AEM_3e_Chapter02

AEM_3e_Chapter02 - 2 First-Order Differential Equations...

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Unformatted text preview: 2 First-Order Differential Equations EXERCISES 2.1 Solution Curves Without the Solution y 3 1. 2. y 10 2 5 1 -3 -2 -1 1 2 x 0 x 3 -1 -5 -2 -10 -3 3. y -5 10 y 4. 4 5 0 4 2 2 x 0 x 0 -2 -2 -4 -4 5. -2 0 2 4 -4 y 6. 4 2 0 2 4 y 4 2 x 0 x 0 -2 -4 -2 -2 -2 0 2 4 -4 22 -2 0 2 4 2.1 7. y 8. 4 Solution Curves Without the Solution y 4 2 2 x 0 x 0 -2 -2 -4 -4 9. -2 0 2 4 -4 y 10. 4 2 x 2 4 y 4 x 0 -2 -2 -4 -2 0 2 4 -4 y 12. 4 2 -2 0 2 4 y 4 2 x 0 x 0 -2 -2 -4 13. 0 2 0 11. -2 -2 0 2 4 -4 y -2 2 4 y 14. 3 0 4 2 2 1 x 0 x 0 -1 -2 -2 -4 -3 -3 -2 -1 0 1 2 3 -4 23 -2 0 2 4 2.1 Solution Curves Without the Solution 15. (a) The isoclines have the form y = −x + c, which are straight y 3 lines with slope −1. 2 1 -3 -2 -1 1 3 x 2 -1 -2 -3 (b) The isoclines have the form x2 + y 2 = c, which are circles centered at the origin. y 2 1 -2 -1 1 x 2 -1 -2 16. (a) When x = 0 or y = 4, dy/dx = −2 so the lineal elements have slope −2. When y = 3 or y = 5, dy/dx = x−2, so the lineal elements at (x, 3) and (x, 5) have slopes x − 2. (b) At (0, y0 ) the solution curve is headed down. If y → ∞ as x increases, the graph must eventually turn around and head up, but while heading up it can never cross y = 4 where a tangent line to a solution curve must have slope −2. Thus, y cannot approach ∞ as x approaches ∞. y = x2 − 2y is positive and the portions of solution curves “outside” the nullcline parabola are increasing. When y > 1 x2 , 2 17. When y < 1 2 2x , y = x2 − 2y is negative and the portions of the solution curves “inside” the nullcline parabola are decreasing. y 3 2 1 x 0 -1 -2 -3 -3 -2 -1 0 1 2 3 18. (a) Any horizontal lineal element should be at a point on a nullcline. In Problem 1 the nullclines are x2 −y 2 = 0 or y = ±x. In Problem 3 the nullclines are 1 − xy = 0 or y = 1/x. In Problem 4 the nullclines are (sin x) cos y = 0 or x = nπ and y = π/2 + nπ, where n is an integer. The graphs on the next page show the nullclines for the differential equations in Problems 1, 3, and 4 superimposed on the corresponding direction field. 24 2.1 y Solution Curves Without the Solution y y 4 3 4 2 2 2 1 x 0 x 0 x 0 -1 -2 -2 -2 -4 -3 -3 -2 -1 0 1 2 Problem 1 -4 -4 3 4 -2 0 2 Problem 3 -4 -2 0 2 Problem 4 4 (b) An autonomous first-order differential equation has the form y = f (y). Nullclines have the form y = c where f (c) = 0. These are the graphs of the equilibrium solutions of the differential equation. 19. Writing the differential equation in the form dy/dx = y(1 − y)(1 + y) we see that critical points are located at y = −1, y = 0, and y = 1. The phase portrait is shown at the right. 1 y (a) y (b) 5 4 0 1 3 -1 2 1 -2 1 2 y (c) -1 1 (d) -1 1 2 x y 1 -2 2 x x 2 x -1 -2 -3 -4 -1 -5 20. Writing the differential equation in the form dy/dx = y 2 (1 − y)(1 + y) we see that critical points are located at y = −1, y = 0, and y = 1. The phase portrait is shown at the right. y (a) 1 y (b) 5 4 0 1 3 2 -1 1 -2 1 y 2 -1 1 2 x y (c) (d) -2 -2 -1 1 2 x x -1 -1 -2 -3 -4 -1 -5 25 x 2.1 Solution Curves Without the Solution 21. Solving y 2 − 3y = y(y − 3) = 0 we obtain the critical points 0 and 3. From the phase portrait we see that 0 is asymptotically stable (attractor) and 3 is unstable (repeller). 3 0 22. Solving y 2 − y 3 = y 2 (1 − y) = 0 we obtain the critical points 0 and 1. From the phase portrait we see that 1 is asymptotically stable (attractor) and 0 is semi-stable. 1 0 23. Solving (y − 2)4 = 0 we obtain the critical point 2. From the phase portrait we see that 2 is semi-stable. 2 24. Solving 10 + 3y − y 2 = (5 − y)(2 + y) = 0 we obtain the critical points −2 and 5. From the phase portrait we see that 5 is asymptotically stable (attractor) and −2 is unstable (repeller). 5 -2 26 2.1 Solution Curves Without the Solution 25. Solving y 2 (4 − y 2 ) = y 2 (2 − y)(2 + y) = 0 we obtain the critical points −2, 0, and 2. From the phase portrait we see that 2 is asymptotically stable (attractor), 0 is semi-stable, and −2 is unstable (repeller). 2 0 -2 26. Solving y(2 − y)(4 − y) = 0 we obtain the critical points 0, 2, and 4. From the phase portrait we see that 2 is asymptotically stable (attractor) and 0 and 4 are unstable (repellers). 4 2 0 27. Solving y ln(y + 2) = 0 we obtain the critical points −1 and 0. From the phase portrait we see that −1 is asymptotically stable (attractor) and 0 is unstable (repeller). 0 -1 -2 28. Solving yey − 9y = y(ey − 9) = 0 we obtain the critical points 0 and ln 9. From the phase portrait we see that 0 is asymptotically stable (attractor) and ln 9 is unstable (repeller). ln 9 0 29. The critical points are 0 and c because the graph of f (y) is 0 at these points. Since f (y) > 0 for y < 0 and y > c, the graph of the solution is increasing on (−∞, 0) and (c, ∞). Since f (y) < 0 for 0 < y < c, the graph of the solution is decreasing on (0, c). 27 2.1 Solution Curves Without the Solution y c c x 0 30. The critical points are approximately at −2, 2, 0.5, and 1.7. Since f (y) > 0 for y < −2.2 and 0.5 < y < 1.7, the graph of the solution is increasing on (−∞, −2.2) and (0.5, 1.7). Since f (y) < 0 for −2.2 < y < 0.5 and y > 1.7, the graph is decreasing on (−2.2, 0.5) and (1.7, ∞). y 2 1.7 1 0.5 -2 -1 1 2 x -1 -2 -2.2 31. From the graphs of z = π/2 and z = sin y we see that (π/2)y − sin y = 0 has only three solutions. By inspection 1 we see that the critical points are −π/2, 0, and π/2. Π From the graph at the right we see that 2 y − sin y π 2 y − sin y π Π 2 Π 2 y Π -1 <0 >0 for y < −π/2 for y > π/2 > 0 for < 0 for − π/2 < y < 0 0 < y < π/2. Π 2 0 Π 2 This enables us to construct the phase portrait shown at the right. From this portrait we see that π/2 and −π/2 are unstable (repellers), and 0 is asymptotically stable (attractor). 32. For dy/dx = 0 every real number is a critical point, and hence all critical points are nonisolated. 33. Recall that for dy/dx = f (y) we are assuming that f and f are continuous functions of y on some interval I. Now suppose that the graph of a nonconstant solution of the differential equation crosses the line y = c. If the point of intersection is taken as an initial condition we have two distinct solutions of the initial-value problem. This violates uniqueness, so the graph of any nonconstant solution must lie entirely on one side of any equilibrium solution. Since f is continuous it can only change signs at a point where it is 0. But this is a critical point. Thus, f (y) is completely positive or completely negative in each region Ri . If y(x) is oscillatory 28 2.1 Solution Curves Without the Solution or has a relative extremum, then it must have a horizontal tangent line at some point (x0 , y0 ). In this case y0 would be a critical point of the differential equation, but we saw above that the graph of a nonconstant solution cannot intersect the graph of the equilibrium solution y = y0 . 34. By Problem 33, a solution y(x) of dy/dx = f (y) cannot have relative extrema and hence must be monotone. Since y (x) = f (y) > 0, y(x) is monotone increasing, and since y(x) is bounded above by c2 , limx→∞ y(x) = L, where L ≤ c2 . We want to show that L = c2 . Since L is a horizontal asymptote of y(x), limx→∞ y (x) = 0. Using the fact that f (y) is continuous we have f (L) = f ( lim y(x)) = lim f (y(x)) = lim y (x) = 0. x→∞ x→∞ x→∞ But then L is a critical point of f . Since c1 < L ≤ c2 , and f has no critical points between c1 and c2 , L = c2 . 35. Assuming the existence of the second derivative, points of inflection of y(x) occur where y (x) = 0. From dy/dx = f (y) we have d2 y/dx2 = f (y) dy/dx. Thus, the y-coordinate of a point of inflection can be located by solving f (y) = 0. (Points where dy/dx = 0 correspond to constant solutions of the differential equation.) 36. Solving y 2 − y − 6 = (y − 3)(y + 2) = 0 we see that 3 and −2 are critical points. Now d2 y/dx2 = (2y −1) dy/dx = (2y −1)(y −3)(y +2), so the only possible point y 5 of inflection is at y = 1 , although the concavity of solutions can be different on 2 either side of y = −2 and y = 3. Since y (x) < 0 for y < −2 and 1 < y < 3, 2 and y (x) > 0 for −2 < y < 1 and y > 3, we see that solution curves are 2 concave down for y < −2 and 1 < y < 3 and concave up for −2 < y < 1 and 2 2 y > 3. Points of inflection of solutions of autonomous differential equations will -5 have the same y-coordinates because between critical points they are horizontal translates of each other. 5 x -5 37. If (1) in the text has no critical points it has no constant solutions. The solutions have neither an upper nor lower bound. Since solutions are monotonic, every solution assumes all real values. 38. The critical points are 0 and b/a. From the phase portrait we see that 0 is an attractor and b/a is a repeller. Thus, if an initial population satisfies P0 > b/a, the population becomes unbounded as t increases, most probably in finite time, i.e. P (t) → ∞ as t → T . If 0 < P0 < b/a, then the population eventually dies out, that is, P (t) → 0 as t → ∞. Since population P > 0 we do not consider the case P0 < 0. b a 0 39. (a) Writing the differential equation in the form dv k = dt m mg −v k we see that a critical point is mg/k. From the phase portrait we see that mg/k is an asymptotically stable critical point. Thus, limt→∞ v = mg/k. 29 mg k 2.1 Solution Curves Without the Solution (b) Writing the differential equation in the form dv k = dt m k mg − v2 = k m mg −v k mg +v k we see that the only physically meaningful critical point is From the phase portrait we see that critical point. Thus, limt→∞ v = mg/k. mg/k. mg/k is an asymptotically stable mg k 40. (a) From the phase portrait we see that critical points are α and β. Let X(0) = X0 . If X0 < α, we see that X → α as t → ∞. If α < X0 < β, we see that X → α as t → ∞. If X0 > β, we see that X(t) increases in an unbounded manner, but more specific behavior of X(t) as t → ∞ is Β not known. Α (b) When α = β the phase portrait is as shown. If X0 < α, then X(t) → α as t → ∞. If X0 > α, then X(t) increases in an unbounded manner. This could happen in a finite amount of time. That is, the phase portrait does not indicate that X becomes unbounded as t → ∞. Α (c) When k = 1 and α = β the differential equation is dX/dt = (α − X)2 . For X(t) = α − 1/(t + c) we have dX/dt = 1/(t + c)2 and (α − X)2 = α − α − 1 t+c 2 = 1 dX = . 2 (t + c) dt For X(0) = α/2 we obtain X(t) = α − 1 . t + 2/α X(t) = α − 1 . t − 1/α For X(0) = 2α we obtain 30 2.2 X X 2α α α −2/α Separable Variables α/2 t 1/α t For X0 > α, X(t) increases without bound up to t = 1/α. For t > 1/α, X(t) increases but X → α as t→∞ EXERCISES 2.2 Separable Variables In many of the following problems we will encounter an expression of the form ln |g(y)| = f (x) + c. To solve for g(y) we exponentiate both sides of the equation. This yields |g(y)| = ef (x)+c = ec ef (x) which implies g(y) = ±ec ef (x) . Letting c1 = ±ec we obtain g(y) = c1 ef (x) . 1. From dy = sin 5x dx we obtain y = − 1 cos 5x + c. 5 2. From dy = (x + 1)2 dx we obtain y = 1 (x + 1)3 + c. 3 3. From dy = −e−3x dx we obtain y = 1 e−3x + c. 3 1 1 1 = x + c or y = 1 − . dy = dx we obtain − (y − 1)2 y−1 x+c 1 4 5. From dy = dx we obtain ln |y| = 4 ln |x| + c or y = c1 x4 . y x 1 1 1 6. From 2 dy = −2x dx we obtain − = −x2 + c or y = 2 . y y x + c1 4. From 7. From e−2y dy = e3x dx we obtain 3e−2y + 2e3x = c. 1 8. From yey dy = e−x + e−3x dx we obtain yey − ey + e−x + e−3x = c. 3 1 y2 x3 1 9. From y + 2 + dy = x2 ln x dx we obtain + 2y + ln |y| = ln |x| − x3 + c. y 2 3 9 1 1 1 2 10. From = + c. dy = dx we obtain (2y + 3)2 (4x + 5)2 2y + 3 4x + 5 1 1 11. From dy = − 2 dx or sin y dy = − cos2 x dx = − 1 (1 + cos 2x) dx we obtain 2 csc y sec x 1 1 − cos y = − 2 x − 4 sin 2x + c or 4 cos y = 2x + sin 2x + c1 . 12. From 2y dy = − sin 3x dx or 2y dy = − tan 3x sec2 3x dx we obtain y 2 = − 1 sec2 3x + c. 6 cos3 3x 31 2.2 Separable Variables 13. From 14. From ey 2 (ey + 1) y 1/2 (1 + y 2 ) −ex −1 −2 y = 1 (ex + 1) 3 dx we obtain − (e + 1) 2 (ex + 1) x 1/2 dy = dx we obtain 1 + y 2 = 1 + x2 1/2 (1 + x2 ) dy = + c. 1/2 + c. 1 dS = k dr we obtain S = cekr . S 1 16. From dQ = k dt we obtain ln |Q − 70| = kt + c or Q − 70 = c1 ekt . Q − 70 15. From 17. From 1 dP = P − P2 1 1 + P 1−P dP = dt we obtain ln |P | − ln |1 − P | = t + c so that ln P = t + c or 1−P P c1 et . = c1 et . Solving for P we have P = 1−P 1 + c1 et 1 t+2 t+2 18. From dN = tet+2 − 1 dt we obtain ln |N | = tet+2 − et+2 − t + c or N = c1 ete −e −t . N y−2 x−1 5 5 19. From dy = dx or 1 − dy = 1 − dx we obtain y − 5 ln |y + 3| = x − 5 ln |x + 4| + c y+3 x+4 y+3 x+4 5 x+4 or = c1 ex−y . y+3 y+1 x+2 dy = dx or y−1 x−3 (y − 1)2 or = c1 ex−y . (x − 3)5 20. From 21. From x dx = 22. From 1 1 − y2 1+ 2 y−1 dy = 1+ 5 x−3 dx we obtain y + 2 ln |y − 1| = x + 5 ln |x − 3| + c dy we obtain 1 x2 = sin−1 y + c or y = sin 2 x2 + c1 . 2 1 1 1 ex 1 . dx we obtain − = tan−1 ex + c or y = − dy = x dx = x 2 y2 e + e−x (e ) + 1 y tan−1 ex + c 1 dx = 4 dt we obtain tan−1 x = 4t + c. Using x(π/4) = 1 we find c = −3π/4. The solution of the +1 3π 3π initial-value problem is tan−1 x = 4t − or x = tan 4t − . 4 4 23. From x2 1 1 1 dy = 2 dx or y2 − 1 x −1 2 1 1 − dx we obtain x−1 x+1 y−1 c(x − 1) ln |y − 1| − ln |y + 1| = ln |x − 1| − ln |x + 1| + ln c or = . Using y(2) = 2 we find y+1 x+1 24. From 1 1 − y−1 y+1 c = 1. A solution of the initial-value problem is dy = 1 2 y−1 x−1 = or y = x. y+1 x+1 1 1 1 1−x 1 dx = − dy = dx we obtain ln |y| = − − ln |x| = c or xy = c1 e−1/x . Using y(−1) = −1 2 2 y x x x x we find c1 = e−1 . The solution of the initial-value problem is xy = e−1−1/x or y = e−(1+1/x) /x. 25. From 1 dy = dt we obtain − 1 ln |1 − 2y| = t + c or 1 − 2y = c1 e−2t . Using y(0) = 5/2 we find c1 = −4. 2 1 − 2y The solution of the initial-value problem is 1 − 2y = −4e−2t or y = 2e−2t + 1 . 2 26. From 27. Separating variables and integrating we obtain √ dx − 1 − x2 dy 1 − y2 =0 and 32 sin−1 x − sin−1 y = c. 2.2 sin 28. From −1 x − sin √ 3/2 we obtain c = −π/3. Thus, an implicit solution of the initial-value problem is y = π/3. Solving for y and using an addition formula from trigonometry, we get √ √ 3 1 − x2 π π π x y = sin sin−1 x + = x cos + 1 − x2 sin = + . 3 3 3 2 2 Setting x = 0 and y = −1 Separable Variables −x 1 dy = 2 dx we obtain 2 1 + (2y) 1 + (x2 ) 1 1 tan−1 2y = − tan−1 x2 + c or 2 2 tan−1 2y + tan−1 x2 = c1 . Using y(1) = 0 we find c1 = π/4. Thus, an implicit solution of the initial-value problem is −1 −1 2 tan 2y + tan x = π/4 . Solving for y and using a trigonometric identity we get π 2y = tan − tan−1 x2 4 1 π y = tan − tan−1 x2 2 4 1 tan π − tan(tan−1 x2 ) 4 = 2 1 + tan π tan(tan−1 x2 ) 4 = 1 1 − x2 . 2 1 + x2 29. (a) The equilibrium solutions y(x) = 2 and y(x) = −2 satisfy the initial conditions y(0) = 2 and y(0) = −2, respectively. Setting x = 1 and y = 1 in y = 2(1 + ce4x )/(1 − ce4x ) we obtain 4 1=2 1 + ce , 1 − ce 1 − ce = 2 + 2ce, −1 = 3ce, and c = − 1 . 3e The solution of the corresponding initial-value problem is y=2 1 − 1 e4x−1 3 − e4x−1 3 1 4x−1 = 2 3 + e4x−1 . 1 + 3e (b) Separating variables and integrating yields 1 1 ln |y − 2| − ln |y + 2| + ln c1 = x 4 4 ln |y − 2| − ln |y + 2| + ln c = 4x ln c(y − 2) = 4x y+2 y−2 c = e4x . y+2 Solving for y we get y = 2(c + e4x )/(c − e4x ). The initial condition y(0) = −2 implies 2(c + 1)/(c − 1) = −2 which yields c = 0 and y(x) = −2. The initial condition y(0) = 2 does not correspond to a value of c, and it must simply be recognized that y(x) = 2 is a solution of the initial-value problem. Setting x = 1 and y = 1 in y = 2(c + e4x )/(c − e4x ) leads to c = −3e. Thus, a solution of the 4 initial-value problem is −3e + e4x 3 − e4x−1 y=2 =2 . −3e − e4x 3 + e4x−1 30. Separating variables, we have dy dx = y2 − y x dy = ln |x| + c. y(y − 1) or 33 2.2 Separable Variables Using partial fractions, we obtain 1 1 − y−1 y dy = ln |x| + c ln |y − 1| − ln |y| = ln |x| + c ln y−1 =c xy y−1 = ec = c1 . xy Solving for y we get y = 1/(1 − c1 x). We note by inspection that y = 0 is a singular solution of the differential equation. (a) Setting x = 0 and y = 1 we have 1 = 1/(1 − 0), which is true for all values of c1 . Thus, solutions passing through (0, 1) are y = 1/(1 − c1 x). (b) Setting x = 0 and y = 0 in y = 1/(1 − c1 x) we get 0 = 1. Thus, the only solution passing through (0, 0) is y = 0. (c) Setting x = 1 2 and y = (d) Setting x = 2 and y = 1 2 1 4 we have we have 1 4 1 2 = 1/(1 − 1 2 c1 ), so c1 = −2 and y = 1/(1 + 2x). = 1/(1 − 2c1 ), so c1 = − 3 and y = 1/(1 + 2 3 2 x) = 2/(2 + 3x). 