Section 1.6: Substitution Methods and Exact Equations 75
or
0gy
, or
0gy. Thus the solution is given by sin tan
x
eyxyC.
41. The condition x
FM implies that
222
43
23
,xy xy
Fxy dx gy
yx yx
, and then
42. The condition x
FM implies that
2/3 5/2 2/3 3/2
3
,2
Fxy y x ydx xy x y gy
,
43. Since the dependent variable y is missing, we can substitute yp
and yp
as in
Equation (34) of the text. This leads to
x
pp
, a separable equation for p as a function
44. Since the independent variable x is missing, we can substitute yp
and dp
yp
dy
as in
Equation (36) of the text. This leads to 20
dp
yp p
dy
, or , a separable equation for p as
76 Chapter 1: First-Order Differential Equations
45. Since the independent variable x is missing, we can substitute yp
and dp
yp
dy
as in
Equation (36) of the text. This leads to 40
dp
py
dy
, or 4
p
dp y dy
, or
46. Since the dependent variable y is missing, we can substitute yp
and yp
as in
Equation (34) of the text. This leads to 4
x
pp x
, a linear equation for p as a function
of x which we can rewrite as
4
x
D
xp x (thus, no integrating factor is needed), or
x
47. Since the dependent variable y is missing, we can substitute yp
and yp
as in
Equation (34) of the text. This leads to 2
p
p
, a separable equation for p as a function
of x. Separating variables gives 2
dp
x
dx
p
, or 1
x
B
p
, or 1
p
x
B
, that is,
p
Section 1.6: Substitution Methods and Exact Equations 77
D
48. Since the dependent variable y is missing, we can substitute yp
and yp
as in
Equation (34) of the text. This leads to 232xp xp
, a linear equation for p as a func-
tion of x. We rewrite this equation as 2
32
pp
x
x
, showing that an integrating factor is
x
x
49. Since the independent variable x is missing, we can substitute yp
and dp
yp
dy
as in
Equation (36) of the text. This leads to 2
4
dp
p
ypyp
dy
, or dp
ypy
dy , a linear
equation for p as a function as a function of y which we can rewrite as
y
D
yp y, or
50. Since the dependent variable y is missing, we can substitute yp
and yp
as in
Equation (34) of the text. This leads to
2
p
xp
, a first-order equation for p as a
function of x which is neither linear nor separable. However, the further substitution
78 Chapter 1: First-Order Differential Equations
51. Since the independent variable x is missing, we can substitute yp
and dp
yp
dy
as in
Equation (36) of the text. This leads to 3
2
p
pyp
, or 2
2
p
yp
, or 2
12dp y dy
,
52. Since the independent variable x is missing, we can substitute yp
and dp
yp
dy
as in
Equation (36) of the text. This leads to 31ypp
, or 3
1
pdp dy
, or
53. Since the independent variable x is missing, we can substitute yp
and dp
yp
dy
as in
Equation (36) of the text. This leads to 2pp yp
, or 2dp y dy
, or 2
p
yA,
54. Since the independent variable x is missing, we can substitute yp
and dp
yp
dy
as in
Equation (36) of the text. This leads to 2
3
y
pp p
, or 13
dp dy
p
y
, or
Section 1.6: Substitution Methods and Exact Equations 79
p
55. The proposed substitution vaxbyc implies that
1
y
vaxc
y
, so that
56. The proposed substitution
1n
vy
implies that
11n
yv
and thus that
111
1
11
nn
n
dy dv dv
vv
. Substituting into the given Bernoulli equation
Problems 57-62 illustrate additional substitutions that are helpful in solving certain types of first-
order differential equation.
57. The proposed substitution lnvy implies that v
y
e, and thus that v
dy dv
e
dx dx
. Substi-
58. By Problem 57, substituting lnvy into the given equation yields the linear equation
2
24
x
vvx
, which we rewrite (for 0x) as 24vvx
x
x
. An integrating factor is
80 Chapter 1: First-Order Differential Equations
59. The substitution yvk implies that dy dv
dx dx
, leading to
11
33
xvk xvk
dv
dxxvk xvk
.
Likewise the substitution
x
uh implies that uxh and thus that dv dv du dv
dx du dx du
which means that 1h and 2k . These choices for h and k lead to the homogene-
ous equation
1
1
v
dv u v u
v
du u v
u
,
or
22
21
p
puC
. Back-substituting v
u for p leads to
2
2
21
vv
uC
uu
, or
22
2vuvuC. Finally, back-substituting 1
x
for u and 2y for v gives the implic-
it solution
x
Section 1.6: Substitution Methods and Exact Equations 81
which means that 3h and 2k . These choices for h and k lead to the homogeneous
equation
21
2
43 43
v
dv v u u
v
du u v
u
,
which calls for the further substitution v
pu
, so that vpu and thus dv dp
pu
.
