Slope

We begin with a simple concept from calculus: A derivative dy/dx of a differentiable function y = y(x) gives slopes of tangent lines at points on its graph. Because a solution y = y(x) of a first-order differential equation dy/dx = f(x, y) is necessarily a differentiable function on its interval I of definition, it must also be continuous on I. Thus the corresponding solution curve on I must have no breaks and must possess a tangent line at each point (x, y(x)). The slope of the tangent line at (x, y(x)) on a solution curve is the value of the first derivative dy/dx at this point, and this we know from the differential equation f(x, y(x)). Now suppose that (x, y) represents any point in a region of the xy-plane over which the function f is defined. The value f(x, y) that the function f assigns to the point represents the slope of a line, or as we shall envision it, a line segment called a lineal element. For example, consider the equation dy/dx = 0.2xy, where f(x, y) = 0.2xy. At, say, the point (2, 3), the slope of a lineal element is f(2, 3) = 0.2(2)(3) = 1.2.FIGURE 2.1.1(a) shows a line segment with slope 1.2 passing through (2, 3). As shown in Figure 2.1.1(b), if a solution curve also passes through the point (2, 3), it does so tangent to this line segment; in other words, the lineal element is a miniature tangent line at that point.

Direction Field

If we systematically evaluate f over a rectangular grid of points in the xy-plane and draw a lineal element at each point (x, y) of the grid with slope f(x, y), then the collection of all these lineal elements is called a direction field or a slope field of the differential equation dy/dx = f(x, y). Visually, the direction field suggests the appearance or shape of a family of solution curves of the differential equation, and consequently it may be possible to see at a glance certain qualitative aspects of the solutions—regions in the plane, for example, in which a solution exhibits an unusual behavior. A single solution curve that passes through a direction field must follow the flow pattern of the field; it is tangent to a lineal element when it intersects a point in the grid.

EXAMPLE 1 Direction Field

The direction field for the differential equation dy/dx = 0.2xy shown in FIGURE 2.1.2(a) was obtained using computer software in which a 5 × 5 grid of points (mh, nh), m and n integers, was defined by letting −5 ≤ m ≤ 5, −5 ≤ n ≤ 5 and h = 1. Notice in Figure 2.1.2(a) that at any point along the x-axis (y = 0) and the y-axis (x = 0) the slopes are f(x, 0) = 0 and f(0, y) = 0, respectively, so the lineal elements are horizontal. Moreover, observe in the first quadrant that for a fixed value of x, the values of f(x, y) = 0.2xy increase as y increases; similarly, for a fixed y, the values of f(x, y) = 0.2xy increase as x increases. This means that as both x and y increase, the lineal elements become almost vertical and have positive slope (f(x, y) = 0.2xy > 0 for x > 0, y > 0). In the second quadrant, |f(x, y)| increases as |x| and y increase, and so the lineal elements again become almost vertical but this time have negative slope (f(x, y) = 0.2xy < 0 for x < 0, y > 0). Reading left to right, imagine a solution curve starts at a point in the second quadrant, moves steeply downward, becomes flat as it passes through the y-axis, and then as it enters the first quadrant moves steeply upward—in other words, its shape would be concave upward and similar to a horseshoe. From this it could be surmised that y → as x → ±. Now in the third and fourth quadrants, since f(x, y) = 0.2xy > 0 and f(x, y) = 0.2xy < 0, respectively, the situation is reversed; a solution curve increases and then decreases as we move from left to right.

We saw in (1) of Section 1.1 that is an explicit solution of the differential equation dy/dx = 0.2xy; you should verify that a one-parameter family of solutions of the same equation is given by . For purposes of comparison with Figure 2.1.2(a) some representative graphs of members of this family are shown in Figure 2.1.2(b). ≡

EXAMPLE 2 Direction Field

Use a direction field to sketch an approximate solution curve for the initial-value problem dy/dx = sin y, y(0) = − .

