10.2 SOLUTIONS
11.
(a)
ex +e,x ; then dy
2
dx
If y =
If k = 1; then
k
=
ex ,e,x ; and d2 y2
2
dx
r
1+
dy 2 =
dx
=
=
=
(b)
y = eAx +2Ae,Ax , so
ex +e,x :
=
2
r
r
x
e , e,x 2
1+
r
=
2
e2x + 1 + e,2x
1+
4
2
x
e + e,x 2
r
=
e2x , 1 + e,2x
4
2
4
x
e + e,x ex + e,x
=
2
2
4
2
(since
ex + e,x > 0)
d2 y :
dx2
dy = eAx , e,Ax
dx
2
and
d2y
dx2
Ax ,Ax
A e +2 e
:
=
Therefore we have
1+
This means
k
r
1+
r
dy 2 = k
dx
,
=
Since we want
d2 y
dx2
=
k
dy 2 = 1 + eAx , e,Ax 2 = 1 + 1 ,e2Ax + e,2Ax , 2
dx
2
4
,
,
1 2Ax
,2Ax + 2 = 1 eAx + e,Ax 2 :
=
+e
e
4
4
k
2
1 , Ax
k eAx + e,Ax e
+ e,Ax
=
4
2
Ax
,Ax e
+
e
2
(since
eAx + e,Ax > 0).
q
, 2
1 + dy
dx , we must have A = k.
Solutions for Section 10.2
1.
y
(a)
i
4
,4
4
ii
(b)
(c)
,
,4
x
iii
, ,
The solution through ( 1; 0) appears to be linear, so its equation is y = x 1.
If y = x 1, then y0 = 1 and x + y = x + ( x 1) = 1, so this checks as a solution.
, ,
,
, ,
,
511
512
2.
CHAPTER TEN /SOLUTIONS
y
(a)
6
ii
,6
x
6
i
,6
(b)
3.
(d)
Since y0 = y, the slope is negative above the x-axis (when y is positive) and positive below the x-axis (when y
is negative). The only slope field for which this is true is II.
Since y0 = y, the slope is positive for positive y and negative for negative y. This is true of both I and III. As y get
larger, the slope should get larger, so the correct slope field is I.
Since y0 = x, the slope is positive for positive x and negative for negative x. This corresponds to slope field V.
1
Since y0 = , the slope is positive for positive y and negative for negative y. As y approaches 0, the slope becomes
(e)
larger in magnitude, which correspond to solution curves close to vertical. The correct slope field is III.
Since y0 = y2 , the slope is always positive, so this must correspond to slope field IV.
(a)
(b)
(c)
4.
5.
6.
We can see that the slope lines are flat when y is an integer multiple of . We conclude from the figure that the
solution is y = n in this case.
To check this, we note that if y = n, then (sin x)(sin y) = (sin x)(sin n) = 0 = y0 . Thus y = n is a
solution to y0 = (sin x)(sin y), and it passes through (0;n).
,
(a) II
y
(b) VI
Notice that y0
(c) IV
x+y
x,y
(d) I
(e) III
(f) V
x = ,y and is undefined when x = y. A solution curve will be horizontal
(slope= 0) when passing through a point with x = ,y, and will be vertical (slope undefined) when passing through a
point with x = y. The only slope field for which this is true is slope field (b).
The slope fields in (I) and (II) appear periodic. (I) has zero slope at x = 0, so (I) matches y0 = sin x, whereas (II) matches
y0 = cos x. The slope in (V) tends to zero as x ! 1, so this must match y0 = e,x . Of the remaining slope fields,
only (III) shows negative slopes, matching y0 = xe,x . The slope in (IV) is zero at x = 0, so it matches y0 = x2 e,x .
This leaves field (VI) to match y0 = e,x .
=
is zero when
2
7.
(a), (b)
(vi)
y
(i)
(iii)
x
(iv)
(ii)
(v)
Figure 10.1
10.3 SOLUTIONS
(c)
,
513
Figure 10.1 shows that a solution will be increasing if its y-values fall in the range 1 < y < 2. This makes sense
since if we examine the equation y0 = 0:5(1 + y)(2 y); we will find that y0 > 0 if 1 < y < 2. Notice that if
the y-value ever gets to 2, then y0 = 0 and the function becomes constant, following the line y = 2. (The same is
true if ever y = 1.)
From the graph, the solution is decreasing if y > 2 or y < 1. Again, this also follows from the equation,
since in either case y0 < 0.
The curve has a horizontal tangent if y0 = 0, which only happens if y = 2 or y = 1. This also can be seen
on the graph in Figure 10.1.
,
,
,
,
,
8.
(a)
For y = x2 + 2x + 2, we have y x2 = 2x + 2 and y0 = 2x + 2; thus, y0 = y x2 and the differential equation
is satisfied.
For y = x2 + 2x + 2 2ex, we have y x2 = 2x + 2 2ex and y0 = 2x + 2 2ex and so again y0 = y x2 .
The equation y = x2 + 2x + 2 gives an upward opening parabola, so it corresponds to the top solution shown
on the slope field.
,
,
,
,
,
,
,
y
(b)
I
II
x
III
Figure 10.2
(c)
(d)
(e)
The solution curves in Region I have one critical point each. They are concave up everywhere, decreasing at first,
then increasing very rapidly.
The curves in Region II have two critical points each and an inflection point. They are decreasing and concave up
at first, then they become increasing, eventually becoming concave down and finally decreasing very rapidly.
The curves in Region III are decreasing everywhere. They have no critical points.
Solutions for Section 10.3
1.
(a)
Figure 10.3