Unformatted text preview: 2 FirstOrder 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 ﬁrstorder 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 semistable.
1 0 23. Solving (y − 2)4 = 0 we obtain the critical point 2. From the phase portrait we see that 2 is
semistable. 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 semistable, 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 initialvalue
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 ycoordinate 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 ycoordinates 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
initialvalue 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 initialvalue 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 initialvalue 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 initialvalue 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 initialvalue 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 initialvalue 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 initialvalue 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 initialvalue
problem. Setting x = 1 and y = 1 in y = 2(c + e4x )/(c − e4x ) leads to c = −3e. Thus, a solution of the
4
initialvalue 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 initialvalue problem. y
1.01 1 0.0040.002 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 0.0040.002 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 0.0040.002 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 0.0040.002 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 righthand 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 xaxis the derivative of y(x) is undeﬁned, so no solution curve can cross the xaxis.
Since −x/y is not deﬁned when y = 0, the initialvalue 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 initialvalue 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 cvalues 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 cvalues
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 initialvalue 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 initialvalue 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 initialvalue 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 initialvalue 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 initialvalue 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 initialvalue 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 lefthand derivative of the function at x = 1 is 1/e and the righthand 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 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 5 of y = x3 − 2/x3 . x 3 (c) The ﬁrst two initialvalue 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 initialvalue 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 rootﬁnding
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 rootﬁnding 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 initialvalue 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 initialvalue 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 initialvalue 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
initialvalue 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 initialvalue 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 initialvalue 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 xaxis is vertical and the yaxis 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
1.50.5 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 initialvalue 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 initialvalue 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 initialvalue 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 initialvalue 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 ﬁrstorder 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 initialvalue 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 initialvalue 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 halflife (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 halflives.
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 C14 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 initialvalue 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 initialvalue 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 ﬁrstorder 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 initialvalue 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 initialvalue 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 initialvalue 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
√
initialvalue 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 initialvalue 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 initialvalue 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 initialvalue 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 xaxis 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 initialvalue 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 FirstOrder DEs EXERCISES 2.9
Modeling with Systems of FirstOrder 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 halflife of A is approximately 4.7
days. To determine the halflife 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 halflife 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 FirstOrder 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 FirstOrder 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 FirstOrder 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 FirstOrder 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 FirstOrder 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 initialvalue 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 semistable.
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 semistable.
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 oneparameter
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 initialvalue 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 initialvalue 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 lefthand 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 phaseline
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 righthand 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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This note was uploaded on 05/05/2008 for the course MATH 240 taught by Professor Storm during the Spring '08 term at UPenn.
 Spring '08
 Storm
 Differential Equations, Equations

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