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Solutions Manual for First Course in Differential Equations with Modeling Applications 11th Edition by Zill IBSN 9781305965720 Full clear download (no formatting errors) at: http://downloadlink.org/p/solutions-manual-for-first-course-in-differentialequations-with-modeling-applications-11th-edition-by-zill-ibsn9781305965720/
Chapter 2 First-Order Differential Equations
2.1
Solution Curves Without a Solution
1. x
3
2. x
y
5
2
x
0
–3
–2
–1
1
2
3
x
–1 –2
–10
–5
0
5
10
–3
3. x
y
4
y
4. x 4
2
2 0
x
0
x
–2
–2
–4 –4
5. x
4
–2
0
2
–4
4
y
6. x
2 0 –2
4
–2
0
2
4
y
2 x
0 –2
x
–4
–2
0
2
–4
4
36
–2
0
2
4
37
CHAPTER 2
7. x
FIRST-ORDER DIFFERENTIAL EQUATIONS
y
8. x
4
0
x
–4
–2
0
2
–4
4
y
4
10. x
–2
0
2
4
y
4 2
0
x
0
–2
x
–2 –4
–2
0
2
–4
4
y
4
12. x
2
–2
0
2
4
y
4 2
0
x
x
0
–2
–2 –4
13. x
x
–2
2
11. x
y
4
0
–2
9. x
37
2
2
–4
2.1 Solution Curves Without a Solution
–2
0
2
–4
4
y
–2
2
4
y
14. x
3
0
4
2 2
1 0
x
–1
0
x
–2
–2
–4
–3 –3 –2
–1
0
1
2
3
–4
–2
0
2
4
15. (a) The isoclines have the form y = −x + c, which are straight lines with slope −1.
3 2 1 –3
–2 –1
1 –1 –2 –3
2
3
38
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.1 Solution Curves Without a Solution
38
2 y
(b) The isoclines have the form x2 + y2 = c, which are circles centered at the origin.
1
–2
–1
1
2 x
–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 ∞. 17. When y < 21 x2 , y ′ = x2 − 2y is positive and the portions of solution curves “outside” the nullcline parabola are increasing. When y > 12 x2 , y ′ = x2 − 2y is negative and the portions of the solution curves “inside” the nullcline parabola are decreasing.
y 3 2 1 0
x
–1 –2 –3 –3
18. (a) Any horizontal lineal element should be at a point on a nullcline. nullclines are − y = 1/x. In Problem where n is an integer. in Problems 1, 3, and x2
y2
–2
–1
0
1
2
3
In Problem 1 the
= 0 or y = ±x. In Problem 3 the nullclines are 1 − xy = 0 or 4 the nullclines are (sin x) cos y = 0 or x = nπ and y = π/2 + nπ, The graphs on the next page show the nullclines for the equations 4 superimposed on the corresponding direction field.
y
y
y 4
3
4 2 2
2
1 x
0 –1
0
x
0
x
–2
–2
–2
–4
–3 –3
–2
–1 0 1 Problem 1
2
3
–4 –4
–2
0 Problem 3
2
4
–4
–2
0 2 Problem 4
4
39
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.1 Solution Curves Without a Solution
39
(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 y = −1, y = 0, and y = 1. The phase portrait is shown at the right. 1
(a) x
(b) x
y
y
5
0 1
4 3
–1
2 1
–2 1
(c) x –1
1
2
2
(d) x
y –2
x
–1
1
2
y 1
–1
x
2
–2 –3 –1
–4 –5
20. Writing the differential equation in the form dy/dx = y2 (1 − y)(1 + y) we see that critical points are y = −1, y = 0, and y = 1. The phase portrait is shown at the right. 1
(a) x
(b) x
y 5
y
0 1
4 3
1
2 1
–2
1
(c) x
2
x
(d) x
y
y –2
–2
–1
x
–1
1
2
x
x
–1 –1 –2 –3
–1
–4 –5
1
2
40
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.1 Solution Curves Without a Solution
21. Solving y2 − 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 y2 − y3 = 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
40
41
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.1 Solution Curves Without a Solution
24. Solving 10 + 3y − y2 = (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
25. Solving y 2 (4 − y2 ) = y2 (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
41
42
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.1 Solution Curves Without a Solution
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 (since ey is always positive) 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).
1n 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 the y-intervals (−∞, 0) and (c, ∞). Since f (y) < 0 for 0 < y < c, the graph of the solution is decreasing on the y-interval (0, c). y
c
0
c
x
42
42
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.1 Solution Curves Without a Solution
43
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 the y-intervals (−∞, −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 the y-interval (−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 (2/π)y − sin y = 0 has only three solutions. By inspection we see that the critical points are −π/2, 0, and π/2.
1
–π
–π 2
π 2
π
y
–1
From the graph at the right we see that ( <0 2 y − sin y >0 π
for for
π 2 0
( >0 2 y − sin y π <0
y < −π/2 y > π/2
for for
− π/2 < y < 0 0 < y < π/2
–
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
44
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.1 Solution Curves Without a Solution
44
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 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)
x→∞
= lim f (y(x)) = lim y ′ (x) = 0. 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 y2 − 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 of inflection is at y = 21 , although the concavity of solutions can be different on either side of y = −2 and y = 3. Since y ′′ (x) < 0 for y < −2 and 21 < y < 3, and y ′′ (x) > 0 for −2 < y < 12 and y > 3, we see that solution curves are concave down for y < −2 and 12 < y < 3 and concave up for −2 < y < 12 and y > 3. Points of inflection of solutions of autonomous differential equations will have the same y-coordinates because between critical points they are horizontal translations of each other.
y 5
x –5
5
–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.
45
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.1 Solution Curves Without a Solution
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. From the equation dP /dt = k (P − h/k) we see that the only critical point of the autonomous differential equationis the positive number h/k. A phase portrait shows that this point is unstable, that is, h/k is a repeller. For any initial condition P (0) = P0 for which 0 < P0 < h/k, dP /dt < 0 which means P (t) is monotonic decreasing and so the graph of P (t) must cross the t-axis or the line P − 0 at some time t1 > 0. But P (t1 ) = 0 means the population is extinct at time t1 .
40. Writing the differential equation in the form dv k mg = −v dt m k
mg 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, lim v = mg/k. t→∞
41. Writing the differential equation in the form r k mg dv k mg 2 −v = −v = dt m k m k
r
mg +v k p we see that the only physically meaningful critical point is mg/k. p From the phase portrait we see that mg/k is an asymptotically stable critical p point. Thus, lim v = mg/k.
√
mg k
t→∞
42. (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 not known.
β
α
45
46
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.1 Solution Curves Without a Solution
(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 → ∞.
(c) When k = 1 and α = β the differential equation is dX/dt = (α − X )2 . X (t) = α − 1/(t + c) we have dX/dt = 1/(t + c)2 and (α − X )2 = α − α −
1 t+c
2
46
α
For
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
x
x 2α
α –2 / α
α 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 → ∞.
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 = − 15 cos 5x + c. 2. From dy = (x + 1)2 dx we obtain y = 13 (x + 1)3 + c.
47
CHAPTER 2
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
3. From dy = −e−3x dx we obtain y = 13 e−3x + c. 1 1 1 . 4. From dy = dx we obtain − = x + c or y = 1 − 2 (y − 1) y−1 x +c 5. From
4 1 dy = dx we obtain ln |y| = 4 ln |x| + c or y = c1 x4 . x y
6. From
1 1 1 . dy = −2x dx we obtain − = −x2 + c or y = 2 2 x + c1 y y
7. From e−2y dy = e3x dx we obtain 3e−2y + 2e3x = c. 8. From yey dy = e−x + e−3x dx we obtain yey − ey + e−x +
1 −3x e = c. 3
1 y2 x3 + 2y + ln |y| = ln |x| − x3 + c. 2 3 9 1 1 1 2 = 10. From dy = dx we obtain + c. (2y + 3)2 (4x + 5)2 2y + 3 4x + 5 9. From
y+2+
1 y
dy = x2 ln x dx we obtain
1 1 dy = − 2 dx or sin y dy = − cos2 x dx = − 12 (1 + cos 2x) dx we obtain sec x csc y − cos y = − 12 x − 14 sin 2x + c or 4 cos y = 2x + sin 2x + c1 .
11. From
12. From 2y dy = −
sin 3x dx or 2y dy = − tan 3x sec 2 3x dx we obtain y2 = − 61 sec2 3x + c. cos3 3x
13. From
−ex ey dy = dx we obtain − (ey + 1)−1 = (ex + 1)3 (ey + 1)2
1 2
14. From
x y dy = dx we obtain 1 + y2 1/2 2 (1 + y ) (1 + x2 )1/2
= 1 + x2
15. From
1 dS = k dr we obtain S = cekr . S
16. From
1 dQ = k dt we obtain ln |Q − 70| = kt + c or Q − 70 = c1 ekt . Q − 70
1/2
(ex + 1)−2 + c. 1/2
+ c.