31. Singular solutions of dy/dx = x 1 − y 2 are y = −1 and y = 1. A singular solution of (ex + e−x )dy/dx = y 2 is y = 0. 32. Differentiating ln(x2 + 10) + csc y = c we get 2x dy − csc y cot y = 0, x2 + 10 dx or 2x 1 cos y dy − · = 0, x2 + 10 sin y sin y dx 2x sin2 y dx − (x2 + 10) cos y dy = 0. Writing the differential equation in the form dy 2x sin2 y = 2 dx (x + 10) cos y we see that singular solutions occur when sin2 y = 0, or y = kπ, where k is an integer. 33. The singular solution y = 1 satisfies the initial-value problem. y 1.01 1 -0.004-0.002 0.98 0.97 34 0.002 0.004 x 2.2 34. Separating variables we obtain − dy = dx. Then (y − 1)2 Separable Variables y 1.02 1 x+c−1 = x + c and y = . y−1 x+c 1.01 Setting x = 0 and y = 1.01 we obtain c = −100. The solution is y= x − 101 . x − 100 -0.004-0.002 0.002 0.004 x 0.99 0.98 dy 35. Separating variables we obtain = dx. Then (y − 1)2 + 0.01 y 1.0004 1 x+c tan . 10 10 Setting x = 0 and y = 1 we obtain c = 0. The solution is 10 tan−1 10(y − 1) = x + c and y = 1 + y =1+ 1 x tan . 10 10 1.0002 -0.004-0.002 0.002 0.004 x 0.9998 0.9996 dy = dx. Then, from (11) in (y − 1)2 − 0.01 1 this section of the manual with u = y − 1 and a = 10 , we get 36. Separating variables we obtain 5 ln y 1.0004 10y − 11 = x + c. 10y − 9 1.0002 Setting x = 0 and y = 1 we obtain c = 5 ln 1 = 0. The solution is 5 ln -0.004-0.002 10y − 11 = x. 10y − 9 0.002 0.004 x 0.9998 Solving for y we obtain 0.9996 y= 11 + 9ex/5 . 10 + 10ex/5 Alternatively, we can use the fact that dy y−1 1 =− tanh−1 = −10 tanh−1 10(y − 1). (y − 1)2 − 0.01 0.1 0.1 (We use the inverse hyperbolic tangent because |y − 1| < 0.1 or 0.9 < y < 1.1. This follows from the initial condition y(0) = 1.) Solving the above equation for y we get y = 1 + 0.1 tanh(x/10). 37. Separating variables, we have dy dy = = 3 y−y y(1 − y)(1 + y) 1 1/2 1/2 + − y 1−y 1+y Integrating, we get ln |y| − 1 1 ln |1 − y| − ln |1 + y| = x + c. 2 2 35 dy = dx. 2.2 Separable Variables When y > 1, this becomes 1 1 y ln(y − 1) − ln(y + 1) = ln = x + c. 2 2 y2 − 1 √ √ √ Letting x = 0 and y = 2 we find c = ln(2/ 3 ). Solving for y we get y1 (x) = 2ex / 4e2x − 3 , where x > ln( 3/2). ln y − When 0 < y < 1 we have 1 1 y = x + c. ln(1 − y) − ln(1 + y) = ln 2 2 1 − y2 √ √ we find c = ln(1/ 3 ). Solving for y we get y2 (x) = ex / e2x + 3 , where −∞ < x < ∞. ln y − 1 2 Letting x = 0 and y = When −1 < y < 0 we have 1 1 −y = x + c. ln(1 − y) − ln(1 + y) = ln 2 2 1 − y2 √ √ we find c = ln(1/ 3 ). Solving for y we get y3 (x) = −ex / e2x + 3 , where ln(−y) − Letting x = 0 and y = − 1 2 −∞ < x < ∞. When y < −1 we have 1 1 −y = x + c. ln(1 − y) − ln(−1 − y) = ln 2 2 y2 − 1 √ √ Letting x = 0 and y = −2 we find c = ln(2/ 3 ). Solving for y we get y4 (x) = −2ex / 4e2x − 3 , where √ x > ln( 3/2). ln(−y) − y y y y 4 4 4 4 2 2 2 2 1 2 3 4 5x -4 -2 2 4 x -4 -2 2 4 x 1 -2 -2 -2 -4 -4 -4 2 3 5x 4 -2 -4 38. (a) The second derivative of y is y 8 d y dy/dx 1/(y − 3) 1 =− =− =− . 2 2 2 dx (y − 1) (y − 3) (y − 3)3 2 2 6 2 The solution curve is concave down when d y/dx < 0 or y > 3, and concave up when d2 y/dx2 > 0 or y < 3. From the phase portrait we see that the solution curve is decreasing when y < 3 and increasing when y > 3. 4 3 2 -4 -2 2 x 4 -2 (b) Separating variables and integrating we obtain y 8 (y − 3) dy = dx 1 2 y − 3y = x + c 2 y 2 − 6y + 9 = 2x + c1 6 4 2 (y − 3)2 = 2x + c1 √ y = 3 ± 2x + c1 . -1 1 -2 36 2 3 4 5 x 2.2 Separable Variables The initial condition dictates whether to use the plus or minus sign. √ When y1 (0) = 4 we have c1 = 1 and y1 (x) = 3 + 2x + 1 . √ When y2 (0) = 2 we have c1 = 1 and y2 (x) = 3 − 2x + 1 . √ When y3 (1) = 2 we have c1 = −1 and y3 (x) = 3 − 2x − 1 . √ When y4 (−1) = 4 we have c1 = 3 and y4 (x) = 3 + 2x + 3 . 39. (a) Separating variables we have 2y dy = (2x + 1)dx. Integrating gives y 2 = x2 + x + c. When y(−2) = −1 we √ find c = −1, so y 2 = x2 + x − 1 and y = − x2 + x − 1 . The negative square root is chosen because of the initial condition. y 2 (b) From the figure, the largest interval of definition appears to be approximately (−∞, −1.65). 1 -5 -4 -3 -2 -1 -1 1 2 x -2 -3 -4 -5 √ (c) Solving x + x − 1 = 0 we get x = ± 5 , so the largest interval of definition is (−∞, − 1 − 1 5 ). 2 2 √ The right-hand endpoint of the interval is excluded because y = − x2 + x − 1 is not differentiable at this point. 2 −1 2 1 2 √ 40. (a) From Problem 7 the general solution is 3e−2y + 2e3x = c. When y(0) = 0 we find c = 5, so 3e−2y + 2e3x = 5. Solving for y we get y = − 1 ln 1 (5 − 2e3x ). 2 3 y (b) The interval of definition appears to be approximately (−∞, 0.3). 2 1 x -2 -1.5 -1 -0.5 -1 -2 (c) Solving 1 (5 − 2e3x ) = 0 we get x = 1 ln( 5 ), so the exact interval of definition is (−∞, 1 ln 5 ). 3 3 2 3 2 √ 2 is defined at x = −5 and x = 5, y (x) is not defined at these values, and so the 41. (a) While y2 (x) = − 25 − x 2 interval of definition is the open interval (−5, 5). (b) At any point on the x-axis the derivative of y(x) is undefined, so no solution curve can cross the x-axis. Since −x/y is not defined when y = 0, the initial-value problem has no solution. 42. (a) Separating variables and integrating we obtain x2 − y 2 = c. For c = 0 the graph is a hyperbola centered at the origin. All four initial conditions imply c = 0 and y = ±x. Since the differential equation is not defined for y = 0, solutions are y = ±x, x < 0 and y = ±x, x > 0. The solution for y(a) = a is y = x, x > 0; for y(a) = −a is y = −x; for y(−a) = a is y = −x, x < 0; and for y(−a) = −a is y = x, x < 0. (b) Since x/y is not defined when y = 0, the initial-value problem has no solution. √ (c) Setting x = 1 and y = 2 in x2 − y 2 = c we get c = −3, so y 2 = x2 + 3 and y(x) = x2 + 3 , where the positive square root is chosen because of the initial condition. The domain is all real numbers since x2 + 3 > 0 for all x. 37 2.2 Separable Variables 1 + y 2 sin2 y = dx which is not readily integrated (even by a CAS). We note that dy/dx ≥ 0 for all values of x and y and that dy/dx = 0 when y = 0 and y = π, which are equilibrium solutions. 43. Separating variables we have dy/ y 3.5 3 2.5 2 1.5 1 0.5 -6 -4 -2 2 4 6 8 x √ √ √ 44. Separating variables we have dy/( y + y) = dx/( x + x). To integrate dx/( x + x) we substitute u2 = x and get √ 2u 2 du = du = 2 ln |1 + u| + c = 2 ln(1 + x ) + c. u + u2 1+u Integrating the separated differential equation we have √ √ √ √ 2 ln(1 + y ) = 2 ln(1 + x ) + c or ln(1 + y ) = ln(1 + x ) + ln c1 . √ Solving for y we get y = [c1 (1 + x ) − 1]2 . 45. We are looking for a function y(x) such that dy dx y2 + 2 = 1. Using the positive square root gives dy = dx 1 − y 2 =⇒ dy 1 − y2 = dx =⇒ sin−1 y = x + c. Thus a solution is y = sin(x + c). If we use the negative square root we obtain y = sin(c − x) = − sin(x − c) = − sin(x + c1 ). Note that when c = c1 = 0 and when c = c1 = π/2 we obtain the well known particular solutions y = sin x, y = − sin x, y = cos x, and y = − cos x. Note also that y = 1 and y = −1 are singular solutions. y 46. (a) 3 −3 3 x −3 (b) For |x| > 1 and |y| > 1 the differential equation is dy/dx = √ y 2 − 1 / x2 − 1 . Separating variables and integrating, we obtain dy y2 −1 =√ dx x2 − 1 and cosh−1 y = cosh−1 x + c. Setting x = 2 and y = 2 we find c = cosh−1 2 − cosh−1 2 = 0 and cosh−1 y = cosh−1 x. An explicit solution is y = x. 47. Since the tension T1 (or magnitude T1 ) acts at the lowest point of the cable, we use symmetry to solve the problem on the interval [0, L/2]. The assumption that the roadbed is uniform (that is, weighs a constant ρ 38 2.2 Separable Variables pounds per horizontal foot) implies W = ρx, where x is measured in feet and 0 ≤ x ≤ L/2. Therefore (10) becomes dy/dx = (ρ/T1 )x. This last equation is a separable equation of the form given in (1) of Section 2.2 in the text. Integrating and using the initial condition y(0) = a shows that the shape of the cable is a parabola: y(x) = (ρ/2T1 )x2 + a. In terms of the sag h of the cable and the span L, we see from Figure 2.22 in the text that y(L/2) = h + a. By applying this last condition to y(x) = (ρ/2T1 )x2 + a enables us to express ρ/2T1 in terms of h and L: y(x) = (4h/L2 )x2 + a. Since y(x) is an even function of x, the solution is valid on −L/2 ≤ x ≤ L/2. 48. (a) Separating variables and integrating, we have (3y 2 +1)dy = −(8x+5)dx and y 3 + y = −4x2 − 5x + c. Using a CAS we show various contours of f (x, y) = y 3 + y + 4x2 + 5x. The plots shown on [−5, 5] × [−5, 5] correspond to c-values of 0, ±5, ±20, ±40, ±80, and ±125. y 4 2 0 x -2 -4 -4 (b) The value of c corresponding to y(0) = −1 is f (0, −1) = −2; to y(0) = 2 is f (0, 2) = 10; to y(−1) = 4 is f (−1, 4) = 67; and to y(−1) = −3 is −31. -2 0 2 4 y 4 2 x 0 -2 -4 -4 -2 0 2 4 49. (a) An implicit solution of the differential equation (2y + 2)dy − (4x3 + 6x)dx = 0 is y 2 + 2y − x4 − 3x2 + c = 0. The condition y(0) = −3 implies that c = −3. Therefore y 2 + 2y − x4 − 3x2 − 3 = 0. (b) Using the quadratic formula we can solve for y in terms of x: y= −2 ± 4 + 4(x4 + 3x2 + 3) . 2 The explicit solution that satisfies the initial condition is then y = −1 − x4 + 3x3 + 4 . (c) From the graph of the function f (x) = x4 + 3x3 + 4 below we see that f (x) ≤ 0 on the approximate interval −2.8 ≤ x ≤ −1.3. Thus the approximate domain of the function y = −1 − x4 + 3x3 + 4 = −1 − f (x) is x ≤ −2.8 or x ≥ −1.3. The graph of this function is shown below. 39 2.2 Separable Variables 1 f x f x -4 4 -2 x 2 -2 2 -4 -4 x -2 -6 -2 -8 -4 -10 (d) Using the root finding capabilities of a CAS, the zeros of f are found to be −2.82202 and −1.3409. The domain of definition of the solution y(x) is then x > −1.3409. The 1 f x x 2 equality has been removed since the derivative dy/dx does not exist at the points where -2 f (x) = 0. The graph of the solution y = φ(x) is given on the right. -4 -6 -8 -10 50. (a) Separating variables and integrating, we have y (−2y + y 2 )dy = (x − x2 )dx 4 and 1 1 1 −y 2 + y 3 = x2 − x3 + c. 3 2 3 Using a CAS we show some contours of 2 -2 f (x, y) = 2y 3 − 6y 2 + 2x3 − 3x2 . The plots shown on [−7, 7]×[−5, 5] correspond to c-values of −450, −300, −200, −120, −60, −20, −10, −8.1, −5, −0.8, 20, 60, and 120. -4 is f 0, 3 = − 27 . 2 4 The portion of the graph between the dots corresponds to the solution curve satisfying the intial condition. To determine the (b) The value of c corresponding to y(0) = -6 -4 3 2 -2 0 2 4 6 y 4 2 interval of definition we find dy/dx for 2y 3 − 6y 2 + 2x3 − 3x2 = − x 0 x 0 27 . 4 -2 Using implicit differentiation we get y = (x − x2 )/(y 2 − 2y), which is infinite when y = 0 and y = 2. Letting y = 0 in -4 -2 0 2 4 6 2y 3 − 6y 2 + 2x3 − 3x2 = − 27 and using a CAS to solve for x 4 we get x = −1.13232. Similarly, letting y = 2, we find x = 1.71299. The largest interval of definition is approximately (−1.13232, 1.71299). 40 2.3 (c) The value of c corresponding to y(0) = −2 is f (0, −2) = −40. The portion of the graph to the right of the dot corresponds to the solution curve satisfying the initial condition. To determine the interval of definition we find dy/dx for Linear Equations y 4 2 x 0 -2 2y 3 − 6y 2 + 2x3 − 3x2 = −40. -4 Using implicit differentiation we get y = (x − x )/(y − 2y), which is infinite when y = 0 and y = 2. Letting y = 0 in 2y 3 − 6y 2 + 2x3 − 3x2 = −40 and using a CAS to solve for x 2 2 -6 -8 -4 -2 0 2 4 6 8 10 we get x = −2.29551. The largest interval of definition is approximately (−2.29551, ∞). EXERCISES 2.3 Linear Equations d −5x y = 0 and y = ce5x for −∞ < x < ∞. e dx d 2x so that e y = 0 and y = ce−2x for −∞ < x < ∞. dx 1. For y − 5y = 0 an integrating factor is e− 5 dx = e−5x so that 2. For y + 2y = 0 an integrating factor is e 2 dx = e2x The transient term is ce−2x . 3. For y +y = e3x an integrating factor is e dx = ex so that The transient term is ce−x . 4. For y + 4y = 4 3 4 dx an integrating factor is e d x [e y] = e4x and y = 1 e3x +ce−x for −∞ < x < ∞. 4 dx d 4x e y = dx = e4x so that −∞ < x < ∞. The transient term is ce−4x . 3x2 dx 5. For y + 3x2 y = x2 an integrating factor is e 3 = ex so that −∞ < x < ∞. The transient term is ce−x . 3 6. For y + 2xy = x3 an integrating factor is e 2x dx 2 = ex so that for −∞ < x < ∞. The transient term is ce−x . 2 7. For y + 1 1 y = 2 an integrating factor is e x x (1/x)dx 8. For y − 2y = x2 + 5 an integrating factor is e− y = − 1 x2 − 1 x − 2 2 11 4 4 4x 3e and y = 3 3 d ex y = x2 ex and y = dx 1 3 1 3 + ce−4x for + ce−x for 2 2 d ex y = x3 ex and y = 1 x2 − 2 dx 3 1 2 + ce−x 2 d 1 1 c [xy] = and y = ln x + for 0 < x < ∞. dx x x x d −2x = e−2x so that y = x2 e−2x + 5e−2x and e dx = x so that 2 dx + ce2x for −∞ < x < ∞. d 1 1 1 y = x sin x an integrating factor is e− (1/x)dx = so that y = sin x and y = cx − x cos x for x x dx x 0 < x < ∞. 3 2 d 10. For y + y = an integrating factor is e (2/x)dx = x2 so that x2 y = 3x and y = 3 +cx−2 for 0 < x < ∞. 2 x x dx 4 d 11. For y + y = x2 −1 an integrating factor is e (4/x)dx = x4 so that x4 y = x6 −x4 and y = 1 x3 − 1 x+cx−4 7 5 x dx 9. For y − for 0 < x < ∞. 41 2.3 Linear Equations x d y = x an integrating factor is e− [x/(1+x)]dx = (x+1)e−x so that (x + 1)e−x y = x(x+1)e−x (1 + x) dx 2x + 3 cex and y = −x − + for −1 < x < ∞. x+1 x+1 12. For y − 13. For y + y= 1+ 2 x ce−x 1 ex + 2 2 x2 x ex d 2 x [x e y] = e2x and an integrating factor is e [1+(2/x)]dx = x2 ex so that 2 x dx ce−x for 0 < x < ∞. The transient term is . x2 y = 1 −x d e sin 2x an integrating factor is e [1+(1/x)]dx = xex so that [xex y] = sin 2x and x dx 1 ce−x y = − e−x cos 2x + for 0 < x < ∞. The entire solution is transient. 2x x 1 x 14. For y + 1 + 15. For y= dx 4 − x = 4y 5 an integrating factor is e− dy y (4/y)dy = eln y −4 = y −4 so that d −4 y x = 4y and x = 2y 6 +cy 4 dy for 0 < y < ∞. dx 2 + x = ey an integrating factor is e dy y c for 0 < y < ∞. The transient term is 2 . y 16. For (2/y)dy = y 2 so that tan x dx 17. For y + (tan x)y = sec x an integrating factor is e d 2 2 2 c y x = y 2 ey and x = ey − ey + 2 ey + 2 dy y y y = sec x so that y = sin x + c cos x for −π/2 < x < π/2. 18. For y +(cot x)y = sec2 x csc x an integrating factor is e cot x dx d [(sec x)y] = sec2 x and dx = eln | sin x| = sin x so that d [(sin x) y] = sec2 x dx and y = sec x + c csc x for 0 < x < π/2. 19. For y + y= x+2 2xe−x y= an integrating factor is e x+1 x+1 [(x+2)/(x+1)]dx = (x + 1)ex , so d [(x + 1)ex y] = 2x and dx x2 −x c e + e−x for −1 < x < ∞. The entire solution is transient. x+1 x+1 4 d 5 an integrating factor is e [4/(x+2)]dx = (x + 2)4 so that y= (x + 2)4 y = 5(x + 2)2 x+2 (x + 2)2 dx 5 and y = (x + 2)−1 + c(x + 2)−4 for −2 < x < ∞. The entire solution is transient. 3 dr 21. For + r sec θ = cos θ an integrating factor is e sec θ dθ = eln | sec x+tan x| = sec θ + tan θ so that dθ d [(sec θ + tan θ)r] = 1 + sin θ and (sec θ + tan θ)r = θ − cos θ + c for −π/2 < θ < π/2 . dθ 2 dP d t2 −t 2 22. For P = (4t − 2)et −t and + (2t − 1)P = 4t − 2 an integrating factor is e (2t−1) dt = et −t so that e dt dt 2 2 P = 2 + cet−t for −∞ < t < ∞. The transient term is cet−t . 