642 Solutions to exercises
- (i)
(ii)
- 6 x
2
- 5 e
− 2 x1 − 11
Section 11.3
13.y
2
1 = 12 x
3
1 + 1 c 14.
- 16.y
3
1 = 13 e
x1 + 1 c
- 18.y 1 = 1 cx
19.y 1 = 111 − 1 x 20.
21.y 1 = 1 x 1 − 11 22.y 1 = 1 −ln(2 1 − 1 e
x)
23.y 1 = 1 x(ln 1 x 1 + 1 2) 24.
- 27.y 1 = 1 ae
x1 − 12 x 1 − 15
28.(x 1 − 1 y)
2
1 − 14 y 1 = 1 c
Section 11.4
Section 11.5
33.y 1 = 121 + 1 ce
− 2 x35.y 1 = 1 e
− 3 x(x 1 + 1 c)
37.y 1 = 1 ce
− 12 x1 − 14 38.y 1 = 1 cos 1 x 1 + 1 c 1 cos
2
1 x
Section 11.6
−
−
−+
1
1
23
23kk
e
kkt()
()
[]C ( )
ak k
kkkk
e
kt=
+−
−
−
12
2311
1
1
1[]C
ak k
kk
ee
kk
ee
kt kt kt=
−
−
−
−
−
−− − −
12
21 31
13 2kktkk
332
−
y
x
na
cx
na=
++
+
−
1
1
y
b
a
ce
ax nn=+
−+
+ 1() 1
y
x
=+xxxx xc
−+
2
22
2
2
sin cos sin
yce
x=−
2
2
14
xt
ka
kk
e
kkt()
()
=
−
−
−+
−1
11
1
1111
1
11
xa
nkt
nn−−−=−()
τ1 nn
k
=
ln
yx
x
444
2
=
ln
y
x
x
22
1
3
1
=−
yy
x
x
2
3
3
+= −
y
ce
x=
1
1
yae
x=
3y
xc
=
−
1
2
2
2
4
2e
−xxt
t
m
() t
cos
=
−
12
4
2
π
π
m
dx
dt
t
2
2
=cos 2 π
Section 11.7
- (i)I(t) 1 = 1 Ae
−t 2 RC(ii)
Chapter 12
Section 12.2
4.y 1 = 1 ae
− 2 x1 + 1 be
2 x 235.y 1 = 1 (a 1 + 1 bx)e
3 x6.y 1 = 1 a 1 cos 12 x 1 + 1 b 1 sin 12 x
Section 12.3
7.y 1 = 1 ae
3 x1 + 1 be
− 2 x9.y 1 = 1 (a 1 + 1 bx)e
4 x10.y 1 = 1 (a 1 + 1 bx)e
− 3 x 2211.y 1 = 1 e
− 2 x(A 1 cos 1 x 1 + 1 B 1 sin 1 x)
Section 12.4
- 14.x 1 = 1 e
−3(t−1)(t 1 − 1 1)
- 16.x 1 = 1 (cos 1 t 1 − 1 sin 1 t)e
t17.y 1 = 1 e
π− 2 x(cos 12 x 1 − 1 e
π 221 sin 12 x)
18.y 1 = 1 −sin 13 x 19.y 1 = 1 xe
4(1−x)20.y 1 = 12 e
− 2 x21.θ
n(t) 1 = 1 Ae
int 2τ1 + 11 Be
−int 2τ,n 1 = 10 ,± 1 ,± 2 ,=
Section 12.5
- (i)
(ii)
23.x(t) 1 = 1 (U
0
2 ω)sin 1 ωt
Section 12.6
- (i)x 2 l 1 = 10 , 124 , 122 , 324 , 1
(ii)x 2 l 1 = 10 , 125 , 225 , 325 , 425 , 1
- (i) ifnodd,
ifneven
ψnx
l
nx
l
()= sin
2 π
ψ
nx
l
nx
l
()= cos
2 π
A=,= 24 δ π
Aab=+,= ba
22 − 1
δ tan ( )
xt=
1
3
sin 3
xe e
tt=+
−
2
3
1
3
2
y e ae be
xix
ix=+
−− 32
11 2
11 2
()
yae be e
xxx=+
−
()
25 25 2
×+
cosωωtRC tsinω
It ce
EC
RC
tRC()
()
=+
− 0
2
1
ω
ω
+−
RtL tsinωω ωcos
It
E
RL
Le
Rt L()=
0 −
222
ω
ω