chapter 8 mathematical modeling with di erential...

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404

CHAPTER 8

Mathematical Modeling with DifferentialEquations

EXERCISE SET 8.1

1. y′ = 9x2ex3

= 3x2y and y(0) = 3 by inspection.

2. y′ = x3 − 2 sinx, y(0) = 3 by inspection.

3. (a) first order;dy

dx= c; (1 + x)

dy

dx= (1 + x)c = y

(b) second order; y′ = c1 cos t− c2 sin t, y′′ + y = −c1 sin t− c2 cos t+ (c1 sin t+ c2 cos t) = 0

4. (a) first order; 2dy

dx+ y = 2

(− c

2e−x/2 + 1

)+ ce−x/2 + x− 3 = x− 1

(b) second order; y′ = c1et − c2e−t, y′′ − y = c1e

t + c2e−t −

(c1e

t + c2e−t) = 0

5. False. It is a first-order equation, because it involves y and dy/dx, but not dny/dxn for n > 1.

6. True. y = −1/2 is a solution.

7. True. As mentioned in the marginal note after equation (2), the general solution of an n’th orderdifferential equation usually involves n arbitrary constants.

8. False. Every solution of the first order differential equation y′ = y has the form y = Aex = Aex+b

with b = 0.

9. (a) If y = e−2x then y′ = −2e−2x and y′′ = 4e−2x, soy′′ + y′ − 2y = 4e−2x + (−2e−2x)− 2e−2x = 0.

If y = ex then y′ = ex and y′′ = ex, so y′′ + y′ − 2y = ex + ex − 2ex = 0.(b) If y = c1e

−2x + c2ex then y′ = −2c1e−2x + c2e

x and y′′ = 4c1e−2x + c2ex, so

y′′ + y′ − 2y = (4c1e−2x + c2ex) + (−2c1e−2x + c2e

x)− 2(c1e−2x + c2ex) = 0.

10. (a) If y = e−2x then y′ = −2e−2x and y′′ = 4e−2x, soy′′ − y′ − 6y = 4e−2x − (−2e−2x)− 6e−2x = 0.

If y = e3x then y′ = 3e3x and y′′ = 9e3x, so y′′ − y′ − 6y = 9e3x − 3e3x − 6e3x = 0.(b) If y = c1e

−2x + c2e3x then y′ = −2c1e−2x + 3c2e3x and y′′ = 4c1e−2x + 9c2e3x, so

y′′ − y′ − 6y = (4c1e−2x + 9c2e3x)− (−2c1e−2x + 3c2e3x)− 6(c1e−2x + c2e3x) = 0.

11. (a) If y = e2x then y′ = 2e2x and y′′ = 4e2x, so y′′ − 4y′ + 4y = 4e2x − 4(2e2x) + 4e2x = 0.If y = xe2x then y′ = (2x+ 1)e2x and y′′ = (4x+ 4)e2x, soy′′ − 4y′ + 4y = (4x+ 4)e2x − 4(2x+ 1)e2x + 4xe2x = 0.

(b) If y = c1e2x + c2xe

2x then y′ = 2c1e2x + c2(2x + 1)e2x and y′′ = 4c1e2x + c2(4x + 4)e2x, soy′′ − 4y′ + 4y = (4c1e2x + c2(4x+ 4)e2x)− 4(2c1e2x + c2(2x+ 1)e2x) + 4(c1e2x + c2xe

2x) = 0.

12. (a) If y = e4x then y′ = 4e4x and y′′ = 16e4x, so y′′ − 8y′ + 16y = 16e4x − 8(4e4x) + 16e4x = 0.If y = xe4x then y′ = (4x+ 1)e4x and y′′ = (16x+ 8)e4x, soy′′ − 8y′ + 16y = (16x+ 8)e4x − 8(4x+ 1)e4x + 16xe4x = 0.

(b) If y = c1e4x + c2xe

4x then y′ = 4c1e4x + c2(4x+ 1)e4x and y′′ = 16c1e4x + c2(16x+ 8)e4x, soy′′−8y′+16y = (16c1e4x+c2(16x+8)e4x)−8(4c1e4x+c2(4x+1)e4x)+16(c1e4x+c2xe4x) = 0.


Exercise Set 8.1 405

13. (a) If y = sin 2x then y′ = 2 cos 2x and y′′ = −4 sin 2x, so y′′ + 4y = −4 sin 2x+ 4 sin 2x = 0.

If y = cos 2x then y′ = −2 sin 2x and y′′ = −4 cos 2x, so y′′ + 4y = −4 cos 2x+ 4 cos 2x = 0.(b) If y = c1 sin 2x+ c2 cos 2x then y′ = 2c1 cos 2x− 2c2 sin 2x and y′′ = −4c1 sin 2x− 4c2 cos 2x,

so y′′ + 4y = (−4c1 sin 2x− 4c2 cos 2x) + 4(c1 sin 2x+ c2 cos 2x) = 0.

14. (a) If y = e−2x sin 3x then y′ = e−2x(−2 sin 3x+3 cos 3x) and y′′ = e−2x(−5 sin 3x−12 cos 3x), soy′′+ 4y′+ 13y = e−2x(−5 sin 3x−12 cos 3x) + 4e−2x(−2 sin 3x+ 3 cos 3x) + 13e−2x sin 3x = 0.

If y = e−2x cos 3x then y′ = e−2x(−3 sin 3x− 2 cos 3x) and y′′ = e−2x(12 sin 3x− 5 cos 3x), soy′′ + 4y′ + 13y = e−2x(12 sin 3x− 5 cos 3x) + 4e−2x(−3 sin 3x− 2 cos 3x) + 13e−2x cos 3x = 0.

(b) If y = e−2x(c1 sin 3x+ c2 cos 3x) then y′ = e−2x[−(2c1 + 3c2) sin 3x+ (3c1 − 2c2) cos 3x] andy′′ = e−2x[(−5c1 + 12c2) sin 3x− (12c1 + 5c2) cos 3x], soy′′ + 4y′ + 13y = e−2x[(−5c1 + 12c2) sin 3x− (12c1 + 5c2) cos 3x]

+ 4e−2x[−(2c1 + 3c2) sin 3x+ (3c1 − 2c2) cos 3x]+ 13e−2x(c1 sin 3x+ c2 cos 3x) = 0.