20. For y + 23. For y + 3 + 1 x y= e−3x an integrating factor is e x [3+(1/x)]dx = xe3x so that d xe3x y = 1 and dx ce−3x for 0 < x < ∞. The transient term is ce−3x /x. x 2 2 x−1 x+1 24. For y + 2 y = an integrating factor is e [2/(x −1)]dx = so that x −1 x−1 x+1 (x − 1)y = x(x + 1) + c(x + 1) for −1 < x < 1. y = e−3x + 42 d x−1 y dx x + 1 = 1 and 2.3 Linear Equations 1 d 1 c 1 y = ex an integrating factor is e (1/x)dx = x so that [xy] = ex and y = ex + for 0 < x < ∞. x x dx x x 1 2−e If y(1) = 2 then c = 2 − e and y = ex + . x x dx 1 1 d 1 For − x = 2y an integrating factor is e− (1/y)dy = so that x = 2 and x = 2y 2 +cy for 0 < y < ∞. dy y y dy y 49 If y(1) = 5 then c = −49/5 and x = 2y 2 − y. 5 d Rt/L E di R E i = eRt/L and For + i= an integrating factor is e (R/L) dt = eRt/L so that e dt L L dt L E E E i= + ce−Rt/L for −∞ < t < ∞. If i(0) = i0 then c = i0 − E/R and i = + i0 − e−Rt/L . R R R dT d −kt For −kT = −Tm k an integrating factor is e (−k)dt = e−kt so that [e T ] = −Tm ke−kt and T = Tm +cekt dt dt for −∞ < t < ∞. If T (0) = T0 then c = T0 − Tm and T = Tm + (T0 − Tm )ekt . 1 d ln x For y + y = an integrating factor is e [1/(x+1)]dx = x + 1 so that [(x + 1)y] = ln x and x+1 x+1 dx x x c x x 21 y= ln x − + for 0 < x < ∞. If y(1) = 10 then c = 21 and y = ln x − + . x+1 x+1 x+1 x+1 x+1 x+1 d For y + (tan x)y = cos2 x an integrating factor is e tan x dx = eln | sec x| = sec x so that [(sec x) y] = cos x dx and y = sin x cos x + c cos x for −π/2 < x < π/2. If y(0) = −1 then c = −1 and y = sin x cos x − cos x. 25. For y + 26. 27. 28. 29. 30. 31. For y + 2y = f (x) an integrating factor is e2x so that ye2x = 1 + c1 , 0 ≤ x ≤ 3 1 2x 2e c2 , y x > 3. If y(0) = 0 then c1 = −1/2 and for continuity we must have c2 = 1 e6 − 1 2 2 so that y= 1 2 (1 − e−2x ), 1 6 2 (e − 1)e−2x , x > 3. 32. For y + y = f (x) an integrating factor is ex so that 1 0≤x≤1 e + c1 , x −e + c2 , x > 1. x yex = y 5 If y(0) = 1 then c1 = 0 and for continuity we must have c2 = 2e so that y= x 5 0≤x≤3 x 3 x -1 1, 0≤x≤1 1−x 2e − 1, x > 1. 2 33. For y + 2xy = f (x) an integrating factor is ex so that 2 yex = 1 x2 2e 2 + c1 , 0 ≤ x ≤ 1 c2 , y x > 1. If y(0) = 2 then c1 = 3/2 and for continuity we must have c2 = 1 e + 2 3 2 so that y= 1 2 + 3 e−x , 2 2 1 2e + 3 2 0≤x≤1 e−x , 2 x > 1. 43 2.3 Linear Equations 34. For y + x 1 + x2 , 0 ≤ x ≤ 1 1 2x y= −x 1 + x2 , x > 1, 1 + x2 5 x -1 an integrating factor is 1 + x2 so that 1 + x2 y = y 1 2 2x 0≤x≤1 + c1 , − 1 x2 + c2 , 2 x > 1. If y(0) = 0 then c1 = 0 and for continuity we must have c2 = 1 so that 1 1 − 2 2 (1 + x2 ) , 0 ≤ x ≤ 1 y= 3 1 − , x > 1. 2 (1 + x2 ) 2 35. We first solve the initial-value problem y + 2y = 4x, y(0) = 3 on the interval [0, 1]. The integrating factor is e 2 dx = e , so d 2x [e y] = 4xe2x dx e2x y = y 20 2x 15 4xe2x dx = 2xe2x − e2x + c1 10 y = 2x − 1 + c1 e−2x . 5 −2x Using the initial condition, we find y(0) = −1+c1 = 3, so c1 = 4 and y = 2x−1+4e , −2 −2 0 ≤ x ≤ 1. Now, since y(1) = 2−1+4e = 1+4e , we solve the initial-value problem y − (2/x)y = 4x, y(1) = 1 + 4e−2 on the interval (1, ∞). The integrating factor is e (−2/x)dx 3 x = e−2 ln x = x−2 , so 4 d −2 [x y] = 4xx−2 = dx x 4 x−2 y = dx = 4 ln x + c2 x y = 4x2 ln x + c2 x2 . (We use ln x instead of ln |x| because x > 1.) Using the initial condition we find y(1) = c2 = 1 + 4e−2 , so y = 4x2 ln x + (1 + 4e−2 )x2 , x > 1. Thus, the solution of the original initial-value problem is y= 2x − 1 + 4e−2x , 0≤x≤1 4x2 ln x + (1 + 4e−2 )x2 , x > 1. See Problem 42 in this section. x 36. For y + ex y = 1 an integrating factor is ee . Thus x d ex e y = ee dx −ex From y(0) = 1 we get c = e, so y = e x et e dt 0 x x and ee y = t ee dt + c. 0 1−ex +e . When y + ex y = 0 we can separate variables and integrate: dy = −ex dx y and 44 ln |y| = −ex + c. 2.3 Linear Equations Thus y = c1 e−e . From y(0) = 1 we get c1 = e, so y = e1−e . x x When y + ex y = ex we can see by inspection that y = 1 is a solution. 37. An integrating factor for y − 2xy = 1 is e−x . Thus 2 2 d −x2 y] = e−x [e dx x e−x y = 2 e−t dt = 2 0 √ y= √ π erf(x) + c 2 2 π x2 e erf(x) + cex . 2 √ √ From y(1) = ( π/2)e erf(1) + ce = 1 we get c = e−1 − 2π erf(1). The solution of the initial-value problem is √ √ 2 π x2 π −1 y= e erf(x) + e − erf(1) ex 2 2 √ π x2 x2 −1 =e + e (erf(x) − erf(1)). 2 38. We want 4 to be a critical point, so we use y = 4 − y. 39. (a) All solutions of the form y = x5 ex − x4 ex + cx4 satisfy the initial condition. In this case, since 4/x is discontinuous at x = 0, the hypotheses of Theorem 1.1 are not satisfied and the initial-value problem does not have a unique solution. (b) The differential equation has no solution satisfying y(0) = y0 , y0 > 0. (c) In this case, since x0 > 0, Theorem 1.1 applies and the initial-value problem has a unique solution given by y = x5 ex − x4 ex + cx4 where c = y0 /x4 − x0 ex0 + ex0 . 0 40. On the interval (−3, 3) the integrating factor is e and so x dx/(x2 −9) d dx = e− x dx/(9−x2 ) 9 − x2 y = 0 1 2 = e 2 ln(9−x and y = √ ) = 9 − x2 c . 9 − x2 41. We want the general solution to be y = 3x − 5 + ce−x . (Rather than e−x , any function that approaches 0 as x → ∞ could be used.) Differentiating we get y = 3 − ce−x = 3 − (y − 3x + 5) = −y + 3x − 2, so the differential equation y + y = 3x − 2 has solutions asymptotic to the line y = 3x − 5. 42. The left-hand derivative of the function at x = 1 is 1/e and the right-hand derivative at x = 1 is 1 − 1/e. Thus, y is not differentiable at x = 1. 43. (a) Differentiating yc = c/x3 we get 3c 3 c 3 =− = − yc 4 3 x x x x so a differential equation with general solution yc = c/x3 is xy + 3y = 0. Now yc = − xyp + 3yp = x(3x2 ) + 3(x3 ) = 6x3 so a differential equation with general solution y = c/x3 + x3 is xy + 3y = 6x3 . This will be a general solution on (0, ∞). 45 2.3 Linear Equations (b) Since y(1) = 13 − 1/13 = 0, an initial condition is y(1) = 0. Since y(1) = 13 + 2/13 = 3, an initial condition is y(1) = 3. In each case the y 3 interval of definition is (0, ∞). The initial-value problem xy + 3y = 6x3 , y(0) = 0 has solution y = x3 for −∞ < x < ∞. In the figure the lower curve is the graph of y(x) = x3 − 1/x3 ,while the upper curve is the graph 5 of y = x3 − 2/x3 . x -3 (c) The first two initial-value problems in part (b) are not unique. For example, setting y(2) = 23 − 1/23 = 63/8, we see that y(2) = 63/8 is also an initial condition leading to the solution y = x3 − 1/x3 . 44. Since e P (x)dx+c c1 e = ec e P (x)dx P (x)dx y = c2 + = c1 e c1 e P (x)dx , we would have P (x)dx f (x) dx and e P (x)dx y = c3 + e P (x)dx f (x) dx, which is the same as (6) in the text. 45. We see by inspection that y = 0 is a solution. 46. The solution of the first equation is x = c1 e−λ1 t . From x(0) = x0 we obtain c1 = x0 and so x = x0 e−λ1 t . The second equation then becomes dy = x0 λ1 e−λ1 t − λ2 y dt or dy + λ2 y = x0 λ1 e−λ1 t dt which is linear. An integrating factor is eλ2 t . Thus d λ2 t [e y ] = x0 λ1 e−λ1 t eλ2 t = x0 λ1 e(λ2 −λ1 )t dt x0 λ1 (λ2 −λ1 )t eλ2 t y = e + c2 λ2 − λ 1 x0 λ1 −λ1 t y= e + c2 e−λ2 t . λ2 − λ 1 From y(0) = y0 we obtain c2 = (y0 λ2 − y0 λ1 − x0 λ1 )/(λ2 − λ1 ). The solution is y= 47. Writing the differential equation as x0 λ1 −λ1 t y0 λ2 − y0 λ1 − x0 λ1 −λ2 t e + e . λ2 − λ 1 λ2 − λ1 dE 1 + E = 0 we see that an integrating factor is et/RC . Then dt RC d t/RC E] = 0 [e dt et/RC E = c E = ce−t/RC . From E(4) = ce−4/RC = E0 we find c = E0 e4/RC . Thus, the solution of the initial-value problem is E = E0 e4/RC e−t/RC = E0 e−(t−4)/RC . 46 2.3 Linear Equations 48. (a) An integrating factor for y − 2xy = −1 is e−x . Thus 2 2 d −x2 y] = −e−x [e dx x √ π erf(x) + c. 2 0 √ √ From y(0) = π/2, and noting that erf(0) = 0, we get c = π/2. Thus √ √ √ √ 2 π π π x2 π x2 y = ex − erf(x) + = e (1 − erf(x)) = e erfc(x). 2 2 2 2 −x2 e y=− −t2 e dt = − y (b) Using a CAS we find y(2) ≈ 0.226339. 5 5 x 49. (a) An integrating factor for y + is x2 . Thus 2 10 sin x y= x x3 d 2 sin x [x y] = 10 dx x x sin t 2 x y = 10 dt + c t 0 y = 10x−2 Si(x) + cx−2 . From y(1) = 0 we get c = −10Si(1). Thus y = 10x−2 Si(x) − 10x−2 Si(1) = 10x−2 (Si(x) − Si(1)). (b) y 2 1 1 2 3 5 4 x -1 -2 -3 -4 -5 (c) From the graph in part (b) we see that the absolute maximum occurs around x = 1.7. Using the root-finding capability of a CAS and solving y (x) = 0 for x we see that the absolute maximum is (1.688, 1.742). x − 50. (a) The integrating factor for y − (sin x2 )y = 0 is e d − [e dx − e x 0 x 0 0 sin t2 dt sin t2 dt . Then y] = 0 sin t2 dt y = c1 x y = c1 e 47 0 sin t2 dt . 2.3 Linear Equations Letting t = π/2 u we have dt = π/2 du and √ x 2/π x π π 2 u du = sin t2 dt = sin 2 0 2 0 2 x π √ √ √ √ so y = c1 e π/2 S( 2/π x) . Using S(0) = 0 and y(0) = c1 = 5 we have y = 5e π/2 S( 2/π x) . (b) π S 2 y 10 5 -10 -5 x 5 10 (c) From the graph we see that as x → ∞, y(x) oscillates with decreasing amplitudes approaching 9.35672. √ Since limx→∞ 5S(x) = 1 , we have limx→∞ y(x) = 5e π/8 ≈ 9.357, and since limx→−∞ S(x) = − 1 , we 2 √ 2 − π/8 have limx→−∞ y(x) = 5e ≈ 2.672. (d) From the graph in part (b) we see that the absolute maximum occurs around x = 1.7 and the absolute minimum occurs around x = −1.8. Using the root-finding capability of a CAS and solving y (x) = 0 for x, we see that the absolute maximum is (1.772, 12.235) and the absolute minimum is (−1.772, 2.044). EXERCISES 2.4 Exact Equations 1. Let M = 2x − 1 and N = 3y + 7 so that My = 0 = Nx . From fx = 2x − 1 we obtain f = x2 − x + h(y), h (y) = 3y + 7, and h(y) = 3 y 2 + 7y. A solution is x2 − x + 3 y 2 + 7y = c. 2 2 2. Let M = 2x + y and N = −x − 6y. Then My = 1 and Nx = −1, so the equation is not exact. 3. Let M = 5x + 4y and N = 4x − 8y 3 so that My = 4 = Nx . From fx = 5x + 4y we obtain f = 5 x2 + 4xy + h(y), 2 h (y) = −8y 3 , and h(y) = −2y 4 . A solution is 5 x2 + 4xy − 2y 4 = c. 2 4. Let M = sin y − y sin x and N = cos x + x cos y − y so that My = cos y − sin x = Nx . From fx = sin y − y sin x we obtain f = x sin y + y cos x + h(y), h (y) = −y, and h(y) = − 1 y 2 . A solution is x sin y + y cos x − 1 y 2 = c. 2 2 5. Let M = 2y 2 x−3 and N = 2yx2 +4 so that My = 4xy = Nx . From fx = 2y 2 x−3 we obtain f = x2 y 2 −3x+h(y), h (y) = 4, and h(y) = 4y. A solution is x2 y 2 − 3x + 4y = c. 6. Let M = 4x3 −3y sin 3x−y/x2 and N = 2y−1/x+cos 3x so that My = −3 sin 3x−1/x2 and Nx = 1/x2 −3 sin 3x. The equation is not exact. 7. Let M = x2 − y 2 and N = x2 − 2xy so that My = −2y and Nx = 2x − 2y. The equation is not exact. 8. Let M = 1 + ln x + y/x and N = −1 + ln x so that My = 1/x = Nx . From fy = −1 + ln x we obtain f = −y + y ln x + h(y), h (x) = 1 + ln x, and h(y) = x ln x. A solution is −y + y ln x + x ln x = c. 48 2.4 Exact Equations 9. Let M = y 3 − y 2 sin x − x and N = 3xy 2 + 2y cos x so that My = 3y 2 − 2y sin x = Nx . From fx = y 3 − y 2 sin x − x we obtain f = xy 3 + y 2 cos x − 1 x2 + h(y), h (y) = 0, and h(y) = 0. A solution is xy 3 + y 2 cos x − 1 x2 = c. 2 2 10. Let M = x3 + y 3 and N = 3xy 2 so that My = 3y 2 = Nx . From fx = x3 + y 3 we obtain f = 1 x4 + xy 3 + h(y), 4 h (y) = 0, and h(y) = 0. A solution is 1 x4 + xy 3 = c. 4 11. Let M = y ln y − e−xy and N = 1/y + x ln y so that My = 1 + ln y + xe−xy and Nx = ln y. The equation is not exact. 12. Let M = 3x2 y + ey and N = x3 + xey − 2y so that My = 3x2 + ey = Nx . From fx = 3x2 y + ey we obtain f = x3 y + xey + h(y), h (y) = −2y, and h(y) = −y 2 . A solution is x3 y + xey − y 2 = c. 13. Let M = y − 6x2 − 2xex and N = x so that My = 1 = Nx . From fx = y − 6x2 − 2xex we obtain f = xy − 2x3 − 2xex + 2ex + h(y), h (y) = 0, and h(y) = 0. A solution is xy − 2x3 − 2xex + 2ex = c. 14. Let M = 1 − 3/x + y and N = 1 − 3/y + x so that My = 1 = Nx . From fx = 1 − 3/x + y we obtain 3 f = x − 3 ln |x| + xy + h(y), h (y) = 1 − , and h(y) = y − 3 ln |y|. A solution is x + y + xy − 3 ln |xy| = c. y 15. Let M = x2 y 3 − 1/ 1 + 9x2 and N = x3 y 2 so that My = 3x2 y 2 = Nx . From fx = x2 y 3 − 1/ 1 + 9x2 obtain f = 1 x3 y 3 − 1 arctan(3x) + h(y), h (y) = 0, and h(y) = 0. A solution is x3 y 3 − arctan(3x) = c. 3 3 we 16. Let M = −2y and N = 5y − 2x so that My = −2 = Nx . From fx = −2y we obtain f = −2xy + h(y), h (y) = 5y, and h(y) = 5 y 2 . A solution is −2xy + 5 y 2 = c. 2 2 17. Let M = tan x − sin x sin y and N = cos x cos y so that My = − sin x cos y = Nx . From fx = tan x − sin x sin y we obtain f = ln | sec x| + cos x sin y + h(y), h (y) = 0, and h(y) = 0. A solution is ln | sec x| + cos x sin y = c. 2 2 18. Let M = 2y sin x cos x − y + 2y 2 exy and N = −x + sin2 x + 4xyexy so that 2 2 My = 2 sin x cos x − 1 + 4xy 3 exy + 4yexy = Nx . 2 2 From fx = 2y sin x cos x − y + 2y 2 exy we obtain f = y sin2 x − xy + 2exy + h(y), h (y) = 0, and h(y) = 0. A 2 solution is y sin2 x − xy + 2exy = c. 19. Let M = 4t3 y − 15t2 − y and N = t4 + 3y 2 − t so that My = 4t3 − 1 = Nt . From ft = 4t3 y − 15t2 − y we obtain f = t4 y − 5t3 − ty + h(y), h (y) = 3y 2 , and h(y) = y 3 . A solution is t4 y − 5t3 − ty + y 3 = c. 2 20. Let M = 1/t + 1/t2 − y/ t2 + y 2 and N = yey + t/ t2 + y 2 so that My = y 2 − t2 / t2 + y 2 = Nt . From 1 t ft = 1/t + 1/t2 − y/ t2 + y 2 we obtain f = ln |t| − − arctan + h(y), h (y) = yey , and h(y) = yey − ey . t y A solution is 1 t ln |t| − − arctan + yey − ey = c. t y 21. Let M = x2 + 2xy + y 2 and N = 2xy + x2 − 1 so that My = 2(x + y) = Nx . From fx = x2 + 2xy + y 2 we obtain f = 1 x3 + x2 y + xy 2 + h(y), h (y) = −1, and h(y) = −y. The solution is 1 x3 + x2 y + xy 2 − y = c. If y(1) = 1 3 3 then c = 4/3 and a solution of the initial-value problem is 1 x3 + x2 y + xy 2 − y = 4 . 3 3 22. Let M = ex + y and N = 2 + x + yey so that My = 1 = Nx . From fx = ex + y we obtain f = ex + xy + h(y), h (y) = 2 + yey , and h(y) = 2y + yey − y. The solution is ex + xy + 2y + yey − ey = c. If y(0) = 1 then c = 3 and a solution of the initial-value problem is ex + xy + 2y + yey − ey = 3. 23. Let M = 4y + 2t − 5 and N = 6y + 4t − 1 so that My = 4 = Nt . From ft = 4y + 2t − 5 we obtain f = 4ty + t2 − 5t + h(y), h (y) = 6y − 1, and h(y) = 3y 2 − y. The solution is 4ty + t2 − 5t + 3y 2 − y = c. If y(−1) = 2 then c = 8 and a solution of the initial-value problem is 4ty + t2 − 5t + 3y 2 − y = 8. 49 2.4 Exact Equations 24. Let M = t/2y 4 and N = 3y 2 − t2 /y 5 so that My = −2t/y 5 = Nt . From ft = t/2y 4 we obtain f = t2 + h(y), 4y 4 3 3 t2 3 , and h(y) = − 2 . The solution is 4 − 2 = c. If y(1) = 1 then c = −5/4 and a solution of the y3 2y 4y 2y t2 3 5 initial-value problem is − 2 =− . 4 4y 2y 4 h (y) = 25. Let M = y 2 cos x − 3x2 y − 2x and N = 2y sin x − x3 + ln y so that My = 2y cos x − 3x2 = Nx . From fx = y 2 cos x − 3x2 y − 2x we obtain f = y 2 sin x − x3 y − x2 + h(y), h (y) = ln y, and h(y) = y ln y − y. The solution is y 2 sin x − x3 y − x2 + y ln y − y = c. If y(0) = e then c = 0 and a solution of the initial-value problem is y 2 sin x − x3 y − x2 + y ln y − y = 0. 26. Let M = y 2 + y sin x and N = 2xy − cos x − 1/ 1 + y 2 so that My = 2y + sin x = Nx . From fx = y 2 + y sin x we −1 obtain f = xy 2 −y cos x+h(y), h (y) = , and h(y) = − tan−1 y. The solution is xy 2 −y cos x−tan−1 y = c. 1 + y2 π If y(0) = 1 then c = −1 − π/4 and a solution of the initial-value problem is xy 2 − y cos x − tan−1 y = −1 − . 4 27. Equating My = 3y 2 + 4kxy 3 and Nx = 3y 2 + 40xy 3 we obtain k = 10. 28. Equating My = 18xy 2 − sin y and Nx = 4kxy 2 − sin y we obtain k = 9/2. 29. Let M = −x2 y 2 sin x + 2xy 2 cos x and N = 2x2 y cos x so that My = −2x2 y sin x + 4xy cos x = Nx . From fy = 2x2 y cos x we obtain f = x2 y 2 cos x + h(y), h (y) = 0, and h(y) = 0. A solution of the differential equation is x2 y 2 cos x = c. 30. Let M = (x2 +2xy−y 2 )/(x2 +2xy+y 2 ) and N = (y 2 +2xy−x2 /(y 2 +2xy+x2 ) so that My = −4xy/(x+y)3 = Nx . 2y 2 From fx = x2 + 2xy + y 2 − 2y 2 /(x + y)2 we obtain f = x + + h(y), h (y) = −1, and h(y) = −y. A x+y solution of the differential equation is x2 + y 2 = c(x + y). 