Chapter 9 Differential equations - scunderwood · PDF fileChapter 9 – Differential...
Transcript of Chapter 9 Differential equations - scunderwood · PDF fileChapter 9 – Differential...
Chapter 9 – Differential equations
Solutions to Exercise 9A 1
a
If y = Ae2t – 2 then dy
dt = 2Ae2t
Given dy
dt = 2y + 4
LHS = 2Ae2t
RHS = 2(Ae2t – 2) + 4
= 2Ae2t – 4 + 4
= 2Ae2t
y = Ae2t – 2 is a solution of dy
dt = 2y + 4
Substituting y(0) = 2 into y = Ae2t – 2
gives:
2 = Ae20 – 2
= A – 2
A = 4
y = 4e2t – 2 is the particular solution.
b
If y = x loge |x| – x + c
then dy
dx = loge |x| + 1 – 1
= loge |x|
y = x loge |x| – x + c is a solution of dy
dx = loge |x|
Substituting y(1) = 3 into
y = x loge |x| – x + c gives:
3 = 1 loge |1| – 1 + c
= –1 + c
c = 4
y = x loge |x| – x + 4 is the particular
solution.
c
If y = 2x + c
then dy
dx = 1
2 2x + c 2
= 1
2x + c
= 1y
y = 2x + c is a solution of dy
dx =
1
y
Substituting y(1) = 9 into
y = 2x + c gives:
9 = 2 1 + c
81 = 2 + c
c = 79
y = 2x + 79 is the particular solution.
d
If y – loge |y + 1| = x + c
then dxdy
= 1 – 1y + 1
= y + 1 1
y + 1
= y
y + 1
dy
dx =
y + 1
y
y – loge |y + 1| = x + c is a solution of dy
dx =
y + 1
y
Substituting y(3) = 0 into
y – loge |y + 1| = x + c gives:
0 – loge |0 + 1| = 3 + c
0 = 3 + c
c = –3
y – loge |y + 1| = x – 3 is the
particular solution.
e
If y = x4
2 + Ax + B
then dy
dx = 2x3 + A
and d2 y
dx2 = 6x2
y = x4
2 + Ax + b is a solution of
d2 y
dx2 = 6x2
Essential Specialist Mathematics Complete Worked Solutions 611
Substituting y(0) = 2 and y(1) = 2
into y = x4
2 + Ax + B gives:
2 = 04
2 + A 0 + B
B = 2
and
2 = 14
2 + A 1 + B
= 12
+ A 1 + 2
A = – 12
y = x4
2 – x
2 + 2 is the particular
solution.
f
If y = Ae2x + Be–2x
then dy
dx = 2Ae2x – 2Be–2x
and d2 y
dx2 = 4Ae2x + 4Be–2x
= 4(Ae2x + Be–2x)
= 4y
y = Ae2x + Be–2x is a solution of
d2 y
dx2 = 4y
Substituting y(0) = 3 and y(loge 2) = 9 into
y = Ae2x + Be–2x gives:
3 = Ae20 + Be–20
3 = A + B 1
and
9 = Ae2 loge 2
+ Be–2 loge 2
= Ae
loge 4 + Be
–loge 4
9 = 4A + 14
B 2
4 1 – 2 gives
3 = 0 + 154
B
B = 1215
= 45
Substituting B = 45
in 1 gives
3 = A + 45
A = 115
y = 115
e2x + 45
e–2x is the particular
solution.
g
If x = A sin 3t + B cos 3t + 2
then dxdt
= 3A cos 3t – 3B sin 3t
and d2 x
dt2 = –9A sin 3t – 9B cos 3t
= –9(A sin 3t + B cos 3t)
Given d2 x
dt2 + 9x = 18
LHS = –9(A sin 3t + B cos 3t)
+ 9(A sin 3t + B cos 3t + 2)
= –9A sin 3t – 9B cos 3t + 9A sin 3t
+ 9B cos 3t + 18
= 18
= RHS
x = A sin 3t + B cos 3t + 2 is a solution
of d2 x
dt2 + 9x = 18
Now x(0) = 4
4 = B + 2
B = 2
and
x(p2
) = –1
–1 = –A + 2
A = 3
x = 3 sin 3t + 2 cos 3t + 2 is the
particular solution.
2
a
y = 4e2x
dy
dx = 8e2x
dy
dx = 2y
b
y = 12
x–2
dy
dx = – x
–3 = –
1
x3
–4 xy2 = –4 x
1
2x2
2
= –4x
4x4 = –
1
x3
dy
dx = –4xy
2
Essential Specialist Mathematics Complete Worked Solutions 612
c
y = x log e |x| + x dy
dx = loge |x| + xx + 1 = loge |x| + 2
yx + 1 = loge |x| + 1 + 1 = loge |x| + 2
dy
dx =
yx + 1
d
y = (3x2 + 27)13
dy
dx =
1
3 (3x
2 + 27 )
–2
3 . (6x)
=
2x
(3x2 + 27)23
2x
y2 =
2x
(3x2 + 27)23
dy
dx = 2x
y2
e
y = e–2x + e3x
dy
dx = –2e–2x + 3e3x
d2 y
dx2 = 4e–2x + 9e3x
d2 y
dx2 –
dy
dx – 6y
= 4e–2 x
+ 9e3x
– (–2e–2 x
+ 3e3x
)
– 6(e–2 x
+ e3x
) = (–6 + 2 + 4)e–2x + (–6 – 3 + 9)e3x
= 0
d
2y
d2x
– dy
dx – 6y = 0
f
y = e4x
(x + 1) = xe4x
+ e4x
dy
dx = 4xe4x + e4x + 4e4x
= 4xe4x
+ 5e4x
d2 y
dx2 = 16xe4x + 4e4x + 20e4x
= 16 xe4x
+ 24 e4x
d2 y
dx2 – 8
dy
dx + 16y
= (16 – 32 + 16)xe4x + (16 – 40 + 24)e4x
= 0
d
2y
d2x
– 8 dy
dx + 16 y = 0
g
y = a sin (nx) dy
dx = na cos(nx)
d2 y
dx2 = –n2a sin(nx) = –n2y
h
y = enx
+ e– nx
dy
dx = nenx – ne–nx
d2 y
dx2 = n2enx + n2e–nx
= n2(enx + e–nx)
= n2y
i
y = x + 1
1 – x dy
dx =
(1 x) (1)(1 + x)
(1 x)2
=
2(1 - x) 2
1 + y2
1 + x2 = 2
2
2
1
)1 (
)1 ( 1
x
x
x
= 22
22
)1 )( 1(
)1 ( )1 (
xx
xx
= 22
2
)1 )(1 (
)1 (2
xx
x
= 2)1 (
2
x
dy
dx =
1 + y2
1 + x2
Essential Specialist Mathematics Complete Worked Solutions 613
j
y = 4
x + 1 = 4(x + 1)
–1
dy
dx =
-4(x + 1) 2
d2 y
dx2 = 8
(x + 1)3
y d2 y
dx2 = 4
x + 1 8
(x + 1)3 = 32
(x + 1)4
2(dy
dx)2
= 2 ´ 16
(x + 1)4 = 32
(x + 1)4
y d2 y
dx2 = 2(
dy
dx)2
3
dx
dy
1
y
dx
dy =
k
y (y > 0)
x = k loge y + c
Substituting y(0) = 2 and y(2) = 4
0 = k loge 2 + c 1
and
2 = k loge 4 + c
2 = 2k loge 2 + c 2
2 – 1
k loge 2 = 2
k = 2loge 2
Substituting into 1 gives
c = –2
x = 2 log e y
log e 2 – 2
When x = 3,
3 = 2log e y
log e 2 – 2
loge y = 52
loge 2
y = 4 2
4
If y = axn
then dy
dx = naxn–1
and d2 y
dx2 = n (n – 1)axn–2
Therefore,
x2 d2 y
dx2 – 2x
dy
dx – 10y
= n (n – 1)axn – 2anxn – 10axn = 0
n(n – 1) – 2n – 10 = 0
n2 – 3n – 10 = 0
(n – 5)(n + 2) = 0
n = –2 or n = 5
5
If y = a + bx + cx2
then dy
dx = b + 2cx
d2 y
dx2 = 2c
as d2 y
dx2 + 2
dy
dx + 4y = 4x2
2c + 2(b + 2cx) + 4(a + bx + cx2) = 4x2
4c = 4 c = 1
4b + 4c = 0 b = –1
2c + 2b + 4a = 0 a = 0
a = 0 , b = –1 and c = 1
6
If x = t (a cos 2t + b sin 2t)
then dxdt
= a cos 2t + b sin 2t
+ t (–2a sin 2t + 2b cos 2t)
= (a + 2bt) cos 2t + (b – 2at) sin 2t
d2 x
dt2 = 2b cos 2t – 2(a + 2bt) sin 2t
– 2a sin 2t + 2(b – 2at) cos 2t
= (4b – 4at) cos 2t – (4a + 4bt) sin 2t
d2 x
dt2 + 4x = 4b cos 2t – 4a sin 2t
and since d
2x
dt2
+ 4x = 2 cos 2 t
4b cos 2t – 4a sin 2t = 2 cos 2t
Essential Specialist Mathematics Complete Worked Solutions 614
4b = 2 b = 1
2 and a = 0
a = 0 , b = 1
2
7
If y = ax3 + bx2 + cx + d,
then dy
dx = 3ax2 + 2bx + c
d2 y
dx2 = 6ax + 2b
d2 y
dx2 + 2
dy
dx + y
= ax3 + (b+6a)x2 + (c +4b+6a)x + (d +2c+2b)
and since d
2y
dx2 + 2
dy
dx + y = x
3
a = 1
b + 6a = 0 b = –6
c + 4b + 6a = 0 c = 18
d + 2c + 2b = 0 d = –24
a = 1, b = –6 , c = 18 , d = –24
Essential Specialist Mathematics Complete Worked Solutions 615
Solutions to Exercise 9B 1
a dy
dx = x2 – 3x + 2
y = 13
x3 – 32
x2 + 2x + c,
c R, is the general solution.
b
dy
dx =
x 2 + 3x - 1x
, x ≠ 0
= x + 3 – 1x
y = 12
x2 + 3x – loge |x| + c,
c R, is the general solution.
c dy
dx = (2x + 1)3
= 8x3 + 12x2 + 6x + 1
y = 2x4 + 4x3 + 3x2 + x + c,
c R, is the general solution.
d dy
dx = 1
x, x > 0
1
2x
y = 2x12 + c
y = 2 x + c ,
c R, is the general solution.
e dy
dt =
12t - 1
, t ≠ 12
y = 12
loge |2t – 1| + c,
c R, is the general solution.
f dy
dt = sin (3t – 2)
y = – 13
cos (3t – 2) + c,
c R, is the general solution.
g dy
dt = tan (2t)
= sin (2 t)cos (2 t)
Let u = cos (2t)
dudt
= –2 sin (2t)
y = – 12u
dudt
dt
= – 12 1u
du
= – 12
loge |u| + c,
y = – 12
loge |cos (2t)| + c,
c R, is the general solution.
h dxdy
= e–3y
x = – 13
e–3y + c,
c R, is the general solution.
i
dx
dy =
1
4 – y2
= 1
22 – y
2
x = sin–1
y
2
+ c ,
c R, is the general solution
j
dx
dy = –
1
(1 – y)2
Let u = 1 – y, then du
dy = –1
x = u–2 (– du)
= u–2 du
= – u–1 + c
x = –
11 - y
+ c
x = 1
y – 1 + c ,
c R, is the general solution.