2
43 1 1 15 1
ln 1 5ln 3 1 ln
3214 1314
pdp dp p p C
pp p p
,
so the solution is given by
ln 1 5ln 3 1 ln 4ln
p
pCu ,
82 Chapter 1: First-Order Differential Equations
61. The expression
x
y appearing on the right-hand side suggests that we try the substitu-
tion vxy, which implies that yxv
, and thus that 1
dy dv
dx dx
. This gives the
the solution is given by 2
sec sec tanvvvdvdx
, or tan sec
x
vvC
. Finally,
back-substituting
x
y for v gives the implicit solution
tan sec
x
xy xy C.
However, for no value of the constant C does this general solution include the “basic” so-
62. First we note that the given differential equation is homogeneous; for 0x we have
3
33
3
33
2
2
221
y
yx y
dy y
x
dx x
xy x y
x
,
and substituting
y
v
x
as usual leads to
Section 1.6: Substitution Methods and Exact Equations 83
63. The substitution 1
1
yy v
, which implies that 12
1dy dv
y
dx v dx
, gives
2
111
2
111dv
y
Ax y Bx y Cx
vdx v v
,
which upon expanding becomes
y
64. Here
1Ax ,
0Bx, and
2
1Cx x . Thus the substitution 1
11
yy x
vv
leads to the linear equation 21
dv xv
dx
. An integrating factor is given by
84 Chapter 1: First-Order Differential Equations
x
65. Here
1Ax ,
2
B
xx , and
2
1Cx x . Thus the substitution
1
11
yy x
vv
yields the trivial linear equation 1
dv
dx , with immediate solution
vx C x. Hence the general solution of our Riccati equation is given by
1
yx x Cx
.
67. First, the line 2
1
4
y
Cx C has slope C and passes through the point
2
11
24
,CC; the
same is true of the parabola 2
y
x at the point
2
11
24
,CC, because
1
2
22
dy
x
CC
y
y
y
. Thus the line is tangent to the parabola at this point. It follows
4
68. Substituting lnCka into
2
ln 1 lnvvkxC
gives
2
ln 1 ln ln ln k
vvkxkaxa
,
Section 1.6: Substitution Methods and Exact Equations 85
69. With 100a and 1
10
k, Equation (19) in the text is
910 1110
50 100 100
xx
y
.
We find the maximum northward displacement of plane by setting
70. With
0
10 1
500 10
w
kv
, Equation (16) in the text gives
21
ln 1 ln
10
vv xC
,
where v denotes
y
x
. Substitution of 200x, 150y, and 3
4
v yields
110
ln 2 200C , which gives
71. Equations (12)-(19) apply to this situation as with the airplane in flight.
(a) With 100a and
0
21
42
w
kv
, the solution given by Equation (19) is
86 Chapter 1: First-Order Differential Equations
72. We note that the dependent variable y is missing in the given differential equation
32
2
1ry y
, leading us to substitute y
, and y
. This results in
32
2
1rp
, a separable first-order differential equation for
as a function of x.
CHAPTER 1 Review Problems
The main objective of this set of review problems is practice in the identification of the different
types of first-order differential equations discussed in this chapter. In each of Problems 1–36 we
identify the type of the given equation and indicate one or more appropriate method(s) of solu-
tion.