1 1 1 dP = dP = dt we obtain ln |P | − ln |1 − P | = t + c so that + P −P2 P 1− P P P c1 e t ln = t + c or = c1 et . Solving for P we have P = . 1−P 1− P 1 + c1 e t
17. From
18. From
1 t+2 t+2 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 dy = dx or y+3 x+4
5 5 dy = 1 − y+3 x +4 x +4 5 y − 5 ln |y + 3| = x − 5 ln |x + 4| + c or = c1 ex−y . y+3
19. From
1−
dx we obtain
47
48
CHAPTER 2
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
y+1 x+2 dy = dx or y−1 x−3
2 5 dy = 1 + y−1 x −3 (y − 1)2 y + 2 ln |y − 1| = x + 5 ln |x − 3| + c or = c1 e x − y . 5 (x − 3)
20. From
21. From x dx = p
1+
dx we obtain
x2 1 + c1 . dy we obtain 12 x2 = sin−1 y + c or y = sin 2 1 − y2
1 ex 1 1 dx we obtain − = tan−1 ex + c or dx = dy = 2 x 2 x −x y (e ) + 1 e +e y 1 y=− . tan−1 ex + c 1 23. From 2 dx = 4 dt we obtain tan−1 x = 4t + c. Using x(π/4) = 1 we find c = −3π/4. The x +1 3π 3π solution of the initial-value problem is tan−1 x = 4t − or x = tan 4t − . 4 4 22. From
24. From
1 1 1 dy = 2 dx or y2 − 1 x −1 2
1 1 − y−1 y+1
dy =
1 2
1 1 − x −1 x +1
dx we obtain
y−1 c(x − 1) = . Using y(2) = 2 we y+1 x +1 x −1 y−1 find c = 1. A solution of the initial-value problem is or y = x. = y+1 x +1 ln |y − 1| − ln |y + 1| = ln |x − 1| − ln |x + 1| + ln c or
1− x 1 1 1 1 − dy = dx = dx we obtain ln |y| = − − ln |x| = c or xy = c1 e−1/x . 2 2 x x y x x Using y(−1) = −1 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. 1 dy = dt we obtain − 12 ln |1 − 2y| = t + c or 1 − 2y = c1 e−2t . Using y(0) = 5/2 we 26. From 1 − 2y find c1 = −4. The solution of the initial-value problem is 1 − 2y = −4e−2t or y = 2e−2t + 12 . 25. From
27. Separating variables and integrating we obtain dx dy √ −p = 0 and 2 1−x 1 − y2
sin−1 x − sin−1 y = c.
√ Setting x = 0 and y = 3/2 we obtain c = −π/3. Thus, an implicit solution of the initialvalue problem is sin−1 x − sin−1 y = π/3. Solving for y and using an addition formula from trigonometry, we get √ √ p 3 1 − x2 π π π x . y = sin sin−1 x + = x cos + 1 − x2 sin = + 3 3 3 2 2
48
49
CHAPTER 2
28. From
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
1 −x dy = dx we obtain 2 1 + (2y) 1 + (x2 )2 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 tan−1 2y + tan−1 x2 = π/4 . Solving for y and using a trigonometric identity we get
2y = tan y=
π − tan−1 x2 4
1 π tan − tan−1 x2 2 4
=
1 tan π4 − tan (tan−1 x2 ) 2 1 + tan π4 tan (tan−1 x2 )
=
1 1 − x2 . 2 1 + x2
29. Separating variables and then proceeding as in Example 5 we get 2 dy = ye−x dx
ˆ
x 4
2 1 dy = e−x y dx ˆ x 1 dy 2 dt = e−t dt y(t) dt 4 ˆ x x 2 e−t dt ln y(t) =
4
4
ˆ
x
ln y(x) − ln y(4) =
e−t dt 2
4
ˆ
x
ln y(x) =
e−t dt 2
4
y(x) = e
´x 4
2
e−t dt
49
50
CHAPTER 2
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
30. Separating variables and then proceeding as in Example 5 we get
dy = y2 sin (x2 ) dx 1 dy = sin (x 2 ) y2 dx ˆ x ˆ x 1 dy dt = sin (t2 ) dt 2 (t) dt y −2 −2 ˆ x −1 x = sin (t2 ) dt y(t) −2 −2 ˆ x 1 −1 + = sin (t2 ) dt y(x) y(−2) −2 ˆ x −1 +3= sin (t2 ) dt y(x) −2 ˆ y(x) = 3 −
x
−1
sin (t2 ) dt
−2
31. Separating variables we get 2x + 1 dy = dx 2y 2y dy = (2x + 1) dx ˆ ˆ 2y dy = (2x + 1) dx y2 = x2 + x + c
√ The condition y(−2) = −1 implies c = −1. Thus y2 = x2 + x − 1 and y = − x2 + x − 1 in order for y to be negative. Moreover√for ! an interval containing −2 for values of x such that 5 1 . x2 + x − 1 > 0 we get −∞, − − 2 2
50
51
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.2 Separable Variables
32. Separating variables we get (2y − 2)
dy = 3x2 + 4x + 2 dx
(2y − 2) dy = 3x2 + 4x + 2 dx ˆ ˆ (2y − 2) dy = 3x2 + 4x + 2 dx ˆ
ˆ 2 (y − 1) dy =
3x2 + 4x + 2 dx
(y − 1)2 = x3 + 2x2 + 2x + c √ The condition y(1) = −2 implies c = 4. Thus y = 1 − x3 + 2x2 + 2x + 4 where the minus sign is indicated by the initial condition. Now x3 + 2x2 + 2x + 4 = (x + 2) x2 + 1 > 0 implies x > −2, so the interval of definition is (−2, ∞). 33. Separating variables we get ey dx − e−x dy = 0 ey dx = e−x dy ex dx = e−y dy ˆ ˆ ex dx = e−y dy ex = −e−y + c The condition y(0) = 0 implies c = 2. Thus e−y = 2 − ex . Therefore y = − ln (2 − ex ). Now we must have 2 − ex > 0 or ex < 2. Since ex is an increasing function this imples x < ln 2 and so the interval of definition is (−∞, ln 2). 34. Separating variables we get ˆ
sin x dx + y dy = 0 ˆ ˆ sin x dx + y dy = 0 dx
1 − cos x + y 2 = c 2 The condition y(0) = 1 implies c = − 12 . Thus − cos x + 12 y2 = − 12 or y2 = 2 cos x − 1. √ Therefore y = 2 cos x − 1 where the positive root is indicated by the initial condition. Now we must have 2 cos x − 1 > 0 or cos x > 12 . This means −π/3 < x < π/3, so the the interval of definition is (−π/3, π/3). 35. (a) The equilibrium solutions y(x) = 2 and y(x) = −2 satisfy the initial conditions y(0) = 2
51
52
CHAPTER 2
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
and y(0) = −2, respectively. Setting x = 14 and y = 1 in y = 2(1 + ce4x )/(1 − ce4x ) we obtain 1 + ce 1 1=2 , 1 − ce = 2 + 2ce, −1 = 3ce, and c = − . 1 − ce 3e The solution of the corresponding initial-value problem is 1 − 13 e4x−1 3 − e4x−1 = 2 . y=2 3 + e4x−1 1 + 13 e4x−1 (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 c
y−2 = 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 = 41 and y = 1 in y = 2(c + e4x )/(c − e4x ) leads to c = −3e. Thus, a solution of the initial-value problem is y=2
3 − e4x−1 −3e + e4x = 2 . 3 + e4x−1 −3e − e4x
36. Separating variables, we have dy dx = y2 − y x Using partial fractions, we obtain ˆ
ˆ or
1 1 − y−1 y
dy = ln |x| + c. y(y − 1)
dy = ln |x | + c
ln |y − 1| − ln |y| = ln |x| + c ln
y−1 =c xy y−1 = ec = c 1 . 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.
52
53
CHAPTER 2
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
(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 =
1 2
we have
(d) Setting x = 2 and y = 14 we have y = 1/(1 + 32 x) = 2/(2 + 3x).
1 2 1 4
= 1/(1 − 12 c1 ), so c1 = −2 and y = 1/(1 + 2x). = 1/(1 − 2c1 ), so c1 = − 32 and
p 37. Singular solutions of dy/dx = x 1 − y2 are y = −1 and y = 1. A singular solution of (ex + e−x )dy/dx = y2 is y = 0. 38. Differentiating ln (x2 + 10) + csc y = c we get 2x dy − csc y cot y = 0, x2 + 10 dx 2x 1 cos y dy = 0, − · x2 + 10 sin y sin y dx or 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. 39. The singular solution y = 1 satisfies the initial-value problem.