31. We note that (My − Nx )/N = 1/x, so an integrating factor is e dx/x = x. Let M = 2xy 2 + 3x2 and N = 2x2 y so that My = 4xy = Nx . From fx = 2xy 2 + 3x2 we obtain f = x2 y 2 + x3 + h(y), h (y) = 0, and h(y) = 0. A solution of the differential equation is x2 y 2 + x3 = c. 32. We note that (My − Nx )/N = 1, so an integrating factor is e dx = ex . Let M = xyex + y 2 ex + yex and N = xex + 2yex so that My = xex + 2yex + ex = Nx . From fy = xex + 2yex we obtain f = xyex + y 2 ex + h(x), h (y) = 0, and h(y) = 0. A solution of the differential equation is xyex + y 2 ex = c. 33. We note that (Nx −My )/M = 2/y, so an integrating factor is e 2dy/y = y 2 . Let M = 6xy 3 and N = 4y 3 +9x2 y 2 so that My = 18xy 2 = Nx . From fx = 6xy 3 we obtain f = 3x2 y 3 + h(y), h (y) = 4y 3 , and h(y) = y 4 . A solution of the differential equation is 3x2 y 3 + y 4 = c. 34. We note that (My −Nx )/N = − cot x, so an integrating factor is e− cot x dx = csc x. Let M = cos x csc x = cot x and N = (1 + 2/y) sin x csc x = 1 + 2/y, so that My = 0 = Nx . From fx = cot x we obtain f = ln(sin x) + h(y), h (y) = 1 + 2/y, and h(y) = y + ln y 2 . A solution of the differential equation is ln(sin x) + y + ln y 2 = c. 35. We note that (My − Nx )/N = 3, so an integrating factor is e 3 dx = e3x . Let M = (10 − 6y + e−3x )e3x = 10e3x − 6ye3x + 1 and N = −2e3x , so that My = −6e3x = Nx . From fx = 10e3x − 6ye3x + 1 we obtain f = 10 3x 10 3x 3x 3x 3 e − 2ye + x + h(y), h (y) = 0, and h(y) = 0. A solution of the differential equation is 3 e − 2ye + x = c. 36. We note that (Nx − My )/M = −3/y, so an integrating factor is e−3 dy/y = 1/y 3 . Let M = (y 2 + xy 3 )/y 3 = 1/y + x and N = (5y 2 − xy + y 3 sin y)/y 3 = 5/y − x/y 2 + sin y, so that My = −1/y 2 = Nx . From fx = 1/y + x we obtain f = x/y + 1 x2 + h(y), h (y) = 5/y + sin y, and h(y) = 5 ln |y| − cos y. A solution of the differential 2 equation is x/y + 1 x2 + 5 ln |y| − cos y = c. 2 50 2.4 Exact Equations 37. We note that (My − Nx )/N = 2x/(4 + x2 ), so an integrating factor is e−2 x dx/(4+x ) = 1/(4 + x2 ). Let M = x/(4 + x2 ) and N = (x2 y + 4y)/(4 + x2 ) = y, so that My = 0 = Nx . From fx = x(4 + x2 ) we obtain f = 1 ln(4+x2 )+h(y), h (y) = y, and h(y) = 1 y 2 . A solution of the differential equation is 1 ln(4+x2 )+ 1 y 2 = c. 2 2 2 2 2 38. We note that (My − Nx )/N = −3/(1 + x), so an integrating factor is e−3 dx/(1+x) = 1/(1 + x)3 . Let M = (x2 + y 2 − 5)/(1 + x)3 and N = −(y + xy)/(1 + x)3 = −y/(1 + x)2 , so that My = 2y/(1 + x)3 = Nx . From fy = −y/(1 + x)2 we obtain f = − 1 y 2 /(1 + x)2 + h(x), h (x) = (x2 − 5)/(1 + x)3 , and h(x) = 2/(1 + x)2 + 2 2/(1 + x) + ln |1 + x|. A solution of the differential equation is − y2 2 2 + ln |1 + x| = c. + + 2(1 + x)2 (1 + x)2 (1 + x) 39. (a) Implicitly differentiating x3 + 2x2 y + y 2 = c and solving for dy/dx we obtain 3x2 + 2x2 dy dy + 4xy + 2y =0 dx dx and dy 3x2 + 4xy =− 2 . dx 2x + 2y By writing the last equation in differential form we get (4xy + 3x2 )dx + (2y + 2x2 )dy = 0. (b) Setting x = 0 and y = −2 in x3 + 2x2 y + y 2 = c we find c = 4, and setting x = y = 1 we also find c = 4. Thus, both initial conditions determine the same implicit solution. y 4 (c) Solving x3 + 2x2 y + y 2 = 4 for y we get y1 (x) = −x2 − 4 − x3 + x4 y2 (x) = −x2 + 4 − x3 + x4 . 2 y2 and -4 -2 2 -2 Observe in the figure that y1 (0) = −2 and y2 (1) = 1. 4 x y1 -4 -6 40. To see that the equations are not equivalent consider dx = −(x/y)dy. An integrating factor is µ(x, y) = y resulting in y dx + x dy = 0. A solution of the latter equation is y = 0, but this is not a solution of the original equation. (3 + cos2 x)/(1 − x2 ) . Since 3 + cos2 x > 0 for all x we must have 1 − x2 > 0 or −1 < x < 1. Thus, the interval of definition is (−1, 1). y y 42. (a) Since fy = N (x, y) = xexy +2xy+1/x we obtain f = exy +xy 2 + +h(x) so that fx = yexy +y 2 − 2 +h (x). x x y Let M (x, y) = yexy + y 2 − 2 . x 1 −1 (b) Since fx = M (x, y) = y 1/2 x−1/2 + x x2 + y we obtain f = 2y 1/2 x1/2 + ln x2 + y + g(y) so that 2 1 2 1 2 −1 −1 −1/2 1/2 −1/2 1/2 x + + g (x). Let N (x, y) = y x + . x +y x +y fy = y 2 2 43. First note that x y d x2 + y 2 = dx + dy. 2 + y2 2 + y2 x x 41. The explicit solution is y = Then x dx + y dy = x2 + y 2 dx becomes x x2 + y2 dx + y x2 + y2 dy = d 51 x2 + y 2 = dx. 2.4 Exact Equations x2 + y 2 and the right side is the total differential of x + c. Thus x2 + y 2 = x + c is a solution of the differential equation. The left side is the total differential of 44. To see that the statement is true, write the separable equation as −g(x) dx+dy/h(y) = 0. Identifying M = −g(x) and N = 1/h(y), we see that My = 0 = Nx , so the differential equation is exact. 45. (a) In differential form we have (v 2 − 32x)dx + xv dv = 0. This is not an exact form, but µ(x) = x is an integrating factor. Multiplying by x we get (xv 2 − 32x2 )dx + x2 v dv = 0. This form is the total differential of u = 1 x2 v 2 − 32 x3 , so an implicit solution is 1 x2 v 2 − 32 x3 = c. Letting x = 3 and v = 0 we find c = −288. 2 3 2 3 Solving for v we get x 9 − 2. 3 x v=8 (b) The chain leaves the platform when x = 8, so the velocity at this time is v(8) = 8 8 9 − ≈ 12.7 ft/s. 3 64 2xy + y 2 )2 and 46. (a) Letting M (x, y) = (x2 N (x, y) = 1 + y 2 − x2 (x2 + y 2 )2 we compute My = 2x3 − 8xy 2 = Nx , (x2 + y 2 )3 so the differential equation is exact. Then we have ∂f 2xy = M (x, y) = 2 = 2xy(x2 + y 2 )−2 ∂x (x + y 2 )2 y f (x, y) = −y(x2 + y 2 )−1 + g(y) = − 2 + g(y) x + y2 ∂f y 2 − x2 y 2 − x2 + g (y) = N (x, y) = 1 + 2 . = 2 2 )2 ∂y (x + y (x + y 2 )2 y Thus, g (y) = 1 and g(y) = y. The solution is y − 2 = c. When c = 0 the solution is x2 + y 2 = 1. x + y2 (b) The first graph below is obtained in Mathematica using f (x, y) = y − y/(x2 + y 2 ) and ContourPlot[f[x, y], {x, -3, 3}, {y, -3, 3}, Axes−>True, AxesOrigin−>{0, 0}, AxesLabel−>{x, y}, Frame−>False, PlotPoints−>100, ContourShading−>False, Contours−>{0, -0.2, 0.2, -0.4, 0.4, -0.6, 0.6, -0.8, 0.8}] The second graph uses x=− y 3 − cy 2 − y c−y and x= y 3 − cy 2 − y . c−y In this case the x-axis is vertical and the y-axis is horizontal. To obtain the third graph, we solve y − y/(x2 + y 2 ) = c for y in a CAS. This appears to give one real and two complex solutions. When graphed in Mathematica however, all three solutions contribute to the graph. This is because the solutions involve the square root of expressions containing c. For some values of c the expression is negative, causing an apparent complex solution to actually be real. 52 2.5 Solutions by Substitutions y 3 2 2 1 -2 y 3 2 -3 x 3 1 1 -1 1 2 3 x y -1.5-0.5 0.511.5 -1 -3 -2 -1 1 -1 -1 -1 -2 -2 -2 -3 -3 -3 EXERCISES 2.5 Solutions by Substitutions 1. Letting y = ux we have (x − ux) dx + x(u dx + x du) = 0 dx + x du = 0 dx + du = 0 x ln |x| + u = c x ln |x| + y = cx. 2. Letting y = ux we have (x + ux) dx + x(u dx + x du) = 0 (1 + 2u) dx + x du = 0 dx du + =0 x 1 + 2u 1 ln |x| + ln |1 + 2u| = c 2 y x2 1 + 2 = c1 x x2 + 2xy = c1 . 53 2 3 2.5 Solutions by Substitutions 3. Letting x = vy we have vy(v dy + y dv) + (y − 2vy) dy = 0 vy 2 dv + y v 2 − 2v + 1 dy = 0 v dv dy + =0 2 (v − 1) y 1 ln |v − 1| − + ln |y| = c v−1 x 1 ln −1 − + ln y = c y x/y − 1 (x − y) ln |x − y| − y = c(x − y). 4. Letting x = vy we have y(v dy + y dv) − 2(vy + y) dy = 0 y dv − (v + 2) dy = 0 dv dy − =0 v+2 y ln |v + 2| − ln |y| = c x + 2 − ln |y| = c y ln x + 2y = c1 y 2 . 5. Letting y = ux we have u2 x2 + ux2 dx − x2 (u dx + x du) = 0 u2 dx − x du = 0 dx du − 2 =0 x u 1 =c u x ln |x| + = c y ln |x| + y ln |x| + x = cy. 6. Letting y = ux and using partial fractions, we have u2 x2 + ux2 dx + x2 (u dx + x du) = 0 x2 u2 + 2u dx + x3 du = 0 dx du + =0 x u(u + 2) ln |x| + 1 1 ln |u| − ln |u + 2| = c 2 2 x2 u = c1 u+2 y y +2 x2 = c1 x x x2 y = c1 (y + 2x). 54 2.5 Solutions by Substitutions 7. Letting y = ux we have (ux − x) dx − (ux + x)(u dx + x du) = 0 u2 + 1 dx + x(u + 1) du = 0 dx u+1 + 2 du = 0 x u +1 ln |x| + 1 ln u2 + 1 + tan−1 u = c 2 ln x2 y2 y + 1 + 2 tan−1 = c1 x2 x ln x2 + y 2 + 2 tan−1 y = c1 . x 8. Letting y = ux we have (x + 3ux) dx − (3x + ux)(u dx + x du) = 0 u2 − 1 dx + x(u + 3) du = 0 dx u+3 + du = 0 x (u − 1)(u + 1) ln |x| + 2 ln |u − 1| − ln |u + 1| = c x(u − 1)2 = c1 u+1 2 y y x − 1 = c1 +1 x x (y − x)2 = c1 (y + x). 9. Letting y = ux we have −ux dx + (x + √ u x)(u dx + x du) = 0 √ (x + x u ) du + xu3/2 dx = 0 2 2 u−3/2 + 1 u du + dx =0 x −2u−1/2 + ln |u| + ln |x| = c ln |y/x| + ln |x| = 2 x/y + c y(ln |y| − c)2 = 4x. 10. Letting y = ux we have ux + x2 − (ux)2 dx − x(udx + xdu) du = 0 x2 − u2 x2 dx − x2 du = 0 x 1 − u2 dx − x2 du = 0, (x > 0) dx du −√ =0 x 1 − u2 ln x − sin−1 u = c sin−1 u = ln x + c1 55 2.5 Solutions by Substitutions sin−1 y = ln x + c2 x y = sin(ln x + c2 ) x y = x sin(ln x + c2 ). See Problem 33 in this section for an analysis of the solution. 11. Letting y = ux we have x3 − u3 x3 dx + u2 x3 (u dx + x du) = 0 dx + u2 x du = 0 dx + u2 du = 0 x 1 ln |x| + u3 = c 3 3x3 ln |x| + y 3 = c1 x3 . Using y(1) = 2 we find c1 = 8. The solution of the initial-value problem is 3x3 ln |x| + y 3 = 8x3 . 12. Letting y = ux we have (x2 + 2u2 x2 )dx − ux2 (u dx + x du) = 0 x2 (1 + u2 )dx − ux3 du = 0 dx u du =0 − x 1 + u2 1 ln |x| − ln(1 + u2 ) = c 2 x2 = c1 1 + u2 x4 = c1 (x2 + y 2 ). Using y(−1) = 1 we find c1 = 1/2. The solution of the initial-value problem is 2x4 = y 2 + x2 . 13. Letting y = ux we have (x + uxeu ) dx − xeu (u dx + x du) = 0 dx − xeu du = 0 dx − eu du = 0 x ln |x| − eu = c ln |x| − ey/x = c. Using y(1) = 0 we find c = −1. The solution of the initial-value problem is ln |x| = ey/x − 1. 14. Letting x = vy we have y(v dy + y dv) + vy(ln vy − ln y − 1) dy = 0 y dv + v ln v dy = 0 dv dy + =0 v ln v y ln |ln |v|| + ln |y| = c y ln 56 x = c1 . y 2.5 Using y(1) = e we find c1 = −e. The solution of the initial-value problem is y ln 15. From y + Solutions by Substitutions x = −e. y 1 dw 1 3 3 y = y −2 and w = y 3 we obtain + w = . An integrating factor is x3 so that x3 w = x3 + c x x dx x x or y 3 = 1 + cx−3 . 16. From y − y = ex y 2 and w = y −1 we obtain or y −1 = − 1 ex + ce−x . 2 dw + w = −ex . An integrating factor is ex so that ex w = − 1 e2x + c 2 dx 17. From y + y = xy 4 and w = y −3 we obtain xe−3x + 1 e−3x + c or y −3 = x + 3 18. From y − 1 + 1 x 1 3 + ce3x . dw − 3w = −3x. An integrating factor is e−3x so that e−3x w = dx y = y 2 and w = y −1 we obtain xex w = −xex + ex + c or y −1 = −1 + dw 1 + 1+ dx x w = −1. An integrating factor is xex so that 1 c + e−x . x x 1 dw 1 1 1 19. From y − y = − 2 y 2 and w = y −1 we obtain + w = 2 . An integrating factor is t so that tw = ln t + c t t dt t t 1 c t or y −1 = ln t + . Writing this in the form = ln t + c, we see that the solution can also be expressed in the t t y form et/y = c1 t. 2 dw −2t 2t 2t w= . An integrating factor is y= y 4 and w = y −3 we obtain − 3 (1 + t2 ) 3 (1 + t2 ) dt 1 + t2 1 + t2 1 w 1 so that = + c or y −3 = 1 + c 1 + t2 . 1 + t2 1 + t2 1 + t2 20. From y + 21. From y − 2 dw 3 6 9 y = 2 y 4 and w = y −3 we obtain + w = − 2 . An integrating factor is x6 so that x x dx x x x6 w = − 9 x5 + c or y −3 = − 9 x−1 + cx−6 . If y(1) = 5 5 1 2 then c = 49 5 and y −3 = − 9 x−1 + 5 49 −6 . 5 x dw 3 3 + w = . An integrating factor is e3x/2 so that e3x/2 w = dx 2 2 = 1 + ce−3x/2 . If y(0) = 4 then c = 7 and y 3/2 = 1 + 7e−3x/2 . 22. From y + y = y −1/2 and w = y 3/2 we obtain e3x/2 + c or y 3/2 du 1 du = dx. Thus tan−1 u = x + c or − 1 = u2 or dx 1 + u2 u = tan(x + c), and x + y + 1 = tan(x + c) or y = tan(x + c) − x − 1. 23. Let u = x + y + 1 so that du/dx = 1 + dy/dx. Then 24. Let u = x + y so that du/dx = 1 + dy/dx. Then and (x + y)2 = 2x + c1 . du 1−u −1 = or u du = dx. Thus 1 u2 = x + c or u2 = 2x + c1 , 2 dx u 25. Let u = x + y so that du/dx = 1 + dy/dx. Then du − 1 = tan2 u or cos2 u du = dx. Thus 1 u + 2 dx 1 4 sin 2u = x + c or 2u + sin 2u = 4x + c1 , and 2(x + y) + sin 2(x + y) = 4x + c1 or 2y + sin 2(x + y) = 2x + c1 . 26. Let u = x + y so that du/dx = 1 + dy/dx. Then (1 − sin u)/(1 − sin u) we have du 1 − 1 = sin u or du = dx. Multiplying by dx 1 + sin u 1 − sin u du = dx or (sec2 u − sec u tan u)du = dx. Thus tan u − sec u = x + c or cos2 u tan(x + y) − sec(x + y) = x + c. 57 2.5 Solutions by Substitutions √ √ du 1 27. Let u = y − 2x + 3 so that du/dx = dy/dx − 2. Then + 2 = 2 + u or √ du = dx. Thus 2 u = x + c and dx u √ 2 y − 2x + 3 = x + c. du 28. Let u = y − x + 5 so that du/dx = dy/dx − 1. Then + 1 = 1 + eu or e−u du = dx. Thus −e−u = x + c and dx −ey−x+5 = x + c. du 1 29. Let u = x + y so that du/dx = 1 + dy/dx. Then − 1 = cos u and du = dx. Now dx 1 + cos u 1 1 − cos u 1 − cos u = csc2 u − csc u cot u = = 1 + cos u 1 − cos2 u sin2 u so we have (csc2 u − csc u cot u)du = dx and − cot u + csc u = x + c. Thus − cot(x + y) + csc(x + y) = x + c. √ Setting x = 0 and y = π/4 we obtain c = 2 − 1. The solution is √ csc(x + y) − cot(x + y) = x + 2 − 1. 30. Let u = 3x + 2y so that du/dx = 3 + 2 dy/dx. Then du 2u 5u + 6 u+2 =3+ = and du = dx. Now by dx u+2 u+2 5u + 6 long division u+2 1 4 = + 5u + 6 5 25u + 30 so we have 1 4 + 5 25u + 30 and 1 5u + 4 25 du = dx ln |25u + 30| = x + c. Thus 1 4 (3x + 2y) + ln |75x + 50y + 30| = x + c. 5 25 Setting x = −1 and y = −1 we obtain c = or 4 25 ln 95. The solution is 1 4 4 (3x + 2y) + ln |75x + 50y + 30| = x + ln 95 5 25 25 5y − 5x + 2 ln |75x + 50y + 30| = 2 ln 95. 31. We write the differential equation M (x, y)dx + N (x, y)dy = 0 as dy/dx = f (x, y) where f (x, y) = − M (x, y) . N (x, y) The function f (x, y) must necessarily be homogeneous of degree 0 when M and N are homogeneous of degree α. Since M is homogeneous of degree α, M (tx, ty) = tα M (x, y), and letting t = 1/x we have M (1, y/x) = Thus 1 M (x, y) xα or M (x, y) = xα M (1, y/x). dy xα M (1, y/x) M (1, y/x) = f (x, y) = − α =− =F dx x N (1, y/x) N (1, y/x) 32. Rewrite (5x2 − 2y 2 )dx − xy dy = 0 as xy dy = 5x2 − 2y 2 dx and divide by xy, so that dy x y =5 −2 . dx y x 58 y . x 2.5 We then identify F y y =5 x x −1 −2 Solutions by Substitutions y . x 33. (a) By inspection y = x and y = −x are solutions of the differential equation and not members of the family y = x sin(ln x + c2 ). (b) Letting x = 5 and y = 0 in sin−1 (y/x) = ln x + c2 we get sin−1 0 = ln 5 + c or c = − ln 5. Then sin−1 (y/x) = ln x − ln 5 = ln(x/5). Because the range of the arcsine function is [−π/2, π/2] we must have π x π − ≤ ln ≤ 2 5 2 x −π/2 e ≤ ≤ eπ/2 5 −π/2 5e ≤ x ≤ 5eπ/2 . y 20 15 10 5 5 10 15 20 x The interval of definition of the solution is approximately [1.04, 24.05]. 34. As x → −∞, e6x → 0 and y → 2x + 3. Now write (1 + ce6x )/(1 − ce6x ) as (e−6x + c)/(e−6x − c). Then, as x → ∞, e−6x → 0 and y → 2x − 3. 35. (a) The substitutions y = y1 + u and dy dy1 du = + dx dx dx lead to dy1 du + = P + Q(y1 + u) + R(y1 + u)2 dx dx 2 = P + Qy1 + Ry1 + Qu + 2y1 Ru + Ru2 or du − (Q + 2y1 R)u = Ru2 . dx This is a Bernoulli equation with n = 2 which can be reduced to the linear equation dw + (Q + 2y1 R)w = −R dx by the substitution w = u−1 . dw + dx 1 4 + w = −1. An integrating x x 2 −1 factor is x3 so that x3 w = − 1 x4 + c or u = − 1 x + cx−3 . Thus, y = + u. 4 4 x 36. Write the differential equation in the form x(y /y) = ln x + ln y and let u = ln y. Then du/dx = y /y and the differential equation becomes x(du/dx) = ln x + u or du/dx − u/x = (ln x)/x, which is first-order and linear. (b) Identify P (x) = −4/x2 , Q(x) = −1/x, and R(x) = 1. Then An integrating factor is e− dx/x − = 1/x, so that (using integration by parts) d 1 ln x u = 2 dx x x and u 1 ln x =− − + c. x x x The solution is ln y = −1 − ln x + cx or y = 37. Write the differential equation as dv 1 + v = 32v −1 , dx x 59 ecx−1 . x 2.5 Solutions by Substitutions and let u = v 2 or v = u1/2 . Then du dv 1 = u−1/2 , dx 2 dx and substituting into the differential equation, we have 1 −1/2 du 1 1/2 = 32u−1/2 u + u 2 dx x or The latter differential equation is linear with integrating factor e du 2 + u = 64. dx x (2/x)dx = x2 , so d 2 [x u] = 64x2 dx and x2 u = 64 3 x +c 3 or v2 = 64 c x+ 2 . 3 x 38. Write the differential equation as dP/dt − aP = −bP 2 and let u = P −1 or P = u−1 . Then dp du = −u−2 , dt dt and substituting into the differential equation, we have −u−2 du − au−1 = −bu−2 dt or The latter differential equation is linear with integrating factor e du + au = b. dt a dt = eat , so d at [e u] = beat dt and b at e +c a b eat P −1 = eat + c a b P −1 = + ce−at a 1 a P = = . −at b/a + ce b + c1 e−at eat u = EXERCISES 2.6 A Numerical Method 1. We identify f (x, y) = 2x − 3y + 1. Then, for h = 0.1, yn+1 = yn + 0.1(2xn − 3yn + 1) = 0.2xn + 0.7yn + 0.1, and y(1.1) ≈ y1 = 0.2(1) + 0.7(5) + 0.1 = 3.8 y(1.2) ≈ y2 = 0.2(1.1) + 0.7(3.8) + 0.1 = 2.98. For h = 0.05, yn+1 = yn + 0.05(2xn − 3yn + 1) = 0.1xn + 0.85yn + 0.1, 60 2.6 and A Numerical Method y(1.05) ≈ y1 = 0.1(1) + 0.85(5) + 0.1 = 4.4 y(1.1) ≈ y2 = 0.1(1.05) + 0.85(4.4) + 0.1 = 3.895 y(1.15) ≈ y3 = 0.1(1.1) + 0.85(3.895) + 0.1 = 3.47075 y(1.2) ≈ y4 = 0.1(1.15) + 0.85(3.47075) + 0.1 = 3.11514. 