. (d) The coefficient of the term dy/dx is y instead of an expression involving only the independent variable x. Hence, this equation is nonlinear in the dependent variable y. (e) This equation is linear. The term x 2 is the coefficient function. (f) For this equation, note that x is the dependent variable; t is the independent variable. This equation, known as the pendulum equation because it models the motion of a simple pendulum, is nonlinear because it involves a nonlinear function of the dependent variable x, sin x. If an ordinary differential equation (ODE) has the form dy/dx = f (x), we can use integration to determine y = y(x), although the result may be in terms of integrals that cannot be evaluated using standard techniques of integration. The following examples of differential equations of this type illustrate some of the typical methods of integration that may be encountered. Example 1.4. Solve the following differential dy dy x equations. (a) , = cos x, (b) = √ 2 dx dx x +1 dy dy dy 1 1 (c) , (d) , = 2 = xex , (e) = dx dx dx x + 16 4 − x2 dy and (f) = sin x tan x. dx Solution: In each case, we integrate the indicated function and graph the solution for several values of the constant of integration. Each solution contains a constant of integration so there are infinitely many solutions to each equation. (a) y = cos x dx = sin x + C (see Fig. 1.4A). (b) To eval x uate y = √ dx, we let u = x 2 + 1 so that x2 + 1
1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS: VOCABULARY
7
FIGURE 1.4 (A) Graph of y = sin x + C for various values of C. (B) Graph of y = x 2 + 1 + C for various values of C.
1 du = x dx. Then, 2 2x 1 1 1 y= dx = √ √ du 2 2 u x2 + 1 1 u−1/2 du = u1/2 + C = 2 (See Fig. 1.4B.) = x 2 + 1 + C.
du = 2x dx or
1 dx we use a 2 x + 16 trigonometric substitution. Letting x = 4 tan θ , −π/2 < θ < π/2, so that dx = 4 sec2 θ dθ gives us 1 4 sec2 θ dθ y= 16 + (4 tan θ )2 1 4 sec2 θ dθ = 16 + 16 tan2 θ 1 1 4 sec2 θ dθ = 16 sec2 θ x 1 1 1 = dθ = θ + C = tan−1 + C. 4 4 4 4 (c) To integrate y =
First, we use the identity 1 + tan2 θ = sec2 θ and then resubstitute: x/4 = tan θ so θ = tan−1 (x/4).
(d) To evaluate y = xex by hand, we use the Integration by Parts formula with u = x and dv =
ex dx. Then, du = dx and v = ex . This gives x x y = xe dx = xe − ex dx = xex − ex + C. (e) Use partial fractions to evaluate this integral. First, find the partial fraction decomposition of 1/(4 − x 2 ), which is determined by finding constants A and B that satisfy the equation 1 A B = + . (2 − x)(2 + x) 2 − x 2 + x These values are A = B = 1/4. (Why?) Thus, 1 1 1 + dx y= 4 2−x 2+x 1 = [− ln |2 − x| + ln |2 + x|] + C 4 1 2 + x = ln + C. (See Fig. 1.5.) 4 2−x The Integration by Parts formula states that u dv = u v − v du.
Find C if y(1) = 2.
Notice that the solutions of dy/dx = 1/(4−x 2 ) are undefined if x = −2 or x = 2. This is because
8
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
we can find only an equation involving the independent and dependent variables that the solution satisfies. In this case, we say that we have found an implicit solution. We will see that given an arbitrary differential equation, constructing an explicit or implicit solution is nearly always impossible. Consequently, although mathematicians were first concerned with finding analytic (explicit or implicit) solutions to differential equations, they have since (frequently) turned their attention to addressing properties of the solution and finding algorithms to approximate solutions. FIGURE 1.5 Graph of y =
1 2 + x ln +C for various val4 2−x
ues of C.
1/(4 − x 2 ) is undefined at these two values of x. Later, we will discuss in greater detail the relationship between the differential equation and its solutions. sin2 x (f) In this case, y = sin x tan x dx = dx becos x cause tan x = sin x/ cos x. Now use the identity sin2 x + cos2 x = 1 or sin2 x = 1 − cos2 x and divide: sin2 x y= dx cos x 1 − cos2 x y= dx cos x y = (sec x − cos x) dx y = ln | sec x + tan x| − sin x + C. For −π/2 < x < π/2, + C.
sec x dx = ln(sec x + tan x)
In Example 1.4, each solution is given as a function y = y(x) of the independent variable. In these cases, the solution is said to be explicit. In solving some differential equations, however,
Example 1.5. Verify that the equation 2x 2 + y 2 − 2xy + 5x = 0 satisfies the differential equation dy 2y − 4x − 5 = . dx 2y − 2x Solution: We use implicit differentiation to compute y = dy/dx if 2x 2 + y 2 − 2xy + 5x = 0: 4x + 2y
dy dy − 2x − 2y + 5 = 0 dx dx dy (2y − 2x) = 2y − 4x − 5 dx dy 2y − 4x − 5 = . dx 2y − 2x
The equation 2x 2 + y 2 − 2xy + 5x = 0 satisfies the 2y − 4x − 5 dy = . differential equation dx 2y − 2x Although we cannot solve 2x 2 + y 2 − 2xy + 5x = 0 for y as a function of x (see Fig. 1.6), we can determine the corresponding y value(s) for a given value of x. For example, if x = −1, then 2 + y 2 + 2y − 5 = y 2 + 2y − 3 = (y + 3)(y − 1) = 0. Therefore, the points (−1, −3) and (−1, 1) lie on the graph of 2x 2 + y 2 − 2xy + 5x = 0. (See Fig. 1.7.)