15. From Exercise 9, y = c1e−2x + c2e

x is a solution of the differential equation, withy′ = −2c1e−2x + c2e

x. Setting y(0) = −1 and y′(0) = −4 gives c1 + c2 = −1 and −2c1 + c2 = −4.So c1 = 1, c2 = −2, and y = e−2x − 2ex.

16. From Exercise 10, y = c1e−2x + c2e

3x is a solution of the differential equation, withy′ = −2c1e−2x + 3c2e3x. Setting y(0) = 1 and y′(0) = 8 gives c1 + c2 = 1 and −2c1 + 3c2 = 8.So c1 = −1, c2 = 2, and y = −e−2x + 2e3x.

17. From Exercise 11, y = c1e2x + c2xe

2x is a solution of the differential equation, withy′ = 2c1e2x + c2(2x+ 1)e2x. Setting y(0) = 2 and y′(0) = 2 gives c1 = 2 and 2c1 + c2 = 2,so c2 = −2 and y = 2e2x − 2xe2x.

18. From Exercise 12, y = c1e4x + c2xe

4x is a solution of the differential equation, withy′ = 4c1e4x + c2(4x+ 1)e4x. Setting y(0) = 1 and y′(0) = 1 gives c1 = 1 and 4c1 + c2 = 1,so c2 = −3 and y = e4x − 3xe4x.

19. From Exercise 13, y = c1 sin 2x+ c2 cos 2x is a solution of the differential equation, withy′ = 2c1 cos 2x− 2c2 sin 2x. Setting y(0) = 1 and y′(0) = 2 gives c2 = 1 and 2c1 = 2,so c1 = 1 and y = sin 2x+ cos 2x.

20. From Exercise 14, y = e−2x(c1 sin 3x+ c2 cos 3x) is a solution of the differential equation, withy′ = e−2x[−(2c1 + 3c2) sin 3x+ (3c1 − 2c2) cos 3x]. Setting y(0) = −1 and y′(0) = −1 givesc2 = −1 and 3c1 − 2c2 = −1, so c1 = −1 and y = −e−2x(sin 3x+ cos 3x).

21. y′ = 2−4x so y =∫

(2−4x) dx = −2x2+2x+C. Setting y(0) = 3 gives C = 3, so y = −2x2+2x+3.

22. (y′)′ = −6x so y′ =∫

(−6x) dx = −3x2 +C. Setting y′(0) = 2 gives C = 2, so y′ = −3x2 + 2 and

y =∫

(−3x2 + 2) dx = −x3 + 2x+D. Setting y(0) = 1 gives D = 1 so y = −x3 + 2x+ 1.

23. If the solution has an inverse function x(y) then, by equation (3) of Section 3.3,dx

dy=

1dy/dx

= y−2. So x =∫y−2 dy = −y−1+C. When x = 1, y = 2, so C =

32

and x =32−y−1.

Solving for y gives y =2

3− 2x. The solution is valid for x <

32

.


406 Chapter 8

24. If the solution has an inverse function x(y) then, by equation (3) of Section 3.3,dx

dy=

1dy/dx

=1

1 + y2. So x =

∫dy

1 + y2= tan−1 y + C. When x = 0, y = 0,

so C = 0, x = tan−1 y, and y = tanx. The solution is valid for −π/2 < x < π/2.

25. By the product rule,d

dx(x2y) = x2y′ + 2xy = 0, so x2y = C and y = C/x2. Setting y(1) = 2

gives C = 2 so y = 2/x2. The solution is valid for x > 0.

26. By the product rule,d

dx(xy) = xy′ + y = ex, so xy =

∫ex dx = ex + C. Setting y(1) = 1 + e

gives C = 1, so xy = ex + 1 and y =ex + 1x

. The solution is valid for x > 0.

27. (a)dy

dt= ky2, y(0) = y0, k > 0 (b)

dy

dt= −ky2, y(0) = y0, k > 0

28. (a) Either y is always zero, or y is positive and increases at a rate proportional to the squareroot of y.

(b) Either y is always zero, or y is positive and decreases at a rate proportional to the cube of y,or y is negative and increases at a rate proportional to the cube of y.

29. (a)ds

dt=

12s (b)

d2s

dt2= 2

ds

dt

30. (a)dv

dt= −2v2 (b)

d2s

dt2= −2

(ds

dt

)2

31. (a) Since k > 0 and y > 0, equation (3) givesdy

dt= ky > 0, so y is increasing.

(b)d2y

dt2=

d

dt(ky) = k

dy

dt= k2y > 0, so y is concave upward.

32. (a) Both y = 0 and y = L satisfy equation (4).

(b) The rate of growth isdy

dt= k

(1− y

L

)y; we wish to find the value of y which maximizes

this. Sinced

dy

[k(

1− y

L

)y]

=k

L(L − 2y), which is positive for y < L/2 and negative for

y > L/2, the maximum growth rate occurs for y = L/2.

33. (a) Both y = 0 and y = L satisfy equation (6).

(b) The rate of growth isdy

dt= ky(L− y); we wish to find the value of y which maximizes this.

Sinced

dy[ky(L − y)] = k(L − 2y), which is positive for y < L/2 and negative for y > L/2,

the maximum growth rate occurs for y = L/2.

34. If T is constant thendT

dt= 0, so equation (7) gives T = Te. Hence T = Te is the unique constant

solution of (7).



35. If x = c1 cos

(√k

mt

)+ c2 sin

(√k

mt

)then

dx

dt= c2

√k

mcos

(√k

mt

)− c1

√k

msin

(√k

mt

)

andd2x

dt2= −c1

k

mcos

(√k

mt

)− c2

k

msin

(√k

mt

)= − k

mx. So m

d2x

dt2= −kx; thus x satisfies

the differential equation for the vibrating string.