1.01
y
1
–0.004 –0.002 0.98
0.97
0.002 0.004
53
54
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
40. Separating variables we obtain −
dy = dx. Then (y − 1)2 y = x +c−1 . x +c
1 = x + c and y−1
Setting x = 0 and y = 1.01 we obtain c = −100. The solution is y= . x − 100
2.2 Separable Variables
54
y
1.02
1.01
–0.004 –0.002
0.002
0.004
x
0.99 0.98
41. Separating variables we obtain
dy = dx. Then (y − 1)2 + 0.01
10 tan −1 10(y − 1) = x + c and
y=1+
1 x +c tan . 10 10
Setting x = 0 and y = 1 we obtain c = 0. The solution is y = 1+
1 x tan . 10 10
y 1.0004 1.0002 x
–0.004 –0.002
0.002 0.004
0.9998 0.9996
dy = dx. Then, (y − 1)2 − 0.01 1 , we get with u = y − 1 and a = 10
42. Separating variables we obtain
10y − 11 5 ln = x + c. 10y − 9 Setting x = 0 and y = 1 we obtain c = 5 ln 1 = 0. The solution is 10y − 11 5 ln = x. 10y − 9 Solving for y we obtain 11 + 9ex/5 y= . 10 + 10ex/5
y 1.0004 1.0002
–0.004 –0.002
x 0.002 0.004
0.9998 0.9996
Alternatively, we can use the fact that ˆ y−1 dy 1 =− tanh−1 = −10 tanh−1 10(y − 1). 2 (y − 1) − 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).
55
CHAPTER 2
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
43. Separating variables, we have dy dy = = y − y3 y(1 − y)(1 + y)
1/2 1 1/2 + − y 1−y 1+y
dy = dx.
Integrating, we get ln |y| −
1 1 ln |1 − y| − ln |1 + y| = x + c. 2 2
When y > 1, this becomes ln y −
1 1 y = x + c. ln (y − 1) − ln (y + 1) = ln p 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). When 0 < y < 1 we have ln y −
Letting x = 0 and y = where −∞ < x < ∞.
1 2
1 1 y = x + c. ln (1 − y) − ln (1 + y) = ln p 2 2 1 − y2 √ √ we find c = ln (1/ 3 ). Solving for y we get y2 (x) = ex / e2x + 3 ,
When −1 < y < 0 we have ln (−y) −
1 1 −y = x + c. ln (1 − y) − ln (1 + y) = ln p 2 2 1 − y2
√ √ Letting x = 0 and y = − 12 we find c = ln (1/ 3 ). Solving for y we get y3 (x) = −ex / e2x + 3 , where −∞ < x < ∞. When y < −1 we have ln (−y) −
1 1 −y = x + c. ln (1 − y) − ln (−1 − y) = ln p 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). y
y
y
y
4
4
4
4
2
2
2
2
1
2
3
4
5
x
–4
–2
2
4
x
–4
–2
2
4
x
1
–2
–2
–2
–2
–4
–4
–4
–4
2
3
4
5
x
55
56
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.2 Separable Variables
y
44. (a) The second derivative of y is dy/dx 1/(y − 3) 1 d2 y =− =− =− . dx2 (y − 1)2 (y − 3)2 (y − 3)3
8 6 3
4
The solution curve is concave down when d2 y/dx2 < 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.
56
2 x –4
–2
2
4
–2
(b) Separating variables and integrating we obtain
8
y
6
(y − 3) dy = dx
4
1 2 y − 3y = x + c 2 y2 − 6y + 9 = 2x + c1 (y − 3)2 = 2x + c1 √ y = 3 ± 2x + c1 .
2 x
–1
1
2
3
4
5
–2
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 where (−1/2, ∞). √ When y2 (0) = 2 we have c1 = 1 and y2 (x) = 3 − 2x + 1 where (−1/2, ∞). √ When y3 (1) = 2 we have c1 = −1 and y3 (x) = 3 − 2x − 1 where (1/2, ∞). √ When y4 (−1) = 4 we have c1 = 3 and y4 (x) = 3 + 2x + 3 where (−3/2, ∞). 45. We separate variables and rationalize the denominator. Then dy =
1 1 − sin x 1 − sin x 1 − sin x · dx dx = dx = 2 1 + sin x 1 − sin x cos2 x 1 − sin x
= sec2 x − tan x sec x dx. Integrating, we have y = tan x − sec x + C . √ √ 46. Separating variables we have y dy = sin x dx. Then ˆ ˆ ˆ √ √ √ 2 3/2 y dy = sin x dx and y = sin x dx. 3 √ √ 1 To integrate sin x we first make the substitution u = x. Then du = √ dx = 2 x ˆ ˆ ˆ √ sin x dx = (sin u) (2u) du = 2 u sin u du.
1 2u
du and
57
CHAPTER 2
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
Using integration by parts we find ˆ √ √ √ u sin u du = −u cos u + sin u = − x cos x + sin x. Thus 2 y= 3
ˆ
√ √ √ √ sin x dx = −2 x cos x + 2 sin x + C
and √ √ √ y = 32/3 − x cos x + sin x + C . √ √ 47. Separating variables we have dy/ y + y = dx/ ( x + x). To integrate
ˆ dx/
we substitute u2 = x and get ˆ ˆ √ 2 2u | | du = 2 ln 1 + u + c = 2 ln 1 + x + c. du = 1+ u u + u2 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 . 48. Separating variables and integrating we have ˆ ˆ dy = dx y 2/3 1 − y 1/3 ˆ
y 2/3 dy = x + c1 1 − y 1/3
−3 ln 1 − y 1/3 = x + c1 ln1 − y 1/3 = −
x + c2 3
1 − y 1/3 = c3 e−x/3 1 − y 1/3 = c4 e−x/3 y 1/3 = 1 + c5 e−x/3 y = 1 + c5 e−x/3
3
.
√ x +x
57
58
CHAPTER 2
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
√ √ 49. Separating variables we have y dy = e x dx. If u = x , then u2 = x and 2u du = dx. Thus, ˆ √ ˆ e x dx = 2ueu du and, using integration by parts, we find ˆ ˆ ˆ √ √ √ √ 1 2 x dx so y = 2ueu du = −2eu + C = 2 x e x − 2e x + C, y dy = e 2
and q√ √ √ xe x − e x +C . y=2 To find C we solve y(1) = 4. q√ √ √ √ 1e 1 − e 1 + C =2 C = 4 2 y(1) =
so
C = 4.
q√ √ √ and the solution of the intial-value problem is y = 2 x e x −e x +4. 50. Seperating variables we have y dy = x tan−1 x dx. Integrating both sides and using integration by parts with u = tan−1 x and dv = x dx we have ˆ y dy = x tan−1 x dx 1 2 1 2 1 1 y = x tan−1 x − x + tan−1 x + C 2 2 2 2 y2 = x2 tan−1 x − x + tan−1 x + C1 p y = x2 tan−1 x − x + tan−1 x + C1 To find C1 we solve y(0) = 3. p y(0) =
02 tan−1 0 − 0 + tan−1 0 + C1 =
and the solution of the initial-value problem is y =
p C1 = 3
√
so
C1 = 9,
x2 tan−1 x − x + tan−1 x + 9 .
√ 51. (a) While y2 (x) = − 25 − x2 is defined at x = −5 and x = 5, y ′2(x) is not defined at these values, and so the 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. 2
52. The derivative of y = 14 x2 − 1 is dy/dx = x 14 x2 − 1 . We note that xy41/2 = x 1 x2 − 1 . We see from the graphs of y (black), dy/dx (red), and xy 1/2 (blue), below that dy/dx = xy 1/2 on (−∞, 2] and [2, ∞).
58
59
CHAPTER 2
59
√ X 2 = |X | we can write
Alternatively, because
√ =x y= x
1 2 4x
x
s xy 1/2
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
1 2 x −1 4
2
1 = x x2 − 1 = 4
1 2 4x
−x x
−1 ,
1 2 4x
−∞ < x ≤ −2
−1 ,
−2 < x < 2
−1 ,
2≤x <∞.
From this we see that dy/dx = xy 1/2 on (−∞, −2] and on [2, ∞). p 53. Separating variables we have dy/ 1 + y2 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.
3.5 3 2.5 2 1.5 1 0.5 –6
–4
–2
2
4
6
8
54. (a) The solution of y ′ = y, y(0) = 1, is y = ex . Using separation of variables we find that the solution of y ′ = y [1 + 1/ (x ln x)], y(e) = 1, is y = ex−e ln x. Solving the two solutions simultaneously we obtain ex = ex−e ln x,
so
ee = ln x
e
and
x = ee e .
e (b) Since y = e(e ) ≈ 2.33 × 101,656,520 , the y-coordinate of the point of intersection of the two solution curves has over 1.65 million digits.
55. We are looking for a function y(x) such that dy2 y + = 1. dx 2
Using the positive square root gives dy p = 1 − y2 dx p
dy = dx 1 − y2
sin−1 y = x + c.
60
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.2 Separable Variables
60
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. 56. (a) x
y 3
–3
3
x
–3
p √ (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 dx p =√ 2 and cosh−1 y = cosh−1 x + c. 2 x − 1 y −1 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. 57. 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 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 2.2.5 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. 58. (a) Separating variables and integrating, we have (3y2 + 1) dy = −(8x + 5) dx and y3 + y = −4x2 − 5x + c. Using a CAS we show various contours of f (x, y) = y3 + 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
–2
0
2
4
61
CHAPTER 2
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
61
y
(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.
4 2 0
x
–2 –4 –4
–2
0
2
4
59. (a) An implicit solution of the differential equation (2y + 2)dy − (4x3 + 6x) dx = 0 is y2 + 2y − x4 − 3x2 + c = 0. The condition y(0) = −3 implies that c = −3. Therefore y2 + 2y − x4 − 3x2 − 3 = 0.