9
1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS: VOCABULARY
Example 1.6 (Duffing’s Equation). Duffing’s equation is the second order nonlinear equation d 2x dx +k − x + x 3 = cos ωt, dt dt 2
(1.2)
where k, , and ω are positive constants. Sources: See texts like Jordan and Smith’s Nonlinear Ordinary Differential Equations, [16].
Write Duffing’s equation as a system of first order equations
FIGURE 1.6 Graph of 2x 2 + y 2 − 2xy + 5x = 0.
Solution: Let y = x . Then, y = x and substituting into Duffing’s equation gives us x + kx − x + x 3 = cos ωt
Find an equation of the line tangent to the graph of 2x 2 + y 2 − 2xy + 5x = 0 at the points (−1, −3) and (−1, 1).
In the same manner that we consider systems of equations in algebra, we can also consider systems of differential equations. For example, if x and y represent functions of t , we will learn in Chapter 6 to solve the system of linear equations
dx/dt = ax + by , dy/dt = cx + dy where a, b, c, and d represent constants and differentiation is with respect to t . We will see that systems of differential equations arise naturally in many physical situations that are modeled with more than one equation and involve more than one dependent variable. In addition, we will see that it is often useful to write a differential equation of order greater than one as a system of first order equations, especially when the original equation is nonlinear.
y + ky − x + x 3 = cos ωt y = x − x 3 − ky + cos ωt. Thus, Duffing’s equation is equivalent to the nonlinear system
x = y y = x − x 3 − ky + cos ωt Note that a system of differential equations can consist of more than two equations. For example, the basic equations that describe the competition between two organisms, with population densities x1 and x2 , respectively, in a chemostat are Sources: See Smith and Waltman’s The Theory of the Chemostat, [27], for a detailed discussion of chemostat models.
⎧ m1 S m2 S ⎪ x1 − x2 S = 1 − S − ⎪ ⎪ ⎪ a + S a 1 2+S ⎪ ⎨ m1 S x1 = x1 −1 a ⎪ 1+S ⎪ ⎪ ⎪ m2 S ⎪ ⎩x2 = x2 −1 a2 + S
,
(1.3)
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1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.7 Notice that near the points (−1, −3) and (−1, 1) the implicit solution looks like a function. In fact, when we zoom in near the points (−1, −3) and (−1, 1), we see what appears to be the graph of a function.
where denotes differentiation with respect to t ; S = S(t), x1 = x1 (t), and x2 = x2 (t). For Eqs. (1.3), we remark that S denotes the concentration of the nutrient available to the competitors with population densities x1 and x2 . We investigate chemostat models in more detail in Chapter 7.
7. 8. 9. 10.
EXERCISES 1.1 Determine if each of the following equations is an ordinary differential equation or a partial differential equation. If the equation is an ordinary differential equation, then determine (a) the order of the ordinary differential equation and (b) if the equation is linear or nonlinear. 1. 2. 3. 4. 5.
d 2y
dy + − 2y = x 3 dx 2 dx dy y + y 4 = sin x dx 2 ∂ 2y 2 ∂ y , c > 0 constant = c ∂t 2 ∂x 2 y − 2y + 5y + y = ex ( = d/dx; y = y(x)) 2 dy +y =0 dx
d 2y dy +t + 2y = 0 2 dt dt 2 2 1 ∂ z ∂ z ∂ 2z = + , (z = z(t, x, y)) c2 ∂t 2 ∂x 2 ∂y 2 u ux + ut = 0, (u = u(t, x)) 2 2 d y + 2y = 2x x dx 2 d 2x + 2 sin x = sin 2t (x = x(t)) dt 2 ut + u ux = σ uxx , σ constant
6. t 2
11.
12. (2x − 1) dx − dy = 0 13. (2t − y) dt − dy = 0 ∂u ∂u 14. = u (u = u(x, y)) ∂x ∂y 15. (2x − y) dx − y dy = 0 16. Write each of the following second order equations as a system of first order equations. d 2 x dx (a) − − 6x = 0 dt dt 2 d 2x dx (b) 4 2 + 4 + 37x = 0 dt dt d 2x (c) L 2 + g sin x = 0, L, g positive condt stants, x = x(t) dx d 2x − μ(1 − x 2 ) + x = 0, μ > 0 con(d) 2 dt dt stant
11
EXERCISES 1.1
d 2x dx +(b −t) −ax = 0, a, b constants dt dt 2 In Exercises 17–28, verify that each of the given functions is a solution to the corresponding differential equation. (A, B, and C represent constants.) (e) t
17. dy/dx + 2y = 0, y(x) = e−2x , y(x) = 5e−2x 2 18. dy/dx + xy = 0, y(x) = e−x /2 19. dy/dx + y = sin x, y(x) = e−x − 12 cos x + 1 2 sin x d 2 y dy 20. − − 12y = 0, y(t) = e4t , y(t) = e−3t dt dt 2 21. y + 9y = 0, y = A + Be−9t ( denotes d/dt) 22. x + 3x − 10x = 0, x(t) = Ae2t + Be−5t 23. x + x = t cos t − cos t, x = A cos t + B sin t + 1 2 1 1 4 t sin t − 2 t sin t + 4 t cos t 24. y − 12y + 40y = 0, y = e6x cos 2x, y = e6x sin 2x ( denotes d/dx) 25. y − 4y = 0, y = A + Be2x + Ce−2x 26. y − 2y = 0, y = A + Bt + Ce2t 27. x 2 y − 12x y + 42y = 0, y = Ax 6 + Bx 7 28. t 2 y + 3t y + 5y = 0, y = t −1 (A cos(2 ln t) + B sin(2 ln t)) In Exercises 29–33, verify that the given equation satisfies the differential equation. Use the equation to determine y for the given value of x (or t). 29. dy/dx = −x/y, x 2 + y 2 = 16, x = 0 30. 3y(t 2 + y)dt + t (t 2 + 6y)dy = 0, t 3 y + 3ty 2 = 8, t = 2 31. dy/dx = −2y/x − 3, x 3 + x 2 y = 100, x = 1 32. y cos t dt + (2y + sin t) dy = 0, y 2 + y sin t = 1, t =0 33. (y/x + cos y) dx + (ln x − x sin y) dy = 0, y ln x + x cos y = 0, x = 1 In Exercises 34–43, use integration to find a solution to the differential equation. 34. 35. 36. 37. 38. 39.