Essential Specialist Mathematics Complete Worked Solutions 616
2
a
d2 y
dx2 = 5x3
dy
dx =
54
x4 + c
y = 14
x5 + cx + d,
where c, d R, is the general solution.
b
d2 y
dx2 = 1 – x
Let u = 1 – x, then dudx
= –1
dy
dx = u
12 (– du)
= – u12 du
= – 23
u32 + c
y = – 23
u32 dx + c dx
dxcduu )(3
22
3
= 23 u
32 dudx
dx + c dx
= 23 u
32 du + c dx
= 415
u52 + cx + d
= 415
(1 – x)52 + cx + d,
where c, d R
y = 415
(1 – x)52 + cx + d
is the general solution.
c
d2 y
dx2 = sin (2x +
p4
)
dy
dx = – 1
2 cos (2x +
p4
) + c
y = – 14
sin (2x + p4
) + cx + d,
where c, d R, is the general solution.
d
d2 y
dx2 = e
x2
dy
dx = 2e
x2 + c
y = 4ex2 + cx + d,
where c, d R, is the general solution.
e
d2 y
dx2 = 1
cos2 x
= sec2 x
dy
dx = tan x + c
= sin xcos x
+ c
y = sin xcos x
+ c dx
Let u = cos x dudx
= – sin x
y = – 1u
dudx
dx + c dx
dxcduu
1
= – loge |u| + cx + d,
= – loge |cos x| + cx + d,
where c, d R
y = – loge |cos x| + cx + d,
is the general solution.
f
d2 y
dx2 =
1
(x + 1)2
= (x + 1)–2
dy
dx = – (x + 1)–1 + c
=
-1x + 1
+ c
y = – loge |x + 1| + cx + d,
where c, d R is the general solution.
Essential Specialist Mathematics Complete Worked Solutions 617
3
a dy
dx = 1
x2
y = dxx2
y = – 1x + c
3Initial condition: (4)
4y
34
= – 14
+ c
c = 1
y = –1
x + 1 =
x – 1
x
b dy
dx = e–x
y = e–x dx
y = – e–x + c
Initial condition: (0) 0y
0 = –1 + c
c = 1
y = 1 – e–x
c
dy
dx =
x2 – 4
x
y =
x 2 - 4x dx
= (x – 4x ) dx
= x2
2 – 4 loge |x| + c
3Initial condition: (1)
2y
32
= 12
+ c (Note: loge 1 = 0)
c = 1
y = x2
2 – 4 loge |x| + 1
d
Using partial fractions
x
x2 – 4
= A
x – 2 +
B
x + 2
x = A (x + 2) + B (x – 2)
When x = 2, A = 12
y =
x dx
x 2 - 4
= 12
dxx - 2
+ 12
dxx + 2
= 12
loge |x2 – 4| + c
y = 12
loge |x2 – 4| + c
Initial condition: (2 2 ) log 2e
y
loge 2 = loge ((2 2)2 – 4) + c
loge 2 = loge 4 + c
loge 2 = loge 2 + c
c = 0
y = 12
loge |x2 – 4|
e
dy
dx = x x
2 – 4
Let x2 – 4 = u, then du
dx = 2x
y = x x2 – 4 dx
= 12
u du
y = 13
u32 + c
1Initial condition: (4)
4 3y
But when x = 4, u = 12
1
4 3 =
1
3 (12 12 ) + c
c = 3
12 – 4 12
c = 3
12 – 8 3
c = 12
395
y = 13
(x2 – 4)32 –
95 312
f
dy
dx =
2 4
1
x
y =
2 4 x
dx
y = sin–1
x
2
+ c
Initial condition: (1)3
y
Essential Specialist Mathematics Complete Worked Solutions 618
When x = –2, B = 12
p3
= sin–1 12
+ c
c = p6
y = sin–1
x
2
+
6
g
2
1
4
1
(2 )(2 )
1 1 1
4 2 2
using partial fractions.
1log 2 log 2
4
1 2log
4 2
Initial condition: (0) 2
12 log 1
4
2
1 2log 2
4 2
e e
e
e
e
dy
dx x
x x
x x
y x x c
xc
x
y
c
c
xy
x
h dy
dx = 1
4 + x2
y = dx
4 + x2
y = 12
tan–1 x2
+ c
Initial conditions: y(2) = 3
8
3p8
= 12
tan–1 1 + c
3p8
= p8
+ c
c = p4
y = 12
tan–1 x2
+ p4
i
dy
dx = x 4 – x
Let u = 4 – x, du
dx = –1
x = 4 – u
y = x 4 – x dx
= – (4 – u)u12 du
= – 4u12 – u
32 du
= – (83
u32 – 2
5u
52 ) + c
= – (83
(4 – x)32 – 2
5(4 – x)
52 ) + c
3 5
8 (4 ) 2 (4 )
3 5
x xc
Initial conditions: y(0) = –8
15
–8
15 = – 8 43
3 + 2 45
5 + c
–8
15 = –
643
+ 645
+ c
–8
15 = –
128
15 + c
c = 8
3 58 (4 ) 2 (4 )
83 5
x xy
j dy
dx = ex
e x + 1
Let u = ex + 1 dudx
= ex
y = 1u du
y = loge (ex + 1) + c (as ex > 0)
Initial condition: (0) 0y
0 = loge 2 + c
c = – loge 2
y = loge (ex + 1) – loge 2
y = log e
e
x+ 1
2
Essential Specialist Mathematics Complete Worked Solutions 619
4
a
d2y
dx2 = e
– x – e
x
dy
dx = (e–x – ex) dx
dy
dx = – e–x – ex + c1
Initial condition: 0, 0dy
xdx
0 = –1 – 1 + c1
c1 = 2
dy
dx = – e–x – ex + 2
y = (– e–x – ex + 2) dx
y = e–x – ex + 2x + c2
Initial condition: (0) 0y 0 = 1 – 1 + 0 + c2
c2 = 0
y = e–x – ex + 2x
b
d2y
dx2 = 2 – 12 x
dy
dx = (2 – 12x) dx
dy
dx = 2x – 6x2 + c1
Initial condition: 0, 0dy
xdx
c1 = 0
dy
dx = 2x – 6x
2
y = (2x – 6x2) dx
y = x2 – 2x3 + c2
Initial condition: (0) 0y c2 = 0
y = x2 – 2x3
c
d2y
dx2 = 2 – sin 2x
dy
dx = (2 – sin 2x) dx
dy
dx = 2x +
1
2 cos 2x + c1
1
Initial condition: , 02
dyx
dx
dy
dx = 2x –
1
2 cos 2x
y = (2x + 1
2 cos 2x) dx
y = x2 + 14
sin 2x + c2
Initial condition: (0) 1y c2 = – 1
y = x2 + 14
sin 2x – 1
d
d2y
dx2 = 1 –
1
x2
dy
dx = (1 –
1
x2 ) dx
= x + 1x + c1
Initial condition: 0, 0dy
xdx
0 = 1 + 1 + c1
c1 = – 2
dy
dx = x +
1
x – 2
y = (x + 1
x – 2) dx
y = x2
2 + loge |x| – 2x + c2
3Initial condition: (1)
2y
2
3 = 1
2 – 2 + c2 (Note: loge 1 = 0)
c2 = 3
y = x2
2 + loge |x| – 2x + 3
e
d2y
dx2 =
2x
(1 + x2)
2
dy
dx = 2x dx
(1 + x 2 )2
= dww 2
, where w = 1 + x2
= – 1w + c1
= – 1
1 + x2 + c1
Initial condition: 0, 0dy
xdx
c1 = 1
dy
dx = –
1
1 + x2 + 1
Essential Specialist Mathematics Complete Worked Solutions 620
c1 = 0
21
1
dxy dx
x
= – tan–1 x + x + c2
Initial condition: (1) 1y
1 = – p4
+ 1 + c2
c2 = p4
y = x – tan–1 x + p4
f
d2y
dx2 = 24 (2x + 1)
dy
dx = 24(2x + 1) dx
= 24(x2 + x) + c1
Initial condition: 6, 1dy
xdx
c1 = 6
dy
dx = 24(x2 + x) + 6
y = 24(x3
3 +
x2
2) + 6x + c2
= 8x3 + 12x2 + 6x + c2
Initial condition: ( 1) 2y – 2 = – 8 + 12 – 6 + c2
c2 = 0
y = 8x3 + 12x2 + 6x
g
d2y
dx2 =
x
(4 – x2)
3
2
Let 4 – x2 = u
then dudx
= – 2x
dy
dx =
x
(4 – x2)
3
2
dx
= –1
2
1
u
3
2
du
= u
- 1
2 + c1
= 2
4
1
x + c1
1Initial condition: , 0
2
dyx
dx
dy
dx =
1
4 – x2
y = sin–1 x
2 + c2
Initial condition: ( 2)2
y
–p2
= sin–1 (–1) + c2
c2 = 0
y = sin–1 x2
5
a dy
dx = 3x + 4
y = 32
x2 + 4x + c,
c R, represents the family of curves.
b
d2 y
dx2 = – 2x
dy
dx = – x2 + c
y = – 13
x3 + cx + d,
c, d R, represents the family of curves.
c dy
dx =
1x - 3
y = loge |x – 3| + c,
c R, represents the family of curves.