36. (a) From Exercise 35 we have x = c1 cos

(√k

mt

)+ c2 sin

(√k

mt

)and

x′ = c2

√k

mcos

(√k

mt

)− c1

√k

msin

(√k

mt

). Setting x(0) = x0 and x′(0) = 0 gives

c1 = x0 and c2

√k

m= 0, so c2 = 0 and x = x0 cos

(√k

mt

).

(b) From the discussion before Example 2 in Section 0.3, the amplitude is |x0|, the period is2π√k/m

= 2π√m

k, and the frequency is

√k/m

2π. The amplitude is the maximum displacement

of the mass from its rest position. The period is the length of time the mass takes to moveback and forth once. The frequency tells how often the mass moves back and forth in oneunit of time.

EXERCISE SET 8.2

1.1ydy =

1xdx, ln |y| = ln |x|+ C1, ln

∣∣∣yx

∣∣∣ = C1,y

x= ±eC1 = C, y = Cx

including C = 0 by inspection.

2.dy

1 + y2= 2x dx, tan−1 y = x2 + C, y = tan

(x2 + C

)

3.dy

1 + y= − x√

1 + x2dx, ln |1 + y| = −

√1 + x2 + C1, 1 + y = ±e−

√1+x2

eC1 = Ce−√

1+x2,

y = Ce−√

1+x2 − 1, C 6= 0

4. y dy =x3 dx

1 + x4,y2

2=

14

ln(1 + x4) + C1, 2y2 = ln(1 + x4) + C, y = ±√

[ln(1 + x4) + C]/2

5.2(1 + y2)

ydy = exdx, 2 ln |y|+ y2 = ex + C; by inspection, y = 0 is also a solution.

6.dy

y= −x dx, ln |y| = −x2/2 + C1, y = ±eC1e−x

2/2 = Ce−x2/2, including C = 0 by inspection.

7. eydy =sinx

cos2 xdx = secx tanx dx, ey = secx+ C, y = ln(secx+ C)

8.dy

1 + y2= (1 + x) dx, tan−1 y = x+

x2

2+ C, y = tan(x+ x2/2 + C)


408 Chapter 8

9.dy

y2 − y=

dx

sinx,

∫ [−1y

+1

y − 1

]dy =

∫cscx dx, ln

∣∣∣∣y − 1y

∣∣∣∣ = ln | cscx− cotx|+ C1,

y − 1y

= ±eC1(cscx− cotx) = C(cscx− cotx), y =1

1− C(cscx− cotx), C 6= 0;

by inspection, y = 0 is also a solution, as is y = 1.

10.1ydy = cosx dx, ln |y| = sinx+ C, y = C1e

sin x

11. (2y + cos y) dy = 3x2 dx, y2 + sin y = x3 + C, π2 + sinπ = C,C = π2, y2 + sin y = x3 + π2

12.dy

dx= (x+ 2)ey, e−ydy = (x+ 2)dx, −e−y =

12x2 + 2x+ C, −1 = C,

−e−y =12x2 + 2x− 1, e−y = −1

2x2 − 2x+ 1, y = − ln

(1− 2x− 1

2x2

)

13. 2(y − 1) dy = (2t+ 1) dt, y2 − 2y = t2 + t+ C, 1 + 2 = C,C = 3, y2 − 2y = t2 + t+ 3

14.dy

dxcosh2 x = y cosh 2x, y−1 dy =

cosh 2xcosh2 x

dx =2 cosh2 x− 1

cosh2 xdx = (2− sech2 x) dx,

ln y = 2x− tanhx+ C, y = eCe2x−tanh x, 3 = eCe2·0−tanh 0 = eC , y = 3e2x−tanh x

15. (a)dy

y=dx

2x, ln |y| = 1

2ln |x|+ C1,

|y| = C2|x|1/2, y2 = Cx;

by inspection y = 0 is also a solu-tion.

(b) 12 = C · 2, C = 1/2, y2 = x/2

–2 2

–2

2

x

yx = –0.5y2

x = –1.5y2x = y2 x = 2y2

x = 2.5y2

y = 0x = –3y2

16. (a) y dy = −x dx, y2

2= −x

2

2+ C1, y = ±

√C2 − x2

(b) y =√

25− x2

–3 3

–3

3

x

yy = √9 – x2

y = –√6.25 – x2

y = –√4 – x2

y = –√1 – x2

y = √2.25 – x2

y = √0.25 – x2

17.dy

y= − x dx

x2 + 4

ln |y| = −12

ln(x2 + 4) + C1

y =C√x2 + 4

1.5

–1

–2 2C = –1

C = 1C = 0C = 2

C = –2



18. cos y dy = cosx dxsin y = sinx+ Cy = 2nπ + sin−1(sinx+ C) or y = (2n+ 1)π − sin−1(sinx+ C) for some integer n

For C = 0, the integral curves are lines of the form y =2nπ+x and y = (2n+1)π−x for integers n. These dividethe xy-plane into squares rotated 45◦ from the axes. ForC 6= 0, −2 < C < 2, each integral curve stays within eitherthe top half or the bottom half of one of these squares. Thefigure shows 5 such curves in the square x − 2π < y < x,π − x < y < 3π − x. From top to bottom, their equationsare y = π − sin−1(sinx+ 0.5), y = π − sin−1(sinx+ 1.5),y = sin−1(sinx + 1.7), y = sin−1(sinx + 1), and y =sin−1(sinx+ 0.3).

x

y

! 2!

!

19. (1− y2) dy = x2 dx,

y − y3

3=x3

3+ C1, x

3 + y3 − 3y = C

–2 2

–2

2

x

y

20.(

1y

+ y

)dy = dx, ln |y|+ y2

2= x+ C1,

yey2/2 = ±eC1ex = Cex including C = 0.

–5 5

–3

3

x

y

21. True. The equation can be rewritten as1

f(y)dy

dx= 1, which has the form (1).

22. False. The equation can be rewritten as1

g(y)dy

dx=

1h(x)

, which has the form (1).