SOLUTION

Before proceeding, recall that from the continuity of f(x, y) = sin y and ∂f/∂y = cos y, Theorem 1.2.1 guarantees the existence of a unique solution curve passing through any specified point (x₀, y₀) in the plane. Now we set our computer software again for a 5 × 5 rectangular region and specify (because of the initial condition) points in that region with vertical and horizontal separation of unit—that is, at points (mh, nh), h = , m and n integers such that −10 ≤ m ≤ 10, −10 ≤ n ≤ 10. The result is shown in FIGURE 2.1.3. Since the right-hand side of dy/dx = sin y is 0 at y = 0 and at y = −π, the lineal elements are horizontal at all points whose second coordinates are y = 0 or y = −π. It makes sense then that a solution curve passing through the initial point (0, − ) has the shape shown in color in the figure.≡

Increasing/Decreasing

Interpretation of the derivative dy/dx as a function that gives slope plays the key role in the construction of a direction field. Another telling property of the first derivative will be used next, namely, if dy/dx > 0 (or dy/dx < 0) for all x in an interval I, then a differentiable function y = y(x) is increasing (or decreasing) on I.

DEs Free of the Independent Variable

In Section 1.1 we divided the class of ordinary differential equations into two types: linear and nonlinear. We now consider briefly another kind of classification of ordinary differential equations, a classification that is of particular importance in the qualitative investigation of differential equations. An ordinary differential equation in which the independent variable does not appear explicitly is said to be autonomous. If the symbol x denotes the independent variable, then an autonomous first-order differential equation can be written in general form as F(y, y′) = 0 or in normal form as

(1)

We shall assume throughout the discussion that follows that f in (1) and its derivative f′ are continuous functions of y on some interval I. The first-order equations

are autonomous and nonautonomous, respectively.

Many differential equations encountered in applications, or equations that are models of physical laws that do not change over time, are autonomous. As we have already seen in Section 1.3, in an applied context, symbols other than y and x are routinely used to represent the dependent and independent variables. For example, if t represents time, then inspection of

where k, n, and T_m are constants, shows that each equation is time-independent. Indeed, all of the first-order differential equations introduced in Section 1.3 are time-independent and so are autonomous.

Critical Points

The zeros of the function f in (1) are of special importance. We say that a real number c is a critical point of the autonomous differential equation (1) if it is a zero of f, that is, f(c) = 0. A critical point is also called an equilibrium point or stationary point. Now observe that if we substitute the constant function y(x) = c into (1), then both sides of the equation equal zero. This means

A constant solution y(x) = c of (1) is called an equilibrium solution; equilibria are the only constant solutions of (1).

As already mentioned, we can tell when a nonconstant solution y = y(x) of (1) is increasing or decreasing by determining the algebraic sign of the derivative dy/dx; in the case of (1) we do this by identifying the intervals on the y-axis over which the function f(y) is positive or negative.

EXAMPLE 3 An Autonomous DE

The differential equation

where a and b are positive constants, has the normal form dP/dt = f(P), which is (1) with t and P playing the parts of x and y, respectively, and hence is autonomous. From f(P) = P(a − bP) = 0, we see that 0 and a/b are critical points of the equation and so the equilibrium solutions are P(t) = 0 and P(t) = a/b. By putting the critical points on a vertical line, we divide the line into three intervals defined by − < P < 0, 0 < P < a/b, a/b < P < . The arrows on the line shown in FIGURE 2.1.4 indicate the algebraic sign of f(P) = P(a − bP) on these intervals and whether a nonconstant solution P(t) is increasing or decreasing on an interval. The following table explains the figure.