6
a dy
dx = 2 – e–x
y = 2x + e–x + c
Now y(0) = 1
1 = 2 0 + e–0 + c
c = 0
y = 2x + e–x
Essential Specialist Mathematics Complete Worked Solutions 621
c1 = 0 b dy
dx = x + sin 2x
y = 12
x2 – 12
cos 2x + c
Now y(0) = 4
4 = 12
(0)2 – 12
cos (2 0) + c
= 0 – 12
+ c
c = 92
y = 12
x2 – 12
cos 2x + 92
c
dy
dx =
x 2
1
y = – loge |2 – x| + c
Now y(3) = 2
2 = – loge |2 – 3| + c
= – 0 + c
c = 2
y = 2 – loge |2 – x|
Essential Specialist Mathematics Complete Worked Solutions 622
Solutions to Exercise 9C 1
a dy
dx = 3y – 5
dxdy
=
13y - 5
x = 13
loge |3y – 5| + c, c R,
x – c = 13
loge |3y – 5|
3(x – c) = loge |3y – 5|
|3y – 5| = e3(x–c)
|3y – 5| = Ae3x
where A = e–3 c
3y – 5 = Ae3x
or 3y – 5 = – Ae3x
y = 1
3 (Ae
3x+ 5) or y =
1
3 (5 – Ae
3x)
y = 1
3 (Ae
3x + 5) for y >
5
3
or y = 1
3 (5 – Ae
3x) for y <
5
3
b dy
dx = 1 – 2y
dxdy
=
11 - 2y
=
-12y - 1
x = – 12
loge |2y – 1| + c, c R,
x – c = – 12
loge |2y – 1|
– 2(x – c) = loge |2y – 1|
|2y – 1| = e–2(x–c)
|2y – 1| = Ae–2 x
where A = e2c
2y – 1 = Ae–2 x
or 2y – 1 = – Ae–2 x
y = 1
2 (Ae
–2 x+ 1) or y =
1
2 (1 – Ae
–2 x)
y = 1
2 (Ae
–2 x + 1) for y >
1
2
or y = 1
2 (1 – Ae
–2 x) for y <
1
2
c dy
dx = e2y–1
dxdy
= e1–2y
x = – 12
e1–2y + c, c R
x – c = – 12
e1–2y
– 2(x – c) = e1–2y
1 – 2y = loge |– 2(x – c)|
2y = 1 – loge |– 2(x – c)|
y = 12
(1 – loge |– 2(x – c)|)
y = 12
– 12
loge |2c – 2x|
d dy
dx = cos2 y
dxdy
= sec2 y
x = tan y + c
x – c = tan y
y = tan–1 (x – c)
e dy
dx = cot y
dxdy
= tan y
= sin y
cos y
Let u = cos y dudy
= – sin y
dxdy
= – 1u
dudy
x = – 1u
dudy
dy
= – 1u
du
= – loge |u| + c, c R
= – loge |cos y| + c
x – c = – loge |cos y|
– (x – c) = loge |cos y|
|cos y| = ec–x
y = cos–1
(ec – x
) for cos y > 0
or y = cos–1
( – ec – x
) for cos y < 0
Essential Specialist Mathematics Complete Worked Solutions 623
f dy
dx = y2 – 1
dxdy
=
1
y 2 - 1
=
1(y + 1)(y - 1)
Let )1 )(1 (
1
yy A
y + 1 +
By - 1
A (y – 1) + B (y + 1) = 1
When y = –1,
– 2A = 1
A = – 12
When y = 1,
2B = 1
B = 12
1(y + 1)(y - 1)
12(y - 1)
– 12(y + 1)
dxdy
=
12(y - 1)
– 12(y + 1)
x = 12
loge |y – 1| – 12
loge |y + 1| + c,
c R,
=
1
2 log e
y – 1
y + 1
+ c
x – c = 1
2 log e
y – 1
y + 1
2(x – c) = log e y – 1
y + 1
y – 1
y + 1
= e2(x – c)
y – 1
y + 1
= Ae2x
where A = e–2 c
For y > 1 or y < – 1:
y – 1 = Ae2x
(y + 1)
y – Aye2x
= Ae2x
+ 1
y (1 – Ae2x
) = Ae2x
+ 1
y = 1 + Ae
2x
1 – Ae2x
For – 1 < y < 1 :
y = 1 – Ae
2x
1 + Ae2x
g dy
dx = 1 + y2
dxdy
= 1
1 + y2
x = tan–1 y + c, c R
x – c = tan–1 y
y = tan (x – c)
h dy
dx = 1
5y2 + 2y
dxdy
= 5y2 + 2y
x = 53
y3 + y2 + c, c R
i dy
dx = y = y
12
dxdy
= y
- 1
2
x = 2y 12 + c, c R
x – c = 2y 12
12
(x – c) = y 12
y = 14
(x – c)2
j dy
dx = y2 + 4y
dxdy
= 1
y2 + 4y
= 1y(y + 4)
Let 1y(y + 4)
Ay
+ By + 4
A (y + 4) + By = 1
When y = – 4,
– 4B = 1
B = – 14
When y = 0,
4A = 1
A = 14
1y(y + 4)
14y
– 14(y + 4)
dxdy
= 14y
– 14(y + 4)
x = 14
loge |y| – 14
loge |y + 4| + c,
c R
Essential Specialist Mathematics Complete Worked Solutions 624
x = 1
4 log e
y
y + 4
+ c
x – c = 1
4 log e
y
y + 4
4(x – c) = log e y
y + 4
e4(x – c)
= y
y + 4
y
y + 4
= Ae4x
where A = e–4 c
For y > 0 or y < – 4:
y = Ae4x
(y + 4)
y (1 – Ae4x
) = 4Ae4x
y = 4Ae
4x
1 – Ae4x
For – 4 < y < 0:
y = – 4Ae
4x
1 + Ae4x
2
a
dy
dx = y
x = dyy
x + c = log e |y|
When x = 0, y = e
c = 1 x + 1 = log e |y|
ex + 1
= |y|
y = ex + 1
for y > 0
b dy
dx = y + 1
x = dy
y + 1
loge |y + 1| = x + c
When x = 4, y = 0
c = –4
|y + 1| = ex – 4
y = ex – 4
– 1 for y > 1
c
dy
dx = 2y
x = 12 dyy
loge |y| = 2x + c
When x = 1, y = 1
c = –2
|y| = e2x – 2
y = e2x – 2
for y > 0
d dy
dx = 2y + 1
x = dy
2y + 1
x = 1
2 log e |2y + 1| + c
When x = 0, y = – 1
c = 0 2x = log e |2y + 1|
|2y + 1| = e2x
21 1( 1) for
2 2
xy e y
e
dy
dx =
ey
ey + 1
x = ey + 1
ey dy
x = (1 + e–y) dy
x = y – e–y + c
When x = 0, y = 0
c = 1 x = y – e–y + 1
f
dy
dx = 9 – y
2
x =
dy
9 – y2
x = sin–1 y
3 + c
Essential Specialist Mathematics Complete Worked Solutions 625
When x = 0, y = 3
c = –
2
y = 3 sin
x +
2
y = 3 cos x
Also –
2 < x +
2 <
2 – < x < 0 y = 3 cos x , – < x < 0
g
dy
dx = 9 – y
2
x =
1
9 – y2 dy
1
9 – y2
A
3 – y +
B
3 + y
1 = A (3 + y) + B (3 – y)
When y = –3, B = 16
When y = 3, A = 16
x = 16
dy
3 - y + 1
6
dy
3 + y
x = –1
6 log e |3 – y| +
1
6 log e |3 + y | + c
x = 1
6 log e
3 + y
3 – y
+ c
When x = 7
6 , y = 0
c = 7
6
6
x – 7
6
= log e 3 + y
3 – y
3 + y
3 – y
= e6x – 7
For – 3 < y < 3:
3 + y = e6x – 7
(3 – y)
y (1 + e6x – 7
) = 3e6x – 7
– 3
y = 3(e
6x – 7 – 1)
e6x – 7
+ 1
h
dy
dx = 1 + 9y
2
x =
1
1 + 9y2 dy
x =
1
1 + (3y)2 dy
x = 1
3 tan–1 3y + c
When x = –
12 , y = –
1
3
13
tan–1 (–1) + c = –p
12
c – p
12 = –
p12
c = 0
tan–1 3y = 3x
3y = tan 3x
y = 13
tan 3x
Also –
2 < 3x <
2
–
6 < x <
6
y = 1
3 tan 3x , –
6 < x <
6
i
dy
dx =
y2 + 2y
2
x =
2
y2 + 2y
dy
x =
2
y(y + 2) dy
2
y2 + 2y = A
y + B
y + 2
2 = A (y + 2) + By
When y = 0, A = 1
When y = –2, B = –1
x = 2 dy
y2 + 2y
x = dy
y –
dy
y + 2
x = log e y
y + 2
+ c
When x = 0, y = – 4
c = – log e 2
x = log e y
2(y + 2)
Essential Specialist Mathematics Complete Worked Solutions 626
For y < – 2
y = 2ex (y + 2)
y(1 – 2ex) = 4e
x
y = 4e
x
1 – 2ex
y = 4e
x
1 – 2ex
e– x
e– x
y = 4
e– x
– 2
3
a dy
dx = 1
y2
dxdy
= y2
x = 13
y3 + c
x – c = 13
y3
3(x – c) = y3
y = [3(x – c)]
1
3
is the equation for the family of curves.
b dy
dx = 2y – 1
dxdy
=
12y - 1
x = 12
loge |2y – 1| + c, c R,
x – c = 12
loge |2y – 1|
2(x – c) = loge |2y – 1|
|2y – 1| = e2(x–c)
|2y – 1| = Ae2x where A = e–2c
21( 1)
2
xy Ae
Essential Specialist Mathematics Complete Worked Solutions 627
Solutions to Exercise 9D 1
a
From the table, dxdt
= 2t + 1
x = t2 + t + c
Now x(0) = 3
3 = c
x = t2 + t + 3
b
From the table, dxdt
= 3t – 1
x = 32
t2 – t + c
Now x(1) = 1
1 = 32
– 1 + c
c = 12
x = 32
t2 – t + 12
c
From the table, dxdt
= – 2t + 8
x = – t2 + 8t + c
Now x(2) = – 3
– 3 = – (2)2 + 8(2) + c
= – 4 + 16 + c
= 12 + c
c = –15
x = – t2 + 8t – 15
2
a
dy
dx =
1
y , y 0
b
dy
dx =
1
y2 , y 0
c
dN
dt
1
N2 , N 0
dN
dt =
k
N2 , N 0 and k > 0
since the population is increasing.
d
dx
dt
1
x , x 0
dx
dt =
k
x , x 0 and k > 0
e
dm
dt – m
dm
dt = – km , k > 0
or alternatively,
dm
dt = km , k < 0
f
The gradient of the tangent at the point (x, y) is
y
x . Three times this is,
3y
x
Therefore the gradient of the normal at the point
(x, y) is
dy
dx =
–1
3y
x
dy
dx =
– x
3y , y 0
3
a
i.
dP
dt P
dP
dt = kP , k > 0
Essential Specialist Mathematics Complete Worked Solutions 628
ii. dtdP
= 1kP
t = 1k
1P
dP
t = 1k
loge P + c, P > 0
b
i.
When t = 0, P = 1000
0 = 1k
loge 1000 + c 1
When t = 2, P = 1100
2 = 1k
loge 1100 + c 2
2 – 1 yields
2 = 1k
loge 1100 – 1k
loge 1000
= 1k
loge (11001000
)
= 1k
loge 1.1
k = 12
loge 1.1 3
Substituting 3 in 1 yields
0 = 1
12
loge 1×1 loge 1000 + c
= 2 loge 1000
loge 1×1 + c
c =
-2 loge
1000
loge
1 ×1
t = 1
12
loge 1×1 loge P +
-2 loge
1000
loge
1 ×1
= 2loge 1×1
(loge P – loge 1000)
t = 2loge 1×1
loge ( P1000
)
Rearranging to make P the subject of the
formula: loge (1×1) t
2 = loge ( P
1000)
P = 1000e12 t loge (1·1)
P = 1000 (1.1)t2
When t = 5,
P = 1000 (1.1)52
= 1269.05870…
After five years, the population is 1269.
ii.
P = 1000 (1.1)t2 , t ≥ 0
4
a
i.
dP
dt – P , P > 0
dP
dt = k P , k < 0 and P > 0
ii. dtdP
= 1
k P
t = 1k
P
- 1
2 dP
= 1k
2P12 + c
t = 2 Pk
+ c, k < 0
b
i.
When t = 0, P = 15 000
0 = 2 15000
k + c
0 = 100 6k
+ c 1
When t = 5, P = 13 500
5 = 2 13500
k + c
5 = 20 135k
+ c 2
2 – 1 yields
5 = 20 135k
– 100 6k
5 = 20k
( 135 – 5 6 )
k = 4 135 – 20 6 3
1000
t
P
0
Essential Specialist Mathematics Complete Worked Solutions 629
Substituting 3 in 1 yields
0 = 620 1354
6100
+ c
= 65 135
625
+ c
c = 65135
625
t = 620 1354
2
P –
65135
625
= )65 135(2
P –
)65 135(2
)625(2
= )65 135(2
650
P 4
Rearranging to make P the subject of the
formula:
2t ( 135 – 5 6 ) + 50 6 = P
P = [2t ( 135 – 5 6 ) + 50 6 ]2
When t = 10,
P = (20 135 – 100 6 + 50 6 )2
= (20 135 – 50 6 )2
= 400135 + 25006 – 220 135 50 6
= 69 000 – 2000 810
= 69 000 – 18 000 10
= 12 079.00212…
The population after 10 years is 12 079.
ii.