23. True. After t minutes there will be 32 · (1/2)t grams left; when t = 5 there will be 32 · (1/2)5 = 1gram.

24. True. The population will quadruple in twice the doubling time.

25. Of the solutions y =1

2x2 − C, all pass through the point

(0,− 1

C

)and thus never through (0, 0).

A solution of the initial value problem with y(0) = 0 is (by inspection) y = 0. The method ofExample 1 fails in this case because it starts with a division by y2 = 0.

26. If y0 6= 0 then, proceeding as before, we get C = 2x2 − 1y, C = 2x2

0 −1y0

, and

y =1

2x2 − 2x20 + 1/y0

, which is defined for all x provided 2x2 is never equal to 2x20 − 1/y0; this

last condition will be satisfied if and only if 2x20−1/y0 < 0, or 0 < 2x2

0y0 < 1. If y0 = 0 then y = 0is, by inspection, also a solution for all real x.

27.dy

dx= xe−y, ey dy = x dx, ey =

x2

2+ C, x = 2 when y = 0 so 1 = 2 + C,C = −1, ey = x2/2− 1


410 Chapter 8

28.dy

dx=

3x2

2y, 2y dy = 3x2 dx, y2 = x3 + C, 1 = 1 + C,C = 0,

y2 = x3, y = x3/2 passes through (1, 1).

2

00 1.6

29. (a)dy

dt= 0.02y, y0 = 10,000 (b) y = 10,000et/50

(c) T =1

0.02ln 2 ≈ 34.657 h (d) 45,000 = 10,000et/50,

t = 50 ln45,00010,000

≈ 75.20 h

30. k =1T

ln 2 =120

ln 2

(a)dy

dt= ((ln 2)/20)y, y(0) = 1 (b) y(t) = et(ln 2)/20 = 2t/20

(c) y(120) = 26 = 64 (d) 1,000,000 = 2t/20,

t = 20ln 106

ln 2≈ 398.63 min

31. (a)dy

dt= −ky, y(0) = 5.0× 107; 3.83 = T =

1k

ln 2, so k =ln 23.83

≈ 0.1810

(b) y = 5.0× 107e−0.181t

(c) y(30) = 5.0× 107e−0.1810(30) ≈ 219,000

(d) y(t) = (0.1)y0 = y0e−kt, −kt = ln 0.1, t = − ln 0.1

0.1810= 12.72 days

32. (a) k =1T

ln 2 =1

140ln 2 ≈ 0.0050, so

dy

dt= −0.0050y, y0 = 10.

(b) y = 10e−0.0050t

(c) 10 weeks = 70 days so y = 10e−0.35 ≈ 7 mg.

(d) 0.3y0 = y0e−kt, t = − ln 0.3

0.0050≈ 240.8 days

33. 100e0.02t = 10,000, e0.02t = 100, t =1

0.02ln 100 ≈ 230 days

34. y = 10,000ekt, but y = 12,000 when t = 5 so 10,000e5k = 12,000, k =15

ln 1.2. y = 20,000 when

2 = ekt, t =ln 2k

= 5ln 2

ln 1.2≈ 19, in the year 2017.

35. y(t) = y0e−kt = 10.0e−kt, 3.5 = 10.0e−k(5), k = −1

5ln

3.510.0

≈ 0.2100, T =1k

ln 2 ≈ 3.30 days

36. y = y0e−kt, 0.7y0 = y0e

−5k, k = −15

ln 0.7 ≈ 0.07

(a) T =ln 2k≈ 9.90 yr



(b) y(t) ≈ y0e−0.07t,y

y0≈ e−0.07t, so e−0.07t × 100 percent will remain.

38. (a) None; the half-life is independent of the initial amount.(b) kT = ln 2, so T is inversely proportional to k.

39. (a) T =ln 2k

; and ln 2 ≈ 0.6931. If k is measured in percent, k′ = 100k,

then T =ln 2k≈ 69.31

k′≈ 70k′

.

(b) 70 yr (c) 20 yr (d) 7%

40. Let y = y0ekt with y = y1 when t = t1 and y = 3y1 when t = t1 + T ; then y0e

kt1 = y1 (i) and

y0ek(t1+T ) = 3y1 (ii). Divide (ii) by (i) to get ekT = 3, T =

1k

ln 3.

41. From (19), y(t) = y0e−0.000121t. If 0.27 =

y(t)y0

= e−0.000121t then t = − ln 0.270.000121

≈ 10,820 yr, and

if 0.30 =y(t)y0

then t = − ln 0.300.000121

≈ 9950, or roughly between 9000 B.C. and 8000 B.C.

42. (a) 1

00 50000

(b) t = 1988 yields

y/y0 = e−0.000121(1988) ≈ 79%.

43. (a) Let T1 = 5730− 40 = 5690, k1 =ln 2T1≈ 0.00012182;T2 = 5730 + 40 = 5770, k2 ≈ 0.00012013.

With y/y0 = 0.92, 0.93, t1 = − 1k1

lny

y0= 684.5, 595.7; t2 = − 1

k2ln(y/y0) = 694.1, 604.1; in

1988 the shroud was at most 695 years old, which places its creation in or after the year 1293.

(b) Suppose T is the true half-life of carbon-14 and T1 = T (1 + r/100) is the false half-life. Then

with k =ln 2T, k1 =

ln 2T1

we have the formulae y(t) = y0e−kt, y1(t) = y0e

−k1t. At a certain

point in time a reading of the carbon-14 is taken resulting in a certain value y, which in thecase of the true formula is given by y = y(t) for some t, and in the case of the false formulais given by y = y1(t1) for some t1.

If the true formula is used then the time t since the beginning is given by t = −1k

lny

y0. If

the false formula is used we get a false value t1 = − 1k1

lny

y0; note that in both cases the

value y/y0 is the same. Thus t1/t = k/k1 = T1/T = 1 + r/100, so the percentage error inthe time to be measured is the same as the percentage error in the half-life.

44. If y = y0ekt and y = y1 = y0e

kt1 then y1/y0 = ekt1 , k =ln(y1/y0)

t1; if y = y0e

−kt and

y = y1 = y0e−kt1 then y1/y0 = e−kt1 , k = − ln(y1/y0)

t1.


