Chapter 9 Differential equations - scunderwood · PDF fileChapter 9 – Differential equations Solutions to Exercise 9A 1 a If y 2= Ae t – 22 then dy dt = 2Ae t Given dy dt = 2y

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


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


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


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


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


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


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 dy

dx = (2x + 1)3

= 8x3 + 12x2 + 6x + 1

y = 2x4 + 4x3 + 3x2 + x + c,


d dy

dx = 1

x, x > 0

1

2x

y = 2x12 + c

y = 2 x + c ,


e dy

dt =

12t - 1

, t ≠ 12

y = 12

loge |2t – 1| + c,


f dy

dt = sin (3t – 2)

y = – 13

cos (3t – 2) + c,


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,


h dxdy

= e–3y

x = – 13

e–3y + c,


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 ,



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,


d

d2 y

dx2 = e

x2

dy

dx = 2e

x2 + c

y = 4ex2 + cx + d,


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.


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


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


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


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)


0 = loge 2 + c

c = – loge 2

y = loge (ex + 1) – loge 2

y = log e

e

x+ 1

2


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


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


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


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


xdx

c1 = 1

dy

dx = –

1

1 + x2 + 1


c1 = 0

21

1

dxy dx

x

= – tan–1 x + x + c2


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


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


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|


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


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


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


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)


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


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


= 3t – 1

x = 32

t2 – t + c

Now x(1) = 1

1 = 32

– 1 + c

c = 12

x = 32

t2 – t + 12

c


= – 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


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



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


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



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


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


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


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


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


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


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


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


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


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


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


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


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


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 .


0

4

1

cos (x) dx + 1

c

dy

dx = log e (x

2) , y(1) = 2

Find y when x = e.


1

e1

ln (x2) dx + 2

d

dy

dx = log e x , y(1) = 2

Find y when x = e.


1

e1

ln (x) dx + 2


2

Ensure your calculator is set to Exact/Standard

mode.

a

dy

dx = tan

–1 x


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


deSolve (y' = x3 ln (x), x, y)


y = 1

4 x

4log e(x) –

x4

16 + c

c

dy

dx = e

3x sin 2x


deSolve (y' = e3x

sin (2x), x, y)


y = e3x

3

13 sin 2x –

2

13 cos 2x

+ c

d

dy

dx = e

3x cos (2x)


deSolve (y' = e3x

cos (2x), x, y)


y = e3x

3

13 cos 2x +

2

13 sin 2x

+ c

e

dy

dx = 2

x


deSolve (y' = 2x, x, y)


y = 2

x

log e 2 + c


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


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


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


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


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


When x = 2,

y = 1 – 1

2 =

1

2 = 0.5

ii.


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.


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,


y = log e 3

2 = 0.4055

ii.


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





iii.

step size = 0.01




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






6

dy

dx =

1

cos (x2) , with y(0) = 0



Therefore the solution at x = 0.3 using the

Euler program and a step size of 0.01 is



7

dy

dx =

1

2

e

–x

2

2 , with y(0) =

1

2

a



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:


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.



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.



Therefore an approximation to Pr(Z ≤ 1) using

the Euler program and a step size of 0.01 is



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


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


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


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


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


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 )


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


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


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


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


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)


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


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.


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


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


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


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.


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.


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)


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.


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)


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


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


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


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


= 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

Documents

Chapter 9 Differential equations - scunderwood · PDF fileChapter 9 – Differential equations Solutions to Exercise 9A 1 a If y 2= Ae t – 22 then dy dt = 2Ae t Given dy dt = 2y