P = [2t( 135 – 5 6 ) + 50 6 ]2, t ≥ 0
From 4 in part b i.
When P = 0, t = 135 65
625
5
a
i.
dP
dt
1
P
dP
dt =
k
P , k > 0 and P > 0
ii. dtdP
= Pk
t = 1k
P dP
= 1k
12
P2 + c
t = 12k
P2 + c
b
i.
When t = 0, P = 1 000 000
0 = 12k
(1 000 000)2 + c 1
When t = 4, P = 1 100 000
4 = 12k
(1 100 000)2 + c 2
2 – 1 yields
4 = 12k
(1 100 000)2 – 12k
(1 000 000)2
= 12k
((1 100 000)2 – (1 000 000)2)
= 12k
(2.1 1011)
= 1×05 ´ 1011
k
k = 14
(1.05 1011)
= 2.625 1010 3
5 6 – 135
25 6
15 000
t
P
0
Essential Specialist Mathematics Complete Worked Solutions 630
Substituting 3 in 1 yields
0 = 1
2(2×625 ´ 1010)(1 000 000)2 + c
= 1 ´ 1012
5×25 ´ 1010 + c
= 100
5.25
= 40021
+ c
c = –40021
t = 1
2(2×625 ´ 1010)P2 –
40021
= 1
5×25 ´ 1010P2 – 40021
Rearranging to make P the subject of the
formula:
P2 = 5.25 1010 (t + 40021
)
P = 5×25 ´ 1010(t + 40021
)
(The negative square root is not appropriate as
P ≥ 0.)
= 50 000 21(t + 40021
)
P = 50 000 21t + 400, t ≥ 0
ii.
P = 50 000 21t + 400, t ≥ 0
6 dy
dx = 1
10y
x = 10 1y dy
x = 10 loge y + c , y > 0
y = 10 when x = 0
c = – 10 loge 10
x = 10 loge y
10
y = 10ex
1 0
7 dqdt
= 0.01
dtdq
= 100
q
t = 100 loge + c , > 0
= 300 when t = 0
c = –100 loge 300
t = 100 loge (q
300)
= 300e0·01t
When t = 10,
= 300e0·1
≈ 331.55° K
8 dQdt
= – kQ
dtdQ
=
1-kQ
t =
-1k
loge Q + c , Q > 0
When t = 0, Q = 50
c = 1k
loge 50
When t = 10, Q = 25
10 = –1
k log e 25 +
1
k log e 50
10 = 1
k log e 2
k = 1
10 loge 2
t = 10
loge 2 loge ( 50
Q)
t10
loge 2 = loge 50Q
When Q = 10,
t = 10 loge 5
loge 2 ≈ 23.22
9 dmdt
= – km
dtdm
=
1-km
t = – 1k
loge m + c, m > 0
Let m = m0 initially.
c = 1k
loge m0
t = 1k
loge m0m
When m = m0
2, t = 1
k loge 2
1000000
t
P
0
Essential Specialist Mathematics Complete Worked Solutions 631
10
a dxdt
=
20 - 3x30
dtdx
=
3020 - 3x
t = –30
3 log e(20 – 3x) + c , x <
20
3 = –10 loge (20 – 3x) + c
When t = 0, x = 2
c = 10 loge (14)
t = 10 loge (
1420 - 3x
)
loge
1420 - 3x
= t10
1
1420 - 3x
= et
1 0
20 - 3x14
= e
- t
1 0
20 – 3x = 14e
- t10
3x = 20 – 14e
- t
1 0
x =
20 - 14e- t
1 0
3
1014203
1t
ex
b
From 1 ,
t = 10 loge
1420 - 3x
Therefore when x = 6,
t = 10 loge
1420 - 18
t = 10 loge 7 ≈ 19 min
11 dy
dx = 10 –
y
10
dy
dx =
100 - y
10
dxdy
=
10100 - y
x = –10 loge (100 – y) + c, y < 100
When x = 0, y = 10
c = 10 loge (90)
x = 10 loge (
90100 - y
)
90100 - y
= ex
1 0
100 - y
90 = e
-x
1 0
100 – y = 90e
-x
1 0
y = 100 – 90e
-x
1 0
12 dndt
= kn
dtdn
= 1kn
t = 1k
loge n + c , n > 0
When t = 0, n = 4000
c = – 1k
loge 4000
When t = 4, n = 8000
4 = 1
k log e 8000 –
1
k log e 4000
4 = 1
k log e 2
k = 1
4 loge 2
t = 4loge 2
loge n4000
After 3 more days, t = 7
7
4 log e 2 = log e
n
4000
loge n = 74
loge 2 + loge 4000
n = 4000 274
≈ 13 454
≈ 13 500 (to the nearest hundred)
10
Essential Specialist Mathematics Complete Worked Solutions 632
13
a
dN
dt N
dN
dt = kN, k > 0
dtdN
= 1kN
t = 1k
loge N + c, N > 0
Let year 1990 be t = 0, then 2000 is t = 10
and 2010 is t = 20.
When t = 0, N = 10000:
c = – 1k
loge 10000
When t = 10, N = 12 000:
10 = 1
k log e 12000 –
1
k log e 10000
10 = 1
k log e
6
5
k = 110
loge 65
t = 10loge 1×2
loge N10000
N = 10 000e0·1t l oge 1·2
0.110000(1.2)
t
For t = 20,
N = 10 000 (1.2)2
= 14 400
b
dN
dt
1
N
dNdt
= kN
, k > 0
dtdN
= Nk
t = N 2
2k + c
When t = 0, N = 10 000
c =
-108
2k
When t = 10, N = 12 000
10 = 1
k 144 10
6
2 –
108
2
k = (144 ´ 106
2 –
108
2) ÷ 10
= 22 105
t = N2
44 ´ 105 – 108
44 ´ 105
For t = 20,
N = 8510)1044(20 ≈ 13 711
c
dN
dt N
dNdt
= k N , k > 0
dtdN
= 1
k N =
1kN
-1
2
t = 2k
N12 + c
When t = 0, N = 10 000
c =
-200k
t = 2k
N 12 –
200k
When t = 10, N = 12 000
10 = 2
k 12000 –
200
k
10 = 1
k (2 12000 – 200 )
k = 1
512 000 – 20
= 4 30 – 20
For t = 20,
N
1
2 =
1
2
(4 30 – 20 )
20 +
200
4 30 – 20
= 1
2 (80 30 – 400 + 200 )
= 40 30 – 100
N = (40 30 – 100 )2
≈ 14 182
14
a
rate of inflow = 0.3
rate of outflow = 0.2 V
dVdt
= 0.3 – 0.2 V , V > 0
b
rate of inflow = 5 10 = 50
rate of outflow = volume
m 12
dV
dt = rate in – rate out
= 10 – 12
= –2
Essential Specialist Mathematics Complete Worked Solutions 633
V = –2 t + c , c is a constant When t = 0, V = 200:
c = 200 V = 200 – 2t
rate of outflow = 12 m
200 – 2t
=
6m
100 – t
dm
dt = 50 –
6m
100 – t , 0 t < 100
c
rate of inflow = 0 6
= 0
rate of outflow = volume
x 5
dV
dt = rate in – rate out
= 6 – 5
= 1 V = t + c , c is a constant When t = 0, V = 200:
c = 200 V = 200 + t
rate of outflow = 5x
200 + t
dx
dt = 0 –
5x
200 + t = –
5x
200 + t where t 0
15
a
rate of outflow = m
volume rate out
= 100
m 1
= m100
The sugar is being removed at m100
kg/min at
time t minutes.
b
rate of inflow = 0 1 = 0 dmdt
= rate of inflow – rate of outflow
= 0 – m100
dmdt
= –m
100
c dtdm
=
-100m
t =
-100m
dm
= – 100 loge m + c, m > 0
When t = 0, m = 20:
0 = – 100 loge 20 + c
c = 100 loge 20
t = – 100 loge m + 100 loge 20
t = 100 loge ( 20m )
t100
= loge (20m )
20m = e
t100
m = 20e
- t
100, t ≥ 0
d
m = 20e
- t
100, t ≥ 0
16
a
rate of inflow = 0.25 1
= 0.25
The sugar is being added at a rate of 0.25
kg/min at time t.
20
t
m
0
Essential Specialist Mathematics Complete Worked Solutions 634
b
rate of outflow = m100
1
= m100
The sugar is being removed at a rate of m100
kg/min at time t.
c dmdt
= rate of inflow – rate of outflow
= 0.25 – m100
d dmdt
=
25 - m100
dtdm
=
10025 - m
t = 100
125 - m
dm
= – 100 loge (25 – m) + c,
where 0 < m < 25
When t = 0, m = 0
0 = – 100 loge 25 + c
c = 100 loge 25
t = – 100 loge (25 – m) + 100 loge 25
t = 100 loge (
2525 - m
) 1
t100
= loge (
2525 - m
)
2525 - m
= et
100
25 = (25 – m) et
100
= 25et
100 – met
100
met
100 = 25et
100 – 25
= 25(et
100 – 1)
m = 25(et
100 – 1) e
- t
100
m = 25(1 – e
- t
100), t ≥ 0
e
When the concentration is 0.1 kg per litre,
m = 0.1 100 = 10
Substitute m = 10 in 1 from d.
t = 100 loge (
2525 - 10
)
= 100 loge (2515
)
= 100 loge (53
)
t = 51.08256…
It will take 51 minutes (to the nearest minute)
for the concentration in the tank to reach 0.1 kg
per litre.
f
m = 25(1 – e
- t
100), t ≥ 0
When t = 0, m = 25(1 – e0)
= 25(1 – 1)
= 0
As t , e
- t
100 0 m 25
m = 25 is a horizontal asymptote.
17
a
If x L is the amount of pure serum in the tank at
time t, then with 2 L of solution drawn off we
lose 0.02x L of pure serum, but we add at the
same time 0.2 L each minute.
dxdt
= 0.2 – 0.02x
dxdt
= 100
2 20 x =
50
10 x
b
dt
dx =
50
10 – x = –
50
x – 10 t = –50 loge (x – 10) + c
When t = 0, x = 20
c = 50 loge 10
t = 50 log e 10
x – 10
When x = 18,
t = 50 loge 108
≈ 11.16 min
Essential Specialist Mathematics Complete Worked Solutions 635
18
a dxdt
= 0.4 – 2x400
,
where 0.4 is a constant rate added and 2x400
is
the rate of solution drawn off.