412 Chapter 8

45. (a) If y = y0ekt, then y1 = y0e

kt1 , y2 = y0ekt2 , divide: y2/y1 = ek(t2−t1), k =

1t2 − t1

ln(y2/y1),

T =ln 2k

=(t2 − t1) ln 2

ln(y2/y1). If y = y0e

−kt, then y1 = y0e−kt1 , y2 = y0e

−kt2 ,

y2/y1 = e−k(t2−t1), k = − 1t2 − t1

ln(y2/y1), T =ln 2k

= − (t2 − t1) ln 2ln(y2/y1)

.

In either case, T is positive, so T =∣∣∣∣ (t2 − t1) ln 2

ln(y2/y1)

∣∣∣∣.(b) In part (a) assume t2 = t1 + 1 and y2 = 1.25y1. Then T =

ln 2ln 1.25

≈ 3.1 h.

46. (a) In t years the interest will be compounded nt times at an interest rate of r/n each time. Thevalue at the end of 1 interval is P + (r/n)P = P (1 + r/n), at the end of 2 intervals it isP (1 + r/n) + (r/n)P (1 + r/n) = P (1 + r/n)2, and continuing in this fashion the value at theend of nt intervals is P (1 + r/n)nt.

(b) Let x = r/n, then n = r/x andlim

n→+∞P (1 + r/n)nt = lim

x→0+P (1 + x)rt/x = lim

x→0+P [(1 + x)1/x]rt = Pert.

(c) The rate of increase is dA/dt = rPert = rA.

47. (a) A = 1000e(0.08)(5) = 1000e0.4 ≈ $1, 491.82

(b) Pe(0.08)(10) = 10, 000, Pe0.8 = 10, 000, P = 10, 000e−0.8 ≈ $4, 493.29

(c) From (11), with k = r = 0.08, T = (ln 2)/0.08 ≈ 8.7 years.

48. Let r be the annual interest rate when compounded continuously and r1 the effective annualinterest rate. Then an amount P invested at the beginning of the year is worth Per = P (1 + r1)at the end of the year, and r1 = er − 1.

49. (a) Givendy

dt= k

(1− y

L

)y, separation of variables yields

(1y

+1

L− y

)dy = k dt so that

lny

L− y= ln y − ln(L − y) = kt + C. The initial condition gives C = ln

y0L− y0

so

lny

L− y= kt+ ln

y0L− y0

,y

L− y= ekt

y0L− y0

, and y(t) =y0L

y0 + (L− y0)e−kt.

(b) If y0 > 0 then y0 + (L − y0)e−kt = Le−kt + y0(1 − e−kt) > 0 for all t ≥ 0, so y(t) exists for

all such t. Since limt→+∞

e−kt = 0, limt→+∞

y(t) =y0L

y0 + (L− y0) · 0= L.

(Note that for y0 < 0 the solution “blows up” at t = −1k

ln−y0L− y0

, so limt→+∞

y(t) is undefined.)

50. The differential equation for the spread of disease can be rewritten asdy

dt= kL

(1− y

L

)y, which is

the logistic equation with k replaced by kL. Making this replacement in the solution from Exercise

49 gives y(t) =y0L

y0 + (L− y0)e−kLt.



51. (a) k = L = 1, y0 = 2

2

00 2

(b) k = L = y0 = 1

2

00 2

(c) k = y0 = 1, L = 2

2

00 5

(d) k = 1, y0 = 0.5, L = 10

00

10

10

52. y0 ≈ 400, L ≈ 1000; since the curve y =400,000

400 + 600e−ktpasses through the point (200, 600),

600 =400,000

400 + 600e−200k, 600e−200k =

8003, k =

1200

ln 2.25 ≈ 0.00405.

53. y0 ≈ 2, L ≈ 8; since the curve y =2 · 8

2 + 6e−ktpasses through the point (2, 4), 4 =

162 + 6e−2k

,

6e−2k = 2, k =12

ln 3 ≈ 0.5493.

54. This is the logistic equation (Equation (4) of Section 8.1) withk = 0.98, L = 5, and y0 = 1. From Exercise 49, the solution is

given by y(t) =5

1 + 4e−0.98t.

5

00 12

55. (a) y0 = 5 (b) L = 12 (c) k = 1

(d) L/2 = 6 =60

5 + 7e−t, 5 + 7e−t = 10, t = − ln(5/7) ≈ 0.3365

(e)dy

dt=

112y(12− y), y(0) = 5

56. (a) y0 = 1 (b) L = 1000 (c) k = 0.9

(d) 750 =1000

1 + 999e−0.9t, 3(1 + 999e−0.9t) = 4, t =

10.9

ln(3 · 999) ≈ 8.8949

(e)dy

dt=

0.91000

y(1000− y), y(0) = 1


414 Chapter 8

57. (a) Assume that y(t) students have had the flu t days after the break. If the disease spreadsas predicted by equation (6) of Section 8.1 and if nobody is immune, then Exercise 50 gives

y(t) =y0L

y0 + (L− y0)e−kLt, where y0 = 20 and L = 1000. So y(t) =

2000020 + 980e−1000kt

=

10001 + 49e−1000kt

. Using y(5) = 35 we find that k = − ln(193/343)5000

. Hence y =1000

1 + 49(193/343)t/5.

(b) ty(t)

0

20

1

22

2

25

3

28

4

31

5

35

6

39

7

44

8

49

9

54

10

61

11

67

12

75

13

83

14

93

(c)

3 6 9 12

25

50

75

100

t

y

58. If T0 < Ta thendT

dt= k(Ta − T ) where k > 0. If T0 > Ta then

dT

dt= −k(T − Ta) where k > 0;

both cases yield T (t) = Ta + (T0 − Ta)e−kt with k > 0.

59. (a) From Exercise 58 with T0 = 95 and Ta = 21, we have T = 21 + 74e−kt for some k > 0.