Figure 2.1.4 is called a one-dimensional phase portrait, or simply phase portrait, of the differential equation dP/dt = P(a − bP). The vertical line is called a phase line. ≡

Solution Curves

Without solving an autonomous differential equation, we can usually say a great deal about its solution curves. Since the function f in (1) is independent of the variable x, we can consider f defined for − < x < or for 0 ≤ x < . Also, since f and its derivative f′ are continuous functions of y on some interval I of the y-axis, the fundamental results of Theorem 1.2.1 hold in some horizontal strip or region R in the xy-plane corresponding to I, and so through any point (x₀, y₀) in R there passes only one solution curve of (1). See FIGURE 2.1.5 (a). For the sake of discussion, let us suppose that (1) possesses exactly two critical points, c₁ and c₂, and that c₁ < c₂. The graphs of the equilibrium solutions y(x) = c₁ and y(x) = c₂ are horizontal lines, and these lines partition the region R into three subregions R₁, R₂, and R₃ as illustrated in Figure 2.1.5(b). Without proof, here are some conclusions that we can draw about a nonconstant solution y(x) of (1):

• If (x₀, y₀) is in a subregion R_i, i = 1, 2, 3, and y(x) is a solution whose graph passes through this point, then y(x) remains in the subregion R_i for all x. As illustrated in Figure 2.1.5(b), the solution y(x) in R₂ is bounded below by c₁ and above by c₂; that is, c₁ < y(x) < c₂ for all x. The solution curve stays within R₂ for all x because the graph of a nonconstant solution of (1) cannot cross the graph of either equilibrium solution y(x) = c₁ or y(x) = c₂. See Problem 33 in Exercises 2.1.
• By continuity of f we must then have either f(y) > 0 or f(y) < 0 for all x in a subregion R_i, i = 1, 2, 3. In other words, f(y) cannot change signs in a subregion. See Problem 33 in Exercises 2.1.
• Since dy/dx = f(y(x)) is either positive or negative in a subregion R_i, i = 1, 2, 3, a solution y(x) is strictly monotonic—that is, y(x) is either increasing or decreasing in a subregion R_i. Therefore y(x) cannot be oscillatory, nor can it have a relative extremum (maximum or minimum). See Problem 33 in Exercises 2.1.
• If y(x) is bounded above by a critical point c₁ (as in subregion R₁ where y(x) < c₁ for all x), then the graph of y(x) must approach the graph of the equilibrium solution y(x) = c₁ either as x → or as x → −. If y(x) is bounded, that is, bounded above and below by two consecutive critical points (as in subregion R₂ where c₁ < y(x) < c₂ for all x), then the graph of y(x) must approach the graphs of the equilibrium solutions y(x) = c₁ and y(x) = c₂, one as x → and the other as x → −. If y(x) is bounded below by a critical point (as in subregion R₃ where c₂ < y(x) for all x), then the graph of y(x) must approach the graph of the equilibrium solution y(x) = c₂ either as x → or as x → −. See Problem 34 in Exercises 2.1.

With the foregoing facts in mind, let us reexamine the differential equation in Example 3.

EXAMPLE 4 Example 3 Revisited

The three intervals determined on the P-axis or phase line by the critical points P = 0 and P = a/b now correspond in the tP-plane to three subregions:

R₁: − < P < 0,        R₂: 0 < P < a/b,        R₃: a/b < P < ,

where − < t < . The phase portrait in Figure 2.1.4 tells us that P(t) is decreasing in R₁, increasing in R₂, and decreasing in R₃. If P(0) = P₀ is an initial value, then in R₁, R₂, and R₃, we have, respectively, the following:

1. For P₀ < 0, P(t) is bounded above. Since P(t) is decreasing, P(t) decreases without bound for increasing t and so P(t) → 0 as t → −. This means the negative t-axis, the graph of the equilibrium solution P(t) = 0, is a horizontal asymptote for a solution curve.
2. For 0 < P₀ < a/b, P(t) is bounded. Since P(t) is increasing, P(t) → a/b as t → and P(t) → 0 as t → −. The graphs of the two equilibrium solutions, P(t) = 0 and P(t) = a/b, are horizontal lines that are horizontal asymptotes for any solution curve starting in this subregion.
3. For P₀ > a/b, P(t) is bounded below. Since P(t) is decreasing, P(t) → a/b as t → . The graph of the equilibrium solution P(t) = a/b is a horizontal asymptote for a solution curve.