dxdt
= 0.4 – x200
=
80 - x200
dtdx
=
20080 - x
t = –200 loge (80 – x) + c
Initially, x = 10
c = 200 loge 70
t = 200 loge (
7080 - x
)
7080 - x
= et
200
70 = e
t
200 (80 – x)
xe
t
200 = 80 e
t
200 – 70
x = 80 – 70e
-t
200
b
rate of outflow = x
volume 1
dV
dt = 2 – 1 = 1
V = t + c , c is a constant When t = 0, V = 400:
c = 400 V = 400 + t
rate of outflow = x
400 + t
dx
dt = 0.4 –
x
400 + t
19
rate of inflow = dxdV
dVdt
= 0.5 10
= 5
rate of outflow = dxdV
dVdt
= x100
10
= x10
dxdt
= 5 – x10
=
50 - x10
dtdx
=
1050 - x
t = 10
150 - x
dx
= –10 loge (50 – x) + c, 0 < x < 50
When t = 0, x = 0
0 = –10 loge 50 + c
c = 10 loge 50
t = –10 loge (50 – x) + 10 loge 50
= 10 loge (
5050 - x
)
t10
= loge (
5050 - x
)
5050 - x
= et
1 0
50 = (50 – x) et
1 0
= 50et
1 0 – xet
1 0
xet
1 0 = 50(et
1 0 – 1)
x = 50(1 – e
-t
1 0 ), t ≥ 0
20
a
rate of inflow = 0 2 = 0
rate of outflow = x
20 2 =
x
10
dx
dt = 0 –
x
10 = –
x
10
Essential Specialist Mathematics Complete Worked Solutions 636
b dtdx
= –10x
t = –10 loge (x) + c
When t = 0, t = 10:
c = 10 loge 10
t = 10 log e
10
x
1
x = 10e
-t
1 0 , t ≥ 0
c
d
From 1 in part b
When x = 5,
t = 10 loge 2 ≈ 6.93 min
21
a dNdt
= 0.1N – 5000
dtdN
=
10 ×1N - 5000
t =
10 ×1N - 5000
dN
= 10×1
loge (0.1N – 5000) + c,
where N > 50 000
= 10 loge (0.1N – 5000) + c
When t = 0, N = 5 000 000:
0 = 10 loge (0.1 5 000 000 – 5000) + c
= 10 loge (495 000) + c
c = – 10 loge (495 000)
t = 10 loge (
0 ×1N - 5000495 000
) 1
t10
= loge (
0 ×1N - 5000495 000
)
et
1 0 =
0 ×1N - 5000495 000
0.1N = 495 000 et
1 0 + 5000
N = 4 950 000 et
1 0 + 50 000
N = 50 000 (99et
1 0 + 1), t ≥ 0
b
From 1 in a, when N = 10 000 000
t = 10 loge (
0 ×1 ́ 10 000000 - 5000495 000
)
= 10 loge (19999
)
= 6.98184…
The country will have a population of 10 million
at the end of 2006.
10
t
x
0
Essential Specialist Mathematics Complete Worked Solutions 637
Solutions to Exercise 9E 1
a
25cm
h
50cmr
Let V cm3 be the volume at time t minutes.
min/cm500
min/cm10005.0L/min5.0
3
3
500dt
dV
Volume of cone hrV2
3
1
Using similar triangles 25025
hr
hr
1223
132
hh
hV
So, 4
2h
dh
dV
0,2000
5004
2
2
hh
h
dt
dV
dV
dh
dt
dh
b
h
hcQdt
dV out ratein rate
Volume of tank
area theis where, AhAV
Adh
dV
And so,
0 where,1
hhcQA
dt
dV
dV
dh
dt
dh
c
Vdt
dV2.03.0out rate in rate
Volume of tank hV 6
6dh
dV
Hence,
0 where,2360
1
2.03.06
1
VV
V
dt
dV
dV
dh
dt
dh
1
(3 2 6 ), w here 060
h h
Essential Specialist Mathematics Complete Worked Solutions 638
d
hdt
dV
Volume of cylinder hrV2
With 5.1r ;
4
9
2
32
hhV
4
9
dh
dV
And so,
0 where,9
4
9
4
hh
h
dt
dV
dV
dh
dt
dh
2
a
hdt
dV5
Volume of cone hyV2
3
1
Using similar triangles; r
y
r
h
hy
Hence,
3
3
1hV
2h
dh
dV
And so,
0 where,5
51
3
2
h
h
hh
dt
dV
dV
dh
dt
dh
ch
t
cht
dhht
dhh
t
h
dh
dt
25
2
5
2
5
5
5
5
5
3
3
2
5
2
3
When 0t , 25h :
250
25
31252
c
c
025
2250
5
hh
t
W hen 0, 250
785.40m in
13hrs 5m in
h t
t
t
1.5 cm
4 m
h
h cm
r
r
y
Essential Specialist Mathematics Complete Worked Solutions 639
b
Tank is empty when h = 0:
3
a
When y = 2 – h,
x2 + (2 – h)
2 = 4
x2 + h
2 – 4h + 4 = 4
x = ± 4h – h2
Therefore the width of the rectangular water
surface = 2 4h – h2
Thus the area of rectangular water surface
= 6 2 4h – h2
= 12 4h – h2
Changing h to x, the area of the rectangular
water surface A = 12 4x – x2
dx
dt =
–0.025 x
12 4x – x2
= –x
480 4x – x2
1
x
1
x
dx
dt = –
1
480 4 – x
b
cxt
cx
t
dxxt
xdx
dt
2
3
2
1
4320
12
3
4480
4480
4480
2
3
When :4,0 xt
2
3
4320
0
xt
c
c
Tank is empty when .0x
So,
hrs 42min 2560832043203
22
3
t
min 40 hrs 42 t
4
a
Given: 2
2 Adt
dV
dt
dV
dV
dr
dt
dr
Volume of sphere 3
3
4rV
24 r
dr
dV
And so,
dr
dt =
1
4r2 –2 A
2
= –2 A
A
where 4r2 is the surface area of a sphere.
dr
dt = –2 A
dr
dt = –2 4r
2
y
x
x2 + y
2 = 4
2-2
2
-2
h
2 – h
Essential Specialist Mathematics Complete Worked Solutions 640
b
cr
t
crt
drrt
drr
t
drr
t
rdr
dt
8
1
8
1
8
1
1
8
1
8
1
8
1
1
2
2
2
2
When :2,0 rt
16
1 c
16
1
8
1
rt
116
2
2
1161
2
18
1
16
1
8
1
tr
t
r
tr
tr
dr
dt = –8r
2
c
radius-time graph
surface area-time graph
Surface area , A = 4r2
r = + A
4
A
4 =
2
16 t + 1
A
4 =
4
(16 t + 1)2
A = 16
(16 t + 1)2
t
r
2
t
A
Essential Specialist Mathematics Complete Worked Solutions 641
5
a
h
Let V cm
3 be the volume at time t minutes.
khQdt
dV out ratein rate
(Q – kh) L/min = (Q – kh) 1000 cm3 / min
= 1000 (Q – kh) cm3 / min
dV
dt = 1000 (Q – kh) cm
3 / min
Volume of tank
area theis where, AhAV
Adh
dV
And so,
0 where,1000
hkhQA
dt
dV
dV
dh
dt
dh
b
dt
dh =
A
1000 (Q – kh)
t = A
1000
1
Q – kh dh
t = –A
1000 k log e (Q – kh) + c
When t = 0, h = h0:
c = A
1000 k log e (Q – kh0) , (Q > kh0)
t = A
1000 k log e
Q – kh0
Q – kh
, Q > kh0
c
When h = Q + kh0
2k ,
t = A
1000 k log e
Q – kh 0
Q – k Q + kh 0
2k
t = A
1000 k log e
Q – kh 0
Q – Q + kh 0
2
t = A
1000 k log e
Q – kh 0
Q – kh 0
2
t = A
1000 k log e
(Q – kh0) 2
Q – kh0
t = A
1000 k log e 2
minutes
Essential Specialist Mathematics Complete Worked Solutions 642
Solutions to Exercise 9F 1
TI: Set Calculation Mode to Approximate and
Display Digits to Fix4
CP: Set to Decimal mode and change the
Number Format to Fix4
a
dy
dx = cos x , y(0) = 1
Find y when x =
4 .
On your calculator type:
0
4
cos (x) dx + 1
b
dy
dx =
1
cos x , y(0) = 1
Find y when x =
4 .
On your calculator type:
0
4
1
cos (x) dx + 1
c
dy
dx = log e (x
2) , y(1) = 2
Find y when x = e.
On your calculator type:
1
e1
ln (x2) dx + 2
d
dy
dx = log e x , y(1) = 2
Find y when x = e.
On your calculator type:
1
e1
ln (x) dx + 2
Essential Specialist Mathematics Complete Worked Solutions 643
2
Ensure your calculator is set to Exact/Standard
mode.
a
dy
dx = tan
–1 x
On your calculator type:
deSolve (y ' = tan–1
(x) , x, y)
Therefore the general solution is
y = –1
2 ln (x
2 + 1) + x tan
–1 x + c
b
dy
dx = x
3 log e x
On your calculator type:
deSolve (y' = x3 ln (x), x, y)
Therefore the general solution is
y = 1
4 x
4log e(x) –
x4
16 + c
c
dy
dx = e
3x sin 2x
On your calculator type:
deSolve (y' = e3x
sin (2x), x, y)
Therefore the general solution is
y = e3x
3
13 sin 2x –
2
13 cos 2x
+ c
d
dy
dx = e
3x cos (2x)
On your calculator type:
deSolve (y' = e3x
cos (2x), x, y)
Therefore the general solution is
y = e3x
3
13 cos 2x +
2
13 sin 2x
+ c
e
dy
dx = 2
x
On your calculator type:
deSolve (y' = 2x, x, y)
Therefore the general solution is
y = 2
x
log e 2 + c
Essential Specialist Mathematics Complete Worked Solutions 644
Solutions to Exercise 9G 1
a
dy
dx = cos x and y(0) = 1
Using Euler’s method:
yn + 1 = yn + 0.1 [cos (xn)] 1
with x0 = 0 , y0 = 1 , h = 0.1
Put n = 0 into 1 :
y1 = y0 + 0.1 cos (x0) = 1 + 0.1 1 y1 = 1.1 and x1 = 0 + 0.1 = 0.1
Put n = 1 into 1 :
y2 = y1 + 0.1 [cos (x1)] = 1.1 + 0.1 cos (0.1 ) = 1.19950041 . . . y2 = 1.1995 and x2 = 0.1 + 0.1 = 0.2
Put n = 2 into 1 :
y3 = y2 + 0.1 [cos (x2)] = 1.1995 + 0.1 cos (0.2 )
= 1.29750707 . . . y3 = 1.2975 and x3 = 0.3
b
dy
dx =
1
x2 and y(1) = 0
Using Euler’s method:
y n + 1 = y n + 0.01
1
x n 2
2
with x0 = 1 , y0 = 0 , h = 0.01
Put n = 0 into 2 :
y 1 = y 0 + 0.01
1
x 0 2
= 0 + 0.01 (1) y1 = 0.01 and x1 = 1 + 0.01 = 1.01
Put n = 1 into 2 :
y 2 = y 1 + 0.01
1
x 1 2
= 0.01 + 0.01
1
1.012
=
20201
1020100
= 0.01980296 . . .
y2 = 20201
1020100 and x2 = 1.02
Put n = 2 into 2 :
y 3 = y 2 + 0.01
1
x 2 2
=
20201
1020100 + 0.01
1
1.022
=
78045301
2653280100
= 0.029414648 . . .
y3 = 78045301
2653280100 and x3 = 1.03
Put n = 3 into 2 :
y 4 = 78045301
2653280100 + 0.01
1
1.032
= 0.038840607 . . .
y4 = 0.0388 and x4 = 1.04
c
dy
dx = x and y(1) = 1
Using Euler’s method:
yn + 1 = yn + 0.1 [ xn ] 3
with x0 = 1 , y0 = 1 , h = 0.1
Put n = 0 into 3 :
y1 = y0 + 0.1 x0
= 1 + 0.1 1
y1 = 1.1 and x1 = 1.1
Essential Specialist Mathematics Complete Worked Solutions 645
Put n = 1 into 3 :
y2 = y1 + 0.1 x1
= 1.1 + 0.1 1.1 = 1.20488088 . . . y2 = 1.20488 . . . and x2 = 1.2
Put n = 2 into 3 :
y3 = y2 + 0.1 x2
= 1.20488 . . . + 0.1 1.2 = 1.31442539 . . . y3 = 1.3144 and x3 = 1.3
d
dy
dx =
1
x2 + 3x + 2
and y(0) = 0
Using Euler’s method:
y n + 1 = y n + 0.01
1
x n 2 + 3x n + 2
4
with x0 = 0 , y0 = 0 , h = 0.01
Put n = 0 into 4 :
y 1 = y0 + 0.01
1
x0 2 + 3x 0 + 2
= 0 + 0.01
1
2
y1 = 0.005 and x1 = 0.01
Put n = 1 into 4 :
y2 = 0.005 + 0.01 1
2.0301
y2 = 0.009925865 . . . and x2 = 0.02
Put n = 2 into 4 :
y3 = 0.009925 . . . + 0.01 1
2.0604
= 0.01477929 . . . y3 = 0.0148 and x3 = 0.03
2
a
i.
dy
dx = cos x with y(0) = 1
y = sin x + c When x = 0, y = 1:
c = 1 y = sin x + 1 When x = 1,
y = sin (1) + 1 1.8415
ii.