(b) Using the quadratic formula we can solve for y in terms of x: p −2 ± 4 + 4(x4 + 3x2 + 3) y= . 2 The explicit solution that satisfies the initial condition is then y = −1 −
p 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 −
p
x4 + 3x3 + 4 = −1 −
p f (x)
is x ≤ −2.8 or x ≥ −1.3. The graph of this function is shown below. –1 – √f(x) f(x) –4
4
–2
2 –2
2 –4 –4
x
–2 –2 –4
–6 –8 –10
x
62
CHAPTER 2
2.2 Separable Variables
FIRST-ORDER DIFFERENTIAL EQUATIONS
62
–1 – √f(x)
(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 equality has been removed since the derivative dy/dx does not exist at the points where f (x) = 0. The
2 –2 –4 –6 –8 –10
60. (a) Separating variables and integrating, we have (−2y + y2 ) dy = (x − x2 ) dx
y 4 2
and −y 2 +
1 3 1 2 1 3 y = x − x +c 3 2 3
Using a CAS we show some contours of
0
x
–2 –4
f (x, y) = 2y3 − 6y2 + 2x3 − 3x2 .
–6
–4
–2
0
2
4
6
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.
(b) The value of c corresponding to y(0) = 32 is f 0, 32 = − 274 . The portion of the graph between the dots corresponds to the solution curve satisfying the intial condition. To determine the interval of definition we find dy/dx for 2y3 − 6y2 + 2x3 − 3x2 = −
27 . 4
y 4 2 0
x
–2 –4 –2
0
2
4
6
Using implicit differentiation we get y ′ = (x − x2 )/(y 2 − 2y), which is infinite when y = 0 and y = 2. Letting y = 0 in 2y3 − 6y2 + 2x3 − 3x2 = − 274 and using a CAS to solve for x 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).
63
CHAPTER 2
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
63
y
(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
4 2 0
x
–2 –4 –6 –8
2y3 − 6y2 + 2x3 − 3x2 = −40.
–4
–2
0
2
4
6
8
10
Using implicit differentiation we get y ′ = (x − x2 )/(y 2 − 2y), which is infinite when y = 0 and y = 2. Letting y = 0 in 2y3 − 6y2 + 2x3 − 3x2 = −40 and using a CAS to solve for x we get x = −2.29551. The largest interval of definition is approximately (−2.29551, ∞).
2.3
Linear Equations ´
1. For y ′ − 5y = 0 an integrating factor is e−
5 dx
= e−5x so that
d
e−5x y = 0 and y = ce5x
dx
for −∞ < x < ∞. 2. For y ′ + 2y = 0 an integrating factor is e
´
2 dx
= e2x so that
d
e2x y = 0 and y = ce−2x for
dx
−∞ < x < ∞. The transient term is ce−2x . 3. For y ′ + y = e3x an integrating factor is e
´
dx
d [ex y] = e4x and y = 1 e3x + ce−x 4 dx
= ex so that
for −∞ < x < ∞. The transient term is ce−x . 4. For y ′ +4y =
4 3
´
an integrating factor is e
4 dx
d 4x e y = 43 e4x and y = 13 +ce−4x dx
= e4x so that
for −∞ < x < ∞. The transient term is ce−4x . 5. For y=
y′ 1 3
+
3x2 y
+ ce
−x3
=
x2
´
an integrating factor is e
−
1 2
=
3 ex
so that
−x3
for −∞ < x < ∞. The transient term is ce
6. For y ′ + 2xy = x3 an integrating factor is e 1 x2 2
3x2 dx
´
2x dx
2 + ce−x
d
h
i 3 3 ex y = x2 ex and
dx
.
2
= ex so that −x 2
d
h
i 2 2 ex y = x3 ex and
dx
for −∞ < x < ∞. The transient term is ce . ´ 1 1 d 1 1 c 7. For y ′ + y = 2 an integrating factor is e (1/x) dx = x so that [xy] = and y = ln x+ x x dx x x x y=
for 0 < x < ∞. The entire solution is transient.
64
CHAPTER 2
´
8. For y ′ −2y = x2 +5 an integrating factor is e− and y = − 12 x2 − 12 x − 9. For y ′ −
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
11 4
2 dx
= e−2x so that
d
e−2x y = x2 e−2x +5e−2x
dx
+ ce2x for −∞ < x < ∞. There is no transient term.
´ 1 1 d 1 y = x sin x an integrating factor is e− (1/x) dx = so that y = sin x and x x dx x
y = cx − x cos x for 0 < x < ∞. There is no transient term. ´ 3 d 2 2 10. For y ′ + y = an integrating factor is e (2/x)dx = x2 so that x y = 3x and y = 3 +cx−2 x dx x 2
for 0 < x < ∞. The trancient term is cx−2 . 11. For y ′ +
− ´ 4 d 4 y = x2 − 1 an integrating factor is e (4/x)dx = x4 so that x y = x6 x4 and x dx
y = 17 x3 − 15 x + cx−4 for 0 < x < ∞. The transient term is cx−4 . ´ x y = x an integrating factor is e− [x/(1+x)]dx = (x + 1)e−x so that (1 + x) d 2x + 3 ce x (x + 1)e−x y = x(x + 1)e−x and y = −x − + for −1 < x < ∞. There dx x +1 x +1
12. For y ′ −
is no transient term. 13. For y ′ + 1 + and y =
2 x
y=
1 ex ce−x + 2 2 2x x
´ ex [1+(2/x)]dx = x2 ex so that d [x2 ex y] = e2x an integrating factor is e 2 dx x −x ce for 0 < x < ∞. The transient term is 2 . x
´ 1 −x e sin 2x an integrating factor is e [1+(1/x)]dx = xex so that x d 1 ce−x for 0 < x < ∞. The entire solution [xex y] = sin 2x and y = − e−x cos 2x + dx 2x x
14. For y ′ +
1+
1 x
y =
is transient. 15. For
´ dx 4 − x = 4y5 an integrating factor is e− dy y
(4/y) dy
= eln y
−4
= y −4 so that d y −4 x = 4y dy
and x = 2y6 + cy4 for 0 < y < ∞. There is no transient term. ´ dx 2 + x = ey an integrating factor is e (2/y) dy = y2 so that dy y 2 2 c c x = ey − ey + 2 ey + 2 for 0 < y < ∞. The transient term is 2 . y y y y
16. For
d 2 y x = y2 ey and dy
64
65
CHAPTER 2
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
17. For y ′ + (tan x)y = sec x an integrating factor is e
´
tan x dx
= sec x so that
d dx
[(sec x)y] = sec2 x
and y = sin x + c cos x for −π/2 < x < π/2. There is no transient term.
18. For y ′ + (cot x)y = sec2 x csc x an integrating factor is e
´
cot x dx
= eln | sin x| = sin x so that
d [(sin x) y] = sec2 x and y = sec x + c csc x for 0 < x < π/2. There is no transient term. dx ´ 2xe−x x +2 19. For y ′ + y = an integrating factor is e [(x+2)/(x+1)]dx = (x + 1)ex , so x +1 x +1 d c x2 [(x + 1)ex y] = 2x and y = e−x + e−x for −1 < x < ∞. The entire dx x +1 x +1 solution is transient. ´ 5 4 y = an integrating factor is e [4/(x+2)] dx = (x + 2)4 so that 2 x +2 (x + 2) d 5 (x + 2)4 y = 5(x + 2)2 and y = (x + 2)−1 + c(x + 2)−4 for −2 < x < ∞. The dx 3
20. For y ′ +
entire solution is transient. ´ dr + r sec θ = cos θ an integrating factor is e sec θ dθ = eln | sec x+tan x| = sec θ + tan θ so 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. that dθ
21. For
There is no transient term. 22. For
´ dP d h t −t i + (2t − 1)P = 4t − 2 an integrating factor is e (2t−1) dt = et −t so that e P = 2 dt dt 2
(4t − 2)et
2 −t
23. For y ′ + 3 +
2
2
and P = 2 + cet−t for −∞ < t < ∞. The transient term is cet−t . 1 x
and y = e−3x +
´ e−3x d xe3x y = 1 an integrating factor is e [3+(1/x)]dx = xe3x so that x dx 3x
y= ce− x
for 0 < x < ∞. The transient term is ce−3x /x.
´ x +1 2 [2/(x2 −1)]dx = x − 1 an integrating factor is e y = 2 x −1 x −1 x +1 so that d x − 1 y = 1 and (x − 1)y = x(x + 1) + c(x + 1) for −1 < x < 1. There is no dx x + 1
24. For y ′ +
transient term.