In FIGURE 2.1.6, the phase line is the P-axis in the tP-plane. For clarity, the original phase line from Figure 2.1.4 is reproduced to the left of the plane in which the subregions R₁, R₂, and R₃ are shaded. The graphs of the equilibrium solutions P(t) = a/b and P(t) = 0 (the t-axis) are shown in the figure as blue dashed lines; the solid graphs represent typical graphs of P(t) illustrating the three cases just discussed. ≡

In a subregion such as R₁ in Example 4, where P(t) is decreasing and unbounded below, we must necessarily have P(t) → −. Do not interpret this last statement to mean P(t) → − as t → ; we could have P(t) → − as t → T, where T > 0 is a finite number that depends on the initial condition P(t₀) = P₀. Thinking in dynamic terms, P(t) could “blow up” in finite time; thinking graphically, P(t) could have a vertical asymptote at t = T > 0. A similar remark holds for the subregion R₃.

The differential equation dy/dx = sin y in Example 2 is autonomous and has an infinite number of critical points since sin y = 0 at y = nπ, n an integer. Moreover, we now know that because the solution y(x) that passes through (0, −) is bounded above and below by two consecutive critical points (−π < y(x) < 0) and is decreasing (sin y < 0 for −π < y < 0), the graph of y(x) must approach the graphs of the equilibrium solutions as horizontal asymptotes: y(x) → −π as x → and y(x) → 0 as x → −.

EXAMPLE 5 Solution Curves of an Autonomous DE

The autonomous equation dy/dx = (y − 1)² possesses the single critical point 1. From the phase portrait in FIGURE 2.1.7(a), we conclude that a solution y(x) is an increasing function in the subregions defined by − < y < 1 and 1 < y < , where − < x < . For an initial condition y(0) = y₀ < 1, a solution y(x) is increasing and bounded above by 1, and so y(x) → 1 as x → ; for y(0) = y₀ > 1, a solution y(x) is increasing and unbounded.

Now y(x) = 1 − 1/(x + c) is a one-parameter family of solutions of the differential equation. (See Problem 4 in Exercises 2.2.) A given initial condition determines a value for c. For the initial conditions, say, y(0) = − 1 < 1 and y(0) = 2 > 1, we find, in turn, that y(x) = 1 − 1/(x + ) and so y(x) = 1 − 1/(x − 1). As shown in Figure 2.1.7(b) and 2.1.7(c), the graph of each of these rational functions possesses a vertical asymptote. But bear in mind that the solutions of the IVPs

    and    

are defined on special intervals. The two solutions are, respectively,

The solution curves are the portions of the graphs in Figures 2.1.7(b) and 2.1.7(c) shown in blue. As predicted by the phase portrait, for the solution curve in Figure 2.1.7(b), y(x) → 1 as x → ; for the solution curve in Figure 2.1.7(c), y(x) → as x → 1 from the left.

Attractors and Repellers

Suppose y(x) is a nonconstant solution of the autonomous differential equation given in (1) and that c is a critical point of the DE. There are basically three types of behavior y(x) can exhibit near c. In FIGURE 2.1.8 we have placed c on four vertical phase lines. When both arrowheads on either side of the dot labeled c point toward c, as in Figure 2.1.8(a), all solutions y(x) of (1) that start from an initial point (x₀, y₀) sufficiently near c exhibit the asymptotic behavior lim_x→y(x) = c. For this reason the critical point c is said to be asymptotically stable. Using a physical analogy, a solution that starts near c is like a charged particle that, over time, is drawn to a particle of opposite charge, and so c is also referred to as an attractor. When both arrowheads on either side of the dot labeled c point away from c, as in Figure 2.1.8(b), all solutions y(x) of (1) that start from an initial point (x₀, y₀) move away from c as x increases. In this case the critical point c is said to be unstable. An unstable critical point is also called a repeller, for obvious reasons. The critical point c illustrated in Figures 2.1.8(c) and 2.1.8(d) is neither an attractor nor a repeller. But since c exhibits characteristics of both an attractor and a repeller—that is, a solution starting from an initial point (x₀, y₀) sufficiently near c is attracted to c from one side and repelled from the other side—we say that the critical point c is semi-stable. In Example 3, the critical point a/b is asymptotically stable (an attractor) and the critical point 0 is unstable (a repeller). The critical point 1 in Example 5 is semi-stable.