TI: Use the leonhard_euler program as outlied
in the textbook.
CP: Use the Spreadsheet instructions as outlied
in the textbook.
y(1) = 1.8438 using Euler
b
i.
dy
dx =
1
x2 with y(1) = 0
y =
x–2
dx
y = –1
x + c
When x = 1, y = 0:
c = 1
y = 1 – 1
x
Essential Specialist Mathematics Complete Worked Solutions 646
When x = 2,
y = 1 – 1
2 =
1
2 = 0.5
ii.
y(2) = 0.5038 using Euler
c.
i.
dy
dx = x with y(1) = 1
y =
x
1
2 dx
y = 2
3 x
3
2 + c
When x = 1, y = 1:
c = 1
3
y = 2
3 x
3
2 +
1
3 When x = 2,
y = 2
3 (2 2 ) +
1
3 = 2.2190
ii.
y(2) = 2.2169 using Euler
d
i.
dy
dx =
1
x2 + 3x + 2
with y(0) = 0
= 1
(x + 1)(x + 2)
1
(x + 1)(x + 2)
A
x + 1 +
B
x + 2 1 = A(x + 2) + B(x + 1) When x = – 2,
B = –1 When x = – 1,
A = 1
y =
1
x2 + 3x + 2
dx
=
1
x + 1 –
1
x + 2 dx
y = log e(x + 1) – log e(x + 2) + c
y = log e
x + 1
x + 2
+ c
When x = 0, y = 0:
c = – log e
1
2
= log e 2
y = log e
x + 1
x + 2
+ log e 2
y = log e
2x + 2
x + 2
When x = 2,
Essential Specialist Mathematics Complete Worked Solutions 647
y = log e 3
2 = 0.4055
ii.
y(2) = 0.4076 using Euler
3
dy
dx = sec
2 x with y(0) = 2
a
Recall:
sec2 kx =
1
k tan kx + c
y =
sec2 x dx
y = tan x + c When x = 0, y = 2,
c = 2 y = tan x + 2 When x = 1,
y = tan (1) + 2 ( 3.5574 )
b
i.
step size = 0.1
Therefore the solution at x = 1 using the Euler
program and a step size of 0.1 is 3.444969502
correct to 9 decimal places.
ii.
step size = 0.05
Therefore the solution at x = 1 using the Euler
program and a step size of 0.05 is 3.498989223
correct to 9 decimal places.
Essential Specialist Mathematics Complete Worked Solutions 648
iii.
step size = 0.01
Therefore the solution at x = 1 using the Euler
program and a step size of 0.01 is 3.545369041
correct to 9 decimal places.
4
dy
dx = cos
–1(x) , with y(0) = 0
For Euler’s method the method described in
2a(ii.) will be used.
Therefore the solution at x = 0.5 using the
Euler program and a step size of 0.01 is
0.66019008 correct to 8 decimal places.
Note: the ODE was inputted as arccos(x) on
the TI-nspire CAS.
5
dy
dx = sin ( x ) , with y(0) = 0
For Euler’s method the method described in
2a(ii.) will be used.
Therefore the solution at x = 3 using the Euler
program and a step size of 0.1 is 2.474287
correct to 6 decimal places.
6
dy
dx =
1
cos (x2) , with y(0) = 0
For Euler’s method the method described in
2a(ii.) will be used.
Therefore the solution at x = 0.3 using the
Euler program and a step size of 0.01 is
0.30022359 correct to 8 decimal places.
Essential Specialist Mathematics Complete Worked Solutions 649
7
dy
dx =
1
2
e
–x
2
2 , with y(0) =
1
2
a
For Euler’s method the method described in
2a(ii.) will be used.
Tabulating these results (correct to 8 decimal
places) gives:
Pr(Z < z)
z Euler’s Method
0 0.5
0.1 0.53989423
0.2 0.57958948
0.3 0.61869375
0.4 0.65683253
0.5 0.69365955
0.6 0.72886608
0.7 0.76218854
0.8 0.79341393
0.9 0.82238309
1 0.84899161
b
TI: In a Lists & Spreadsheet page, input the
numbers 0,0.1,0.2,0.3,…,1 into column A. In
cell B1 type =normCdf(–∞,A1,0,1). Press
Menu3:DataFill and scroll down to cell
11 to copy the formula into the remaining cells.
Use the down arrow key to view all results.
CP: In a Spreadsheet page, input the numbers
0,0.1,0.2,0.3,…,1 into column A. In cell B1
type =normCDf(–∞,A1,1,0). Select cell B1
through to B11. Tap EditFill Range then
OK.
Tabulating these results against Euler’s method
gives:
Essential Specialist Mathematics Complete Worked Solutions 650
z Pr(Z < z)
Euler’s method From tables
0 0.5 0.5
0.1 0.53989423 0.53983
0.2 0.57958948 0.57926
0.3 0.61869375 0.61791
0.4 0.65683253 0.65542
0.5 0.69365955 0.69146
0.6 0.72886608 0.72575
0.7 0.76218854 0.75804
0.8 0.79341393 0.78814
0.9 0.82238309 0.81594
1 0.84899161 0.84134
c
i.
For Euler’s method the method described in
2a(ii.) will be used.
Therefore an approximation to Pr(Z ≤ 0.5)
using the Euler program and a step size of 0.01
is 0.69169538 correct to 8 decimal places.
ii.
For Euler’s method the method described in
2a(ii.) will be used.
Therefore an approximation to Pr(Z ≤ 1) using
the Euler program and a step size of 0.01 is
0.84212759 correct to 8 decimal places.
Essential Specialist Mathematics Complete Worked Solutions 651
Solutions to Exercise 9H 1 Refer to comment in textbook
2
a
dy
dx = 3x
2 with y(1) = 0
y =
3x2 dx
y = x3
+ c
Using y(1) = 0:
0 = 13
+ c
c = –1
y = x3
– 1
b
dy
dx= sin (x) with y(0) = 0
y =
sin (x) dx
y = – cos (x) + c
using y(0) = 0:
0 = – cos (0) + c
c = 1
y = 1 – cos (x)
c
dy
dx= e
–2 x with y(0) = 1
y =
e–2 x
dx
y = –1
2 e
–2 x+ c
using y(0) = 1:
1 = –1
2 e
0+ c
c =3
2
y = 3
2–
1
2 e
–2 x =
1
2 3 – e
–2 x
Essential Specialist Mathematics Complete Worked Solutions 652
d
dy
dx= y
2 with y(1) = 1
dx
dy =
1
y2
x =
1
y2 dy
x =
y–2
dy
x = –1
y+ c
using y(1) = 1:
1 =–1
1+ c
c = 2
x = –1
y+ 2
x – 2 = –1
y
y = –1
x – 2 , x < 2
y =1
– (x – 2)
y =1
2 – x , x < 2
e dy
dx= y
2 with y(1) = –1
dx
dy =
1
y2
x =
1
y2 dy
x =
y–2
dy
x = –1
y+ c
using y(1) = –1 :
1 =–1
–1+ c
c = 0
x = –1
y y = –
1
x and x > 0
Essential Specialist Mathematics Complete Worked Solutions 653
f
dy
dx= y(y – 1) with y(0) = –1
dx
dy=
1
y(y – 1)
dx
dy=
1
y – 1–
1
y using partial fractions
x =
1
y – 1–
1
y
dy
x = log e|y – 1| – log e|y | + c
x = log e
y – 1
y
+ c
using y(0) = –1 :
0 = log e
–1 – 1
–1
+ c
c = – log e(2)
x = log e
y – 1
y
– log e(2) , y > 1
x = log e
y – 1
2y
ex
=y – 1
2y
ex
=1
2–
1
2y
ex
–1
2= –
1
2y
1 – 2ex =
1
y
y =1
1 – 2ex
g
dy
dx= y(y – 1) with y(0) = 2
from part f.
x = log e
y – 1
y
+ c
using y(0) = 2:
0 = log e
2 – 1
2
+ c
c = – log e
1
2
= log e(2)
x = log e
y – 1
y
+ log e(2) , y > 1
x = log e
2(y – 1)
y
ex
=2(y – 1)
y
ex
= 2 –2
y
ex
– 2 = –2
y
y = –2
ex
– 2
y =2
2 – ex
Essential Specialist Mathematics Complete Worked Solutions 654
h
dy
dx= tan (x) with y(0) = 0
y =
tan (x) dx
y =
sin x
cos x dx
Let u = cos x du
dx= – sin x
y =
–du
dx
u dx
y =
–1
u du
y = – log e(u ) + c , u > 0
y = – log e(cos x) + c
using y(0) = 0:
0 = – log e(1) + c
c = 0
y = – log e(cos x)
3
a
b
Essential Specialist Mathematics Complete Worked Solutions 655
Chapter review: multiple-choice questions 1 a = sin (2t) , when t = 0, v = 4
v =
0
t
sin (2x) dt + 4
Answer is C
2
f ' (x) = x2 – 1 and f(1) = 3
Using Euler’s Method:
yn + 1 = yn + 0.2 [(xn)2 – 1] 1
with x0 = 1 , y0 = 3
Put n = 0 into 1 :
y1 = y0 + 0.2 [(x0)2 – 1]
= 3 + 0.2 [12 – 1]
y1 = 3 and x1 = 1.2
Put n = 1 into 1 :
y2 = y1 + 0.2 [(x1)2 – 1]
= 3 + 0.2 [(1.2 )2 – 1]
= 3 + 0.088 y2 = 3.088 and x = 1.4
Answer is D
3
dy
dx = x log e(x) and y(2) = 2
Using Euler’s Method:
yn + 1 = yn + 0.1 [xn log e(xn)] 2
with x0 = 2 , y0 = 2
Put n = 0 into 2 :
y1 = y0 + 0.1 [x0 log e(x0)] = 2 + 0.1 [2 log e(2)] y1 = 2 + 0.2 log e(2) and x = 2.1
Put n = 1 into 2 :
y2 = y1 + 0.1 [x1 log e(x1)]
= 2 + 0.2 log e(2) + 0.1 [2.1 log e(2.1 )]
= 2 + 0.2 log e(2) + 0.21 log e(2.1 )
= 2.294436 . . .