(b) 85 = T (1) = 21 + 74e−k, k = − ln6474

= − ln3237, T = 21 + 74et ln(32/37) = 21 + 74

(3237

)t,

T = 51 when3074

=(

3237

)t, t =

ln(30/74)ln(32/37)

≈ 6.22 min

60.dT

dt= k(70− T ), T (0) = 40; − ln(70− T ) = kt+ C, 70− T = e−kte−C , T = 40 when t = 0, so

30 = e−C , T = 70− 30e−kt; 52 = T (1) = 70− 30e−k, k = − ln70− 52

30= ln

53≈ 0.5,

T ≈ 70− 30e−0.5t

61. (a)dv

dt=

ck

m0 − kt− g, v = −c ln(m0 − kt)− gt+ C; v = 0 when t = 0 so 0 = −c lnm0 + C,

C = c lnm0, v = c lnm0 − c ln(m0 − kt)− gt = c lnm0

m0 − kt− gt.

(b) m0 − kt = 0.2m0 when t = 100 so

v = 2500 lnm0

0.2m0− 9.8(100) = 2500 ln 5− 980 ≈ 3044 m/s.

62. (a) By the chain rule,dv

dt=dv

dx

dx

dt=dv

dxv so m

dv

dt= mv

dv

dx.

(b)mv

kv2 +mgdv = −dx, m

2kln(kv2 +mg) = −x+ C; v = v0 when x = 0 so

C =m

2kln(kv2

0 +mg),m

2kln(kv2 +mg) = −x+

m

2kln(kv2

0 +mg), x =m

2klnkv2

0 +mg

kv2 +mg.



(c) x = xmax when v = 0 so

xmax=m

2klnkv2

0 +mg

mg=

3.56× 10−3

2(7.3× 10−6)ln

(7.3× 10−6)(988)2 + (3.56× 10−3)(9.8)(3.56× 10−3)(9.8)

≈1298 m

63. (a) A(h) = π(1)2 = π, πdh

dt= −0.025

√h,

π√hdh = −0.025dt, 2π

√h = −0.025t + C;h = 4 when

t = 0, so 4π = C, 2π√h = −0.025t+ 4π,

√h = 2− 0.025

2πt, h ≈ (2− 0.003979 t)2.

(b) h = 0 when t ≈ 2/0.003979 ≈ 502.6 s ≈ 8.4 min.

64. (a) A(h) = 6[2√

4− (h− 2)2]

= 12√

4h− h2,

12√

4h− h2dh

dt= −0.025

√h, 12

√4− h dh = −0.025dt,

−8(4− h)3/2 = −0.025t+ C; h = 4 when t = 0 so C = 0,

(4− h)3/2 = (0.025/8)t, 4− h = (0.025/8)2/3t2/3,

h ≈ 4− 0.021375t2/3 ft

h − 22

2√4 − (h − 2)2

h

(b) h = 0 when t =8

0.025(4− 0)3/2 = 2560 s ≈ 42.7 min

65.dv

dt= − 1

32v2,

1v2dv = − 1

32dt,−1

v= − 1

32t+ C; v = 128 when t = 0 so − 1

128= C,

−1v

= − 132t− 1

128, v =

1284t+ 1

cm/s. But v =dx

dtso

dx

dt=

1284t+ 1

, x = 32 ln(4t+ 1) + C1;

x = 0 when t = 0 so C1 = 0, x = 32 ln(4t+ 1) cm.

66.dv

dt= −0.02

√v,

1√vdv = −0.02dt, 2

√v = −0.02t+ C; v = 9 when t = 0 so 6 = C,

2√v = −0.02t+ 6, v = (3− 0.01t)2 cm/s. But v =

dx

dtso

dx

dt= (3− 0.01t)2,

x = −1003

(3− 0.01t)3 + C1; x = 0 when t = 0 so C1 = 900, x = 900− 1003

(3− 0.01t)3 cm.

67. Suppose that H(y) = G(x) + C. ThendH

dy

dy

dx= G′(x). But

dH

dy= h(y) and

dG

dx= g(x), hence

y(x) is a solution of (1).

68. Suppose that y(x) satisfies y(x0) = y0 and∫ y(x)

y0

h(r) dr =∫ x

x0

g(s) ds for all x in some interval

containing x0. Differentiate with respect to x; by the Fundamental Theorem of Calculus (Theorem

5.6.3) and the chain rule, we have h(y(x))dy

dx= g(x).

69. If h(y) = 0 then (1) implies that g(x) = 0, so h(y) dy = 0 = g(x) dx. Otherwise the slope of L

isdy

dx=g(x)h(y)

. Since (x1, y1) and (x2, y2) lie on L, we havey2 − y1x2 − x1

=g(x)h(y)

. So h(y)(y2 − y1) =

g(x)(x2 − x1); i.e. h(y) dy = g(x) dx.

70. It is true that the method may not give a formula for y as a function of x. But sometimes such aformula (in terms of familiar functions) does not exist. In such cases the method at least gives arelationship between x and y, from which we can find as good an approximation to y as we want.


416 Chapter 8

71. Suppose that y = f(x) satisfies h(y)dy

dx= g(x). Integrating both sides of this with respect to x

gives∫h(y)

dy

dxdx =

∫g(x) dx, so

∫h(f(x))f ′(x) dx =

∫g(x) dx. By equation (2) of Section

5.3 with f replaced by h, g replaced by f , and F replaced by∫h(y) dy, the left side equals

F (f(x)) = F (y). Thus∫h(y) dy =

∫g(x) dx.

EXERCISE SET 8.3

1.

x

y

-2 -1 1 2

-2

-1

1

22.

x

y

1 2 3 4

1

2

3

43.

5

–1

2

x

y

y(0) = 1

y(0) = 2

y(0) = –1

4.dy

dx+ y = 1, µ = e

∫dx = ex,

d

dx[yex] = ex, yex = ex + C, y = 1 + Ce−x

(a) −1 = 1 + C,C = −2, y = 1− 2e−x

(b) 1 = 1 + C,C = 0, y = 1

(c) 2 = 1 + C,C = 1, y = 1 + e−x

5. limx→+∞

y = 1

6. (a) IV, since the slope is positive for x > 0 and negative for x < 0.

