88 Chapter 2. First Order Differential Equations. Problems 1 through 6 involve equations of the form dy/dt = f (y). In each problem sketch

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1 88 Chapter 2. First Order Differential Equations place to permit successful breeding, and the population rapidl declined to extinction. The last survivor died in The precipitous decline in the passenger pigeon population from huge numbers to extinction in a few decades was one of the earl factors contributing to a concern for conservation in this countr. Problems 1 through 6 involve s of the form d/dt = f (). In each problem sketch the graph of f () versus, determine the critical (equilibrium) points, and classif each one as asmptoticall stable or unstable. Draw the phase line, and sketch several graphs of solutions in the t-plane. 1. d/dt = a + b 2, a > 0, b > 0, d/dt = a + b 2, a > 0, b > 0, < 0 < 3. d/dt = ( 1)( 2), d/dt = e 1, < 0 < 5. d/dt = e 1, < 0 < 6. d/dt = 2(arctan )/(1 + 2 ), < 0 < 7. Semistable Equilibrium Solutions. Sometimes a constant equilibrium solution has the propert that solutions ling on one side of the equilibrium solution tend to approach it, whereas solutions ling on the other side depart from it (see Figure 2.5.9). In this case the equilibrium solution is said to be semistable. φ(t) = k k k φ(t) = k t t (a) (b) FIGURE In both cases the equilibrium solution φ(t) = k is semistable. (a) d/dt 0; (b) d/dt 0. (a) Consider the d/dt = k(1 ) 2, (i) where k is a positive constant. Show that = 1 is the onl critical point, with the corresponding equilibrium solution φ(t) = 1. (b) Sketch f () versus. Show that is increasing as a function of t for < 1 and also for > 1. The phase line has upward-pointing arrows both below and above = 1. Thus solutions below the equilibrium solution approach it, and those above it grow farther awa. Therefore φ(t) = 1 is semistable. (c) Solve Eq. (i) subject to the initial condition (0) = 0 and confirm the conclusions reached in part (b). Problems 8 through 13 involve s of the form d/dt = f (). In each problem sketch the graph of f () versus, determine the critical (equilibrium) points, and classif each one

2 2.5 Autonomous Equations and Population Dnamics 89 as asmptoticall stable, unstable, or semistable (see Problem 7). Draw the phase line, and sketch several graphs of solutions in the t-plane. 8. d/dt = k( 1) 2, k > 0, < 0 < 9. d/dt = 2 ( 2 1), < 0 < 10. d/dt = (1 2 ), < 0 < 11. d/dt = a b, a > 0, b > 0, d/dt = 2 (4 2 ), < 0 < 13. d/dt = 2 (1 ) 2, < 0 < 14. Consider the d/dt = f () and suppose that 1 is a critical point, that is, f ( 1 ) = 0. Show that the constant equilibrium solution φ(t) = 1 is asmptoticall stable if f ( 1 )<0 and unstable if f ( 1 )> Suppose that a certain population obes the logistic d/dt = r[1 (/K)]. (a) If 0 = K/3, find the time τ at which the initial population has doubled. Find the value of τ corresponding to r = per ear. (b) If 0 /K = α, find the time T at which (T)/K = β, where 0 < α, β < 1. Observe that T as α 0orasβ 1. Find the value of T for r = per ear, α = 0.1, and β = Another that has been used to model population growth is the Gompertz 14 d/dt = r ln(k/), where r and K are positive constants. (a) Sketch the graph of f () versus, find the critical points, and determine whether each is asmptoticall stable or unstable. (b) For 0 K, determine where the graph of versus t is concave up and where it is concave down. (c) For each in 0 < K, show that d/dt as given b the Gompertz is never less than d/dt as given b the logistic. 17. (a) Solve the Gompertz d/dt = r ln(k/), subject to the initial condition (0) = 0. Hint: You ma wish to let u = ln(/k). (b) For the data given in Example 1 in the text (r = 0.71 per ear, K = kg, 0 /K = 0.25), use the Gompertz model to find the predicted value of (2). (c) For the same data as in part (b), use the Gompertz model to find the time τ at which (τ) = 0.75K. 18. A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k and is lost through evaporation at a rate proportional to the surface area. (a) Show that the volume V(t) of water in the pond at time t satisfies the differential dv/dt = k απ(3a/πh) 2/3 V 2/3, where α is the coefficient of evaporation. 14 Benjamin Gompertz ( ) was an English actuar. He developed his model for population growth, published in 1825, in the course of constructing mortalit tables for his insurance compan.