y2 2.294 and x = 2.2
Answer is B
4
dy
dx =
2 – y
4
dx
dy =
4
2 – y
x =
1
1
2
4
2 – t dt + 3
Answer is A
5 dy
dx =
2x + 1
4 , y(2) = 0
y = 1
4
(2x + 1) dx
y = 1
4 (x
2 + x) + c
When x = 2, y = 0:
c = –3
2
y = 1
4 (x
2 + x) –
3
2
y = 1
4 (x
2 + x – 6)
Answer is E
Essential Specialist Mathematics Complete Worked Solutions 656
6
dy
dx =
(y – 1)2
5 , y(0) = 0
dx
dy =
5
(y – 1)2 = 5(y – 1)
–2
x = 5
(y – 1)–2
dy
x = –5
y – 1 + c
When x = 0, y = 0:
c = –5
x + 5 = –5
y – 1
y – 1 = –5
x + 5
y = 1 – 5
x + 5
y = x
x + 5
Using CAS:
Answer is C
7
dy
dx = e
– x2
, y(1) = 4
y =
1
x
e– u
2
du + 4
Answer is D
8
y = 2xe2x
then dy
dx = 2e
2x + 4xe
2x
and d
2y
dx2 = 4e
2x + 4e
2x + 8xe
2x
= 8e2x
+ 8xe2x
Response A:
dy
dx – 2y = 2e
2x 0
Response B:
d2y
dx2 – 2
dy
dx = 4e
2x 0
Response C: dy
dx + 2y
dy
dx
= 8xe4x
(2x + 1) + e2x
(4x + 2) 0
Response D:
d2y
dx2 – 4y = 8e
2x e
2x
Response E:
d2y
dx2 – 4y = 8e
2x = RHS
Answer is E
9
Given: dV
dt = –
5 h
2h + 45
V = (15 h2 + 225 h)
dV
dh = (10 h + 225 )
= 5 (2h + 45 )
Essential Specialist Mathematics Complete Worked Solutions 657
dh
dt =
dh
dV
dV
dt
dh
dt =
1
5 (2h + 45 )
–5 h
2h + 45
=
– h
(2h + 45 )2
Answer is A
10
dy
dx = y
dx
dy =
1
y x = log e(y) + c , for y > 0 When x = 0, y = 2:
c = – log e(2)
x = log e
y
2
ex =
y
2
y = 2ex , for y > 0
Answer is C
Essential Specialist Mathematics Complete Worked Solutions 658
Chapter review: short-answer questions
1 a dy
dx =
x2 + 1
x2
= 1 + 1
x2
y = (1 + x –2 )
1
1dx
= x – x –1 + c
= x – 1
x + c
b 1
y
dy
dx = 10
dy
dx = 10y
dx
dy =
1
10y
x = 1
10y dy
= 1
10 loge y + d, since y 0
e10(x – d ) = y
y = ce10x
c d
2y
dt 2 =
1
2 (sin 3t + cos 2t), t ≥ 0
dy
dt =
1
2 (sin 3t + cos 2t)
1
1dt
= 1
6 cos 3t +
1
4 sin 2t + c1
y =
1
6 cos 3t +
1
4 sin 2t + c1 dt
= 1
18 sin 3t
1
8 cos 2t + c1t + c2
d d
2y
dx2 = e –3x + e –x
dy
dx =(e –3x + e –x )
1
1dx
= 1
3 e –3x e –x + c1
y =
1
3 e –3x e –x + c1 dx
= 1
9 e –3x + e –x + c1 x + c2
Essential Specialist Mathematics Complete Worked Solutions 659
e dy
dx =
3 y
2
dx
dy =
2
3 y
x = 2
3 y dy
x = 2 loge (3 y) + c, y < 3
3 – y = e
x c
2
y = 3 e
x c
2
f dy
dx =
3 x
2
y =
3 x
2 dx
= 3
2 x –
x2
4 + c
2 a dy
dx = π cos (2πx)
y = πcos (2πx)
1
1dx
=
2 sin (2πx) + c
When y = 1, x = 5
2 , and –1 =
1
2 sin 5π + c
–1 = 1
2 + c
c = 1
2
y = 1
2 sin (2πx)
1
2
b dy
dx =
cos 2x
sin 2x
Let sin 2x = u, then du
dx = 2 cos 2x
y = 1
2
du
u
= 1
2 loge | |u + c
= 1
2 loge | |sin 2x + c
When y = 0, x =
4 , and 0 =
1
2 loge
sin
2 + c
0 = 1
2 loge 1 + c
c = 0
y = 1
2 loge | |sin 2x
Essential Specialist Mathematics Complete Worked Solutions 660
c dy
dx =
1 + x2
x2
y =
1
x + x dx
= loge | |x + x2
2 + c
When y = 0, x = 1, and 0 = loge 1 + 1
2 + c
c = 1
2
y = loge | |x + x2
2
1
2
d dy
dx =
x
1 + x2
Let 1 + x2 = u, 2x = du
dx
y = 1
2
du
u
= 1
2 loge | |u + c
= 1
2 loge (1 + x2 ) + c
When y = 1, x = 0, and 1 = 1
2 loge 1 + c
c = 1
y = 1
2 loge (1 + x2 ) + 1
e dy
dx =
1
2 y
dx
dy =
2
y
x =
2
y dy
= –2 loge | |y + c
When y = e1
, x = 2, and 2 = –2 –1 + c
c = 0
x = –2 loge y
y = e
x
2
f d
2x
dt 2 = –10
dx
dt = –10t + c1
Since dx
dt = 4 when x = 0, 4 = 10 0 + c1
c1 = 4
dx
dt = –10t + 4
Essential Specialist Mathematics Complete Worked Solutions 661
x = –10t + 4
1
1dt
= –5t2 + 4t + c2
When x = 0, t = 4, and 0 = –5 16 + 16 + c2
c2 = 64
x = 64 + 4t – 5t2
3 a dy
dx = sin x + x cos x product rule
d
2y
dx2 = cos x + cos x – x sin x
= 2 cos x – x sin x
x2 d
2y
dx2 = 2x2 cos x – x3 sin x
and kx dy
dx = kx sin x + kx2 cos x
and (x2 – m) y = (x2 – m) x sin x since y = x sin x
= x3 sin x – mx sin x
x2 d
2y
dx2 kx dy
dx + (x2 – m) y = 0
becomes 2x2 cos x – x3 sin x – kx sin x – kx2 cos x + x3 sin x – mx sin x = 0
(2x2 – kx2) cos x + (– kx – mx) sin x = 0
(2 – k) x2 cos x + (– k – m)x sin x = 0
Equating coefficients, 2 – k = 0
k = 2
and – k – m = 0
m = – k
= – 2
So k = 2 and m = 2.
b dy
dx = e2x + 2xe2x
d 2y
dx2 = 2e2x + 2e2x + 4xe2x
= 4e2x + 4xe2x
d 2y
dx2 – dy
dx – 3e2x = 4e2x + 4xe2x – e2x – 2xe2x – 3e2x
= 2xe2x, as required.
4 dV
dt =
dV
dx
dx
dt
Given dV
dt = 3 cm3/s,
dx
dt = 3
dV
dx
Now dV
dx =
d
dx
π
3 (18x2 – x3 )
= π
3 (36x – 3x2 ) = π (12x – x2 )
dx
dt =
3
(12x x2 )
= 3
x (12 x)
Essential Specialist Mathematics Complete Worked Solutions 662
5 Now C = 2πr,
r = C
2
Also A = πr2,
A = π
C
2
2
= C
2
4
dA
dC =
C
2
Now dA
dt =
dA
dC
dC
dt
Given dA
dt = 4,
dC
dt = 4
C
2
= 8
C
6 Each minute, the amount of soap in the solution decreases by the same proportion, i.e. 40
1000 =
1
25, since the volume of water remains constant.
dS
dt = –
S
25
dt
dS = –
25
S
t = –
25
S dS
= –25 loge S + c, S 0
When t = 0, S = 3, and 0 = –25 loge 3 + c
c = 25 loge 3
t = 25 loge S + 25 loge 3
= 25 loge 3
S
t
25 = loge
3
S
S = 3e
t
25
7 dx
dt = –
x
100
dt
dx = –
100
x
t = –
100
x dx
= 100 loge x + c, x 0
When t = 0, x = x0 and 0 = 100 loge x0 + c
c = 100 loge x0
t = 100 loge
x0
x
When x = x0
2 , t = 100 loge 2 69
Essential Specialist Mathematics Complete Worked Solutions 663
It takes approximately 69 days.
8 a d
dt =
30
20
dt
d =
20
30
t = 20
30 d
= –20 loge (30 – ) + c, 30
At t = 0, = 10, 0 = –20 loge 20 + c
c = 20 loge 20
t = 20 loge
20
30 –
= 30 – 20e
t
20
e
t
20 = 20
30 –
b = 30 – 20e
60
20
29
So temperature is 29C approximately.
c t = 20 loge
20
30 –
At = 20, t = 20 loge 2 14
So it takes 14 minutes approximately.
9 a The rate of change is a constant proportion of the area, 2%,
dA
dt = 0.02A
b dA
dt = 0.02A
= A
50
dt
dA =
50
A
t =
50
A dA
= 50 loge A + c, A 0
When t = 0, A = 1
2 , and 0 = 50 loge
1
2 + c
c = 50 loge 1
2
= 50 loge 2
t = 50 loge (2A)
2A = e0.02t
A = 1
2 e0.02t
After 10 hours, the area is A = 1
2 e0.2 0.61 hectares.
Essential Specialist Mathematics Complete Worked Solutions 664
c 3 = 1
2 e0.02t
6 = e0.02t
loge 6 = 0.02t
t = log
e 6
0.02
89.59
So 3 hectares have been covered at 89 1
2 hours.
10 dy
dx =
1
16(L – 3x)
1
1dx
= Lx
16 –
3x2
32 + c
The rate of change of the deflection is zero at the point of the support,
i.e., dy
dx = 0 at x = 0.
c = 0
To find where the deflection has its greatest magnitude, we need to find x for which dy
dx = 0
(x > 0).
Lx
16 –
3x2
32 = 0
x = 2L
3
Now y =
Lx
16 –
3x2
32 dx
= Lx2
32 –
x3
32 + d
The deflection itself is zero at the point of the support, i.e., y = 0 when x = 0,
d = 0
y = Lx2
32 –
x3
32
When x = 2L
3, y =
L 4L2
32 9 –
8L3
32 27
= L3
72
L3
108
= L3
216
So the magnitude is greater at x = 2L
3 where the deflection is
L3
216 .