1. We first rewrite the differential equation for 0x as 2
3
y
yx
x
, showing that the
Review Problems 87
2. We first rewrite the differential equation for ,0xy as 22
3yx
yx
, showing that the
3. Rewriting the differential equation for 0x as
2
2
2
x
yy y y
y
x
xx
shows that the
4. Rewriting the differential equation in differential form gives
322
23sin0
x
Mdx Ndy xy e dx x y y dy
,
and because
2
,6 ,
M
xy xy N xy
yx
, the given equation is exact. Thus we ap-
x
5. We first rewrite the differential equation for ,0xy as 4
23yx
yx
, showing that the
6. We first rewrite the differential equation for ,0xy as 22
12yx
yx
, showing that the
x
88 Chapter 1: First-Order Differential Equations
x
7. We first rewrite the differential equation for 0x as 3
21
yy
x
x
, showing that the
x
x
x
8. We first rewrite the differential equation for 0x as
2
2
dy y y
dx x x
, showing that is it
homogeneous. Substituting y
v
x
then gives 22
dv
vx v v
dx
, or 23
dv
x
vv
dx . Sep-
y
x
y
x
y
x
Alternatively, writing the given equation as 2
2
21dy
y
y
dx x x
shows that it is a Bernoulli
equation with 2n. The substitution 12 1
vy y
implies that 1
y
v
and thus that
2
y
vv
. Substituting gives 212
2
21
vv v v
xx
, or 2
21
vv
x
x
, a linear equa-
x
Review Problems 89
y
9. We first rewrite the differential equation for ,0xy as 12
26
y
yxy
x
, showing that it
is a Bernoulli equation with 12n. The substitution 12
vy implies that 2
y
v and
thus that 2yvv
. Substituting gives 2
2
26vv v xv
x
, or 13vvx
x
, a linear equa-
x
y
x
10. Factoring the right-hand side gives
22
11
dy
x
y
dx , showing that the equation is
y
11. We first rewrite the differential equation for ,0xy as
2
3
dy y y
dx x x
, showing that it
is homogeneous. Substituting
y
v
x
then gives 2
3
dv
vx v v
dx
, or 2
3
dv
x
v
dx . Sepa-
rating variables leads to 2
13
dv dx
vx
, or 13ln
x
C
v
, or 1
3ln
vCx
. Back-
y
x
x
y
y
x
90 Chapter 1: First-Order Differential Equations
x
x
x
12. Rewriting the differential equation in differential form gives
34 22 3
62 9 8 0xy y dx x y xy dy
,
and because
34 23 22 3
62188 9 8
x
yy xyy xyxy
yx
, the given equation is
13. We first rewrite the differential equation for 0y as 4
254
y
x
x
y
, showing that the
x
14. We first rewrite the differential equation for ,0xy as
3
dy y y
dx x x
, showing that it is
homogeneous. Substituting y
v
x
then gives 3
dv
vx vv
dx
, or 3
dv
x
v
dx . Separat-
x
x
Review Problems 91
15. This is a linear differential equation. An integrating factor is given by
3
exp 3
x
dx e
, and multiplying the equation by
gives 332
33
xx
ey ey x
, or
16. Rewriting the differential equation as
2
yxy
suggests the substitution vxy,
which implies that yxv
, and thus that 1yv
. Substituting gives 2
1vv
, or
2
1vv
, a separable equation for v as a function of x. Separating variables gives
17. Rewriting the differential equation in differential form gives
0
xxy yxy
eyedxexedy
,
and because
x
xy xy y xy
e ye xye e xe
yx
, the given equation is exact. We ap-
92 Chapter 1: First-Order Differential Equations
18. We first rewrite the differential equation for ,0xy as
3
2
dy y y
dx x x
, showing that it
is homogeneous. Substituting
y
v
x
then gives 3
2
dv
vx vv
dx
, or 3
dv
x
vv
dx .
y
x
y
Alternatively, rewriting the differential equation for 0x as 3
3
21
y
yy
xx
shows
that it is Bernoulli with 3n. The substitution 13 2
vy y
implies that 12
y
v
, and
thus that 32
1
2
y
vv
. Substituting gives 32 12 32
3
12 1
2vv v v
xx
, or
3
42
vv
x
x
, a linear equation for v as a function of x. An integrating factor is given by
4
4
exp dx x
x
, and multiplying the differential equation by
gives
y
19. We first rewrite the differential equation for ,0xy as 32
223
y
x
x
y
, showing that
Review Problems 93
20. We first rewrite the differential equation for 0x as 52
33
y
yx
x
, showing that the
y
21. We first rewrite the differential equation for 1
x
as 2
11
11
yy
xx
, showing that
1
x
.
22. Writing the given equation for 0x as 323
612
dy
y
xy
dx x
shows that it is a Bernoulli
equation with 23n. The substitution 123 13
vy y
implies that 3
y
v and thus that
2
3
y
vv
. Substituting gives 2332
6
312vv v xv
x
, or 3
24vvx
x
, a linear equation
y
y
23. Rewriting the differential equation in differential form gives
cos sin 0
yy
ey xdxxe xdy
,
94 Chapter 1: First-Order Differential Equations
24. We first rewrite the differential equation for ,0xy as 32 12
29
y
x
x
y
, showing that
25. We first rewrite the differential equation for 1x as 23
1
yy
x
, showing that the
equation is linear. An integrating factor is given by
2
2
exp 1
, and
26. Rewriting the differential equation in differential form gives
12 43 15 32 3/2 1/3 6/5 1/2
912 815 0xy xy dx xy xy dy
,
and because
12 43 15 32 12 43 15 12 3/2 1/3 6/5 1/2
912 1218 815
x
yxy xyxy xyxy
yx
,
27. Writing the given equation for 0x as
2
4
1
3
dy x
yy
dx x
shows that it is a Bernoulli
equation with 4n. The substitution 3
vy
implies that 13
y
v
and thus that