dy/dx = (x 2 − 1)(x 3 − 3x)3 2 dy/dx = x sin √x 2 dy/dx = x/ x − 16 dy/dx = 1/(x ln x) dy/dx = x ln x dy/dx = xe−x
40. 41. 42. 43. 44. 45. 46.
−2(x + 5) dy = dx (x + 2)(x − 4) x − x2 dy = 2 dx (x + √ 1)(x + 1) 2 dy/dx = x − 16/x dy/dx = (4 − x 2 )3/2 dy/dx = 1/(x 2 − 16) dy/dx = cos x cot x dy/dx = sin3 x tan x
In Exercises 45–54, use the indicated conditions with the indicated solution to determine the solution to the given problem. 47. dy/dx + 2y = 0, y(0) = 2, y(x) = Ae−2x 48. dy/dt + y = sin t, y(0) = −1, y(t) = Ae−t − 1 1 2 cos t + 2 sin t 49. y − y − 12y = 0, y(0) = 1, y (0) = −1, y = Ae4x + Be−3x 50. y + 9y = 0, y(0) = 2, y (0) = −1, y = A + Be−9x 51. y − 2y = 0, y(0) = 0, y (0) = 1, y (0) = 3, y = A + Bx + Ce2x 52. y − 4y = 0, y(0) = 1, y (0) = −1, y (0) = 0, y = A + Be2x + Ce−2x 53. t 2 y − 12t y + 42y = 0, y(1) = 0, y (1) = −1, y(t) = At 6 + Bt 7 54. x 2 y + 3xy + 5y = 0, y(1) = 0, y (1) = 1, y(x) = x −1 (A cos(2 ln x) + B sin(2 ln x)) In Exercises 55–58, solve the initial value problem. 55. 56. 57. 58. 59.
dy/dx = 4x 3 − x + 2, y(0) = 1 dy/dt = sin 2t − cos y(0) = 0 2t, dy/dx = x −2 cos x −1 , y(2/π) = 1 dy/dx = (ln x)/x, y(1) = 0 The velocity of a falling object with mass m that is subjected to air resistance proportional the instantaneous velocity v of the object is found by solving the initial value
m dv/dt = mg − cv problem , where c > v(0) = v0 0 is the proportionality constant. (a) Given that a general solution to m dv/dt = mg − cv is v(t) = mg/c +
12
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
Ke−ct/m , find the solution of this initial value problem. (b) Determine limt→∞ v(t). 60. The number of cells in a bacteria colony after t hours is determined by solving the ini
dP /dt = kP . tial value problem P (0) = P0 (a) Given that a general solution of dP /dt = kP is P (t) = Cekt , use the initial condition to find C. (b) Find the value of k so that the population doubles in 8 hours. 61. In 1840, the Belgian mathematicianbiologist Pierre F. Verhulst (1804–1849) developed the logistic equation, dP /dt = rP − aP 2 , where r and a are positive constants, to predict the population P (t) in certain countries. (a) Given that a general solution to this r equation is P (t) = , find the a + Ce−rt solution that satisfies P (0) = P0 , where P0 > 0 is constant. (b) Determine limt→∞ P (t).
63.
64. 65.
66.
67. 68. 69.
Pierre Francois Verhulst (1804–1849)
62. The differential equation dS/dt + 3S/(t + 100) = 0, where S(t) is the number of pounds of salt in a particular tank at time t, is used to approximate the amount of salt in the tank containing a salt-water mixture in
70.
71.
which pure water is allowed to flow into the tank while the mixture is allowed to flow out of the tank. If S(t) = 15,000,000/(t + 100)3 , show that S satisfies dS/dt + 3S/(t + 100) = 0. What is the initial amount of salt in the tank? As t → ∞, what happens to the amount of salt in the tank? The displacement (measured from x = 0) of a mass attached to the end of a spring at time t is given by x(t) = 3 cos 4t + 94 sin 4t. Show that x satisfies the ordinary differential equation x + 16x = 0. What is the initial position of the mass? What is the initial velocity of the mass? Show that u(x, y) = ln x 2 + y 2 satisfies Laplace’s equation, uxx + uyy = 0. The temperature in a thin rod of length 2π after t minutes at a position x between 0 and 2π is given by u(x, t) = 3 − e−16kt cos 4x. Show that u satisfies ut = k uxx . What is the initial temperature (t = 0) at x = π ? What happens to the temperature at each point in the wire as t → ∞? The displacement u of a string of length 1 at time t position x, where x is measured from x = 0 is given by u(x, t) = sin πx cos t. Show that u satisfies π 2 utt = uxx . What is the value of u at the endpoints x = 0 and x = 1 for all values of t? Find the value(s) of m so that y = x m is a solution of x 2 y − 2xy + 2y = 0. Find the value(s) of k so that y = ekt is a solution of y − 3y − 18y = 0. d 2x dy Use the fact that e y = e2x + 2e2x y dx dx dy + 2e2x y = ex . and integration to solve e2x dx dy d x + ex y and Use the fact that (e y) = ex dx dx dy integration to solve ex + ex y = xex . dx The time-independent Schrödinger equation is given by −
h2 d 2 ψ(x) + U (x)ψ(x) = Eψ(x). 2m dx 2
13
EXERCISES 1.1
If U (x) = 0, find conditions on E so that ψ(x) = A sin(nπx/L) is a solution of the time-independent Schrödinger equation.