(b) VI, since the slope is positive for y > 0 and negative for y < 0.

(c) V, since the slope is always positive.

(d) II, since the slope changes sign when crossing the lines y = ±1.

(e) I, since the slope can be positive or negative in each quadrant but is not periodic.

(f) III, since the slope is periodic in both x and y.

7. y0 = 1, yn+1 = yn + 12y

1/3n

nxnyn

0

0

1

1

0.5

1.50

2

1

3 4

2

5

2.07

1.5

2.71 3.41

2.5

4.16

6

3

4.96

7

3.5

5.81

8

4

6.71

2 41 3

9

x

y



8. y0 = 1, yn+1 = yn + (xn − y2n)/4

nxn

yn

0

0

1

1

0.25

0.75

2 3 4 5

0.50

0.67

0.75

0.68

1.00

0.75

1.25

0.86

6

1.50

0.99

7

1.75

1.12

8

2.00

1.24

0.5 1 1.5 2

0.5

1

1.5

2

x

y

9. y0 = 1, yn+1 = yn +12

cos yn

ntnyn

0

0

1

1

0.5

1.27

2

1

3 4

2

1.42

1.5

1.49 1.53

3

3

t

y

10. y0 = 0, yn+1 = yn + e−yn/10

ntnyn

0

0

0

1

0.1

0.10

2 3 4 5

0.2

0.19

0.3

0.27

0.4

0.35

0.5

0.42

6

0.6

0.49

7

0.7

0.55

8

0.8

0.60

9

0.9

0.66

10

1.0

0.71

1

1

t

y

11. h = 1/5, y0 = 1, yn+1 = yn +15

sin(πn/5)

ntnyn

0

0

0

1

0.2

0.12

0.4

0.31

2 3 4

0.6

0.50

0.8

0.62

1.0

0.62

5

12. False. This is only true if the slope at (x, y) does not depend on y.

13. True.dy

dx= exy > 0 for all x and y. So, for any integral curve, y is an increasing function of x.

14. True.d2y

dx2=

d

dx(ey) = ey

dy

dx= e2y > 0 for all y.

15. True. Every cubic polynomial has at least one real root. If p(y0) = 0 then y = y0 is an integralcurve that is a horizontal line.

16. (a) By inspection,dy

dx= e−x

2and y(0) = 0.

(b) yn+1 = yn + e−x2n/20 = yn + e−(n/20)2/20 and y20 = 0.7625. From a CAS, y(1) = 0.7468.

17. (b) y dy = −x dx, y2/2 = −x2/2 + C1, x2 + y2 = C; if y(0) = 1 then C = 1 so y(1/2) =

√3/2.


418 Chapter 8

18. (a) Yes.(b) y = 0 is an integral curve of y′ = 2xy which is a horizontal line.

19. (b) The equation y′ = 1− y is separable:dy

1− y= dx, so

∫dy

1− y=∫

dx, − ln |1− y| = x+C.

Substituting x = 0 and y = −1 gives C = − ln 2, so x = ln 2−ln |1−y| = ln∣∣∣∣ 21− y

∣∣∣∣. Since the

integral curve stays below the line y = 1, we can drop the absolute value signs: x = ln2

1− yand y = 1− 2e−x. Solving y = 0 shows that the x-intercept is ln 2 ≈ 0.693.

20. (a) y0 = 1, yn+1 = yn + (√yn/2)∆x

∆x = 0.2 : yn+1 = yn +√yn/10; y5 ≈ 1.5489

∆x = 0.1 : yn+1 = yn +√yn/20; y10 ≈ 1.5556

∆x = 0.05 : yn+1 = yn +√yn/40; y20 ≈ 1.5590

(b)dy√y

=12dx, 2

√y = x/2 + C, 2 = C,

√y = x/4 + 1, y = (x/4 + 1)2,

y(1) = 25/16 = 1.5625

21. (a) The slope field does not vary with x, hence along a given parallel line all values are equalsince they only depend on the height y.

(b) As in part (a), the slope field does not vary with x; it is independent of x.

(c) From G(y)− x = C we obtaind

dx(G(y)− x) =

1f(y)

dy

dx− 1 =

d

dxC = 0, i.e.

dy

dx= f(y).

22. (a) Separate variables:dy√y

= dx, 2√y = x + C, y = (x/2 + C1)2 is a parabola that opens up,

and is therefore concave up.(b) A curve is concave up if its derivative is increasing, and y′ =

√y is increasing.

23. (a) By implicit differentiation, y3 + 3xy2 dy

dx− 2xy − x2 dy

dx= 0,

dy

dx=

2xy − y3

3xy2 − x2.

(b) If y(x) is an integral curve of the slope field in part (a), thend

dx{x[y(x)]3−x2y(x)} = [y(x)]3 + 3xy(x)2y′(x)− 2xy(x)−x2y′(x) = 0, so the integral curve

must be of the form x[y(x)]3 − x2y(x) = C.(c) x[y(x)]3 − x2y(x) = 2

24. (a) By implicit differentiation, ey + xeydy

dx+ ex

dy

dx+ yex = 0,

dy

dx= − e

y + yex

xey + ex

(b) If y(x) is an integral curve of the slope field in part (a), thend

dx{xey(x) + y(x)ex} = ey(x) + xy′(x)ey(x) + y′(x)ex + y(x)ex = 0 from part (a). Thus

xey(x) + y(x)ex = C.(c) Any integral curve y(x) of the slope field above satisfies xey(x) + y(x)ex = C; if it passes

through (1, 1) then e+ e = C, so xey(x) + y(x)ex = 2e defines the curve implicitly.

25. (a) For any n, yn is the value of the discrete approximation at the right endpoint, that, is anapproximation of y(1). By increasing the number of subdivisions of the interval [0, 1] onemight expect more accuracy, and hence in the limit y(1).



(b) For a fixed value of n we have, for k = 1, 2, . . . , n, yk = yk−1 +yk−11n

=n+ 1n

yk−1. In partic-

ular yn =n+ 1n

yn−1 =(n+ 1n

)2

yn−2 = . . . =(n+ 1n

)ny0 =

(n+ 1n

)n. Consequently,

limn→+∞

yn = limn→+∞

(n+ 1n

)n= e, which is the (correct) value y = ex

∣∣∣∣x=1

.