2. We identify f (x, y) = x + y 2 . Then, for h = 0.1, 2 2 yn+1 = yn + 0.1(xn + yn ) = 0.1xn + yn + 0.1yn , and y(0.1) ≈ y1 = 0.1(0) + 0 + 0.1(0)2 = 0 y(0.2) ≈ y2 = 0.1(0.1) + 0 + 0.1(0)2 = 0.01. For h = 0.05, 2 2 yn+1 = yn + 0.05(xn + yn ) = 0.05xn + yn + 0.05yn , and y(0.05) ≈ y1 = 0.05(0) + 0 + 0.05(0)2 = 0 y(0.1) ≈ y2 = 0.05(0.05) + 0 + 0.05(0)2 = 0.0025 y(0.15) ≈ y3 = 0.05(0.1) + 0.0025 + 0.05(0.0025)2 = 0.0075 y(0.2) ≈ y4 = 0.05(0.15) + 0.0075 + 0.05(0.0075)2 = 0.0150. 3. Separating variables and integrating, we have dy = dx y ln |y| = x + c. and Thus y = c1 ex and, using y(0) = 1, we find c = 1, so y = ex is the solution of the initial-value problem. h=0.1 xn 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 h=0.05 yn 1.0000 1.1000 1.2100 1.3310 1.4641 1.6105 1.7716 1.9487 2.1436 2.3579 2.5937 Actual Value 1.0000 1.1052 1.2214 1.3499 1.4918 1.6487 1.8221 2.0138 2.2255 2.4596 2.7183 % Rel . Abs . Error Error 0.0000 0.00 0.0052 0.47 0.0114 0.93 0.0189 1.40 0.0277 1.86 0.0382 2.32 0.0506 2.77 0.0650 3.23 0.0820 3.68 0.1017 4.13 0.1245 4.58 xn 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 yn 1.0000 1.0500 1.1025 1.1576 1.2155 1.2763 1.3401 1.4071 1.4775 1.5513 1.6289 1.7103 1.7959 1.8856 1.9799 2.0789 2.1829 2.2920 2.4066 2.5270 2.6533 61 Actual Value 1.0000 1.0513 1.1052 1.1618 1.2214 1.2840 1.3499 1.4191 1.4918 1.5683 1.6487 1.7333 1.8221 1.9155 2.0138 2.1170 2.2255 2.3396 2.4596 2.5857 2.7183 % Rel . Abs . Error Error 0.0000 0.00 0.0013 0.12 0.0027 0.24 0.0042 0.36 0.0059 0.48 0.0077 0.60 0.0098 0.72 0.0120 0.84 0.0144 0.96 0.0170 1.08 0.0198 1.20 0.0229 1.32 0.0263 1.44 0.0299 1.56 0.0338 1.68 0.0381 1.80 0.0427 1.92 0.0476 2.04 0.0530 2.15 0.0588 2.27 0.0650 2.39 2.6 A Numerical Method 4. Separating variables and integrating, we have dy = 2x dx and y ln |y| = x2 + c. Thus y = c1 ex and, using y(1) = 1, we find c = e−1 , so y = ex 2 2 h=0.1 xn 1.00 1.10 1.20 1.30 1.40 1.50 5. 7. yn 1.0000 1.2000 1.4640 1.8154 2.2874 2.9278 Actual Value 1.0000 1.2337 1.5527 1.9937 2.6117 3.4903 Abs . Error 0.0000 0.0337 0.0887 0.1784 0.3243 0.5625 xn 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 xn 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 xn yn 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 yn 0.0000 0.0500 0.0976 0.1429 0.1863 0.2278 0.2676 0.3058 0.3427 0.3782 0.4124 1.0000 1.1000 1.2155 1.3492 1.5044 1.6849 1.8955 2.1419 2.4311 2.7714 3.1733 yn 0.5000 0.5125 0.5232 0.5322 0.5395 0.5452 0.5496 0.5527 0.5547 0.5559 0.5565 % Rel . Abs . Error Error 0.0000 0.00 0.0079 0.72 0.0182 1.47 0.0314 2.27 0.0483 3.11 0.0702 4.00 0.0982 4.93 0.1343 5.90 0.1806 6.92 0.2403 7.98 0.3171 9.08 h=0.05 yn 1.0000 1.1000 1.2220 1.3753 1.5735 1.8371 h=0.1 xn 0.00 0.10 0.20 0.30 0.40 0.50 62 Actual Value 1.0000 1.1079 1.2337 1.3806 1.5527 1.7551 1.9937 2.2762 2.6117 3.0117 3.4903 h=0.1 xn 0.00 0.10 0.20 0.30 0.40 0.50 8. h=0.05 yn 0.5000 0.5250 0.5431 0.5548 0.5613 0.5639 % Rel . Error 0.00 2.73 5.71 8.95 12.42 16.12 6. h=0.05 yn 0.0000 0.1000 0.1905 0.2731 0.3492 0.4198 h=0.1 xn 0.00 0.10 0.20 0.30 0.40 0.50 is the solution of the initial-value problem. h=0.05 h=0.1 xn 0.00 0.10 0.20 0.30 0.40 0.50 −1 xn 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 yn 1.0000 1.0500 1.1053 1.1668 1.2360 1.3144 1.4039 1.5070 1.6267 1.7670 1.9332 h=0.05 yn 1.0000 1.1000 1.2159 1.3505 1.5072 1.6902 xn 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 yn 1.0000 1.0500 1.1039 1.1619 1.2245 1.2921 1.3651 1.4440 1.5293 1.6217 1.7219 2.6 9. h=0.1 10. h=0.05 xn 1.00 1.10 1.20 1.30 1.40 1.50 yn 1.0000 1.0000 1.0191 1.0588 1.1231 1.2194 xn 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 h=0.1 yn 1.0000 1.0000 1.0049 1.0147 1.0298 1.0506 1.0775 1.1115 1.1538 1.2057 1.2696 xn 0.00 0.10 0.20 0.30 0.40 0.50 A Numerical Method h=0.05 yn 0.5000 0.5250 0.5499 0.5747 0.5991 0.6231 xn 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 yn 0.5000 0.5125 0.5250 0.5375 0.5499 0.5623 0.5746 0.5868 0.5989 0.6109 0.6228 11. Tables of values were computed using the Euler and RK4 methods. The resulting points were plotted and joined using ListPlot in Mathematica. h=0.25 h=0.1 y h=0.05 y 7 6 5 4 3 2 1 Euler 2 4 6 8 y 7 6 5 4 3 2 1 RK4 10 7 6 5 4 3 2 1 RK4 Euler x 2 4 6 8 10 RK4 Euler x 2 4 6 8 10 x 12. See the comments in Problem 11 above. h=0.25 h=0.1 y h=0.05 y 6 y 6 RK4 5 4 6 RK4 5 Euler 4 Euler 4 3 Euler 3 RK4 5 3 2 2 2 1 1 1 1 2 3 4 5 x 1 2 3 4 5 x 1 2 3 4 5 x 13. Using separation of variables we find that the solution of the differential equation is y = 1/(1 − x2 ), which is undefined at x = 1, where the graph has a vertical asymptote. Because the actual solution of the differential equation becomes unbounded at x approaches 1, very small changes in the inputs x will result in large changes in the corresponding outputs y. This can be expected to have a serious effect on numerical procedures. The graphs below were obtained as described above in Problem 11. 63 2.6 A Numerical Method h=0.25 h=0.1 y 10 y 10 RK4 8 RK4 8 6 6 4 4 Euler 2 Euler 2 0.2 0.4 0.6 0.8 1 x 0.2 0.4 0.6 0.8 1 x EXERCISES 2.7 Linear Models 1. Let P = P (t) be the population at time t, and P0 the initial population. From dP/dt = kP we obtain P = P0 ekt . Using P (5) = 2P0 we find k = 1 ln 2 and P = P0 e(ln 2)t/5 . Setting P (t) = 3P0 we have 3 = e(ln 2)t/5 , so 5 ln 3 = (ln 2)t 5 and t= 5 ln 3 ≈ 7.9 years. ln 2 Setting P (t) = 4P0 we have 4 = e(ln 2)t/5 , so ln 4 = (ln 2)t 5 and t ≈ 10 years. 2. From Problem 1 the growth constant is k = 1 ln 2. Then P = P0 e(1/5)(ln 2)t and 10,000 = P0 e(3/5) ln 2 . Solving 5 for P0 we get P0 = 10,000e−(3/5) ln 2 = 6,597.5. Now P (10) = P0 e(1/5)(ln 2)(10) = 6,597.5e2 ln 2 = 4P0 = 26,390. The rate at which the population is growing is P (10) = kP (10) = 1 (ln 2)26,390 = 3658 persons/year. 5 3. Let P = P (t) be the population at time t. Then dP/dt = kP and P = cekt . From P (0) = c = 500 we see that P = 500ekt . Since 15% of 500 is 75, we have P (10) = 500e10k = 575. Solving for k, we get 1 1 k = 10 ln 575 = 10 ln 1.15. When t = 30, 500 P (30) = 500e(1/10)(ln 1.15)30 = 500e3 ln 1.15 = 760 years and P (30) = kP (30) = 1 (ln 1.15)760 = 10.62 persons/year. 10 4. Let P = P (t) be bacteria population at time t and P0 the initial number. From dP/dt = kP we obtain P = P0 ekt . Using P (3) = 400 and P (10) = 2000 we find 400 = P0 e3k or ek = (400/P0 )1/3 . From P (10) = 2000 we then have 2000 = P0 e10k = P0 (400/P0 )10/3 , so 2000 −7/3 = P0 40010/3 and P0 = 64 2000 40010/3 −3/7 ≈ 201. 2.7 Linear Models 5. Let A = A(t) be the amount of lead present at time t. From dA/dt = kA and A(0) = 1 we obtain A = ekt . Using A(3.3) = 1/2 we find k = 1 3.3 ln(1/2). When 90% of the lead has decayed, 0.1 grams will remain. Setting A(t) = 0.1 we have et(1/3.3) ln(1/2) = 0.1, so t 1 ln = ln 0.1 3.3 2 and t= 3.3 ln 0.1 ≈ 10.96 hours. ln(1/2) 6. Let A = A(t) be the amount present at time t. From dA/dt = kA and A(0) = 100 we obtain A = 100ekt . Using A(6) = 97 we find k = 1 ln 0.97. Then A(24) = 100e(1/6)(ln 0.97)24 = 100(0.97)4 ≈ 88.5 mg. 6 7. Setting A(t) = 50 in Problem 6 we obtain 50 = 100ekt , so kt = ln 1 2 and t= ln(1/2) ≈ 136.5 hours. (1/6) ln 0.97 8. (a) The solution of dA/dt = kA is A(t) = A0 ekt . Letting A = T = −(ln 2)/k. 1 2 A0 and solving for t we obtain the half-life (b) Since k = −(ln 2)/T we have A(t) = A0 e−(ln 2)t/T = A0 2−t/T . (c) Writing 1 A0 = A0 2−t/T as 2−3 = 2−t/T and solving for t we get t = 3T . Thus, an initial amount A0 will 8 decay to 1 A0 in three half-lives. 8 9. Let I = I(t) be the intensity, t the thickness, and I(0) = I0 . If dI/dt = kI and I(3) = 0.25I0 , then I = I0 ekt , k = 1 ln 0.25, and I(15) = 0.00098I0 . 3 10. From dS/dt = rS we obtain S = S0 ert where S(0) = S0 . (a) If S0 = $5000 and r = 5.75% then S(5) = $6665.45. (b) If S(t) =$10,000 then t = 12 years. (c) S ≈ $6651.82 11. Assume that A = A0 ekt and k = −0.00012378. If A(t) = 0.145A0 then t ≈15,600 years. 12. From Example 3 in the text, the amount of carbon present at time t is A(t) = A0 e−0.00012378t . Letting t = 660 and solving for A0 we have A(660) = A0 e−0.0001237(660) = 0.921553A0 . Thus, approximately 92% of the original amount of C-14 remained in the cloth as of 1988. 13. Assume that dT /dt = k(T − 10) so that T = 10 + cekt . If T (0) = 70◦ and T (1/2) = 50◦ then c = 60 and k = 2 ln(2/3) so that T (1) = 36.67◦ . If T (t) = 15◦ then t = 3.06 minutes. 14. Assume that dT /dt = k(T − 5) so that T = 5 + cekt . If T (1) = 55◦ and T (5) = 30◦ then k = − 1 ln 2 and 4 c = 59.4611 so that T (0) = 64.4611◦ . 15. Assume that dT /dt = k(T − 100) so that T = 100 + cekt . If T (0) = 20◦ and T (1) = 22◦ , then c = −80 and k = ln(39/40) so that T (t) = 90◦ , which implies t = 82.1 seconds. If T (t) = 98◦ then t = 145.7 seconds. 16. The differential equation for the first container is dT1 /dt = k1 (T1 − 0) = k1 T1 , whose solution is T1 (t) = c1 ek1 t . Since T1 (0) = 100 (the initial temperature of the metal bar), we have 100 = c1 and T1 (t) = 100ek1 t . After 1 minute, T1 (1) = 100ek1 = 90◦ C, so k1 = ln 0.9 and T1 (t) = 100et ln 0.9 . After 2 minutes, T1 (2) = 100e2 ln 0.9 = 100(0.9)2 = 81◦ C. The differential equation for the second container is dT2 /dt = k2 (T2 − 100), whose solution is T2 (t) = 100 + c2 ek2 t . When the metal bar is immersed in the second container, its initial temperature is T2 (0) = 81, so T2 (0) = 100 + c2 ek2 (0) = 100 + c2 = 81 65 2.7 Linear Models and c2 = −19. Thus, T2 (t) = 100 − 19ek2 t . After 1 minute in the second tank, the temperature of the metal bar is 91◦ C, so T2 (1) = 100 − 19ek2 = 91 9 ek2 = 19 9 k2 = ln 19 and T2 (t) = 100 − 19et ln(9/19) . Setting T2 (t) = 99.9 we have 100 − 19et ln(9/19) = 99.9 0.1 et ln(9/19) = 19 ln(0.1/19) t= ≈ 7.02. ln(9/19) Thus, from the start of the “double dipping” process, the total time until the bar reaches 99.9◦ C in the second container is approximately 9.02 minutes. 17. Using separation of variables to solve dT /dt = k(T − Tm ) we get T (t) = Tm + cekt . Using T (0) = 70 we find c = 70 − Tm , so T (t) = Tm + (70 − Tm )ekt . Using the given observations, we obtain T 1 = Tm + (70 − Tm )ek/2 = 110 2 T (1) = Tm + (70 − Tm )ek = 145. Then, from the first equation, ek/2 = (110 − Tm )/(70 − Tm ) and ek = (ek/2 )2 = 2 110 − Tm 70 − Tm = 145 − Tm 70 − Tm (110 − Tm )2 = 145 − Tm 70 − Tm 2 2 12100 − 220Tm + Tm = 10150 − 250Tm + Tm Tm = 390. The temperature in the oven is 390◦ . 18. (a) The initial temperature of the bath is Tm (0) = 60◦ , so in the short term the temperature of the chemical, which starts at 80◦ , should decrease or cool. Over time, the temperature of the bath will increase toward 100◦ since e−0.1t decreases from 1 toward 0 as t increases from 0. Thus, in the long term, the temperature of the chemical should increase or warm toward 100◦ . (b) Adapting the model for Newton’s law of cooling, we have dT = −0.1(T − 100 + 40e−0.1t ), dt T 100 T (0) = 80. 90 Writing the differential equation in the form 80 dT + 0.1T = 10 − 4e−0.1t dt we see that it is linear with integrating factor e 70 0.1 dt 66 = e0.1t . 10 20 30 40 50 t 2.7 Linear Models Thus d 0.1t [e T ] = 10e0.1t − 4 dt e0.1t T = 100e0.1t − 4t + c and T (t) = 100 − 4te−0.1t + ce−0.1t . Now T (0) = 80 so 100 + c = 80, c = −20 and T (t) = 100 − 4te−0.1t − 20e−0.1t = 100 − (4t + 20)e−0.1t . The thinner curve verifies the prediction of cooling followed by warming toward 100◦ . The wider curve shows the temperature Tm of the liquid bath. 19. From dA/dt = 4 − A/50 we obtain A = 200 + ce−t/50 . −t/50 A = 200 − 170e If A(0) = 30 then c = −170 and . 20. From dA/dt = 0 − A/50 we obtain A = ce−t/50 . If A(0) = 30 then c = 30 and A = 30e−t/50 . 21. From dA/dt = 10 − A/100 we obtain A = 1000 + ce−t/100 . If A(0) = 0 then c = −1000 and A(t) = 1000 − 1000e−t/100 . 22. From Problem 21 the number of pounds of salt in the tank at time t is A(t) = 1000 − 1000e−t/100 . The concentration at time t is c(t) = A(t)/500 = 2 − 2e−t/100 . Therefore c(5) = 2 − 2e−1/20 = 0.0975 lb/gal and limt→∞ c(t) = 2. Solving c(t) = 1 = 2 − 2e−t/100 for t we obtain t = 100 ln 2 ≈ 69.3 min. 23. From dA 10A 2A = 10 − = 10 − dt 500 − (10 − 5)t 100 − t 1 we obtain A = 1000 − 10t + c(100 − t)2 . If A(0) = 0 then c = − 10 . The tank is empty in 100 minutes. 24. With cin (t) = 2 + sin(t/4) lb/gal, the initial-value problem is dA 1 t + A = 6 + 3 sin , dt 100 4 The differential equation is linear with integrating factor e d t/100 A(t)] = [e dt 6 + 3 sin t 4 et/100 A(t) = 600et/100 + and A(t) = 600 + A(0) = 50. dt/100 = et/100 , so et/100 150 t/100 t t 3750 t/100 sin − cos + c, e e 313 4 313 4 150 t 3750 t sin − cos + ce−t/100 . 313 4 313 4 Letting t = 0 and A = 50 we have 600 − 3750/313 + c = 50 and c = −168400/313. Then A(t) = 600 + 150 t 3750 t 168400 −t/100 . sin − cos − e 313 4 313 4 313 The graphs on [0, 300] and [0, 600] below show the effect of the sine function in the input when compared with the graph in Figure 2.38(a) in the text. 67 2.7 Linear Models A t 600 A t 600 500 500 400 400 300 300 200 200 100 100 50 100 25. From 150 200 250 300 t 100 200 300 400 500 600 t dA 4A 2A =3− =3− dt 100 + (6 − 4)t 50 + t we obtain A = 50 + t + c(50 + t)−2 . If A(0) = 10 then c = −100,000 and A(30) = 64.38 pounds. 26. (a) Initially the tank contains 300 gallons of solution. Since brine is pumped in at a rate of 3 gal/min and the mixture is pumped out at a rate of 2 gal/min, the net change is an increase of 1 gal/min. Thus, in 100 minutes the tank will contain its capacity of 400 gallons. (b) The differential equation describing the amount of salt in the tank is A (t) = 6 − 2A/(300 + t) with solution A(t) = 600 + 2t − (4.95 × 107 )(300 + t)−2 , 0 ≤ t ≤ 100, as noted in the discussion following Example 5 in the text. Thus, the amount of salt in the tank when it overflows is A(100) = 800 − (4.95 × 107 )(400)−2 = 490.625 lbs. (c) When the tank is overflowing the amount of salt in the tank is governed by the differential equation dA A = (3 gal/min)(2 lb/gal) − lb/gal (3 gal/min) dt 400 3A =6− , A(100) = 490.625. 400 Solving the equation, we obtain A(t) = 800 + ce−3t/400 . The initial condition yields c = −654.947, so that A(t) = 800 − 654.947e−3t/400 . When t = 150, A(150) = 587.37 lbs. (d) As t → ∞, the amount of salt is 800 lbs, which is to be expected since (400 gal)(2 lb/gal)= 800 lbs. (e) A 800 600 400 200 200 400 600 t 27. Assume L di/dt + Ri = E(t), L = 0.1, R = 50, and E(t) = 50 so that i = and limt→∞ i(t) = 3/5. 68 3 5 + ce−500t . If i(0) = 0 then c = −3/5 2.7 Linear Models 28. Assume L di/dt + Ri = E(t), E(t) = E0 sin ωt, and i(0) = i0 so that i= E0 R 2 ω 2 + R2 L E0 Lω 2 ω 2 + R2 L sin ωt − cos ωt + ce−Rt/L . E0 Lω . L2 ω 2 + R2 29. Assume R dq/dt + (1/C)q = E(t), R = 200, C = 10−4 , and E(t) = 100 so that q = 1/100 + ce−50t . If q(0) = 0 then c = −1/100 and i = 1 e−50t . 2 Since i(0) = i0 we obtain c = i0 + 1 30. Assume R dq/dt + (1/C)q = E(t), R = 1000, C = 5 × 10−6 , and E(t) = 200. Then q = 1000 + ce−200t and 1 i = −200ce−200t . If i(0) = 0.4 then c = − 500 , q(0.005) = 0.003 coulombs, and i(0.005) = 0.1472 amps. We have q→ 1 1000 as t → ∞. 31. For 0 ≤ t ≤ 20 the differential equation is 20 di/dt + 2i = 120. An integrating factor is et/10 , so (d/dt)[et/10 i] = 6et/10 and i = 60 + c1 e−t/10 . If i(0) = 0 then c1 = −60 and i = 60 − 60e−t/10 . For t > 20 the differential equation is 20 di/dt + 2i = 0 and i = c2 e−t/10 . At t = 20 we want c2 e−2 = 60 − 60e−2 so that c2 = 60 e2 − 1 . Thus 60 − 60e−t/10 , 0 ≤ t ≤ 20 i(t) = 2 −t/10 60 e − 1 e , t > 20. 32. Separating variables, we obtain dq dt = E0 − q/C k1 + k2 t q 1 −C ln E0 − ln |k1 + k2 t| + c1 = C k2 (E0 − q/C)−C = c2 . (k1 + k2 t)1/k2 Setting q(0) = q0 we find c2 = (E0 − q0 /C)−C /k1 1/k2 , so (E0 − q/C)−C (E0 − q0 /C)−C = 1/k2 1/k (k1 + k2 t) k1 2 E0 − q C −C E0 − = E0 − q0 C q q0 = E0 − C C −C k1 k + k2 t k1 k + k2 t q = E0 C + (q0 − E0 C) −1/k2 1/Ck2 k1 k + k2 t 1/Ck2 . 