11 r = h tan 30°
= h
3
V = 1
3 πr2h
= 1
3 π
h
3
2
h
= h3
9
30
r
h
Essential Specialist Mathematics Complete Worked Solutions 665
dV
dh =
h2
3 and
dV
dt = 2 – 0.05 h
dh
dt =
dh
dV
dV
dt
= 3
h2 (2 – 0.05 h )
2
6 0.15 h
h
Essential Specialist Mathematics Complete Worked Solutions 666
Chapter review: extended-response questions
1 a i dx
dt = –kx, k > 0
ii Now dt
dx =
1
k
1
x
t = 1
k
1
x dx
= 1
k loge x + c, x 0
x = 100 when t = 0, since the initial amount counts as 100%.
c = 1
k loge 100
t = 1
k loge
100
x
e kt = x
100
x = 100e kt
Now x = 50 when t = 5760,
k = 1
5760 loge 2
x = 100e
t
5760 loge 2
b t = 5760
loge 2 loge
100
45.1
6617 years
The eruption occurred 6617 years ago.
c x = 100e
t loge 2
5760, t ≥ 0
x
0
100
t
2 a Unreacted amount of A after x minutes
= 2 1
4 x
= 8 x
4
Unreacted amount of B after x minutes
= 3 3
4 x
= 3(4 x)
4)
dx
dt =
3k(8 x)(4 x)
16
Essential Specialist Mathematics Complete Worked Solutions 667
b Now dt
dx =
16
3k
1
(8 x)(4 x)
and 1
(8 x)(4 x) =
A
8 x +
B
4 x
A(4 – x) + B(8 – x) = 1
When x = 4, 4B = 1,
B = 1
4
When x = 8, –4A = 1,
A = 1
4
t = 16
3k
1
4 1
4 x –
1
8 x dx
= 4
3k (loge | |8 – x – loge | |4 – x + c)
= 4
3k loge
8 – x
4 x + c
x = 0 when t = 0 (no reaction yet),
c = 4
3k loge 2
t = 4
3k loge
8 – x
2(4 x)
x = 1 when t = 1,
1 = 4
3k loge
8 – 1
2(4 1)
k = 4
3 loge
7
6
t = 1
loge 7
6
loge
8 – x
2(4 x)
c At x = 2, t = 1
loge 7
6
loge 6
4
2.633
2 min 38 s
It takes 2 minutes 38 seconds to form 2 kg of X.
d 8 – x
2(4 x) = et loge
7
6
x (2et loge 7
6 – 1) = 8 (et loge76 – 1)
x = 8((
7
6 )t 1)
2( 7
6 )t 1
When t = 2, x = 8(
49
36 1)
2 49
36 1
= 52
31
The mass of X formed after two minutes is 52
31 kg.
Essential Specialist Mathematics Complete Worked Solutions 668
3 a dT
dt = k (T – Ts ), k < 0
b dT
dt = k(T 22)
dt
dT =
1
k
1
T 22
t = 1
k 1
T 22 dT
= 1
k loge (T 22) + c, T 22
When T = 72, t = 0,
0 = 1
k loge (72 22) + c
c = 1
k loge 50
t = 1
k loge (T 22)
1
k loge 50
= 1
k loge
T 22
50
i k = 1
t loge
T 22
50
When T = 65, t = 5,
k = 1
5 loge
65 22
50
= 1
5 loge 0.86
t = 5
loge 0.86
loge
T 22
50
When T = 50, t = 5
loge 0.86
loge
50 22
50
= 5 log
e 0.56
loge 0.86
19.2
The coffee remains drinkable for 19.2 minutes.
ii Now at t = 30, 30 = 5
loge 0.86
loge
T 22
50
30
5 loge 0.86 = loge
T 22
50
loge (0.86)6 = loge
T 22
50
T 22
50 = (0.86)6
T = 50 (0.86)6 + 22
42.2
The temperature of the coffee at the end of 30 minutes is 42.2°C.
Essential Specialist Mathematics Complete Worked Solutions 669
4 a dp
dt = rate of increase – rate of decrease
= kp – 1000, k > 0
b dt
dp =
1
kp 1000
t = 1
kp 1000 dp
= 1
k loge (kp – 1000) + c, kp – 1000 > 0
When t = 0, p = 5000,
0 = 1
k loge (5000k – 1000) + c
c = 1
k loge (5000k – 1000)
t = 1
k loge (kp – 1000)
1
k loge (5000k – 1000)
= 1
k loge
kp 1000
5000k 1000
c i When t = 5, p = 6000,
5 = 1
k loge
6000k 1000
5000k 1000
5k = loge
1000(6k 1)
1000(5k 1)
5k = loge
6k 1
5k 1
ii TI: Type solve(5×k=ln((6×k-1)/(5×k-1)),k)
Interpreting these results gives k = 0 or k = 0.22183565…
CP: Sketch the graphs of y1=5x and y2=ln((6x–1)/(5x–1)). Tap AnalysisG-
SolveIntersect
Thus an approximation for the value of k of 0.221 835 66.
d t = 1
k loge
kp 1000
5000k 1000
kt = loge
kp 1000
5000k 1000
kp 1000
5000k 1000 = ekt
kp – 1000 = ekt (5000k – 1000)
Essential Specialist Mathematics Complete Worked Solutions 670
p = 1
k (ekt (5000k – 1000) + 1000)
p
t
5000
0
5 a dN
dt = 100 – kN, k > 0
b dt
dN =
1
100 kN
t = 1
100 kN dN
= 1
k loge (100 – kN) + c, 100 – kN > 0
When t = 0, N = 1000,
0 = 1
k loge (100 – 1000k) + c
c = 1
k loge (100 – 1000k)
t = 1
k loge (100 – kN) +
1
k loge (100 – 1000k)
= 1
k loge
100 1000k
100 kN
c When t = 10, N = 700,
10 = 1
k loge
100 1000k
100 700k
10k = loge
1 10k
1 7k
TI: Type solve(10×k=ln((1-10×k)/(1-7×k)),k)
Interpreting these results gives k = 0 or k = 0.16018368…
CP: Sketch the graphs of y1=10x and y2=ln((1–10x)/(1–7x)). Tap AnalysisG-
SolveIntersect
Thus an approximation for the value of k of 0.160 183 68.
Essential Specialist Mathematics Complete Worked Solutions 671
d t = 1
k loge
100 1000k
100 kN
kt = loge
100 1000k
100 kN
ekt = 100 1000k
100 kN
100 – kN = e – kt (100 – 1000k)
N = 1
k (100 – e – kt (100 – 1000k))
When k 0.16, N 1
0.16 (100 – e –0.16t (100 – 1000 0.16))
25
4 (100 + 60
e –0.16t )
1000
0
100
k
t
N
e As t +, N 100
k
The eventual trout population in the lake will be 100
k.
When k 0.16, 100
k 625
So the trout population approaches 625.
6 a dy
dx =
9
40L2 (3x – L)
1
1dx
= 9
40L2
3x2
2 – Lx + c
When x = 0 (at A), dy
dx = 0, c = 0
dy
dx =
9
40L2 ( 3x2
2 – Lx)
dy
dx = 0 when
3x2
2 = Lx (x ≠ 0)
x = 2L
3
The maximum deflection occurs 2L
3 cm from the end A.
b y = 9
40L2
3x2
2 – Lx .dx
= 9
40L2
x3
2 –
Lx2
2 + c
When x = 0, y = 0, c = 0
y = 9x2
80L2 (x – L)
Essential Specialist Mathematics Complete Worked Solutions 672
when x = 2L
3, y =
9
80L2
2L
3
2
2L
3 L
= 9 4L2 (L)
80L2 9 3
= L
60
The maximum deflection is L
60 cm downwards.
7 a dT
dt = 2 – k (T – T0 )
When T = 60, dT
dt = 1, –1 = –k (60 – T0 )
k = 1
60 T0
dT
dt = 2 –
T 20
60 T0
Given T0 = 20, dT
dt = 2 –
T 20
40
= 100 T
40
b dt
dT =
40
100 T
t = 40
100 T dT
= –40 loge (100 – T ) + c, T < 100
When t = 0, T = 20,
c = 40 loge 80
t = 40 loge
80
100 T
e
t
40 = 80
100 T
100 – T = 80e
t
40
T = 100 – 80e
t
40
c When t = 30, T = 100 – 80e
3
4
= 62.210…
The temperature is 62.2° C after 30 minutes.
d
t
100
0
20
T
Essential Specialist Mathematics Complete Worked Solutions 673
8 a i dW
dt = 0.04W
dt
dW =
1
0.04W
= 25
W
t =
25
W dW
= 25 loge W + c, W > 0
When t = 0, W = 350,
0 = 25 loge 350 + c
c = –25 loge 350
t = 25 loge W – 25 loge 350
= 25 loge
W
350
ii t
25 = loge
W
350
W
350 = e
t
25
W = 350e
t
25
0
350
W
t
iii When t = 50, W = 350e
50
25
= 350e2 2586
b If dW
dt = kW and the population remains constant then
dW
dt = 0.
k = 0 since W > 0
c i dW
dt = (0.04 – 0.00005W )W
dt
dW =
1
(0.04 0.00005W )W =
20 000
(800 W )W
Now 20 000
(800 W )W =
A
800 W +
B
W
AW + B (800 – W) = 20 000
When W = 0, 800B = 20 000, B = 25
When W = 800, 800A = 20 000, A = 25
20 000
(800 W )W =
25
800 W +
25
W
dt
dW =
25
800 W +
25
W
t = 25
800 W +
25
W dW
= –25 loge (800 – W ) + 25 loge W + c, 0 < W < 800
= 25 loge
W
800 W + c
t = 0, W = 350, 0 = 25 loge
350
450 + c
Essential Specialist Mathematics Complete Worked Solutions 674
c = –25 loge 7
9
t = 25 loge
W
800 W – 25 loge
7
9
= 25 loge
9W
7(800 W )
ii t
25 = loge
9W
7(800 W )
9W
7(800 W ) = e
t
25
9W = 7 (800 – W ) e
t
25
= 5600 e
t
25 – 7W e
t
25
9W + 7W e
t
25 = 5600 e
t
25
W ( )9 + 7 e
t
25 = 5600 e
t
25
W = 5600 e
t
25
9 + 7 e
t
25
0
350
W
800
t
iii When t = 50, W = 5600 e
50
25
9 + 7 e
50
25
= 5600e2
9 + 7e2
= 681.429 55…
The population after 50 years is approximately 681 iguanas.
9 a i dx
dt = rate of input – rate of output
= R – kx, k > 0
ii dt
dx =
1
R kx
t = 1
R kx dx
t = 1
k loge (R – kx) + c, R – kx > 0
When t = 0, x = 0,
0 = 1
k loge R + c
c = 1
k loge R
t = 1
k loge (R – kx) +
1
k loge R
= 1
k loge
R
R kx
Essential Specialist Mathematics Complete Worked Solutions 675
kt = loge
R
R kx
ekt = R
R kx
(R – kx) ekt = R
kx ekt = R (ekt – 1)
x = R(ekt 1)
kekt
= R
k (1 – e– k t )
b i If R = 50 and k = 0.05,
x = 50
0.05 (1 – e –0.05t )
= 1000 ( )1 – e
t
20
t 0
x
1000
ii t = 1
k loge
R
R kx
When R = 50 and k = 0.05,
t = 20 loge
50
50 0.05x
= 20 loge
1000
1000 x
When x = 200, t = 20 loge
1000
1000 200
= 20 loge 5
4 = 4.4628…
There are 200 mg of the drug in the patient after 4.46 hours, correct to two decimal
places.
c i When t > 20 loge 5
4 ,
dx
dt = –kx and k = 0.05 =
1
20 ,
dx
dt =
x
20
dt
dx =
20
x
t =
20
x dx
= –20 loge x + c, x > 0
When t = 20 loge 5
4 , x = 200,
20 loge 5
4 = –20 loge 200 + c
c = 20 loge 5
4 + 20 loge 200
t = 20 loge 5
4 + 20 loge 200 – 20 loge x
Essential Specialist Mathematics Complete Worked Solutions 676
= 20 loge 250
x
When x = 100, t = 20 loge 5
2
= 18.325 81…
The amount of drug falls to 100 mg after 18.33 hours, correct to two decimal places,
a further 13.86 hours after the drip was disconnected.
ii t = 20 loge
250
x
t
20 = loge
250
x
e
t
20 = 250
x
x = 250 e
t
20
0
x
100
200
20 loge 5
4 t 20 loge
5
2
x =
1000( )1 – e
t
20 0 t 20 loge
5
4
250 e
t
20 t 20 loge 5
4