equation, and (b) graph the solution on the indicated interval(s). Throughout Introductory Differential Equations, we use graphs of solutions of differential equations. In some cases, we are able to predict what the graph of a solution should look like. If the graph of our proposed solution does not appear as predicted we know that either we made a mistake in constructing our proposed solution or our conjecture about the general shape of the graph of the solution is wrong. In other cases, we will find that it is easier to examine the graph of a solution than it is to examine the solution (if we are able to construct one in the first place).
The great physicist Erwin Schrödinger (1887–1961) received a Nobel prize for his work in 1933.
72. A singular solution of a differential equation is a solution that cannot be derived from the general solution of the differential equation. Use implicit differentiation to show that −1/x + 2/x 2 + 1/y − 1/y 2 = C is a general (implicit) solution of the differential equation dy/dx = (x − 4)y 3 /[x 3 (y − 2)]. Is y = 0 a solution of this differential equation? Is y = 0 a singular solution? 73. Show that x + x 2 /y = C is a general (implicit) solution of the differential equation dy/dx = (y 2 + 2xy)/x 2 . Is y = 0 a solution of this differential equation? Is y = 0 a singular solution? 74. The current I (t) in an L-R circuit, which contains a resistor, an inductor, and a voltage source, satisfies the differential equation RI + LdI /dt = E(t), where R and L are constants representing the resistance and the inductance and E(t) is the voltage source. Is this equation linear or nonlinear? Determine the order of the equation. In Exercises 75–77, (a) verify that the indicated function is a solution of the given differential
75. xy + y = cos x, y = (sin x)/x; [−2π, 0) ∪ (0, 2π] 76. 16y + 24y + 153y = 0, y = e−3t/4 cos 3t; [0, 3π/2] 77. x 3 y + x 2 y + xy − 40y = 0, y = x −1 sin(3 ln x); (0, π] 78. (a) Verify that ⎧ √ √ √ ⎨x = e−t 100 3 sin 3t + 20 cos 3t 3 √ √ √ ⎩y = e−t − 40 3 sin 3t + 20 cos 3t 3 is a solution
of the system of differential dx/dt = 4y equations . dy/dt = −x − 2y (b) Graph x(t),
y(t), and the parametric x = x(t) equations for 0 ≤ t ≤ 2π . y = y(t) 79. (a) Show that (x 2 + y 2 )2 = 5xy is an implicit solution of [4x(x 2 + y 2 ) − 5y] dx + [4y(x 2 + y 2 ) − 5x] dy = 0. (b) Graph (x 2 + y 2 )2 = 5xy on the rectangle [−2, 2] × [−2, 2]. (c) Approximate all points on the graph of (x 2 + y 2 )2 = 5xy with x-coordinate 1.
14
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
(d) Approximate all points on the graph of (x 2 + y 2 )2 = 5xy with y-coordinate −0.319. Often the calculus and algebra encountered in solving differential equations can be quite complicated. Computer algebra systems are capable of performing the integration and algebraic simplification encountered when solving many differential equations. Having access to these tools can be a great advantage: a large number of problems can be solved quickly, we make conjectures as to the general form of a solution to different forms of differential equations, and these tools allow us to check and verify our work.
80. Solve dy/dx = sin4 x, y(0) = 0 and graph the resulting solution on the interval [0, 4π ]. 81. A general solution of y (4) + 25 2 y − 5y + 629 −x/2 (c cos 3x + 1 16 y = 0 is given by y = e x/2 c2 sin 3x) + e (c3 cos 2x + c4 sin 2x), where c1 , c2 , c3 , and c4 are constants. Solve the initial value problem
629 y (4) + 25 2 y − 5y + 16 y = 0 y(0) = 0, y (0) = 1, y (0) = −1, y (0) = 1 and graph the resulting solution. 82. A general solution of the system
dx/dt = 4y dy/dt = −4x
x = −c1 cos 4t + c2 sin 4t , where c1 and y = c2 cos 4t + c1 sin 4t c2 are⎧constants. Solve the initial value prob⎪ ⎨dx/dt = 4y lem dy/dt = −4x and then graph ⎪ ⎩ x(0) = 4, y(0) = 0 x(t), y(t), and the parametric equations
x = x(t) . y = y(t) is
83. A general solution of the system
dx/dt = −5x + 4y dy/dt = 2x + 2y
x = −c1 e3t − 4c2 e−6t , where c1 and c2 y = 2c1 e3t + c2 e−6t are constants. Solve the initial value prob⎧ ⎪ x = −5x + 4y ⎨ lem y = 2x + 2y and then graph ⎪ ⎩ x(0) = 4, y(0) = 0 x(t), y(t), and the parametric equations
x = x(t) . y = y(t)
is
1.2 A GRAPHICAL APPROACH TO SOLUTIONS: SLOPE FIELDS AND DIRECTION FIELDS • Systems of Ordinary Differential Equations and Direction Fields • Relationship Between Systems of First Order and Higher Order Equations Suppose that we are asked to solve the ordi2 nary differential equation dy/dx = e−x . In this case, we do not attempt to solve this equation through integration as we did in the previous section because of the presence of the function 2 f (x) = e−x on the right side of the ordinary differential equation. (See Exercise 21 at the end of this section.) Instead, we can gain insight into the behavior of solutions of this ordinary differential equation through a graphical approach by considering the slope of the tangent line to solutions of the ordinary differential equation. Recall from Section 1.1 that the differential equation gives the slope of the tangent line to solutions of the ordinary differential equation at the given point (x, y) in the xy-plane. Therefore, if we wish to determine the slope of the tangent
15
1.2 A GRAPHICAL APPROACH TO SOLUTIONS
2
FIGURE 1.8 (A) Several line segments in the slope field for dy/dx = e−x . (B) One view of the slope field for the equation. (C) A different view of the slope field for the equation.