65
66
CHAPTER 2
25. For y ′ − 5y = x an integrating factor is e ˆ y=e
5x
´
−5 dx
ˆ −3x
d e−5x y = xe−5x and dx 1 1 =− x− + ce5x . 25 5
1 1 1 76 and y = − x − + e5x . The solution is defined on I = (−∞, ∞). 25 5 25 25
26. For y ′ + 3y = 2x and integrating factor is e
y=e
= e−5x so that
1 −5x 1 5x xe−5x dx = e5x − xe− − e +c 25 5
If y(0) = 3 then c =
If y(0) =
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
2xe3x dx = e−3x
´
3 dx
= e3x so that
2 3x − 2 3x e +c xe 9 3
d
e3x y = 2xe3x and
dx =
2 2 x − + ce−3x . 3 9
5 2 5 1 2 then c = and y = x − + e−3x . THe solution is defined on I = (−∞, ∞). 9 9 3 3 9
´ 1 d 1 y = ex an integrating factor is e (1/x)dx = x so that [xy] = ex and y = 1 ex + c x x dx x x 1 x 2− e for 0 < x < ∞. If y(1) = 2 then c = 2 − e and y = e + . The solution is defined on x x
27. For y ′ +
I = (0, ∞). ´ 1 dx 1 − x = 2y an integrating factor is e− (1/y)dy = so that dy y y
d 1 x = 2 and dy y 49 x = 2y2 + cy for 0 < y < ∞. If y(1) = 5 then c = −49/5 and x = 2y2 − y. The solution is 5
28. For
defined on I = (0, ∞). ´ E d h Rt/L i di R E + i= 29. For an integrating factor is e (R/L) dt = eRt/L so that e i = eRt/L L dt L L dt E E + ce−Rt/L for −∞ < t < ∞. If i(0) = i0 then c = i0 − E/R and i = + and i = R R E −Rt/L e i0 − . The solution is defined on I = (−∞, ∞) R 30. For
´ dT d −kt [e T ] = − Tmke−kt − kT = −Tm k an integrating factor is e (−k) dt = e−kt so that dt dt
and T = Tm +cekt for −∞ < t < ∞. If T (0) = T0 then c = T0 −Tm and T = Tm +(T0 −Tm )ekt . The solution is defined on I = (−∞, ∞) ´ 1 1 d 31. For y ′ + y = 4 + an integrating factor is e (1/x) dx = x so that [xy] = 4x + 1 and x x dx ˆ 1 1 c 2 y= (4x + 1) dx = x x 2x + x + c = 2x + 1 + x .
66
67
CHAPTER 2
If y(1) = 8 then c = 5 and y = 2x + 1 +
5 . The solution is defined on I = (0, ∞). x
32. For y ′ + 4xy = x3 ex an integrating factor is e 2
ˆ 2
y=e
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
2
x e
−2x
2
dx = e
1
2
= e2x so that
1 e
2
2 3x
6
If y(0) = −1 then c = −
4x dx
x e
−2x
3 3x
´
−
1
2
+c
d 2x2 2 [e y] = x3 e3x and dx
=
3x
2 x
18
1
2
x e 6
2
e −
2
+ ce
.
−2x
x
18
17 1 1 x2 17 −2x2 2 . The solution is defined on and y = x 2e x − e − e 18 6 18 18
I = (−∞, ∞). 33. For y ′ +
´ ln x 1 d y= an integrating factor is e [1/(x+1)] dx = x+1 so that [(x+1)y] = ln x x +1 x+ 1 dx
and y=
c x x + ln x − x +1 x +1 x +1
If y(1) = 10 then c = 21 and y =
for 0 < x < ∞.
x x 21 + ln x − . The solution is defined on x +1 x +1 x +1
I = (0, ∞). ´ 1 1 y = an integrating factor is e [1/(x+1)] dx = x + 1 so that x +1 x (x + 1) d 1 [(x + 1) y] = and dx x
34. For y ′ +
1 y= x +1
ˆ
If y(e) = 1 then c = e and y =
1 1 ln x c + dx = (ln x + c) = . x x +1 x +1 x +1 e ln x + . The solution is defined on I = (0, ∞). x +1 x +1
35. For y ′ − (sin x) y = 2 sin x an integrating factor is e
´
(− sin x) dx
= ecos x so that
d [ecos x y] = dx
2 (sin x) ecos x and ˆ y=e
− cos x
2 (sin x) ecos x dx = e− cos x (−2ecos x + c) = −2 + ce− cos x .
If y(π/2) = 1 then c = 3 and y = −2 + 3e− cos x . The solution is defined on I = (−∞, ∞).
67
68
CHAPTER 2
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
36. For y ′ + (tan x)y = cos2 x an integrating factor is e
´
tan x dx
68
= eln | sec x| = sec x so that
d [(sec x) y] = cos x and y = sin x cos x + c cos x for −π/2 < x < π/2. If y(0) = −1 dx then c = −1 and y = sin x cos x − cos x. The solution is defined on I = (−π/2, π/2). 37. For y ′ + 2y = f (x) an integrating factor is e2x so that
ye
2x
=
1 2x e + c1 , 2
0≤x ≤3
c2 ,
x > 3.
y 1
1 (1 − e−2x ), 2 1 6 (e − 1)e−2x , 2
y=
0≤x≤3 x > 3.
38. For y ′ + y = f (x) an integrating factor is ex so that x
yex =
e + c1 ,
0≤x≤1
−ex + c2 ,
x > 1.
1,
0≤x≤1
2e1−x − 1,
x > 1.
y 1
5 x
If y(0) = 1 then c1 = 0 and for continuity we must have c2 = 2e so that y=
x
5
If y(0) = 0 then c1 = −1/2 and for continuity we must have c2 = 12 e6 − 12 so that
–1
2
y
ye
x2
=
1 x2 e + c1 , 2
0≤x≤1
c2 ,
x > 1.
If y(0) = 2 then c1 = 3/2 and for continuity we must have c2 = 12 e + 32 so that
y=
1 3 −x2 + e , 2 2 1 2
0≤x≤1
3 e+
2
x2
e
−
,
x > 1.
2
3
x
69
CHAPTER 2
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
40. For y′ +
2x y= 1 + x2
an integrating factor is 1 + x2 y =
x , 1 + x2
0≤x ≤1
−x , 1 + x2
x > 1,
69
If
y 1
x
5
1 + x2
so that 1 2 x + c1 , 0≤x≤1 2 1 − x 2 + c2 , x > 1. 2
–1
y(0) = 0 then c1 = 0 and for continuity we must have c2 = 1 so that
y=
1 1 − , 2 2 (1 + x2 )
0≤x≤1
3 1 − , 2 (1 + x2 ) 2
x > 1.
41. 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
y 20
= e2x , so d 2x [e y] = 4xe2x dx ˆ e2x y = 4xe2x dx = 2xe2x − e2x + c1
15 10 5
y = 2x − 1 + c1 e
−2x
. 3
x
Using the initial condition, we find y(0) = −1 + c1 = 3, so c1 = 4 and y = 2x − 1 + 4e−2x , 0 ≤ x ≤ 1. Now, since y(1) = 2 − 1 + 4e−2 = 1 + 4e−2 , 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
= e−2 ln x = x−2 , so d −2 4 [x y] = 4xx−2 = dx x ˆ 4 x−2 y = dx = 4 ln x + c 2 x y = 4x2 ln x + c2 x2 .
(We use ln x instead of ln |x| because x > 1.) Using the initial condition we find
70
CHAPTER 2
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
70
y(1) = c2 = 1 + 4e−2 , so y = 4x2 ln x + (1 + 4e−2 )x2 , x > 1. Thus, ⎧ −2x ⎪ ⎨2x − 1 + 4e , y=
⎪ ⎩
0≤x≤1
% & 4x2 ln x + 1 + 4e−2 x2 ,
x > 1.
y(0)´ = 4 on the interval [0, 2]. The integrating factor is e 1 dx = ex , so
y 1
5 x
d dx
–1
ˆ x
e y=
0 dx = c1
y = c1 e−x . Using the initial condition, we find y(0) = c1 = 4, so c1 = 4 and y = 4e−x , 0 ≤ x ≤ 2. Now, since y(2) = 4e−2 , we solve the initial-value problem y ′ + 5y = 0, y(1) = 4e−2 on the interval (2, ∞). The integrating factor is e
´
5 dx
= e5x , so
d ' 5x ( e y =0 dx ˆ 5x e y = 0 dx = c2 y = c2 e−5x .
Using the initial condition we find y(2) = c2 e−10 = 4e−2 , so c2 = 4e8 and y = 4e8 e−5x = 4e8−5x , x > 2. Thus, the solution of the original initial-value problem is
y=
⎧ −x ⎪ ⎨4e ,
0≤x≤2
⎪ ⎩ 8−5x 4x ,
x > 2.
71
CHAPTER 2
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
43. An integrating factor for y ′ − 2xy = 1 is e−x . Thus 2
d −x2 2 [e y] = e−x dx ˆ x √ 2 2 π e−x y = e−t dt = erf (x) + c 2 0 √ π x2 2 y= e erf (x) + cex . 2
√ From y(1) = ( π/2)e erf (1) + ce = 1 we get c = e−1 − initial-value problem is
√
π 2
√ π π x2 −1 erf (1) y= e erf (x) + e − 2 2 √ π x2 x 2 −1 =e + e (erf (x) − erf (1)). 2
erf (1). The solution of the
√
ex
2
44. An integrating factor for y ′ − 2xy = −1 is e−x . Thus 2
d −x2 2 [e y] = −e−x dx ˆ x √ π −x2 −t2 e y=− e dt = − erf (x) + c. 2 0 √ √ π/2, and noting that erf (0) = 0, we get c = π/2. Thus √ √ √ √ 2 π π π x2 x 2 π = e (1 − erf (x)) = e erfc (x). erf (x) + y = ex − 2 2 2 2
From y(0) =
x
45. For y ′ + ex y = 1 an integrating factor is ee . Thus ˆ
x x d ee y = ee dx
From y(0) = 1 we get c = e, so y = e−e
x
x
ex
and
e y=
t
ee dt + c.
