R
7. y 0 + y = x. I(x) = e
x − 1 + Ce−x .
1dx
= ex ⇒ (ex y)0 = xex ⇒ ex y =
R
xex dx = xex − ex + C ⇒ y =
2
− t +C
2
.
13. [(1 + t)u]0 = 1 + t ⇒ (1 + t)u = t + t2 + C ⇒ u = t + 1+t
R
15. (x2 y)0 = ln x ⇒ x2 y = ln xdx = x ln x − x + C ⇒ y =
.
−1 + C ⇒ C = 3 ⇒ y = x ln x−x+3
x2
2
R
3x
2 = y(1) =
3 tan θ
R
dx
x ln x−x+C
.
x2
=
= e x2 +1 = e sec2 θ d(tan θ) = e−3 ln | cos θ| = | sec θ|3 = (1 +
√
R √
x ) ⇒ (1 + x ) y = 3x 1 + x2 ⇒ (1 + x2 )3/2 y = 3x 1 + x2 dx = (1 + x2 )3/2 +
C ⇒ y = 1 + C(1 + x2 )−3/2 .
1 du
−n dy
1−n
−n dy
23. 1−n
+
P
(x)u
=
y
+
P
(x)y
=
y
+
P
(x)y
= y −n Q(x)y n = Q(x).
dx
dx
dx
R
29. 5 dQ
+ 20Q = 60 ⇒ Q0 + 4Q = 12 ⇒ e4t Q = 12e4t dt = 3e4t + C ⇒ Q = 3 + Ce−4t ,
dt
where 3 + C = Q(0) = 0 ⇒ C = −3. Hence Q = 3 − 3e−4t and I = 12e−4t .
R
R
3
3
33. I(t) = e 100+2t dt = e 2 ln |50+t| = (50 + t)3/2 ⇒ (50 + t)3/2 y = 2(50 + t)3/2 dt =
4
(50 + t)5/2 + C ⇒ y = 4(50+t)
+ C(50 + t)−3/2 , where y(0) = 40 + C · 50−3/2 = 0. This
5
5
−3/2
gives C = 5−3−40
= −40·2
= −40000 · 2−3/2 and
10−3
·2−3/2
20.
0
3x
. I(x)
x2 +1
2 3/2 0
y + x23x+1 y
2 3/2
y=
y(20)
100+2·20
≈
35. (a) v 0 +
v =
v=
56−24.147
140
c
v
m
2(100 + 2t)
− 40000(100 + 2t)−3/2
5
≈ 0.2275 kg/L.
= g ⇒ I(t) = e
−ct
mg
+ Ae m . Since
c −ct
mg
1−e m .
c
R
c
dt
m
ct 0
ct
ct
ct
= e m and e m v = e m g ⇒ e m v =
v(0) = 0, we have 0 =
−ct
m
lim
1
−
e
.
(b) lim v = mg
= mg
c t→∞
c
t→∞
h
it
Rt
−cx
mg
m −cx
m
m
(c) 0 mg
1
−
e
dx
=
x
+
=
e
c
c
c
0
36.
dv
dm
=
g
c
1−e
−ct
m
−
mg
c
·
is then sufficient to show
37. (a) z 0 + kz =
−1 0
P
P2
R
kdt
+
k
P
=
mg
c
mg
c
+ A and thus A
h
t+
m −ct
em
c
−ct
−ct
ct
ct −ct
· e m = gc 1 − e m − m
em
m2
−ct
−ct
d dv
dv
> 0: dtd dm
= mg e m − mg e m
dt dm
−kP (1−P/M )
P2
kt 0
+
k
P
=
i
−
m2 g
.
c2
. Note that
+
ct
mg m
e +A ⇒
c
= −mg
. Hence
c
gct
e
m2
−ct
m
=
dv dm t=0
gct −ct
em
m2
= 0. It
> 0.
k
.
M
R kt
kt
kt
(b) I(t) = e
= ekt ⇒ e z = keM ⇒ ekt z = keM dt = eM + C ⇒ z =
M −P0
1
1
M
M
M
+ Ce−kt ⇒ P = 1 +Ce
.
−kt = 1+M Ce−kt , where P0 = 1+M C ⇔ M C = P0 − 1 =
M
P0
Hence P =
M
1+
M
M −P0 −kt
e
P0
.
1