33. (a) From m dv/dt = mg − kv we obtain v = mg/k + ce−kt/m . If v(0) = v0 then c = v0 − mg/k and the solution of the initial-value problem is mg mg −kt/m v(t) = . + v0 − e k k (b) As t → ∞ the limiting velocity is mg/k. (c) From ds/dt = v and s(0) = 0 we obtain s(t) = mg m mg −kt/m m mg t− v0 − e v0 − . + k k k k k 34. (a) Integrating d2 s/dt2 = −g we get v(t) = ds/dt = −gt + c. From v(0) = 300 we find c = 300, and we are given g = 32, so the velocity is v(t) = −32t + 300. 69 2.7 Linear Models (b) Integrating again and using s(0) = 0 we get s(t) = −16t2 + 300t. The maximum height is attained when v = 0, that is, at ta = 9.375. The maximum height will be s(9.375) = 1406.25 ft. 35. When air resistance is proportional to velocity, the model for the velocity is m dv/dt = −mg − kv (using the fact that the positive direction is upward.) Solving the differential equation using separation of variables we obtain v(t) = −mg/k + ce−kt/m . From v(0) = 300 we get mg mg −kt/m v(t) = − + 300 + e . k k Integrating and using s(0) = 0 we find mg m mg s(t) = − t+ 300 + (1 − e−kt/m ). k k k Setting k = 0.0025, m = 16/32 = 0.5, and g = 32 we have s(t) = 1,340,000 − 6,400t − 1,340,000e−0.005t and v(t) = −6,400 + 6,700e−0.005t . The maximum height is attained when v = 0, that is, at ta = 9.162. The maximum height will be s(9.162) = 1363.79 ft, which is less than the maximum height in Problem 34. 36. Assuming that the air resistance is proportional to velocity and the positive direction is downward with s(0) = 0, the model for the velocity is m dv/dt = mg − kv. Using separation of variables to solve this differential equation, we obtain v(t) = mg/k + ce−kt/m . Then, using v(0) = 0, we get v(t) = (mg/k)(1 − e−kt/m ). Letting k = 0.5, m = (125 + 35)/32 = 5, and g = 32, we have v(t) = 320(1 − e−0.1t ). Integrating, we find s(t) = 320t + 3200e−0.1t + c1 . Solving s(0) = 0 for c1 we find c1 = −3200, therefore s(t) = 320t + 3200e−0.1t − 3200. At t = 15, when the parachute opens, v(15) = 248.598 and s(15) = 2314.02. At this time the value of k changes to k = 10 and the new initial velocity is v0 = 248.598. With the parachute open, the skydiver’s velocity is vp (t) = mg/k + c2 e−kt/m , where t is reset to 0 when the parachute opens. Letting m = 5, g = 32, and k = 10, this gives vp (t) = 16 + c2 e−2t . From v(0) = 248.598 we find c2 = 232.598, so vp (t) = 16 + 232.598e−2t . Integrating, we get sp (t) = 16t − 116.299e−2t + c3 . Solving sp (0) = 0 for c3 , we find c3 = 116.299, so sp (t) = 16t − 116.299e−2t + 116.299. Twenty seconds after leaving the plane is five seconds after the parachute opens. The skydiver’s velocity at this time is vp (5) = 16.0106 ft/s and she has fallen a total of s(15) + sp (5) = 2314.02 + 196.294 = 2510.31 ft. Her terminal velocity is limt→∞ vp (t) = 16, so she has very nearly reached her terminal velocity five seconds after the parachute opens. When the parachute opens, the distance to the ground is 15,000 − s(15) = 15,000 − 2,314 = 12,686 ft. Solving sp (t) = 12,686 we get t = 785.6 s = 13.1 min. Thus, it will take her approximately 13.1 minutes to reach the ground after her parachute has opened and a total of (785.6 + 15)/60 = 13.34 minutes after she exits the plane. 37. (a) The differential equation is first-order and linear. Letting b = k/ρ, the integrating factor is e 3b dt/(bt+r0 ) = 3 (r0 + bt) . Then d [(r0 + bt)3 v] = g(r0 + bt)3 dt and (r0 + bt)3 v = g (r0 + bt)4 + c. 4b The solution of the differential equation is v(t) = (g/4b)(r0 + bt) + c(r0 + bt)−3 . Using v(0) = 0 we find 4 c = −gr0 /4b, so that v(t) = 4 4 g gr0 gρ k gρr0 = . (r0 + bt) − r0 + t − 4b 4b(r0 + bt)3 4k ρ 4k(r0 + kt/ρ)3 (b) Integrating dr/dt = k/ρ we get r = kt/ρ + c. Using r(0) = r0 we have c = r0 , so r(t) = kt/ρ + r0 . 70 2.7 Linear Models (c) If r = 0.007 ft when t = 10 s, then solving r(10) = 0.007 for k/ρ, we obtain k/ρ = −0.0003 and r(t) = 0.01 − 0.0003t. Solving r(t) = 0 we get t = 33.3, so the raindrop will have evaporated completely at 33.3 seconds. 38. Separating variables, we obtain dP/P = k cos t dt, so ln |P | = k sin t + c and P = c1 ek sin t . If P (0) = P0 , then c1 = P0 and P = P0 ek sin t . 39. (a) From dP/dt = (k1 − k2 )P we obtain P = P0 e(k1 −k2 )t where P0 = P (0). (b) If k1 > k2 then P → ∞ as t → ∞. If k1 = k2 then P = P0 for every t. If k1 < k2 then P → 0 as t → ∞. 40. (a) Solving k1 (M − A) − k2 A = 0 for A we find the equilibrium solution A = k1 M/(k1 + k2 ). From the phase portrait we see that limt→∞ A(t) = k1 M/(k1 + k2 ). A Since k2 > 0, the material will never be completely memorized and the larger k2 is, the less the amount of material will be memorized over time. M k1 k1 k2 (b) Write the differential equation in the form dA/dt+(k1 +k2 )A = k1 M . Then an integrating factor is e(k1 +k2 )t , and d (k1 +k2 )t A = k1 M e(k1 +k2 )t e dt k1 M (k1 +k2 )t e(k1 +k2 )t A = e +c k1 + k2 A= Using A(0) = 0 we find c = − A→ k1 M + ce−(k1 +k2 )t . k1 + k2 k1 M k1 M 1 − e−(k1 +k2 )t . As t → ∞, and A = k1 + k2 k1 + k2 k1 M . k1 + k2 x 41. (a) Solving r −kx = 0 for x we find the equilibrium solution x = r/k. When x < r/k, dx/dt > 0 and when x > r/k, dx/dt < 0. From the phase portrait we see that limt→∞ x(t) = r/k. r k 71 2.7 Linear Models (b) From dx/dt = r − kx and x(0) = 0 we obtain x = r/k − (r/k)e−kt so that x → r/k as t → ∞. If x(T ) = r/2k then T = (ln 2)/k. x rêk t 42. The bar removed from the oven has an initial temperature of 300◦ F and, after being removed from the oven, approaches a temperature of 70◦ F. The bar taken from the room and placed in the oven has an initial temperature of 70◦ F and approaches a temperature of 300◦ F in the oven. Since the two temperature functions are continuous they must intersect at some time, t∗ . 43. (a) For 0 ≤ t < 4, 6 ≤ t < 10 and 12 ≤ t < 16, no voltage is applied to the heart and E(t) = 0. At the other times, the differential equation is dE/dt = −E/RC. Separating variables, integrating, and solving for e, we get E = ke−t/RC , subject to E(4) = E(10) = E(16) = 12. These intitial conditions yield, respectively, k = 12e4/RC , k = 12e10/RC , k = 12e16/RC , and k = 12e22/RC . Thus 0 ≤ t < 4, 6 ≤ t < 10, 12 ≤ t < 16 0, 12e(4−t)/RC , 4 ≤ t < 6 E(t) = 12e(10−t)/RC , 10 ≤ t < 12 12e(16−t)/RC , 16 ≤ t < 18 12e(22−t)/RC , 22 ≤ t < 24. (b) E 10 5 4 6 10 12 16 18 22 24 t 44. (a) (i) Using Newton’s second law of motion, F = ma = m dv/dt, the differential equation for the velocity v is m dv = mg sin θ dt or dv = g sin θ, dt where mg sin θ, 0 < θ < π/2, is the component of the weight along the plane in the direction of motion. (ii) The model now becomes dv m = mg sin θ − µmg cos θ, dt where µmg cos θ is the component of the force of sliding friction (which acts perpendicular to the plane) along the plane. The negative sign indicates that this component of force is a retarding force which acts in the direction opposite to that of motion. (iii) If air resistance is taken to be proportional to the instantaneous velocity of the body, the model becomes dv = mg sin θ − µmg cos θ − kv, dt where k is a constant of proportionality. m 72 2.7 Linear Models (b) (i) With m = 3 slugs, the differential equation is 3 dv 1 = (96) · dt 2 or dv = 16. dt Integrating the last equation gives v(t) = 16t + c1 . Since v(0) = 0, we have c1 = 0 and so v(t) = 16t. (ii) With m = 3 slugs, the differential equation is √ √ dv 3 3 1 3 = (96) · − · (96) · dt 2 4 2 dv = 4. dt or In this case v(t) = 4t. (iii) When the retarding force due to air resistance is taken into account, the differential equation for velocity v becomes 3 √ √ dv 3 3 1 1 = (96) · − · (96) · − v dt 2 4 2 4 or 3 dv 1 = 12 − v. dt 4 The last differential equation is linear and has solution v(t) = 48 + c1 e−t/12 . Since v(0) = 0, we find c1 = −48, so v(t) = 48 − 48e−t/12 . 45. (a) (i) If s(t) is distance measured down the plane from the highest point, then ds/dt = v. Integrating ds/dt = 16t gives s(t) = 8t2 + c2 . Using s(0) = 0 then gives c2 = 0. Now the length L of the plane is L = 50/ sin 30◦ = 100 ft. The time it takes the box to slide completely down the plane is the solution of s(t) = 100 or t2 = 25/2, so t ≈ 3.54 s. (ii) Integrating ds/dt = 4t gives s(t) = 2t2 + c2 . Using s(0) = 0 gives c2 = 0, so s(t) = 2t2 and the solution of s(t) = 100 is now t ≈ 7.07 s. (iii) Integrating ds/dt = 48 − 48e−t/12 and using s(0) = 0 to determine the constant of integration, we obtain s(t) = 48t + 576e−t/12 − 576. With the aid of a CAS we find that the solution of s(t) = 100, or 100 = 48t + 576e−t/12 − 576 or 0 = 48t + 576e−t/12 − 676, is now t ≈ 7.84 s. (b) The differential equation m dv/dt = mg sin θ − µmg cos θ can be written m dv = mg cos θ(tan θ − µ). dt If tan θ = µ, dv/dt = 0 and v(0) = 0 implies that v(t) = 0. If tan θ < µ and v(0) = 0, then integration implies v(t) = g cos θ(tan θ − µ)t < 0 for all time t. √ (c) Since tan 23◦ = 0.4245 and µ = 3/4 = 0.4330, we see that tan 23◦ < 0.4330. The differential equation √ is dv/dt = 32 cos 23◦ (tan 23◦ − 3/4) = −0.251493. Integration and the use of the initial condition gives v(t) = −0.251493t + 1. When the box stops, v(t) = 0 or 0 = −0.251493t + 1 or t = 3.976254 s. From s(t) = −0.125747t2 + t we find s(3.976254) = 1.988119 ft. (d) With v0 > 0, v(t) = −0.251493t + v0 and s(t) = −0.125747t2 + v0 t. Because two real positive solutions of the equation s(t) = 100, or 0 = −0.125747t2 + v0 t − 100, would be physically meaningless, we use 2 the quadratic formula and require that b2 − 4ac = 0 or v0 − 50.2987 = 0. From this last equality we find v0 ≈ 7.092164 ft/s. For the time it takes the box to traverse the entire inclined plane, we must have 0 = −0.125747t2 + 7.092164t − 100. Mathematica gives complex roots for the last equation: t = 28.2001 ± 0.0124458i. But, for 0 = −0.125747t2 + 7.092164691t − 100, 73 2.7 Linear Models the roots are t = 28.1999 s and t = 28.2004 s. So if v0 > 7.092164, we are guaranteed that the box will slide completely down the plane. 46. (a) We saw in part (b) of Problem 34 that the ascent time is ta = 9.375. To find when the cannonball hits the ground we solve s(t) = −16t2 + 300t = 0, getting a total time in flight of t = 18.75 s. Thus, the time of descent is td = 18.75 − 9.375 = 9.375. The impact velocity is vi = v(18.75) = −300, which has the same magnitude as the initial velocity. (b) We saw in Problem 35 that the ascent time in the case of air resistance is ta = 9.162. Solving s(t) = 1,340,000 − 6,400t − 1,340,000e−0.005t = 0 we see that the total time of flight is 18.466 s. Thus, the descent time is td = 18.466 − 9.162 = 9.304. The impact velocity is vi = v(18.466) = −290.91, compared to an initial velocity of v0 = 300. EXERCISES 2.8 Nonlinear Models 1. (a) Solving N (1 − 0.0005N ) = 0 for N we find the equilibrium solutions N = 0 and N = 2000. When 0 < N < 2000, dN/dt > 0. From the phase portrait we see that limt→∞ N (t) = 2000. A graph of the solution is shown in part (b). N 2000 0 (b) Separating variables and integrating we have dN 1 1 = − dN = dt N (1 − 0.0005N ) N N − 2000 N 2000 1500 1000 and 500 ln N − ln(N − 2000) = t + c. 5 10 15 20 t Solving for N we get N (t) = 2000ec+t /(1 + ec+t ) = 2000ec et /(1 + ec et ). Using N (0) = 1 and solving for ec we find ec = 1/1999 and so N (t) = 2000et /(1999 + et ). Then N (10) = 1833.59, so 1834 companies are expected to adopt the new technology when t = 10. 2. From dN/dt = N (a − bN ) and N (0) = 500 we obtain N= 500a . 500b + (a − 500b)e−at Since limt→∞ N = a/b = 50,000 and N (1) = 1000 we have a = 0.7033, b = 0.00014, and N = 50,000/(1 + 99e−0.7033t ) . 74 2.8 3. From dP/dt = P 10−1 − 10−7 P Nonlinear Models and P (0) = 5000 we obtain P = 500/(0.0005 + 0.0995e−0.1t ) so that P → 1,000,000 as t → ∞. If P (t) = 500,000 then t = 52.9 months. 4. (a) We have dP/dt = P (a − bP ) with P (0) = 3.929 million. Using separation of variables we obtain 3.929a a/b = 3.929b + (a − 3.929b)e−at 1 + (a/3.929b − 1)e−at c = , 1 + (c/3.929 − 1)e−at P (t) = where c = a/b. At t = 60(1850) the population is 23.192 million, so 23.192 = c 1 + (c/3.929 − 1)e−60a or c = 23.192 + 23.192(c/3.929 − 1)e−60a . At t = 120(1910), 91.972 = c 1 + (c/3.929 − 1)e−120a or c = 91.972 + 91.972(c/3.929 − 1)(e−60a )2 . Combining the two equations for c we get (c − 23.192)/23.192 c/3.929 − 1 2 c c − 91.972 −1 = 3.929 91.972 or 91.972(3.929)(c − 23.192)2 = (23.192)2 (c − 91.972)(c − 3.929). The solution of this quadratic equation is c = 197.274. This in turn gives a = 0.0313. Therefore, P (t) = (b) Year 1790 1800 1810 1820 1830 1840 1850 1860 1870 1880 1890 1900 1910 1920 1930 1940 1950 Census Population 3.929 5.308 7.240 9.638 12.866 17.069 23.192 31.433 38.558 50.156 62.948 75.996 91.972 105.711 122.775 131.669 150.697 Predicted Population 3.929 5.334 7.222 9.746 13.090 17.475 23.143 30.341 39.272 50.044 62.600 76.666 91.739 107.143 122.140 136.068 148.445 Error 0.000 -0.026 0.018 -0.108 -0.224 -0.406 0.049 1.092 -0.714 0.112 0.348 -0.670 0.233 -1.432 0.635 -4.399 2.252 197.274 . 1 + 49.21e−0.0313t % Error 0.00 -0.49 0.24 -1.12 -1.74 -2.38 0.21 3.47 -1.85 0.22 0.55 -0.88 0.25 -1.35 0.52 -3.34 1.49 The model predicts a population of 159.0 million for 1960 and 167.8 million for 1970. The census populations for these years were 179.3 and 203.3, respectively. The percentage errors are 12.8 and 21.2, respectively. 75 2.8 Nonlinear Models 5. (a) The differential equation is dP/dt = P (5 − P ) − 4. Solving P (5 − P ) − 4 = 0 for P we P obtain equilibrium solutions P = 1 and P = 4. The phase portrait is shown on the right and solution curves are shown in part (b). We see that for P0 > 4 and 1 < P0 < 4 the population approaches 4 as t increases. For 0 < P < 1 the population decreases to 0 in finite time. 4 1 (b) The differential equation is P dP = P (5 − P ) − 4 = −(P 2 − 5P + 4) = −(P − 4)(P − 1). dt Separating variables and integrating, we obtain 4 1 3 dP = −dt (P − 4)(P − 1) 1/3 1/3 − P −4 P −1 t dP = −dt 1 P −4 ln = −t + c 3 P −1 P −4 = c1 e−3t . P −1 Setting t = 0 and P = P0 we find c1 = (P0 − 4)/(P0 − 1). Solving for P we obtain P (t) = 4(P0 − 1) − (P0 − 4)e−3t . (P0 − 1) − (P0 − 4)e−3t (c) To find when the population becomes extinct in the case 0 < P0 < 1 we set P = 0 in P −4 P0 − 4 −3t = e P −1 P0 − 1 from part (a) and solve for t. This gives the time of extinction 1 4(P0 − 1) t = − ln . 3 P0 − 4 6. Solving P (5 − P ) − if P0 < 5 2 25 4 = 0 for P we obtain the equilibrium solution P = 5 2 . For P = 5 2 , dP/dt < 0. Thus, , the population becomes extinct (otherwise there would be another equilibrium solution.) Using separation of variables to solve the initial-value problem, we get P (t) = [4P0 + (10P0 − 25)t]/[4 + (4P0 − 10)t]. To find when the population becomes extinct for P0 < extinction is t = 4P0 /5(5 − 2P0 ). 5 2 we solve P (t) = 0 for t. We see that the time of 7. Solving P (5 − P ) − 7 = 0 for P we obtain complex roots, so there are no equilibrium solutions. Since dP/dt < 0 for all values of P , the population becomes extinct for any initial condition. Using separation of variables to solve the initial-value problem, we get √ √ 5 3 3 2P0 − 5 √ P (t) = + tan tan−1 t . − 2 2 2 3 76 2.8 Solving P (t) = 0 for t we see that the time of extinction is √ √ √ 2 √ t= 3 tan−1 (5/ 3 ) + 3 tan−1 (2P0 − 5)/ 3 3 Nonlinear Models . P 8. (a) The differential equation is dP/dt = P (1 − ln P ), which has the equilibrium solution P = e. When P0 > e, dP/dt < 0, and when P0 < e, dP/dt > 0. e t (b) The differential equation is dP/dt = P (1 + ln P ), which has the equilibrium solution P = 1/e. When P0 > 1/e, dP/dt > 0, and when P0 < 1/e, dP/dt < 0. P 1êe t −bt (c) From dP/dt = P (a − b ln P ) we obtain −(1/b) ln |a − b ln P | = t + c1 so that P = ea/b e−ce . If P (0) = P0 then c = (a/b) − ln P0 . 9. Let X = X(t) be the amount of C at time t and dX/dt = k(120 − 2X)(150 − X). If X(0) = 0 and X(5) = 10, then 150 − 150e180kt X(t) = , 1 − 2.5e180kt where k = .0001259 and X(20) = 29.3 grams. Now by L’Hˆpital’s rule, X → 60 as t → ∞, so that the amount o of A → 0 and the amount of B → 30 as t → ∞. 