EXAMPLE 6 Classifying Critical Points

Locate and classify all critical points of .

SOLUTION

Rewriting the differential equation as

we see from y(2 − y)(2 + y) = 0 that y = 0, y = 2, and y = − 2 are critical points of the DE.

Now by examining, in turn, the algebraic signs of dy/dx on intervals of the y-axis determined by the critical points, we see from the phase portrait in FIGURE 2.1.9 that:

See Problems 21–28 in Exercises 2.1. ≡

Autonomous DEs and Direction Fields

If a first-order differential equation is autonomous, then we see from the right-hand side of its normal form dy/dx = f(y) that slopes of lineal elements through points in the rectangular grid used to construct a direction field for the DE depend solely on the y-coordinate of the points. Put another way, lineal elements passing through points on any horizontal line must all have the same slope and therefore are parallel; slopes of lineal elements along any vertical line will, of course, vary. These facts are apparent from inspection of the horizontal gray strip and vertical blue strip in FIGURE 2.1.10. The figure exhibits a direction field for the autonomous equation dy/dx = 2(y − 1). The red lineal elements in Figure 2.1.10 have zero slope because they lie along the graph of the equilibrium solution y = 1.

Translation Property

Recall from precalculus mathematics that the graph of a function y = f(x − k), where k is a constant, is the graph of y = f(x) rigidly translated or shifted horizontally along the x-axis by an amount |k|; the translation is to the right if k > 0 and to the left if k < 0.

It turns out that under the assumptions stated after equation (1), solution curves of an autonomous first-order DE are related by the concept of translation. To see this, let’s consider the differential equation dy/dx = y(3 − y), which is a special case of the autonomous equation considered in Examples 3 and 4. Since y = 0 and y = 3 are equilibrium solutions of the DE, their graphs divide the xy-plane into subregions R₁, R₂, and R₃, defined by the three inequalities:

R₁: − < y < 0,        R₂: 0 < y < 3,        R₃: 3 < y < .

In FIGURE 2.1.11 we have superimposed on a direction field of the DE six solution curves. The figure illustrates that all solution curves of the same color, that is, solution curves lying within a particular subregion R_i, all look alike. This is no coincidence but is a natural consequence of the fact that lineal elements passing through points on any horizontal line are parallel. That said, the following translation property of an autonomous DE should make sense:

Hence, if y(x) is a solution of the initial-value problem dy/dx = f(y), y(0) = y₀, then y₁(x) = y(x − x₀) is a solution of the IVP dy/dx = f(y), y(x₀) = y₀. For example, it is easy to verify that y(x) = e^x, − < x < , is a solution of the IVP dy/dx = y, y(0) = 1 and so a solution y₁(x) of, say, dy/dx = y, y(4) = 1 is y(x) = e^x translated 4 units to the right:

y₁(x) = y(x − 4) = e^x−4, − < x < .

2.1 Exercises Answers to selected odd-numbered problems begin on page ANS-2.

2.1.1 Direction Fields

In Problems 1–4, reproduce the given computer-generated direction field. Then sketch, by hand, an approximate solution curve that passes through each of the indicated points. Use different colored pencils for each solution curve.

1. 

   (a) y(−2) = 1

   (b) y(3) = 0

   (c) y(0) = 2

   (d) y(0) = 0

   

2. 

   (a) y(−6) = 0

   (b) y(0) = 1

   (c) y(0) = −4

   (d) y(8) = −4

   

3. 

   (a) y(0) = 0

   (b) y(−1) = 0

   (c) y(2) = 2

   (d) y(0) = −4

   

4. 

   (a) y(0) = 1

   (b) y(1) = 0

   (c) y(3) = 3

   (d) y(0) = −

   

In Problems 5–12, use computer software to obtain a direction field for the given differential equation. By hand, sketch an approximate solution curve passing through each of the given points.