0
´x 0
t
ee dt + e1−e x .
1 y = x. An integrating factor is e1/x . Thus x2 ˆ x d h 1/x i te1/t dt + c. e y = xe1/x and e1/x y = dx 1
46. Dividing by x2 we have y ′ −
From y(1) = 0 we get c = 0, so y = e−1/x
´x 1
te1/t dt.
71
72
CHAPTER 2
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
72
47. An integrating factor for y′ +
10 sin x 2 y= x3 x
is x2 . Thus d dx
x2 y = 10
sin x x ˆ
x
x2 y = 10 0
sin t dt + c t
y = 10x−2 Si (x) + cx−2 . From y(1) = 0 we get c = −10 Si (1). Thus y = 10x−2 Si (x) − 10x−2 Si (1) = 10x−2 (Si (x) − Si (1)) . 48. The integrating factor for y ′ − sin x2 y = 0 is e−
´
x 0
sin t 2 dt .
Then
d h − ´ x sin t2 dt i e 0 y =0 dx e−
´x 0
sin t2 dt
y = c1 ´x
y = c1 e 0 Letting t =
p
p π/2 u we have dt = π/2 du and r ˆ √2/π x r ˆx π π π 2 2 S sin t dt = sin u du = 2 0 2 2 0
√ so y = c1 e
sin t 2 dt
π/2 S
r
! 2 x π √
√
2/π x
. Using S(0) = 0 and y(0) = c1 = 5 we have y = 5e
π/2 S
√
2/π x
.
49. We want 4 to be a critical point, so we use y ′ = 4 − y. 50. (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 1.2.1 are not satisfied and
the
initial-value
problem
does
not
have
a
unique
solution.
73
CHAPTER 2
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
(b) The differential equation has no solution satisfying y(0) = y0 , y0 > 0.
(c) In this case, since x0 > 0, Theorem 1.2.1 applies and the initial-value problem has a unique solution given by y = x5 ex − x4 ex + cx4 where c = y0 /x40 − x0 ex0 + ex0 . 51. On the interval (−3, 3) the integrating factor is ´
e
x dx/(x2 −9)
= e−
´
x dx/(9−x2 )
1
2
= e 2 ln (9−x ) =
p
9 − x2
and so i d hp 9 − x2 y = 0 and dx
y=√
c . 9 − x2
52. 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. 53. 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. 54. (a) Differentiating yc = c/x3 we get y ′c = −
3c x4
=−
3 c
3 = − y x x3 x c
so a differential equation with general solution yc = c/x3 is xy ′ + 3y = 0. Now using yp = x3 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, ∞).
73
74
CHAPTER 2
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL 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 interval of definition is (0, ∞). The initialvalue 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 of y = x3 − 2/x3 .
74
y 3
5
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 . ´
55. Since e ´
c1 e
P (x) dx+c
= ec e
´
P (x) dx
ˆ P (x) dx
y = c2 +
´
c1 e
´
= c1 e P (x) dx
P (x) dx ,
we would have ´
f (x) dx
and
e
ˆ P (x) dx
y = c3 +
´
e
P (x) dx
f (x) dx,
which is the same as (4) in the text.
56. We see by inspection that y = 0 is a solution. 57. 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 h λ2 t i e y = x 0 λ 1 e−λ1 t eλ2 t = x 0λ 1e(λ2 −λ1 )t dt x0 λ1 (λ2 −λ1 )t e + c2 eλ 2 t y = λ2 − λ 1 y=
x0 λ1 −λ1 t + c2 e−λ2 t . λ2 − λ 1 e
75
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.3 Linear Equations
From y(0) = y0 we obtain c2 = (y0 λ2 − y0 λ1 − x0 λ1 ) / (λ2 − λ1 ). The solution is y=
x0 λ1 −λ1 t y0 λ2 − y0 λ1 − x0 λ1 −λ2 t e + e . λ2 − λ 1 λ2 − λ 1
58. Writing the differential equation as
dE 1 + E = 0 we see that an integrating factor is dt RC
et/RC . Then d h t/RC i e E =0 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 .
59. (a) x
(b) Using a CAS we find y(2) ≈ 0.226339.
y 5
5 x
60. (a) x
2
y
1 1 –1 –2 –3 –4 –5
2
3
4
5
x
75
76
CHAPTER 2
2.3 Linear Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
(b) 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).
61. (a) x
y
10
5 –10
–5
5
10
x
(b) From the graph we see that as x → ∞, y(x) oscillates with decreasing amplitudes ap√ 1 proaching 9.35672. Since lim S(x) = , we have lim y(x) = 5e π/8 ≈ 9.357, and x→∞ x→∞ 2 √ 1 since x→−∞ lim S(x) = − , we have x→−∞ lim y(x) = 5e− π/8 ≈ 2.672. 2 (c) 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).
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) = 32 y2 + 7y. A solution is x2 − x + 32y2 + 7y = c. 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 − 8y3 so that My = 4 = Nx . From fx = 5x + 4y we obtain f = 25 x2 + 4xy + h(y), h′ (y) = −8y 3 , and h(y) = −2y 4 . A solution is 52 x2 + 4xy − 2y4 = c. 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) = − 12y2 . A solution is x sin y + y cos x − 12 y 2 = c.
76
77
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.4 Exact Equations
5. Let M = 2y 2 x − 3 and N = 2yx2 + 4 so that My = 4xy = Nx . From fx = 2y2 x − 3 we obtain f = x2 y2 − 3x + h(y), h′ (y) = 4, and h(y) = 4y. A solution is x2 y2 − 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 − y2 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(x), h′ (x) = 1 + ln x, and h(x) = x ln x. A solution is −y + y ln x + x ln x = c. 9. Let M = y3 − y2 sin x − x and N = 3xy2 + 2y cos x so that My = 3y2 − 2y sin x = Nx . From fx = y3 − y2 sin x − x we obtain f = xy3 + y2 cos x − 12x2 + h(y), h′ (y) = 0, and h(y) = 0. A solution is xy 3 + y 2 cos x − 12 x2 = c. 10. Let M = x3 + y3 and N = 3xy2 so that My = 3y2 = Nx . From fx = x3 + y3 we obtain f = 14 x4 + xy3 + h(y), h′ (y) = 0, and h(y) = 0. A solution is 14x4 + xy3 = c. 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 − y2 = 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 3 we obtain f = x − 3 ln |x| + xy + h(y), h′ (y) = 1 − , and h(y) = y − 3 ln |y|. A solution is y x + y + xy − 3 ln |xy| = c. 15. Let M = x2 y3 − 1/ 1 + 9x2 and N = x3 y2 so that My = 3x2 y2 = Nx . From fx = x2 y3 − 1/ 1 + 9x2 we obtain f = 13x3 y3 − 13 arctan (3x) + h(y), h′ (y) = 0, and h(y) = 0. A solution is x3 y3 − arctan (3x) = c. 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) = 52 y2 . A solution is −2xy + 52y2 = c. 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.
77
78
CHAPTER 2
2.4 Exact Equations
FIRST-ORDER DIFFERENTIAL EQUATIONS
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 + 4xy3 exy + 4yexy = Nx . From fx = 2y sin x cos x − y + 2y 2 exy we obtain f = y sin2 x − xy + 2exy + h(y), h′ (y) = 0, 2 and h(y) = 0. A solution is y sin2 x − xy + 2exy = c. 2
2
19. Let M = 4t3 y − 15t2 − y and N = t4 + 3y2 − t so that My = 4t3 − 1 = Nt . From ft = 4t3 y − 15t2 − y we obtain f = t4 y − 5t3 − ty + h(y), h′ (y) = 3y2 , and h(y) = y3 . A solution is t4 y − 5t3 − ty + y3 = c. 20. Let M = 1/t + 1/t2 − y/ t2 + y2 and N = yey + t/ t2 + y2 so that 2 My = y 2 − t2 / t 2 + y 2 = Nt . From ft = 1/t + 1/t2 − y/ t2 + y2 we obtain 1 t f = ln |t| − − arctan + h(y), h′ (y) = yey , and h(y) = yey − ey . A solution is t y ln |t| −
1 − arctan t
t y
+ yey − ey = c.