10. From dX/dt = k(150 − X)2 , X(0) = 0, and X(5) = 10 we obtain X = 150 − 150/(150kt + 1) where k = .000095238. Then X(20) = 33.3 grams and X → 150 as t → ∞ so that the amount of A → 0 and the amount of B → 0 as t → ∞. If X(t) = 75 then t = 70 minutes. √ 11. (a) The initial-value problem is dh/dt = −8Ah h /Aw , h(0) = H. h 10 8 6 4 2 Separating variables and integrating we have √ dh 8A 8A √ = − h dt and 2 h = − h t + c. Aw Aw h √ Using h(0) = H we find c = 2 H , so the solution of the 500 √ initial-value problem is h(t) = (Aw H − 4Ah t)/Aw , where √ Aw H − 4Ah t ≥ 0. Thus, √ h(t) = (Aw H − 4Ah t)2 /A2 for 0 ≤ t ≤ Aw H/4Ah . w 1000 1500 t (b) Identifying H = 10, Aw = 4π, and Ah = π/576 we have h(t) = t2 /331,776 − ( 5/2 /144)t + 10. Solving √ h(t) = 0 we see that the tank empties in 576 10 seconds or 30.36 minutes. 12. To obtain the solution of this differential equation we use h(t) from Problem 13 in Exercises 1.3. Then √ h(t) = (Aw H − 4cAh t)2 /A2 . Solving h(t) = 0 with c = 0.6 and the values from Problem 11 we see that w the tank empties in 3035.79 seconds or 50.6 minutes. 77 2.8 Nonlinear Models 13. (a) Separating variables and integrating gives 6h3/2 dh = −5t 12 5/2 h = −5t + c. 5 and √ √ 2/5 Using h(0) = 20 we find c = 1920 5 , so the solution of the initial-value problem is h(t) = 800 5− 25 t . 12 √ Solving h(t) = 0 we see that the tank empties in 384 5 seconds or 14.31 minutes. √ (b) When the height of the water is h, the radius of the top of the water is r = h tan 30◦ = h/ 3 and Aw = πh2 /3. The differential equation is dh Ah = −c dt Aw 2gh = −0.6 π(2/12)2 √ 2 64h = − 3/2 . πh2 /3 5h Separating variables and integrating gives 5h3/2 dh = −2 dt and 2h5/2 = −2t + c. Using h(0) = 9 we find c = 486, so the solution of the initial-value problem is h(t) = (243 − t)2/5 . Solving h(t) = 0 we see that the tank empties in 24.3 seconds or 4.05 minutes. 14. When the height of the water is h, the radius of the top of the water is 2 5 (20 − h) and Aw = 4π(20 − h) /25. The differential equation is 2 dh Ah = −c dt Aw √ √ h π(2/12)2 5 2gh = −0.6 64h = − . 4π(20 − h)2 /25 6 (20 − h)2 Separating variables and integrating we have √ (20 − h)2 80 2 5 5 √ dh = − dt and 800 h − h3/2 + h5/2 = − t + c. 6 3 5 6 h √ Using h(0) = 20 we find c = 2560 5/3, so an implicit solution of the initial-value problem is √ √ 80 2 5 2560 5 800 h − h3/2 + h5/2 = − t + . 3 5 6 3 √ To find the time it takes the tank to empty we set h = 0 and solve for t. The tank empties in 1024 5 seconds or 38.16 minutes. Thus, the tank empties more slowly when the base of the cone is on the bottom. 15. (a) After separating variables we obtain m dv = dt mg − kv 2 1 dv √ = dt g 1 − ( k v/√mg )2 √ mg k/mg dv √ √ = dt √ k g 1 − ( k v/ mg )2 √ m kv tanh−1 √ =t+c kg mg √ kv kg −1 tanh √ = t + c1 . mg m Thus the velocity at time t is mg kg tanh t + c1 k m √ √ Setting t = 0 and v = v0 we find c1 = tanh−1 ( k v0 / mg ). v(t) = 78 . 2.8 (b) Since tanh t → 1 as t → ∞, we have v → mg/k as t → ∞. (c) Integrating the expression for v(t) in part (a) we obtain an integral of the form s(t) = mg k kg t + c1 m tanh Nonlinear Models dt = m ln cosh k kg t + c1 m du/u: + c2 . Setting t = 0 and s = 0 we find c2 = −(m/k) ln(cosh c1 ), where c1 is given in part (a). 16. The differential equation is m dv/dt = −mg − kv 2 . Separating variables and integrating, we have dv dt =− mg + kv 2 m √ 1 kv 1 √ tan−1 √ =− t+c mg m mgk √ kv gk −1 tan =− t + c1 √ mg m v(t) = Setting v(0) = 300, m = 16 32 = 1 2 mg tan c1 − k gk t . m , g = 32, and k = 0.0003, we find v(t) = 230.94 tan(c1 − 0.138564t) and c1 = 0.914743. Integrating v(t) = 230.94 tan(0.914743 − 0.138564t) we get s(t) = 1666.67 ln | cos(0.914743 − 0.138564t)| + c2 . Using s(0) = 0 we find c2 = 823.843. Solving v(t) = 0 we see that the maximum height is attained when t = 6.60159. The maximum height is s(6.60159) = 823.843 ft. 17. (a) Let ρ be the weight density of the water and V the volume of the object. Archimedes’ principle states that the upward buoyant force has magnitude equal to the weight of the water displaced. Taking the positive direction to be down, the differential equation is m dv = mg − kv 2 − ρV. dt (b) Using separation of variables we have m dv = dt (mg − ρV ) − kv 2 √ m k dv √ √ √ = dt k ( mg − ρV )2 − ( k v)2 √ m kv 1 −1 √ √ tanh √ = t + c. mg − ρV k mg − ρV Thus v(t) = mg − ρV tanh k (c) Since tanh t → 1 as t → ∞, the terminal velocity is 79 √ kmg − kρV t + c1 . m (mg − ρV )/k . 2.8 Nonlinear Models 18. (a) Writing the equation in the form (x − x2 + y 2 )dx + y dy = 0 we identify M = x − x2 + y 2 and N = y. Since M and N are both homogeneous functions of degree 1 we use the substitution y = ux. It follows that x− x2 + u2 x2 dx + ux(u dx + x du) = 0 x 1− 1 + u2 + u2 dx + x2 u du = 0 u du dx √ = x 1 + u2 − 1 + u2 u du dx √ √ = . x 1 + u2 (1 − 1 + u2 ) √ √ Letting w = 1 − 1 + u2 we have dw = −u du/ 1 + u2 so that − − ln 1 − 1 + u2 = ln |x| + c 1− 1 √ = c1 x 1 + u2 1− 1 + u2 = − 1+ 1+ c2 = x c2 x (−c2 = 1/c1 ) y2 x2 1+ 2c2 y2 c2 + 2 =1+ 2 . 2 x x x Solving for y 2 we have c2 c2 x+ 2 2 which is a family of parabolas symmetric with respect to the x-axis with vertex at (−c2 /2, 0) and focus at the origin. y 2 = 2c2 x + c2 = 4 2 (b) Let u = x2 + y 2 so that du dy = 2x + 2y . dx dx Then dy 1 du = −x dx 2 dx and the differential equation can be written in the form y √ 1 du − x = −x + u 2 dx or 1 du √ = u. 2 dx Separating variables and integrating gives du √ = dx 2 u √ u=x+c u = x2 + 2cx + c2 x2 + y 2 = x2 + 2cx + c2 y 2 = 2cx + c2 . 19. (a) From 2W 2 − W 3 = W 2 (2 − W ) = 0 we see that W = 0 and W = 2 are constant solutions. 80 2.8 Nonlinear Models (b) Separating variables and using a CAS to integrate we get dW √ = dx W 4 − 2W and − tanh−1 1√ 4 − 2W 2 = x + c. Using the facts that the hyperbolic tangent is an odd function and 1 − tanh2 x = sech2 x we have 1√ 4 − 2W = tanh(−x − c) = − tanh(x + c) 2 1 (4 − 2W ) = tanh2 (x + c) 4 1 1 − W = tanh2 (x + c) 2 1 W = 1 − tanh2 (x + c) = sech2 (x + c). 2 Thus, W (x) = 2 sech2 (x + c). (c) Letting x = 0 and W = 2 we find that sech2 (c) = 1 and c = 0. W 2 −3 3 x 20. (a) Solving r2 + (10 − h)2 = 102 for r2 we see that r2 = 20h − h2 . Combining the rate of input of water, π, with the rate of output due to evaporation, kπr2 = kπ(20h − h2 ), we have dV /dt = π − kπ(20h − h2 ). Using V = 10πh2 − 1 πh3 , we see also that dV /dt = (20πh − πh2 )dh/dt. Thus, 3 (20πh − πh2 ) dh = π − kπ(20h − h2 ) dt and 1 − 20kh + kh2 dh = . dt 20h − h2 (b) Letting k = 1/100, separating variables and integrating (with the help of a CAS), we get 100h(h − 20) dh = dt (h − 10)2 and 100(h2 − 10h + 100) = t + c. 10 − h h 10 8 6 Using h(0) = 0 we find c = 1000, and solving for h we get h(t) = √ 0.005 t2 + 4000t−t , where the positive square root is chosen because 4 h ≥ 0. 2 t 2000 4000 6000 8000 10000 (c) The volume of the tank is V = 2 π(10)3 feet, so at a rate of π cubic feet per minute, the tank will fill in 3 2 3 3 (10) ≈ 666.67 minutes ≈ 11.11 hours. (d) At 666.67 minutes, the depth of the water is h(666.67) = 5.486 feet. From the graph in (b) we suspect that limt→∞ h(t) = 10, in which case the tank will never completely fill. To prove this we compute the limit of h(t): t2 + 4000t − t2 lim h(t) = 0.005 lim t2 + 4000t − t = 0.005 lim √ t→∞ t→∞ t→∞ t2 + 4000t + t 4000t 4000 = 0.005 lim = 0.005(2000) = 10. = 0.005 t→∞ t 1 + 4000/t + t 1+1 81 2.8 Nonlinear Models 21. (a) t P(t) Q(t) 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 3.929 5.308 7.240 9.638 12.866 17.069 23.192 31.433 38.558 50.156 62.948 75.996 91.972 105.711 122.775 131.669 150.697 179.300 0.035 0.036 0.033 0.033 0.033 0.036 0.036 0.023 0.030 0.026 0.021 0.021 0.015 0.016 0.007 0.014 0.019 (b) The regression line is Q = 0.0348391 − 0.000168222P . Q 0.035 0.03 0.025 0.02 0.015 0.01 0.005 20 40 60 80 100 120 140 P (c) The solution of the logistic equation is given in equation (5) in the text. Identifying a = 0.0348391 and b = 0.000168222 we have P (t) = aP0 . bP0 + (a − bP0 )e−at (d) With P0 = 3.929 the solution becomes P (t) = (e) 0.136883 . 0.000660944 + 0.0341781e−0.0348391t P 175 150 125 100 75 50 25 25 50 75 100 125 150 t (f ) We identify t = 180 with 1970, t = 190 with 1980, and t = 200 with 1990. The model predicts P (180) = 188.661, P (190) = 193.735, and P (200) = 197.485. The actual population figures for these years are 203.303, 226.542, and 248.765 millions. As t → ∞, P (t) → a/b = 207.102. 82 2.8 Nonlinear Models 22. (a) Using a CAS to solve P (1 − P ) + 0.3e−P = 0 for P we see that P = 1.09216 is an equilibrium solution. (b) Since f (P ) > 0 for 0 < P < 1.09216, the solution P (t) of dP/dt = P (1 − P ) + 0.3e−P , f P (0) = P0 , 2 is increasing for P0 < 1.09216. Since f (P ) < 0 for P > 1.09216, the solution P (t) is decreasing for P0 > 1.09216. Thus P = 1.09216 is an attractor. 1 0.5 1 1.5 2 2.5 3 p -1 -2 (c) The curves for the second initial-value problem are thicker. The equilibrium solution for the logic model is P = 1. Comparing 1.09216 and 1, we p 2 1.5 see that the percentage increase is 9.216%. 1 0.5 2 4 6 8 10 t 23. To find td we solve m dv = mg − kv 2 , dt v(0) = 0 using separation of variables. This gives mg tanh k v(t) = kg t. m Integrating and using s(0) = 0 gives s(t) = m ln cosh k kg t . m To find the time of descent we solve s(t) = 823.84 and find td = 7.77882. The impact velocity is v(td ) = 182.998, which is positive because the positive direction is downward. 24. (a) Solving vt = 2 mg/k for k we obtain k = mg/vt . The differential equation then becomes m dv mg = mg − 2 v 2 dt vt or 1 dv = g 1 − 2 v2 . dt vt Separating variables and integrating gives vt tanh−1 v = gt + c1 . vt The initial condition v(0) = 0 implies c1 = 0, so v(t) = vt tanh gt . vt We find the distance by integrating: s(t) = vt tanh gt v2 gt dt = t ln cosh vt g vt 83 + c2 . 2.8 Nonlinear Models The initial condition s(0) = 0 implies c2 = 0, so s(t) = 2 vt gt ln cosh g vt . In 25 seconds she has fallen 20,000 − 14,800 = 5,200 feet. Using a CAS to solve 2 5200 = (vt /32) ln cosh 32(25) vt for vt gives vt ≈ 271.711 ft/s. Then s(t) = 2 vt gt ln cosh g vt = 2307.08 ln(cosh 0.117772t). (b) At t = 15, s(15) = 2,542.94 ft and v(15) = s (15) = 256.287 ft/sec. 25. While the object is in the air its velocity is modeled by the linear differential equation m dv/dt = mg −kv. Using m = 160, k = 1 , and g = 32, the differential equation becomes dv/dt + (1/640)v = 32. The integrating factor 4 dt/640 = et/640 and the solution of the differential equation is et/640 v = 32et/640 dt = 20,480et/640 + c. Using v(0) = 0 we see that c = −20,480 and v(t) = 20,480 − 20,480e−t/640 . Integrating we get s(t) = 20,480t + is e 13,107,200e−t/640 + c. Since s(0) = 0, c = −13,107,200 and s(t) = −13,107,200 + 20,480t + 13,107,200e−t/640 . To find when the object hits the liquid we solve s(t) = 500 − 75 = 425, obtaining ta = 5.16018. The velocity at the time of impact with the liquid is va = v(ta ) = 164.482. When the object is in the liquid its velocity is modeled by the nonlinear differential equation m dv/dt = mg − kv 2 . Using m = 160, g = 32, and k = 0.1 this becomes dv/dt = (51,200 − v 2 )/1600. Separating variables and integrating we have √ √ dv 2 v − 160 2 dt 1 √ = and ln t + c. = 51,200 − v 2 1600 640 1600 v + 160 2 √ Solving v(0) = va = 164.482 we obtain c = −0.00407537. Then, for v < 160 2 = 226.274, √ √ √ √ v − 160 2 v − 160 2 2t/5−1.8443 √ =e √ = e 2t/5−1.8443 . or − v + 160 2 v + 160 2 Solving for v we get √ v(t) = 13964.6 − 2208.29e 2t/5 61.7153 + 9.75937e 2t/5 √ Integrating we find √ s(t) = 226.275t − 1600 ln(6.3237 + e . 2t/5 ) + c. Solving s(0) = 0 we see that c = 3185.78, so √ s(t) = 3185.78 + 226.275t − 1600 ln(6.3237 + e 2t/5 ). To find when the object hits the bottom of the tank we solve s(t) = 75, obtaining tb = 0.466273. The time from when the object is dropped from the helicopter to when it hits the bottom of the tank is ta + tb = 5.62708 seconds. 84 2.9 Modeling with Systems of First-Order DEs EXERCISES 2.9 Modeling with Systems of First-Order DEs 1. The linear equation dx/dt = −λ1 x can be solved by either separation of variables or by an integrating factor. Integrating both sides of dx/x = −λ1 dt we obtain ln |x| = −λ1 t + c from which we get x = c1 e−λ1 t . Using x(0) = x0 we find c1 = x0 so that x = x0 e−λ1 t . Substituting this result into the second differential equation we have dy + λ2 y = λ1 x0 e−λ1 t dt which is linear. An integrating factor is eλ2 t so that d λ2 t e y = λ1 x0 e(λ2 −λ1 )t + c2 dt y= λ1 x0 (λ2 −λ1 )t −λ2 t λ1 x0 −λ1 t e e + c2 e−λ2 t = e + c2 e−λ2 t . λ2 − λ1 λ2 − λ1 Using y(0) = 0 we find c2 = −λ1 x0 /(λ2 − λ1 ). Thus y= λ1 x0 e−λ1 t − e−λ2 t . λ 2 − λ1 Substituting this result into the third differential equation we have dz λ1 λ2 x0 −λ1 t − e−λ2 t . e = dt λ2 − λ1 Integrating we find z=− λ2 x0 −λ1 t λ1 x0 −λ2 t e + e + c3 . λ2 − λ1 λ2 − λ1 Using z(0) = 0 we find c3 = x0 . Thus z = x0 1 − λ2 λ1 e−λ1 t + e−λ2 t . λ2 − λ1 λ 2 − λ1 2. We see from the graph that the half-life of A is approximately 4.7 days. To determine the half-life of B we use t = 50 as a base, since at this time the amount of substance A is so small that it contributes very little to substance B. Now we see from the graph that y(50) ≈ 16.2 and y(191) ≈ 8.1. Thus, the half-life of B is approximately 141 days. x, y, z 20 y(t) 15 10 5 x(t) z(t) 25 50 75 100 125 150 t 3. The amounts x and y are the same at about t = 5 days. The amounts x and z are the same at about t = 20 days. The amounts y and z are the same at about t = 147 days. The time when y and z are the same makes sense because most of A and half of B are gone, so half of C should have been formed. 4. Suppose that the series is described schematically by W =⇒ −λ1 X =⇒ −λ2 Y =⇒ −λ3 Z where −λ1 , −λ2 , and −λ3 are the decay constants for W , X and Y , respectively, and Z is a stable element. Let w(t), x(t), y(t), and 85 2.9 Modeling with Systems of First-Order DEs z(t) denote the amounts of substances W , X, Y , and Z, respectively. A model for the radioactive series is dw dt dx dt dy dt dz dt = −λ1 w = λ1 w − λ2 x = λ2 x − λ3 y = λ3 y. 5. The system is 1 1 2 1 x2 − x1 · 4 = − x1 + x2 + 6 50 50 25 50 1 1 1 2 2 x2 = x1 · 4 − x2 − x2 · 3 = x1 − x2 . 50 50 50 25 25 x1 = 2 · 3 + 6. Let x1 , x2 , and x3 be the amounts of salt in tanks A, B, and C, respectively, so that 1 x2 · 2 − 100 1 x2 = x1 · 6 + 100 1 x2 · 5 − x3 = 100 x1 = 1 1 3 x1 · 6 = x2 − x1 100 50 50 1 1 1 3 7 1 x3 − x2 · 2 − x2 · 5 = x1 − x2 + x3 100 100 100 50 100 100 1 1 1 1 x3 − x3 · 4 = x2 − x3 . 100 100 20 20 7. (a) A model is dx1 x2 x1 =3· −2· , dt 100 − t 100 + t dx2 x1 x2 =2· −3· , dt 100 + t 100 − t x1 (0) = 100 x2 (0) = 50. (b) Since the system is closed, no salt enters or leaves the system and x1 (t) + x2 (t) = 100 + 50 = 150 for all time. Thus x1 = 150 − x2 and the second equation in part (a) becomes dx2 2(150 − x2 ) 3x2 300 2x2 3x2 = − = − − dt 100 + t 100 − t 100 + t 100 + t 100 − t or dx2 + dt 2 3 + 100 + t 100 − t x2 = 300 , 100 + t which is linear in x2 . An integrating factor is e2 ln(100+t)−3 ln(100−t) = (100 + t)2 (100 − t)−3 so d [(100 + t)2 (100 − t)−3 x2 ] = 300(100 + t)(100 − t)−3 . dt Using integration by parts, we obtain (100 + t)2 (100 − t)−3 x2 = 300 1 1 (100 + t)(100 − t)−2 − (100 − t)−1 + c . 2 2 Thus 300 1 1 c(100 − t)3 − (100 − t)2 + (100 + t)(100 − t) (100 + t)2 2 2 300 = [c(100 − t)3 + t(100 − t)]. (100 + t)2 x2 = 86 2.9 Modeling with Systems of First-Order DEs Using x2 (0) = 50 we find c = 5/3000. At t = 30, x2 = (300/1302 )(703 c + 30 · 70) ≈ 47.4 lbs. 8. A model is dx1 = (4 gal/min)(0 lb/gal) − (4 gal/min) dt 1 x1 lb/gal 200 dx2 = (4 gal/min) dt 1 x1 lb/gal − (4 gal/min) 200 1 x2 lb/gal 150 dx3 = (4 gal/min) dt 1 x2 lb/gal − (4 gal/min) 150 1 x3 lb/gal 100 or dx1 1 = − x1 dt 50 dx2 1 2 = x1 − x2 dt 50 75 dx3 1 2 = x2 − x3 . dt 75 25 Over a long period of time we would expect x1 , x2 , and x3 to approach 0 because the entering pure water should flush the salt out of all three tanks. 9. Zooming in on the graph it can be seen that the populations are first equal at about t = 5.6. The approximate periods of x and y are both 45. x,y x 10 y 5 t 50 10. (a) The population y(t) approaches 10,000, while the population x(t) 100 x,y 10 approaches extinction. y 5 x 10 10 20 10 (b) The population x(t) approaches 5,000, while the population y(t) approaches extinction. 