5. y′ = x

   (a) y(0) = 0

   (b) y(0) = −3

6. y′ = x + y

   (a) y(−2) = 2

   (b) y(1) = −3

7. 

   (a) y(1) = 1

   (b) y(0) = 4

8. 

   (a) y(0) = 1

   (b) y(−2) = −1

9. 

   (a) y(0) =

   (b) y(2) = −1

10. 

    (a) y(0) = −2

    (b) y(1) = 2.5

11. 

    (a) y(2) = 2

    (b) y(−1) = 0

12. 

    (a)

    (b)

In Problems 13 and 14, the given figures represent the graph of f(y) and f(x), respectively. By hand, sketch a direction field over an appropriate grid for dy/dx = f(y) (Problem 13) and then for dy/dx = f(x) (Problem 14).

13. 

14. 

15. In parts (a) and (b) sketch isoclines f(x, y) = c (see the Remarks on page 36) for the given differential equation using the indicated values of c. Construct a direction field over a grid by carefully drawing lineal elements with the appropriate slope at chosen points on each isocline. In each case, use this rough direction field to sketch an approximate solution curve for the IVP consisting of the DE and the initial condition y(0) = 1.

    (a) dy/dx = x + y; c an integer satisfying −5 ≤ c ≤ 5

    (b) dy/dx = x² + y²; c = , c = 1, c = , c = 4

Discussion Problems

16. 1. Consider the direction field of the differential equation dy/dx = x(y − 4)² − 2, but do not use technology to obtain it. Describe the slopes of the lineal elements on the lines x = 0, y = 3, y = 4, and y = 5.
    2. Consider the IVP dy/dx = x(y − 4)² − 2, y(0) = y₀, where y₀ < 4. Can a solution y(x) → as x → ? Based on the information in part (a), discuss.
17. For a first-order DE dy/dx = f(x, y), a curve in the plane defined by f(x, y) = 0 is called a nullcline of the equation, since a lineal element at a point on the curve has zero slope. Use computer software to obtain a direction field over a rectangular grid of points for dy/dx = x² − 2y, and then superimpose the graph of the nullcline y = x² over the direction field. Discuss the behavior of solution curves in regions of the plane defined by y < x² and by y > x². Sketch some approximate solution curves. Try to generalize your observations.
18. 1. Identify the nullclines (see Problem 17) in Problems 1, 3, and 4. With a colored pencil, circle any lineal elements in FIGURES 2.1.12, 2.1.14, and 2.1.15 that you think may be a lineal element at a point on a nullcline.
    2. What are the nullclines of an autonomous first-order DE?

2.1.2 Autonomous First-Order DEs

19. Consider the autonomous first-order differential equation dy/dx = y − y³ and the initial condition y(0) = y₀. By hand, sketch the graph of a typical solution y(x) when y₀ has the given values.

    (a) y₀ > 1

    (b) 0 < y₀ < 1

    (c) −1 < y₀ < 0

    (d) y₀ < −1

20. Consider the autonomous first-order differential equation dy/dx = y² − y⁴ and the initial condition y(0) = y₀. By hand, sketch the graph of a typical solution y(x) when y₀ has the given values.

    (a) y₀ > 1

    (b) 0 < y₀ < 1

    (c) −1 < y₀ < 0

    (d) y₀ < −1

In Problems 21–28, find the critical points and phase portrait of the given autonomous first-order differential equation. Classify each critical point as asymptotically stable, unstable, or semi-stable. By hand, sketch typical solution curves in the regions in the xy-plane determined by the graphs of the equilibrium solutions.

In Problems 29 and 30, consider the autonomous differential equation dy/dx = f(y), where the graph of f is given. Use the graph to locate the critical points of each differential equation. Sketch a phase portrait of each differential equation. By hand, sketch typical solution curves in the subregions in the xy-plane determined by the graphs of the equilibrium solutions.