21. Let M = x2 + 2xy + y2 and N = 2xy + x2 − 1 so that My = 2(x + y) = Nx . From fx = x2 + 2xy + y2 we obtain f = 31 x3 + x2 y + xy2 + h(y), h′ (y) = −1, and h(y) = −y. The solution is 13 x3 + x2 y + xy 2 − y = c. If y(1) = 1 then c = 4/3 and a solution of the initial-value problem is 13 x 3 + x 2 y + xy 2 − y = 43 . 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 − ey . 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) = 3y2 − y. The solution is 4ty + t2 − 5t + 3y2 − y = c. If y(−1) = 2 then c = 8 and a solution of the initial-value problem is 4ty + t2 − 5t + 3y2 − y = 8. 24. Let M = t/2y 4 and N = 3y2 − t2 /y 5 so that My = −2t/y 5 = Nt . From ft = t/2y 4 we 3 3 t2 t2 3 − 2 = c. If obtain f = 4 + h(y), h′ (y) = 3 , and h(y) = − 2 . The solution is 4 4y 4y y 2y 2y t2 3 5 y(1) = 1 then c = −5/4 and a solution of the initial-value problem is 4 − 2 = − . 4y 2y 4 25. Let M = y2 cos x − 3x2 y − 2x and N = 2y sin x − x3 + ln y so that My = 2y cos x − 3x2 = Nx . From fx = y2 cos x − 3x2 y − 2x we obtain f = y2 sin x − x3 y − x2 + h(y), h′ (y) = ln y, and h(y) = y ln y − y. The solution is y2 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 y2 sin x − x3 y − x2 + y ln y − y = 0. 26. Let M = y2 + y sin x and N = 2xy − cos x − 1/ 1 + y2 so that My = 2y + sin x = Nx . From −1 f x = y2 + y sin x we obtain f = xy2 − y cos x + h(y), h′ (y) = , and h(y) = − tan−1 y. 1 + y2
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The solution is xy2 − y cos x − tan−1 y = c. If y(0) = 1 then c = −1 − π/4 and a solution of π the initial-value problem is xy2 − y cos x − tan−1 y = −1 − . 4 27. Equating My = 3y2 + 4kxy3 and Nx = 3y2 + 40xy3 we obtain k = 10. 28. Equating My = 18xy2 − sin y and Nx = 4kxy2 − sin y we obtain k = 9/2. 29. Let M = −x2 y2 sin x + 2xy2 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 y2 cos x + h(y), h′ (y) = 0, and h(y) = 0. A solution of the differential equation is x2 y2 cos x = c. 30. Let M = (x2 + 2xy − y 2 )/(x2 + 2xy + y2 ) and N = (y2 + 2xy − x2 )/(y 2 + 2xy + x2 ) so that My = −4xy/(x + y)3 = Nx . From fx = x2 + 2xy + y2 − 2y2 /(x + y)2 we obtain 2y2 f =x+ + h(y), h′ (y) = −1, and h(y) = −y. A solution of the differential equation is x +y x2 + y2 = c(x + y). ´
31. We note that (My − Nx )/N = 1/x, so an integrating factor is e dx/x = x. Let M = 2xy2 + 3x2 and N = 2x2 y so that My = 4xy = Nx . From fx = 2xy2 + 3x2 we obtain f = x2 y2 + x3 + h(y), h′ (y) = 0, and h(y) = 0. A solution of the differential equation is x2 y2 + x3 = c. ´
32. We note that (My − Nx )/N = 1, so an integrating factor is e dx = ex . Let M = xyex + y2 ex + yex and N = xex + 2yex so that My = xex + 2yex + ex = Nx . From fy = xex + 2yex we obtain f = xyex + y2 ex + h(x), h′ (x) = 0, and h(x) = 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 2 dy/y = y2 . Let M = 6xy3 and N = 4y3 + 9x2 y2 so that My = 18xy2 = Nx . From fx = 6xy3 we obtain f = 3x2 y3 + h(y), h′ (y) = 4y3 , and h(y) = y 4 . A solution of the differential equation is 3x2 y3 + y4 = 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 y2 . A solution of the differential equation is ln (sin x) + y + ln y2 = 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 . 3x 3x ′ From fx = 10e3x − 6ye3x + 1 we obtain f = 10 3 e − 2ye + x + h(y), h (y) = 0, and h(y) = 0. A solution of the differential equation is 103 e3x − 2ye3x + x = c. ´
36. We note that (Nx − My )/M = −3/y, so an integrating factor is e−3 dy/y = 1/y3 . Let M = (y2 + xy 3 )/y3 = 1/y + x and N = (5y2 − xy + y3 sin y)/y3 = 5/y − x/y2 + sin y, so that My = −1/y 2 = Nx . From fx = 1/y +x we obtain f = x/y +2 1 x2 +h(y), h′ (y) = 5/y +sin y, and h(y) = 5 ln |y| − cos y. A solution of the differential equation is x/y + 21 x2 + 5 ln |y| − cos y = c.
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2.4 Exact Equations
37. We note that (My − Nx )/N = 2x/(4 + x2 ), so an integrating factor is ´ 2 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 = 12 ln (4 + x2 ) + h(y), h′ (y) = y, and h(y) = 12 y2 . A solution of the differential equation is 12 ln (4 + x2 ) + 12y2 = c. Multiplying both 2
sides by 2 the last equation can be written as ey x2 + 4 = c1 . Using the initial condition 2 y(4) = 0 we see that c1 = 20. A solution of the initial-value problem is ey x2 + 4 = 20. ´
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 + y2 − 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 = −21 y2 /(1 + x)2 + h(x), h′ (x) = (x 2 − 5)/(1 + x) 3, and h(x) = 2/(1 + x) 2 + 2/(1 + x) + ln |1 + x|. A solution of the differential equation is −
y2 2 2 + + + ln |1 + x| = c. 2 2 2(1 + x) (1 + x) (1 + x)
Using the initial condition y(0) = 1 we see that c = 7/2. A solution of the initial-value problem is y2 2 2 7 − + + + ln |1 + x| = 2 2 1+x 2 2 (1 + x) (1 + x) 39. (a) Implicitly differentiating x3 + 2x2 y + y2 = c and solving for dy/dx we obtain dy dy dy 3x 2 + 4xy . 3x2 + 2x2 + 4xy + 2y = 0 and =− 2 dx dx 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 + y2 = c we find c = 4, and setting x = y = 1 we also find c = 4. Thus, both initial conditions determine the same implicit solution. (c) Solving x3 + 2x2 y + y2 = 4 for y we get 4 2
and y2 (x) = −x2 +
p
4 − x3 + x4 .
Observe in the figure that y1 (0) = −2 and y2 (1) = 1.
–4
y2 2
–2
4
–2 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.
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p 41. The explicit solution is y = (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 42. (a) Since fy = N (x, y) = xexy + 2xy + 1/x we obtain f = exy + xy2 + + h(x) so that x y y fx = yexy + y2 − 2 + h′ (x). Let M (x, y) = yexy + y2 − 2 . x x −1
(b) Since fx = M (x, y) = y 1/2 x−1/2 + x x2 + y we obtain 1 1 2 f = 2y 1/2 x1/2 + ln x2 + y + g(y) so that f y = y −1/2 x1/2 + x +y 2 2 1 2 −1 N (x, y) = y −1/2 x1/2 + x +y . 2 43. First note that
p d
Then x dx + y dy =
p
x2 + y 2
= p
−1
+ g′ (y). Let
x y dx + p dy. x2 + y 2 x2 + y 2
x2 + y2 dx becomes
p x y x2 + y 2 = dx. dx + p dy = d x2 + y 2 x2 + y 2 p The left side p is the total differential of x2 + y2 and the right side is the total differential of x + c. Thus x2 + y2 = x + c is a solution of the differential equation. p
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 v2 − 32x dx + xv dv = 0 This is not an exact equation, but µ(x) = x is an integrating factor. The new equation xv2 − 32x2 dx + x2 v dv = 0 is exact and solving yields 12x2 v2 − 323 x3 = c. When x = 3, 32 3
2 2
v = 0 and so c = −288. Solving 12 x v −
x = −288 for v yields the explicit solution r x 9 v(x) = 8 − 2. 3 x 3
(b) The chain leaves the platform when x = 8, and so r 8 9 v(8) = 8 − ≈ 12.7 ft/s 3 64 46. (a) Letting M (x, y) =
(x2
2xy + y 2 )2
and
N (x, y) = 1 +
we compute My =
2x3 − 8xy 2 = Nx , (x2 + y 2 )3
y 2 − x2 (x2 + y 2 )2
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FIRST-ORDER DIFFERENTIAL EQUATIONS
82
so the differential equation is exact. Then we have 2xy ∂f = 2xy(x2 + y 2 )−2 = M (x, y) = 2 ∂x (x + y 2 )2 f (x, y) = −y(x2 + y 2 )−1 + g(y) = −
x2
y + g(y) + y2
∂f y 2 − x2 y 2 − x2 ′ = 2 + g (y) = N (x, y) = 1 + . ∂y (x + y 2 )2 (x2 + y2 )2 Thus, g′ (y) = 1 and g(y) = y. The solution is y − is x2 + y2 = 1.
x2
y = c. When c = 0 the solution + y2
(b) The first graph below is obtained in Mathematica using f (x, y) = y − y/(x2 + y2 ) 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 s x=−
s y3
− cy2
−y
and
c−y
x=
y3 − cy2 − 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. y 3 2
3
3
2
2 1
1 –3
–2
–1
y
1
2
3
–1.5 –1 –0.5
0.5 1
1.5
–3
–2
–1 –1
–1
–1
–2
–2
–2
–3
–3
–3
1
2
3
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FIRST-ORDER DIFFERENTIAL EQUATIONS
2.5 Solutions by Substitutions
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 ln |x| +
1 ln |1 + 2u| = c 2 y x2 1 + 2 = c1 x x2 + 2xy = c1 .