20 20 t x,y 10 x 5 y (c) The population y(t) approaches 10,000, while the population x(t) approaches extinction. t x,y 10 y 5 x 87 t 2.9 Modeling with Systems of First-Order DEs (d) The population x(t) approaches 5,000, while the population y(t) x,y 10 approaches extinction. x 5 y 10 11. (a) x,y 10 5 (b) y x (c) 40 40 40 t x,y 10 y 5 y x 20 x 20 (d) 5 y t x,y 10 t x,y 10 5 20 20 x t 20 40 t In each case the population x(t) approaches 6,000, while the population y(t) approaches 8,000. 12. By Kirchhoff’s first law we have i1 = i2 + i3 . By Kirchhoff’s second law, on each loop we have E(t) = Li1 + R1 i2 and E(t) = Li1 + R2 i3 + q/C so that q = CR1 i2 − CR2 i3 . Then i3 = q = CR1 i2 − CR2 i3 so that the system is Li2 + Li3 + R1 i2 = E(t) 1 −R1 i2 + R2 i3 + i3 = 0. C 13. By Kirchhoff’s first law we have i1 = i2 + i3 . Applying Kirchhoff’s second law to each loop we obtain E(t) = i1 R1 + L1 di2 + i2 R2 dt E(t) = i1 R1 + L2 di3 + i3 R3 . dt and Combining the three equations, we obtain the system di2 + (R1 + R2 )i2 + R1 i3 = E dt di3 L2 + R1 i2 + (R1 + R3 )i3 = E. dt L1 14. By Kirchhoff’s first law we have i1 = i2 + i3 . By Kirchhoff’s second law, on each loop we have E(t) = Li1 + Ri2 and E(t) = Li1 + q/C so that q = CRi2 . Then i3 = q = CRi2 so that system is Li + Ri2 = E(t) CRi2 + i2 − i1 = 0. 15. We first note that s(t) + i(t) + r(t) = n. Now the rate of change of the number of susceptible persons, s(t), is proportional to the number of contacts between the number of people infected and the number who are 88 2.9 Modeling with Systems of First-Order DEs susceptible; that is, ds/dt = −k1 si. We use −k1 < 0 because s(t) is decreasing. Next, the rate of change of the number of persons who have recovered is proportional to the number infected; that is, dr/dt = k2 i where k2 > 0 since r is increasing. Finally, to obtain di/dt we use d d (s + i + r) = n = 0. dt dt This gives di dr ds =− − = −k2 i + k1 si. dt dt dt The system of differential equations is then ds = −k1 si dt di = −k2 i + k1 si dt dr = k2 i. dt A reasonable set of initial conditions is i(0) = i0 , the number of infected people at time 0, s(0) = n − i0 , and r(0) = 0. 16. (a) If we know s(t) and i(t) then we can determine r(t) from s + i + r = n. (b) In this case the system is ds = −0.2si dt di = −0.7i + 0.2si. dt We also note that when i(0) = i0 , s(0) = 10 − i0 since r(0) = 0 and i(t) + s(t) + r(t) = 0 for all values of t. Now k2 /k1 = 0.7/0.2 = 3.5, so we consider initial conditions s(0) = 2, i(0) = 8; s(0) = 3.4, i(0) = 6.6; s(0) = 7, i(0) = 3; and s(0) = 9, i(0) = 1. s,i s,i s,i s,i 10 10 5 5 10 10 5 i i 5 i s i s s s 5 10 t 5 10 t 5 10 t 5 10 t We see that an initial susceptible population greater than k2 /k1 results in an epidemic in the sense that the number of infected persons increases to a maximum before decreasing to 0. On the other hand, when s(0) < k2 /k1 , the number of infected persons decreases from the start and there is no epidemic. 89 2.9 Modeling with Systems of First-Order DEs CHAPTER 2 REVIEW EXERCISES CHAPTER 2 REVIEW EXERCISES 1. Writing the differential equation in the form y = k(y + A/k) we see that the critical point −A/k is a repeller for k > 0 and an attractor for k < 0. 2. Separating variables and integrating we have dy 4 = dx y x ln y = 4 ln x + c = ln x4 + c y = c1 x4 . We see that when x = 0, y = 0, so the initial-value problem has an infinite number of solutions for k = 0 and no solutions for k = 0. dy 3. = (y − 1)2 (y − 3)2 dx 4. dy = y(y − 2)2 (y − 4) dx 5. When n is odd, xn < 0 for x < 0 and xn > 0 for x > 0. In this case 0 is unstable. When n is even, xn > 0 for x < 0 and for x > 0. In this case 0 is semi-stable. When n is odd, −xn > 0 for x < 0 and −xn < 0 for x > 0. In this case 0 is asymptotically stable. When n is even, −xn < 0 for x < 0 and for x > 0. In this case 0 is semi-stable. 6. Using a CAS we find that the zero of f occurs at approximately P = 1.3214. From the graph we observe that dP/dt > 0 for P < 1.3214 and dP/dt < 0 for P > 1.3214, so P = 1.3214 is an asymptotically stable critical point. Thus, limt→∞ P (t) = 1.3214. y 7. x 8. (a) linear in y, homogeneous, exact (b) linear in x (c) separable, exact, linear in x and y (d) Bernoulli in x (e) separable (f ) separable, linear in x, Bernoulli (g) linear in x (h) homogeneous 90 CHAPTER 2 REVIEW EXERCISES (i) Bernoulli (j) homogeneous, exact, Bernoulli (k) linear in x and y, exact, separable, homoge- (l) exact, linear in y neous (m) homogeneous (n) separable 9. Separating variables and using the identity cos2 x = 1 (1 + cos 2x), we have 2 cos2 x dx = y2 y dy, +1 1 1 1 x + sin 2x = ln y 2 + 1 + c, 2 4 2 and 2x + sin 2x = 2 ln y 2 + 1 + c. 10. Write the differential equation in the form y ln x dx = y x ln x − y dy. y This is a homogeneous equation, so let x = uy. Then dx = u dy + y du and the differential equation becomes y ln u(u dy + y du) = (uy ln u − y) dy or y ln u du = −dy. Separating variables, we obtain ln u du = − dy y u ln |u| − u = − ln |y| + c x x x ln − = − ln |y| + c y y y x(ln x − ln y) − x = −y ln |y| + cy. 11. The differential equation dy 2 3x2 −2 + y=− y dx 6x + 1 6x + 1 is Bernoulli. Using w = y 3 , we obtain the linear equation dw 6 9x2 + w=− . dx 6x + 1 6x + 1 An integrating factor is 6x + 1, so d [(6x + 1)w] = −9x2 , dx 3x3 c w=− + , 6x + 1 6x + 1 and (6x + 1)y 3 = −3x3 + c. (Note: The differential equation is also exact.) 12. Write the differential equation in the form (3y 2 + 2x)dx + (4y 2 + 6xy)dy = 0. Letting M = 3y 2 + 2x and N = 4y 2 + 6xy we see that My = 6y = Nx , so the differential equation is exact. From fx = 3y 2 + 2x we obtain 91 CHAPTER 2 REVIEW EXERCISES f = 3xy 2 + x2 + h(y). Then fy = 6xy + h (y) = 4y 2 + 6xy and h (y) = 4y 2 so h(y) = 4 y 3 . A one-parameter 3 family of solutions is 4 3xy 2 + x2 + y 3 = c. 3 13. Write the equation in the form dQ 1 + Q = t3 ln t. dt t An integrating factor is eln t = t, so d [tQ] = t4 ln t dt 1 1 tQ = − t5 + t5 ln t + c 25 5 and Q=− 1 4 1 4 c t + t ln t + . 25 5 t 14. Letting u = 2x + y + 1 we have du dy =2+ , dx dx and so the given differential equation is transformed into u du −2 dx =1 or du 2u + 1 = . dx u Separating variables and integrating we get u du = dx 2u + 1 1 1 1 − du = dx 2 2 2u + 1 1 1 u − ln |2u + 1| = x + c 2 4 2u − ln |2u + 1| = 2x + c1 . Resubstituting for u gives the solution 4x + 2y + 2 − ln |4x + 2y + 3| = 2x + c1 or 2x + 2y + 2 − ln |4x + 2y + 3| = c1 . 15. Write the equation in the form dy 8x 2x + 2 y= 2 . dx x + 4 x +4 4 An integrating factor is x2 + 4 , so d dx x2 + 4 4 x2 + 4 y = 2x x2 + 4 4 3 y= 1 2 x +4 4 y= 1 + c x2 + 4 4 and 92 4 +c −4 . CHAPTER 2 REVIEW EXERCISES 16. Letting M = 2r2 cos θ sin θ + r cos θ and N = 4r + sin θ − 2r cos2 θ we see that Mr = 4r cos θ sin θ + cos θ = Nθ , so the differential equation is exact. From fθ = 2r2 cos θ sin θ + r cos θ we obtain f = −r2 cos2 θ + r sin θ + h(r). Then fr = −2r cos2 θ + sin θ + h (r) = 4r + sin θ − 2r cos2 θ and h (r) = 4r so h(r) = 2r2 . The solution is −r2 cos2 θ + r sin θ + 2r2 = c. 17. The differential equation has the form (d/dx) [(sin x)y] = 0. Integrating, we have (sin x)y = c or y = c/ sin x. The initial condition implies c = −2 sin(7π/6) = 1. Thus, y = 1/ sin x, where the interval π < x < 2π is chosen to include x = 7π/6. 18. Separating variables and integrating we have dy = −2(t + 1) dt y2 1 − = −(t + 1)2 + c y 1 y= , (t + 1)2 + c1 where −c = c1 . The initial condition y(0) = − 1 implies c1 = −9, so a solution of the initial-value problem is 8 y= 1 (t + 1)2 − 9 or y= t2 1 , + 2t − 8 where −4 < t < 2. √ 19. (a) For y < 0, y is not a real number. (b) Separating variables and integrating we have dy √ = dx y and √ 2 y = x + c. √ Letting y(x0 ) = y0 we get c = 2 y0 − x0 , so that √ √ 2 y = x + 2 y0 − x0 and y = 1 √ (x + 2 y0 − x0 )2 . 4 √ √ y > 0 for y = 0, we see that dy/dx = 1 (x + 2 y0 − x0 ) must be positive. Thus, the interval on 2 √ which the solution is defined is (x0 − 2 y0 , ∞). Since 20. (a) The differential equation is homogeneous and we let y = ux. Then (x2 − y 2 ) dx + xy dy = 0 (x2 − u2 x2 ) dx + ux2 (u dx + x du) = 0 dx + ux du = 0 u du = − dx x 1 2 u = − ln |x| + c 2 y2 = −2 ln |x| + c1 . x2 The initial condition gives c1 = 2, so an implicit solution is y 2 = x2 (2 − 2 ln |x|). 93 CHAPTER 2 REVIEW EXERCISES (b) Solving for y in part (a) and being sure that the initial condition is √ still satisfied, we have y = − 2 |x|(1 − ln |x|)1/2 , where −e ≤ x ≤ e so that 1 − ln |x| ≥ 0. The graph of this function indicates that the derivative is not defined at x = 0 and x = e. Thus, √ the solution of the initial-value problem is y = − 2 x(1 − ln x)1/2 , for y 2 1 -2 0 < x < e. -1 1 2 x -1 -2 21. The graph of y1 (x) is the portion of the closed black curve lying in the fourth quadrant. Its interval of definition is approximately (0.7, 4.3). The graph of y2 (x) is the portion of the left-hand black curve lying in the third quadrant. Its interval of definition is (−∞, 0). 22. The first step of Euler’s method gives y(1.1) ≈ 9 + 0.1(1 + 3) = 9.4. Applying Euler’s method one more time √ gives y(1.2) ≈ 9.4 + 0.1(1 + 1.1 9.4 ) ≈ 9.8373. 23. From dP = 0.018P and P (0) = 4 billion we obtain P = 4e0.018t so that P (45) = 8.99 billion. dt 24. Let A = A(t) be the volume of CO2 at time t. From dA/dt = 1.2 − A/4 and A(0) = 16 ft3 we obtain A = 4.8 + 11.2e−t/4 . Since A(10) = 5.7 ft3 , the concentration is 0.017%. As t → ∞ we have A → 4.8 ft3 or 0.06%. 25. Separating variables, we have s2 − y 2 dy = −dx. y Substituting y = s sin θ, this becomes s2 − s2 sin2 θ (s cos θ)dθ = −dx s sin θ cos2 s dθ = − dx sin θ s s 1 − sin2 θ dθ = −x + c sin θ (csc θ − sin θ)dθ = −x + c s ln | csc θ − cot θ| + s cos θ = −x + c s ln s − y s2 − y 2 +s y s2 − y 2 = −x + c. s Letting s = 10, this is 10 ln 10 − y 100 − y 2 + y 100 − y 2 = −x + c. Letting x = 0 and y = 10 we determine that c = 0, so the solution is 10 ln 10 − y 100 − y 2 + y 100 − y 2 = −x. 26. From V dC/dt = kA(Cs − C) and C(0) = C0 we obtain C = Cs + (C0 − Cs )e−kAt/V . 94 CHAPTER 2 REVIEW EXERCISES 27. (a) The differential equation dT = k(T − Tm ) = k[T − T2 − B(T1 − T )] dt = k[(1 + B)T − (BT1 + T2 )] = k(1 + B) T − BT1 + T2 1+B is autonomous and has the single critical point (BT1 + T2 )/(1 + B). Since k < 0 and B > 0, by phase-line analysis it is found that the critical point is an attractor and lim T (t) = t→∞ BT1 + T2 . 1+B Moreover, lim Tm (t) = lim [T2 + B(T1 − T )] = T2 + B T1 − t→∞ t→∞ BT1 + T2 1+B = BT1 + T2 . 1+B (b) The differential equation is dT = k(T − Tm ) = k(T − T2 − BT1 + BT ) dt or dT − k(1 + B)T = −k(BT1 + T2 ). dt This is linear and has integrating factor e− k(1+B)dt = e−k(1+B)t . Thus, d −k(1+B)t T ] = −k(BT1 + T2 )e−k(1+B)t [e dt BT1 + T2 −k(1+B)t e e−k(1+B)t T = +c 1+B BT1 + T2 T (t) = + cek(1+B)t . 1+B Since k is negative, limt→∞ T (t) = (BT1 + T2 )/(1 + B). (c) The temperature T (t) decreases to (BT1 + T2 )/(1 + B), whereas Tm (t) increases to (BT1 + T2 )/(1 + B) as t → ∞. Thus, the temperature (BT1 + T2 )/(1 + B), (which is a weighted average, B 1 T1 + T2 , 1+B 1+B of the two initial temperatures), can be interpreted as an equilibrium temperature. The body cannot get cooler than this value whereas the medium cannot get hotter than this value. 28. (a) By separation of variables and partial fractions, ln T − Tm − 2 tan−1 T + Tm T Tm 3 = 4Tm kt + c. Then rewrite the right-hand side of the differential equation as dT 4 4 = k(T 4 − Tm ) = [(Tm + (T − Tm ))4 − Tm ] dt T − Tm Tm 4 = kTm 1+ 4 = kTm 1+4 4 −1 T − Tm +6 Tm T − Tm Tm 95 2 ··· − 1 ← binomial expansion CHAPTER 2 REVIEW EXERCISES (b) When T − Tm is small compared to Tm , every term in the expansion after the first two can be ignored, giving dT 3 ≈ k1 (T − Tm ), where k1 = 4kTm . dt 29. We first solve (1 − t/10)di/dt + 0.2i = 4. Separating variables we obtain di/(40 − 2i) = dt/(10 − t). Then √ 1 − ln |40 − 2i| = − ln |10 − t| + c or 40 − 2i = c1 (10 − t). 2 √ Since i(0) = 0 we must have c1 = 2/ 10 . Solving for i we get i(t) = 4t − 1 t2 , 0 ≤ t < 10. 5 For t ≥ 10 the equation for the current becomes 0.2i = 4 or i = 20. Thus i(t) = 20 10 10 20 4t − 1 t2 , 0 ≤ t < 10 5 20, t ≥ 10. The graph of i(t) is given in the figure. √ √ 30. From y 1 + (y )2 = k we obtain dx = ( y/ k − y )dy. If y = k sin2 θ then dy = 2k sin θ cos θ dθ, dx = 2k 1 1 − cos 2θ 2 2 dθ, and x = kθ − k sin 2θ + c. 2 If x = 0 when θ = 0 then c = 0. 1 1 31. Letting c = 0.6, Ah = π( 32 · 12 )2 , Aw = π · 12 = π, and g = 32, the differential equation becomes √ √ dh/dt = −0.00003255 h . Separating variables and integrating, we get 2 h = −0.00003255t + c, so h = √ √ (c1 − 0.00001628t)2 . Setting h(0) = 2, we find c = 2 , so h(t) = ( 2 − 0.00001628t)2 , where h is measured in feet and t in seconds. 32. One hour is 3,600 seconds, so the hour mark should be placed at √ h(3600) = [ 2 − 0.00001628(3600)]2 ≈ 1.838 ft ≈ 22.0525 in. up from the bottom of the tank. The remaining marks corresponding to the passage of 2, 3, 4, . . . , 12 hours are placed at the values shown in the table. The marks are not evenly spaced because the water is not draining out at a uniform rate; that is, h(t) is not a linear function of time. 33. In this case Aw = πh2 /4 and the differential equation is dh 1 =− h−3/2 . dt 7680 Separating variables and integrating, we have 1 dt 7680 1 =− t + c1 . 7680 h3/2 dh = − 2 5/2 h 5 96 time seconds 0 1 2 3 4 5 6 7 8 9 10 11 12 height inches 24.0000 22.0520 20.1864 18.4033 16.7026 15.0844 13.5485 12.0952 10.7242 9.4357 8.2297 7.1060 6.0648 CHAPTER 2 REVIEW EXERCISES √ Setting h(0) = 2 we find c1 = 8 2/5, so that √ 1 2 5/2 8 2 =− h t+ , 5 7680 5 √ 1 h5/2 = 4 2 − t, 3072 and √ 4 2− h= 2/5 1 t 3072 . In this case h(4 hr) = h(14,400 s) = 11.8515 inches and h(5 hr) = h(18,000 s) is not a real number. Using a CAS to solve h(t) = 0, we see that the tank runs dry at t ≈ 17,378 s ≈ 4.83 hr. Thus, this particular conical water clock can only measure time intervals of less than 4.83 hours. 34. If we let rh denote the radius of the hole and Aw = π[f (h)]2 , then the √ √ differential equation dh/dt = −k h, where k = cAh 2g/Aw , becomes √ √ 2 2 dh cπrh 2g √ 8crh h =− h=− . dt π[f (h)]2 [f (h)]2 h 2 1 For the time marks to be equally spaced, the rate of change of the height must be a constant; that is, dh/dt = −a. (The constant is negative because the height is decreasing.) Thus √ 2 8crh h −a = − , [f (h)]2 √ 2 8crh h , [f (h)] = a 2 and −1 1 r 2c 1/4 h . a r = f (h) = 2rh Solving for h, we have h= a2 4 4 r . 64c2 rh The shape of the tank with c = 0.6, a = 2 ft/12 hr = 1 ft/21,600 s, and rh = 1/32(12) = 1/384 is shown in the above figure. 35. From dx/dt = k1 x(α − x) we obtain 1/α 1/α + x α−x dx = k1 dt so that x = αc1 eαk1 t /(1 + c1 eαk1 t ). From dy/dt = k2 xy we obtain ln |y| = k2 ln 1 + c1 eαk1 t + c or y = c2 1 + c1 eαk1 t k1 k2 /k1 . 36. In tank A the salt input is 7 gal min 2 lb gal + 1 gal min x2 lb 100 gal = 14 + 1 x2 100 lb . min The salt output is 3 gal min x1 lb 100 gal + 5 gal min x1 lb 100 gal = lb 2 x1 . 25 min In tank B the salt input is 5 gal min x1 lb 100 gal = lb 1 x1 . 20 min The salt output is 1 gal min x2 lb 100 gal + 4 gal min 97 x2 lb 100 gal = lb 1 x2 . 20 min CHAPTER 2 REVIEW EXERCISES The system of differential equations is then dx1 1 2 = 14 + x2 − x1 dt 100 25 dx2 1 1 = x1 − x2 . dt 20 20 37. From y = −x − 1 + c1 ex we obtain y = y + x so that the differential equation of the orthogonal family is dy 1 =− dx y+x or dx + x = −y. dy This is a linear differential equation and has integrating factor e dy = ey , so d y [e x] = −yey dy ey x = −yey + ey + c2 x = −y + 1 + c2 e−y . 38. Differentiating the family of curves, we have y 5 1 1 y =− =− 2. 2 (x + c1 ) y The differential equation for the family of orthogonal trajectories is then y = y 2 . Separating variables and integrating we get -5 dy = dx y2 1 − = x + c1 y 1 y=− . x + c1 5 x -5 98 ...
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