3 2.6 Exact Equations and Integrating Factors 99 Thus, if (µm) is to equal (µn) x, it is necessar that dµ dx = M N x µ. (27) N If (M N x )/N is a function of x onl, then there is an integrating factor µ that also depends onl on x; further, µ(x) can be found b solving Eq. (27), which is both linear and separable. A similar procedure can be used to determine a condition under which Eq. (23) has an integrating factor depending onl on ; see Problem 23. EXAMPLE 4 Find an integrating factor for the (3x + 2 ) + (x 2 + x) = 0 (19) and then solve the. In Example 3 we showed that this is not exact. Let us determine whether it has an integrating factor that depends on x onl. On computing the quantit (M N x )/N, we find that M (x, ) N x (x, ) N(x, ) = 3x + 2 (2x + ) x 2 + x = 1 x. (28) Thus there is an integrating factor µ that is a function of x onl, and it satisfies the differential dµ dx = µ x. (29) Hence µ(x) = x. (30) Multipling Eq. (19) b this integrating factor, we obtain (3x 2 + x 2 ) + (x 3 + x 2 ) = 0. (31) The latter is exact, and its solutions are given implicitl b x x2 2 = c. (32) Solutions ma also be found in explicit form since Eq. (32) is quadratic in. You ma also verif that a second integrating factor for Eq. (19) is µ(x, ) = 1 x(2x + ), and that the same solution is obtained, though with much greater difficult, if this integrating factor is used (see Problem 32). Determine whether each of the s in Problems 1 through 12 is exact. If it is exact, find the solution. 1. (2x + 3) + (2 2) = 0 2. (2x + 4) + (2x 2) = 0 3. (3x 2 2x + 2) dx + (6 2 x 2 + 3) d = 0 4. (2x 2 + 2) + (2x 2 + 2x) = 0

4 100 Chapter 2. First Order Differential Equations 5. d ax + b d ax b = 6. = dx bx + c dx bx c 7. (e x sin 2 sin x) dx + (e x cos + 2 cos x) d = 0 8. (e x sin + 3) dx (3x e x sin ) d = 0 9. (e x cos 2x 2e x sin 2x + 2x) dx + (xe x cos 2x 3) d = (/x + 6x) dx + (ln x 2) d = 0, x > (x ln + x) dx + ( ln x + x) d = 0; x > 0, > xdx (x ) + d 3/2 (x ) = 0 3/2 In each of Problems 13 and 14 solve the given initial value problem and determine at least approximatel where the solution is valid. 13. (2x ) dx + (2 x) d = 0, (1) = (9x 2 + 1) dx (4 x) d = 0, (1) = 0 In each of Problems 15 and 16 find the value of b for which the given is exact, and then solve it using that value of b. 15. (x 2 + bx 2 ) dx + (x + )x 2 d = (e 2x + x) dx + bxe 2x d = Assume that Eq. (6) meets the requirements of Theorem in a rectangle R and is therefore exact. Show that a possible function ψ(x, ) is ψ(x, ) = where (x 0, 0 ) is a point in R. 18. Show that an separable is also exact. x x 0 M(s, 0 ) ds + M(x) + N() = 0 0 N(x, t) dt, In each of Problems 19 through 22 show that the given is not exact but becomes exact when multiplied b the given integrating factor. Then solve the. 19. x x(1 + 2 ) = 0, µ(x, ) = 1/x 3 ( ) ( ) sin cos + 2e x cos x 20. 2e x sin x dx + d = 0, µ(x, ) = e x 21. dx+ (2x e ) d = 0, µ(x, ) = 22. (x + 2) sin dx+ x cos d= 0, µ(x, ) = xe x 23. Show that if (N x M )/M = Q, where Q is a function of onl, then the differential M + N = 0 has an integrating factor of the form µ() = exp Q() d. 24. Show that if (N x M )/(xm N) = R, where R depends on the quantit x onl, then the differential M + N = 0 has an integrating factor of the form µ(x). Find a general formula for this integrating factor.

5 144 Chapter 3. Second Order Linear Equations In each of Problems 1 through 8 find the general solution of the given differential = = = = = = = = 0 In each of Problems 9 through 16 find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior as t increases = 0, (0) = 1, (0) = = 0, (0) = 2, (0) = = 0, (0) = 4, (0) = = 0, (0) = 2, (0) = = 0, (0) = 1, (0) = = 0, (0) = 0, (0) = = 0, (1) = 1, (1) = = 0, ( 2) = 1, ( 2) = Find a differential whose general solution is = c 1 e 2t + c 2 e 3t. 18. Find a differential whose general solution is = c 1 e t/2 + c 2 e 2t. 19. Find the solution of the initial value problem = 0, (0) = 5 4, (0) = 3 4. Plot the solution for 0 t 2 and determine its minimum value. 20. Find the solution of the initial value problem = 0, (0) = 2, (0) = 1 2. Then determine the maximum value of the solution and also find the point where the solution is zero. 21. Solve the initial value problem 2 = 0, (0) = α, (0) = 2. Then find α so that the solution approaches zero as t. 22. Solve the initial value problem 4 = 0, (0) = 2, (0) = β. Then find β so that the solution approaches zero as t. In each of Problems 23 and 24 determine the values of α, if an, for which all solutions tend to zero as t ; also determine the values of α, if an, for which all (nonzero) solutions become unbounded as t. 23. (2α 1) + α(α 1) = (3 α) 2(α 1) = Consider the initial value problem = 0, (0) = 1, (0) = β, where β > 0. (a) Solve the initial value problem. (b) Plot the solution when β = 1. Find the coordinates (t 0, 0 ) of the minimum point of the solution in this case. (c) Find the smallest value of β for which the solution has no minimum point.