2
2
FIGURE 1.9 (A) Slope field for dy/dx = e−x . (B) Using the slope field to sketch the solution to dy/dx = e−x , y(−2) = −1. (C) A different view of using the slope field to sketch the solution to dy/dx = e
line to the solutions to the ordinary differential equation that passes through (0, 1) at this point, we substitute (0, 1) into the right side of 2 dy/dx = e−x . Because the right side only depends on x, we have slope dy 2 = e−(0) = 1. dx In fact, the slope of the line tangent to solutions at√all points of the form (0, y) is 1. At the point ( ln 2, 4), we find that the slope is √ dy 1 2 −1 = e−( ln 2) = e− ln 2 = eln 2 = . dx 2
Again, we obtain the same slope at√all points √ ( ln 2, y) ≈ (0.832555, y) and (− ln 2, y) ≈
−x 2
, y(−2) = −1.
(−0.832555, y). In Fig. 1.8, we draw several short line segments using points of the form (0, y) for y = 0, ±1, ±2, ±3. Notice that we use a triangle with base length 1 and height length 1 to assist in sketching the tangent lines with slope 1 at these points. We use a triangle with base length 2 and height length 1 to help us draw the lines of slope 1/2 √ that are tangent to solutions at the points (− ln 2, y) for y = 0, ±1, ±2, ±3. By drawing a set of short line segments representing the tangent lines to solutions of the ordinary differential equation at numerous points in the xy-plane, we construct the slope field of the ordinary differential equa2 tion. We show the slope field for dy/dx = e−x on the square [−2, 2] × [−2, 2] in Fig. 1.9A. (Note that because constructing a slope field is time
16
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.10 (A) Graphs of x(t) = sin t and y(t) = cos t for 0 ≤ t ≤ 2π . (B) Graph of the parametric equations x(t) = sin t and y(t) = cos t for 0 ≤ t ≤ 2π .
consuming, we usually let a computer algebra system do the work for us.) Observe that at each point along the y-axis, the slope is 1 as we predicted earlier. Notice also that the slope appears to be zero for larger values of |x|. This is because for large values of x (in absolute value), 2 dy/dx ≈ 0 because limx→±∞ e−x = 0. Therefore, we expect solutions to “flatten out” as x increases. We can use the slope field to investigate the solution to an initial value problem such as dy 2 = e−x , dx
y(−2) = −1
by starting at the point (−2, −1) and tracing the solution by following the tangent slopes. This solution is sketched in Fig. 1.9B. Thus, although we were not able to determine explicit formulas for either the general solution of the ODE or the solution to the IVP, we were able to determine some properties of the solutions by using the slope field of the differential equation.
Systems of Ordinary Differential Equations and Direction Fields We can also consider systems of differential equations. In Chapter 6, we will learn how to solve systems of first order ordinary differential equations of the form
dx/dt = ax + by dy/dt = cx + dy
,
where a, b, c, and d are given constants. In the case of this system, we solve for x = x(t) and y = y(t). For example, if we consider the system
dx/dt = y , dy/dt = −x we can verify that the parametric equations
x(t) = sin t y(t) = cos t satisfy the system because d dx = (sin t) = cos t = y and dt dt dy d = (cos t) = − sin t = −x. dt dt We can graph each function separately as we do in Fig. 1.10A. (The graph of x(t) is the solid curve; that of y(t) is dashed.) Another option is to graph them as a pair of parametric equations as in Fig. 1.10B. As we recall, the graph of this pair of parametric equations is a circle of radius 1 centered at the origin because x 2 = sin2 t and y 2 = cos2 t , so that x 2 + y 2 = sin2 t + cos2 t = 1. However, we must indicate the orientation of the curve (the direction of increasing parameter value t ). For this pair of parametric equations, we find that at t = 0, x(0) = sin 0 = 0 and y(0) = cos 0 = 1. Therefore, the point (0, 1) corresponds to t = 0. Similarly, the point (1, 0) corresponds to t = π/2 because x(π/2) = sin π/2 = 1 and y(π/2) = cos π/2 = 0. This means that
1.2 A GRAPHICAL APPROACH TO SOLUTIONS
the solution moves from (0, 1) to (1, 0) as t increases. To determine if the orientation is clockwise or counter-clockwise, we test a t -value between 0 and π/2. Choosing t = π/4, we find the √ 2 and y(π/4) = cos π/4 = x(π/4) = sin π/4 = 1/ √ 1/ 2 so the orientation is clockwise. The parametric equations {x(t) = sin t, y(t) = cos t} satisfy the initial value problem