Discussion Problems

31. Consider the autonomous DE dy/dx = (2/π)y − sin y. Determine the critical points of the equation. Discuss a way of obtaining a phase portrait of the equation. Classify the critical points as asymptotically stable, unstable, or semi-stable.
32. A critical point c of an autonomous first-order DE is said to be isolated if there exists some open interval that contains c but no other critical point. Discuss: Can there exist an autonomous DE of the form given in (1) for which every critical point is nonisolated? Do not think profound thoughts.
33. Suppose that y(x) is a nonconstant solution of the autonomous equation dy/dx = f(y) and that c is a critical point of the DE. Discuss: Why can’t the graph of y(x) cross the graph of the equilibrium solution y = c? Why can’t f(y) change signs in one of the subregions discussed on page 38? Why can’t y(x) be oscillatory or have a relative extremum (maximum or minimum)?
34. Suppose that y(x) is a solution of the autonomous equation dy/dx = f(y) and is bounded above and below by two consecutive critical points c₁ < c₂, as in subregion R₂ of Figure 2.1.5(b). If f(y) > 0 in the region, then lim_x→y(x) = c₂. Discuss why there cannot exist a number L < c₂ such that lim_x→y(x) = L. As part of your discussion, consider what happens to y′(x) as x → .
35. Using the autonomous equation (1), discuss how it is possible to obtain information about the location of points of inflection of a solution curve.
36. Consider the autonomous DE dy/dx = y² − y − 6. Use your ideas from Problem 35 to find intervals on the y-axis for which solution curves are concave up and intervals for which solution curves are concave down. Discuss why each solution curve of an initial-value problem of the form dy/dx = y² − y − 6, y(0) = y₀, where −2 < y₀ < 3, has a point of inflection with the same y-coordinate. What is that y-coordinate? Carefully sketch the solution curve for which y(0) = −1. Repeat for y(2) = 2.
37. Suppose the autonomous DE in (1) has no critical points. Discuss the behavior of the solutions.

Mathematical Models

38. Population Model      The differential equation in Example 3 is a well-known population model. Suppose the DE is changed to

    ,

    where a and b are positive constants. Discuss what happens to the population P as time t increases.

39. Population Model      Another population model is given by

    

    where h > 0 and k > 0 are constants. For what initial values P(0) = P₀ does this model predict that the population will go extinct?

40. Terminal Velocity      The autonomous differential equation

    

    where k is a positive constant of proportionality called the drag coefficient and g is the acceleration due to gravity, is a model for the instantaneous velocity v of a body of mass m that is falling under the influence of gravity. Because the term −kv represents air resistance or drag, the velocity of a body falling from a great height does not increase without bound as time t increases. Use a phase portrait of the differential equation to find the limiting, or terminal, velocity of the body. Explain your reasoning. See page 24.

41. Terminal Velocity      In Problem 17 of Exercises 1.3, we indicated that for high-speed motion of a body, air resistance is taken to be proportional to a power of its instantaneous velocity v. If we take air resistance to be proportional to v², then the mathematical model for the instantaneous velocity of a falling body of mass m in Problem 40 becomes

    

    where k > 0. Use a phase portrait to find the terminal velocity of the body. Explain your reasoning. See page 27.

42. Chemical Reactions      When certain kinds of chemicals are combined, the rate at which a new compound is formed is governed by the differential equation

    

    where k > 0 is a constant of proportionality and β > α > 0. Here X(t) denotes the number of grams of the new compound formed in time t. See pages 21 and 22.

    (a) Use a phase portrait of the differential equation to predict the behavior of X as t → .

    (b) Consider the case when α = β. Use a phase portrait of the differential equation to predict the behavior of X as t → when X(0) < α. When X(0) > α.

    (c) Verify that an explicit solution of the DE in the case when k = 1 and α = β is X(t) = α − 1/(t + c). Find a solution satisfying X(0) = α/2. Find a solution satisfying X(0) = 2α. Graph these two solutions. Does the behavior of the solutions as t → agree with your answers to part (b)?