3. Letting x = vy we have vy(v dy + y dv) + (y − 2vy) dy = 0 vy2 dv + y v2 − 2v + 1 dy = 0 v dv dy + =0 (v − 1)2 y ln |v − 1| −
1 + ln |y| = c v−1
1 x ln − 1 − + ln |y| = c y x/y − 1 (x − y) ln |x − y| − y = c(x − y).
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FIRST-ORDER DIFFERENTIAL EQUATIONS
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 ln + 2 − ln |y| = c y 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 x2 = c 1 +2 x x x2 y = c1 (y + 2x).
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FIRST-ORDER DIFFERENTIAL EQUATIONS
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| + ln x2
1 ln u2 + 1 + tan −1 u = c 2 y2 y + 1 + 2 tan−1 = c1 2 x x
ln x2 + y2 + 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 x
y −1 x
2
= c1
y x
+1
(y − x)2 = c1 (y + x).
9. Letting y = ux we have √ −ux dx + (x + u x)(u dx + x du) = 0 √ (x2 + x2 u ) du + xu3/2 dx = 0 1 dx du + =0 u−3/2 + u x −2u−1/2 + ln |u| + ln |x| = c
p
ln |y/x| + ln |x| = 2 y(ln |y| − c)2 = 4x.
x/y + c
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FIRST-ORDER DIFFERENTIAL EQUATIONS
10. Letting y = ux we have p
x2 − u2 x2 dx − x2 du = 0 p 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 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| + u 3 = c 3 3x3 ln |x| + y3 = 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 ln |x| −
1 ln(1 + u 2) = 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 = y2 + x2 .
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FIRST-ORDER DIFFERENTIAL EQUATIONS
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
x = c1 . y
Using y(1) = e we find c1 = −e. The solution of the initial-value problem is y ln
x = −e. y
1 1 dw 3 3 15. From y ′ + y = y −2 and w = y3 we obtain + w = . An integrating factor is x3 so x x x dx x that x3 w = x3 + c or y3 = 1 + cx−3 . 16. From y ′ − y = ex y2 and w = y −1 we obtain that ex w = − 12 e2x + c or y −1 = − 12 ex + ce−x .
dw + w = −ex . An integrating factor is ex so dx
dw − 3w = −3x. An integrating factor is e−3x so dx = x + 13 + ce3x .
17. From y ′ + y = xy4 and w = y −3 we obtain that e−3x w = xe−3x + 13 e−3x + c or y −3 18. From y ′ − 1 +
1 x
dw 1 + 1+ w = −1. An integrating dx x 1 c = −1 + + e−x . x x
y = y 2 and w = y −1 we obtain
factor is xex so that xex w = −xex + ex + c or y −1
dw 1 1 + 19. From y ′ − y = − 2 y 2 and w = y −1 we obtain dt t t 1 c that tw = ln t + c or y −1 = ln t + . Writing this in t t solution can also be expressed in the form et/y = c1 t.
1 1 w = 2 . An integrating factor is t so t t t the form = ln t + c, we see that the y
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FIRST-ORDER DIFFERENTIAL EQUATIONS
2t dw −2t 2 2t y4 and w = y −3 we obtain − y= w= . An 2 2 2 dt 3 (1 + t ) 3 (1 + t ) 1+ t 1 + t2 1 1 w integrating factor is so that = + c or y −3 = 1 + c 1 + t2 . 1 + t2 1 + t2 1 + t2
20. From y ′ +
2 3 dw 6 9 y = 2 y 4 and w = y −3 we obtain + w = − 2 . An integrating factor x x dx x x is x6 so that x6 w = − 95 x5 + c or y −3 = − 95 x−1 + cx−6 . If y(1) = 12 then c = 49 5 and 49 x−6 . 9 −1 −3 y = − 5x + 5
21. From y ′ −
dw 3 3 + w = . An integrating factor is e3x/2 so dx 2 2 that e3x/2 w = e3x/2 + c or y 3/2 = 1 + ce−3x/2 . If y(0) = 4 then c = 7 and y 3/2 = 1 + 7e−3x/2 . du 1 − 1 = u2 or 23. Let u = x + y + 1 so that du/dx = 1 + dy/dx. Then du = dx. Thus dx 1 + u2 tan−1 u = x + c or u = tan (x + c), and x + y + 1 = tan (x + c) or y = tan (x + c) − x − 1. 1− u du −1 = 24. Let u = x + y so that du/dx = 1 + dy/dx. Then or u du = dx. Thus dx u 1 u 2 = x + c or u2 2 = 2x + c1 , and (x + y) = 2x + c1 . 2 22. From y ′ + y = y −1/2 and w = y 3/2 we obtain
du − 1 = tan2 u or cos2 u du = dx. Thus 25. Let u = x + y so that du/dx = 1 + dy/dx. Then dx 1 u + 1 sin 2u = x + c or 2u + sin 2u = 4x + c , and 2(x + y) + sin 2(x + y) = 4x + c or 1 1 2 4 2y + sin 2(x + y) = 2x + c1 . 1 du − 1 = sin u or 26. Let u = x + y so that du/dx = 1 + dy/dx. Then du = dx. dx 1 + sin u 1 − sin u du = dx or (sec 2 u−sec u tan u)du = dx. Multiplying by (1−sin u)/(1−sin u) we have cos2 u Thus tan u − sec u = x + c or tan (x + y) − sec (x + y) = x + c. √ du 1 27. Let u = y − 2x + 3 so that du/dx = dy/dx − 2. Then + 2 = 2 + u or √ du = dx. Thus dx u √ √ 2 u = x + c and 2 y − 2x + 3 = x + c. 28. Let u = y − x + 5 so that du/dx = dy/dx − 1. Then −e−u = x + c and −e−y+x−5 = x + c. 29. Let u = x + y so that du/dx = 1 + dy/dx. Then
du + 1 = 1 + eu or e−u du = dx. Thus dx
1 du − 1 = cos u and du = dx. Now dx 1 + cos u
1 − cos u 1 1 − cos u = csc2 u − csc u cot u = = 2 1 + cos u 1 − cos 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.
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FIRST-ORDER DIFFERENTIAL EQUATIONS
5u + 6 du 2u = 3+ = 30. Let u = 3x + 2y so that du/dx = 3 + 2 dy/dx. Then and dx u+2 u +2 u +2 du = dx. Now by long division 5u + 6 u +2 1 4 = + 5u + 6 5 25u + 30 ˆ
so we have
and
1u 5
+
4 25
1 4 + 5 25u + 30
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 =
4 25
ln 95. The solution is
1 4 4 (3x + 2y) + ln 95 ln |75x + 50y + 30| = x + 5 25 25 or 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 1 M (1, y/x) = M (x, y) or M (x, y) = xα M (1, y/x). α x Thus dy xα M (1, y/x) M (1, y/x) y =− = f (x, y) = − α N (1, y/x) = F x . dx x N (1, y/x) 32. Rewrite (5x2 − 2y 2 )dx − xy dy = 0 as xy
dy = 5x2 − 2y2 dx
and divide by xy, so that dy x y =5 −2 . dx y x We then identify F
y x
=5
y x
−1
−2
y . x
89
90
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2.5 Solutions by Substitutions
90
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 + c2 or c2 = − 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 e−π/2 ≤ ≤ eπ/2 5 5e
−π/2
≤ x ≤ 5e
y 20 15 10 5 5
π/2
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(y 1 + u) + R(y 1 + u)2 dx dx = P + Qy1 + Ry21 + Qu + 2y1 Ru + Ru2 or
du − (Q + 2y1 R)u = Ru 2. 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 . 1 4 dw + − + w = −1. dx x x −1 An integrating factor is x3 so that x3 w = − 14 x4 + c or u = − 14x + cx−3 . Thus,
(b) Identify P (x) = −4/x2 , Q(x) = −1/x, and R(x) = 1. Then
2 y = +u x
or
1 2 y = + − x + cx−3 x 4
−1
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 =
91
CHAPTER 2
2.5 Solutions by Substitutions
FIRST-ORDER DIFFERENTIAL EQUATIONS
(ln x)/x, which is first-order and linear. An integrating factor is e− (using integration by parts) ln x d 1 u = 2 dx x x
and
´
dx/x
= 1/x, so that
u 1 ln x =− − + c. x x x
The solution is ln y = −1 − ln x + cx
or
y=
ecx−1 . x
37. Write the differential equation as dv 1 + v = 32v −1 , dx x and let u = v2 or v = u1/2 . Then dv 1 du = u−1/2 , dx 2 dx and substituting into the differential equation, we have 1 −1/2 du 1 u + u1/2 = 32u−1/2 2 dx x
du 2 + u = 64. dx x
or
´
-
-