dx/dt = y, x(0) = 0 dy/dt = −x, y(0) = 1 because they satisfy the system of differential equations as well as the two initial conditions. Notice that in the case of an initial value problem involving a system of differential equations, an initial condition is given for each of the variables x and y that depend on t. In the parametric plot, the solution to this initial value problem passes through the point (x(0), y(0)) = (0, 1). Another way to view a system of two ordinary differential equations is through the use of a direction field, which is similar to a slope field. For example, for the first order system
dx/dt = ax + by , dy/dt = cx + dy we first write it as a first order equation with dy/dt cx + dy dy = = . dx dx/dt ax + by Observe that the procedure described here can be used for any system of the form
dx/dt = f (x, y) . dy/dt = g(x, y)
Then, we can consider the slope field associated with this differential equation. For example, if we refer back to the system
dx/dt = y , dy/dt = −x
17
we obtain the first order equation dy/dx = −x/y, so we can determine the slope of tangent lines to solutions at points √ xy-plane. For √ in the example, at the point (1/ 2, 1/ 2), the solution to this system √ that passes through this point has −1/ 2 slope √ = −1. In a similar manner, we can 1/ 2 find the slope at other points in the plane. However, as we mentioned in our earlier discussion, we must indicate the orientation when we graph parametric equations, so we consider the vector dx/dt, dy/dt = dx/dt i + dy/dt j with components from the system of differential equations. In the case of this system, we consider √ dx/dt, dy/dt = y, −x . At the √ point √ (1/ 2, √ 1/ 2), we obtain the vector 1/ 2, −1/√ 2 . This √ 2, 1/ 2) means that the solution√ through (1/ √ has tangent vector 1/ 2, −1/ 2 . The direction field up of tangent vectors such √ √ is made as 1/ 2, −1/ 2 to solutions at points in the plane, so it is similar to the slope field for dy/dy = −x/y shown in Fig. 1.11A, except that vectors are used to indicate the orientation of solutions. In Fig. 1.11B, we show the direction field for this system. The vectors in the direction field indicate that solutions to this system are circles in the xy-plane that are directed clockwise. We graph several solutions along with the direction field in Fig. 1.11C. A collection of solutions in the xy-plane is called the phase portrait of the system. Notice that at points in the first quadrant, where x > 0 and y > 0, dx/dt = y > 0 and dy/dt = −x < 0. This means that x(t) increases and y(t) decreases along solutions in the first quadrant. In the second quadrant, where x < 0 and y > 0, dx/dt = y > 0 and dy/dt = −x > 0. Therefore, x(t) and y(t) increase along solutions in the second quadrant. We can perform a similar analysis for points in the other two quadrants. Generally, we will use standard mathematical notation throughout Introductory Differential Equations so i = 1, 0 and j = 0, 1 .
18
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.11 (A) Slope field for dy/dx = −x/y. (B) Direction field for dx/dt = y, dy/dt = −x. (C) Direction field for dx/dt = y, dy/dt = −x and several solution curves.
Relationship Between Systems of First Order and Higher Order Equations
dx/dt = y . If dy/dt = −x we differentiate the equation dx/dt = y with respect to t , we obtain d 2 x/dt 2 = dy/dt. Therefore, if we equate dy/dt = −x and dy/dt = d 2 x/dt 2 , we have d 2 x/dt 2 = −x or d 2 x/dt 2 + x = 0. We say that the system of two first order ODEs is equivalent to the second order ODE d 2 x/dt 2 + x = 0. Often, however, we begin with a second order ODE of the form d 2 x/dt 2 + b dx/dt + c x = f (t) and would like to write the equation as a system of first order ODEs. We do this by letting dx/dt = y and by differentiating this equation with respect to t to obtain d 2 x/dt 2 = dy/dt. Solving the second order ODE for d 2 x/dt 2 , we find d 2 x/dt 2 = −b dx/dt − c x + f (t) so that dy/dt = −b dx/dt − c x + f (t). Replacing dx/dt in this equation with y, we obtain the following equivalent system of first order ODEs:
dx/dt = y . dy/dt = −b y − c x + f (t)
gree greater than two into a system of first order ODEs.
Again, consider the system
By writing the second order ODE as a system of first order ODEs, we investigate the behavior of the solution of the second order ODE by observing the behavior in the direction field and phase portrait of the corresponding system. A similar procedure is used to transform an ODE of de-
EXERCISES 1.2 In Exercises 1–4, use the slope field to determine if the indicated path is that of a solution to the differential equation.
1.
dy = −y/x dx
2.
dy = x2 + y2 dx
19
EXERCISES 1.2
dy 3. = x/y dx
7.
dy =x −y dx
dy = x2 − y 4. dx
8.
dy = y2 − x2 dx
In Exercises 5–8, use the slope field to sketch the solutions of the differential equation that pass through the given points.
5.
dy = x/y dx
6.
dy = x2 − y dx
9. Graph the slope field for dy/dt = sin y. Determine limt→∞ y(t) if y(0) = y0 where (a) y0 = −3π/2; (b) y0 = −π/2; (c) y0 = π/2; (d) y(0) = 3π/2. 10. Graph the slope field for dy/dx = sin x. Does limx→∞ y(x) exist for any initial condition y(0) = y0 ? Solve the ODE and find limx→∞ y(x). Does this match the graphical result? 11. Graph the slope field for dy/dx = e−x . Does limx→∞ y(x) exist for any initial condition y(0) = y0 ? Solve the ODE and find limx→∞ y(x). Does this match the graphical result? 12. Graph the slope field for dy/dx = 1/(x 2 + 1). Does limx→∞ y(x) exist for any initial condition y(0) = y0 ? Solve the ODE and find limx→∞ y(x). Does this match the graphical result? In Exercises 13–14, use the direction field of the given system to sketch the graph of the solution that satisfies the indicated initial conditions. Determine limt→∞ x(t) and limt→∞ y(t) in each case (if they exist).
20
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.12 Figure for Exercise 23.
dx/dt = y ; (a) x(0) = 0, y(0) = dy/dt = x 2; (b) x(0) = −2, y(0) = 0; (c) x(0) = −2,
13. System:
y(0) = 2;
14. System: y(0) = 5;
dx/dt = 3x − 2y dy/dt = 4x − y
;
x(0) = 0,
d 2 x/dt 2 + 4x = 0 d 2 x/dt 2 − 5 dx/dt = 0 d 2 x/dt 2 + 4 dx/dt + 13x = 0 x − 6x + 7x = 0 ( = d/dt; x = x(t)) x + 16x = sin t x + 4x + 13x = e−t (a) Use a computer algebra system to solve 2 dy/dx = e−x . (b) Graph the solution to 2 the IVP dy/dx = e−x , y(0) = a for a = −2, −1, 0, 1 2. (c) Graph the slope field of the differential equation together with the solutions in (b). (d) Do the solutions appear to match the results described at the beginning of the section? 22. (a) Use a computer algebra system to graph the slope field for dy/dx = sin(2x − y). (b) Use the slope fie to graph the solution y(x) that satisfies the initial condition y(0) = 5. Does limx→∞ y(x) appear to exist? 23. Consider the systems
15. 16. 17. 18. 19. 20. 21.
In Exercises 15–20, write the second order equation as a system of first order equations.
dx/dt = −y dy/dt = x
and
dx/dt = −y dy/dt = −x
.
(a) For any given initial conditions, in a brief paragraph explain why you think that the solutions of the systems are similar or different.
EXERCISES 1.2
21
FIGURE 1.13 Figure for Exercise 24.
(b) Fig. 1.12 shows the direction field associated with each system. Use the direction field to help you graph the solutions that satisfy these initial conditions (i) x(0) = 0.5, y(0) = 0; (ii) x(0) = −0.25, y(0) = 0; (iii) x(0) = 0, y(0) = 0.75; and (iv) x(0) = 0, y(0) = −0.5. (c) How do your graphs affect your conjecture in (a)? 24. Consider the systems
dx/dt = x/2 dy/dt = y
and
dx/dt = −x/2 dy/dt = −y
.
(a) For any given initial conditions, in a brief paragraph explain why you think that the solutions of the systems are similar or different. (b) Fig. 1.13 shows the direction field associated with each system. Use the direction field to help you graph the solutions that satisfy these initial condition (i) x(0) = 0.5, y(0) = 0.25; (ii) x(0) = −0.25, y(0) = −0.5; (iii) x(0) = −0.5, y(0) = 0.75; and (iv) x(0) = 0.75, y(0) = −0.5. (c) How do your graphs affect your conjecture in (a)?
25. (Competing Species) The system of equa
dx/dt = x(a − b1 x − b2 y) where a, tions dy/dt = y(c − d1 x − d2 y) b1 , b2 , c, d1 , and d2 represent positive constants can be used to model the size of the population of two species, represented by x(t) and y(t), competing for a common food supply. (a) Fig. 1.14A shows the direction field for the system if a = 1, b1 = 2, b2 = 1, c = 1, d1 = 0.75, and d2 = 2. (i) Use the direction field to graph various solutions if both x(0) and y(0) are positive. (ii) Use the direction field and your graphs to approximate limt→∞ x(t) and limt→∞ y(t). (b) Fig. 1.14B shows the direction field for the system if a = 1, b1 = 1, b2 = 1, c = 0.67, d1 = 0.75, and d2 = 1. (i) Use the direction field to graph various solutions if both x(0) and y(0) are positive. (ii) Use the direction field and your graphs to determine the fate of the species with population y(t). What happens to the species with population x(t)?
22
1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.14 (A) Figure for Exercise 25 (a). (B) Figure for Exercise 25 (b).
CHAPTER 1 SUMMARY: ESSENTIAL CONCEPTS AND FORMULAS Differential Equation (DE) An equation that contains the derivative or differentials of one or more dependent variables with respect to one or more dependent variables. Ordinary Differential Equation (ODE) If a differential equation contains only ordinary derivatives (of one or more dependent variables) with respect to a single independent variable, the equation is called an ordinary differential equation. Partial Differential Equation (PDE) A differential equation that contains the partial derivatives or differentials of one or more independent variables with respect to more than one independent variable is called a partial differential equation. Linear Ordinary Differential Equation A linear ODE is an equation that can be written in the form an (x)y (n) + an−1 (x)y (n−1) + · · · + a2 (x)y + a1 (x)y + a0 y = f (x), where y = y(x). Order of an Equation The order of the highest order derivative in a differential equation is called the order of the equation.
Solution A solution of a differential equation on a given interval is a function that is continuous on the interval and has all the necessary derivatives that are present in the differential equation such that when substituted into the equation yields an identity for all values on the interval. Explicit Solution A solution given as a function of the independent variable. Implicit Solution A solution given as a relation such as f (x, y) = 0, f (t, x) = 0, or f (t, y) = 0. Trivial Solution y = 0 is always a solution of the nth order linear homogeneous equation. Slope Field A collection of line segments that indicate the slope of the tangent line to the solution(s) of a differential equation. Phase Portrait A collection
of graphs of soludx/dt = f (x, y) tions to the system in dy/dt = g(x, y) the xy-plane. Direction Field A collection of vectors that indicate the slope and direction of the tangent line to the solutions of a system of differential equations.
CHAPTER 1 REVIEW EXERCISES
23
CHAPTER 1 REVIEW EXERCISES
lution to determine the solution(s) to the given problem.
In Exercises 1–5, determine (a) if the equation is an ordinary differential or partial differential equation; (b) the order of the differential equation; and (c) if the equation is linear or nonlinear.
20. dy/dx + 2y = x 2 , y(0) = 1, y = 14 − 12 x + 1 2 −2x 2 x + Ae 21. y + 4y = t , y(0) = 1, y(π/4) = π/16, y = t/4 + A cos 2t + B sin 2t 22. x 2 y + 5xy + 4y = 0, y(1) = 1, y (1) = 0, y = Ax −2 + Bx −2 ln x
dy/dt = y a ux + ut = 0, u = u(x, t), a constant d 2 y/dx 2 + 2 dy/dx + y = 0 m x + kx = sin t, m and k positive constants; x = x(t) ∂φ ∂ 2 φ ∂ 2 φ = 2 5. ∂x ∂x 2 ∂y
1. 2. 3. 4.
In Exercises 6–13, verify that the given function is a solution of the corresponding differential equation. (A and B denote constants.) dy/dx + y cos x = 0, y = e− sin x dy/dx − y = sin x, y = (ex − cos x − sin x)/2 y + 4y − 5y = 0, y = e−5x , y = ex y − 6y + 45y = 0, y = e3x (cos 6x − sin 6x) xy − xy − 16y = 0, y = Ax 5 + Bx −3 x 2 y + 3xy + 2y = 0, y = x −1 (cos(ln x) − sin(ln x)) 12. d 2 y/dx 2 + 2 dy/dx + 2y = x, y = (x − 1)/2 13. y − 7y + 12y = 2, y = Ae3x + Be4x + 1/6 6. 7. 8. 9. 10. 11.
In Exercises 14 and 15, verify that the given implicit function satisfies the differential equation. 14. (2x − 3y) dx + (2y − 3x)dy = 0, x 2 − 3xy + y2 = 1 15. (y cos(xy) + sin x) dx + x cos(xy) dy = 0, sin(xy) − cos x = 0 In Exercises 16–19, find a solution of the differential equation. 16. dy/dx = xe−x 17. dy/dx = x 2 sin x 2x 2 − x + 1 dy = 18. 2 dx (x − 1)(x √ + 1) 2 19. dy/dx = x / x 2 − 1 2
Exercises 20–22, use the indicated initial or boundary conditions with the given general so-
Exercises 23 and 24 solve the initial-value problem. Graph the solution on an appropriate interval. 23. dy/dx = cos2 x sin x, y(0) = 0 24. dy/dx = 13 (4x − 9)(x − 3)−2/3 , y(0) = 0 25. The temperature on the surface of a steel ball at time t is given by u(t) = 70e−kt + 30 (in o F) where k is a positive constant. Show that u satisfies the first order equation du/dt = −k(u − 30). What is the initial temperature (t = 0) on the surface of the ball? What happens to the temperature as t → ∞? 26. The displacement (measured from x = 0) of a mass attached to the end of a spring at √ 1 −t time t is given by x(t) = 4 e (cos 35t + √ √9 sin 35t). Show that x satisfies the ordi35 nary differential equation x +2x +36x = 0. What is the initial displacement of the mass? What is the initial velocity of the mass? 27. For a particular wire of length 1 foot, the temperature at time t hours at a position of x feet from the end (x = 0) of the wire is es2 2 timated by u(x, t) = e−π kt sin πx − e−4π kt × sin 2πx. Show that u satisfies the equation ut = k uxx . What is the initial temperature (t = 0) at x = 1? What happens to the temperature at each point in the wire as t → ∞? 28. The height u of a long string at time t and position x where x is measured from the middle of the string (x = 0) is given by u(x, t) = sin x cos 2t . Show that u satisfies the wave equation utt = 4uxx . What is the initial height (t = 0) at x = 0?
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1. INTRODUCTION TO DIFFERENTIAL EQUATIONS
FIGURE 1.15 (A) Figure for Exercise 30. (B) Figure for Exercise 31.
29. Show that u(x, y) = tan−1 (y/x) satisfies Laplace’s equation uxx + uyy = 0. 30. The slope field of y = 2(y − y 2 ) is shown in Fig. 1.15A. Sketch the graph of the solution that satisfies the indicated initial condition. Also, determine if limx→∞ y(x) exists. (a) y(0) = 0.5; (b) y(0) = 1.5; (c) y(0) = −0.5. 31. The direction field for {dx/dt = x, dy/dt = 2y} is shown in Fig. 1.15B. Is it pos-
sible to select initial conditions so that limt→∞ x(t) = 0 and limt→∞ y(t) = 0? (Briefly explain.) (−1)k × 32. The function J1 (x) = ∞ k=0 k!(k + 1)!22k+1 x 2k+1 is called the Bessel function of order 1. Verify that J1 (x) is a solution of Bessel’s equation of order 1, x 2 y + xy + 2 x − 1 y = 0.