26. Euler’s Method is repeated application of local linear approximation, each step dependent on theprevious step.

27. Visual inspection of the slope field may show where the integral curves are increasing, decreasing,concave up, or concave down. It may also help to identify asymptotes for the integral curves.

For example, in Exercise 3 we see that y = 1 is an integral curve that is an asymptote of all otherintegral curves. Those curves with y < 1 are increasing and concave down; those with y > 1 aredecreasing and concave up.

EXERCISE SET 8.4

1. µ = e∫4 dx = e4x, e4xy =

∫ex dx = ex + C, y = e−3x + Ce−4x

2. µ = e2∫x dx = ex

2,d

dx

[yex

2]

= xex2, yex

2=

12ex

2+ C, y =

12

+ Ce−x2

3. µ = e∫dx = ex, exy =

∫ex cos(ex)dx = sin(ex) + C, y = e−x sin(ex) + Ce−x

4.dy

dx+ 2y =

12

, µ = e∫

2dx = e2x, e2xy =∫

12e2xdx =

14e2x + C, y =

14

+ Ce−2x

5.dy

dx+

x

x2 + 1y = 0, µ = e

∫(x/(x2+1))dx = e

12 ln(x2+1) =

√x2 + 1,

d

dx

[y√x2 + 1

]= 0, y

√x2 + 1 = C, y =

C√x2 + 1

6.dy

dx+y = − 1

1− ex, µ = e

∫dx = ex, exy = −

∫ex

1− exdx = ln(1−ex)+C, y = e−x ln(1−ex)+Ce−x

7.dy

dx+

1xy = 1, µ = e

∫(1/x)dx = eln x = x,

d

dx[xy] = x, xy =

12x2 + C, y =

x

2+C

x,

2 = y(1) =12

+ C,C =32, y =

x

2+

32x

8. Divide by x to put the differential equation in the form (3):dy

dx−x−1y = x. We have p(x) = − 1

xand

q(x) = x, so∫p(x) dx = − ln |x|. So we may take e− ln |x| = |x−1| as an integrating factor. Since

integrating factors are only determined up to a constant factor, we may drop the absolute value

signs and simply take µ = x−1. We haved

dx(x−1y) = x−1 dy

dx− x−2y = x−1

(dy

dx− x−1y

)= 1, so

x−1y = x+ C and y = x2 + Cx. Since y(1) = −1, C = −2 and y = x2 − 2x.


420 Chapter 8

9. µ = e−2∫x dx = e−x

2, e−x

2y =

∫2xe−x

2dx = −e−x

2+ C,

y = −1 + Cex2, 3 = −1 + C, C = 4, y = −1 + 4ex

2

10. µ = e∫dt = et, ety =

∫2et dt = 2et + C, y = 2 + Ce−t, 1 = 2 + C, C = −1, y = 2− e−t

11. False. If y1 and y2 both satisfydy

dx+p(x)y = q(x) then

d

dx(y1 +y2)+p(x)(y1 +y2) = 2q(x). Unless

q(x) = 0 for all x, y1 + y2 is not a solution of the original differential equation.

12. True. If y = C is a solution, then dy/dx = 0, so p(x)C = q(x).

13. True. The concentration in the tank will approach the concentration in the solution flowing intothe tank.

14. False. Although equation (18) implies that vτ =mg

c, where mg is the weight of the object, the

model does not specify how c depends on the object.

15.

-2 2

–10

10

x

y

y(0) = –1

y(–1) = 0y(1) = 1

16.dy

dx− 2y = −x, µ = e−2

∫dx = e−2x,

d

dx

[ye−2x

]= −xe−2x,

ye−2x =14

(2x+ 1)e−2x + C, y =14

(2x+ 1) + Ce2x

(a) 1 = 3/4 + Ce2, C = 1/(4e2), y =14

(2x+ 1) +14e2x−2

(b) −1 = 1/4 + C,C = −5/4, y =14

(2x+ 1)− 54e2x

(c) 0 = −1/4 + Ce−2, C = e2/4, y =14

(2x+ 1) +14e2x+2

17. It appears that limx→+∞

y =

{+∞, if y0 ≥ 1/4;

−∞, if y0 < 1/4.

To confirm this, we solve the equation using the method of integrating factors:dy

dx− 2y = −x, µ = e−2

∫dx = e−2x,

d

dx

[ye−2x

]= −xe−2x, ye−2x =

14

(2x + 1)e−2x + C,

y =14

(2x+ 1) + Ce2x. Setting y(0) = y0 gives C = y0 −14

, so y =14

(2x+ 1) +(y0 −

14

)e2x.

If y0 = 1/4, then y =14

(2x + 1) → +∞ as x → +∞. Otherwise, we rewrite the solution as

y = e2x(y0 −

14

+2x+ 14e2x

); since lim

x→+∞

2x+ 14e2x

= 0, we obtain the conjectured limit.



18. (a) y0 = 1, yn+1 = yn + (2yn − xn)(0.1) = (12yn − xn)/10

n 0 1 2 3 4 5xn 0 0.1 0.2 0.3 0.4 0.5yn 1 1.2 1.43 1.696 2.0052 2.36624

(b) Less. The integral curve appears to be concave up, so each yn is an underestimate of theactual value of y(xn).

(c) From Exercise 16, we have y =14

(2x + 1) + Ce2x for some constant C. Since y(0) = 1 we

find C =34

, so y =3e2x + 2x+ 1

4and y

(12

)=

3e+ 24≈ 2.53871.

19. (a) y0 = 1,yn+1 = yn + (xn + yn)(0.2) = (xn + 6yn)/5

nxn

yn

0

0

1

1

0.2

1.20

2 3 4 5

0.4

1.48

0.6

1.86

0.8

2.35

1.0

2.98

(b) y′ − y = x, µ = e−x,d

dx

[ye−x

]= xe−x,

ye−x = −(x+ 1)e−x + C, 1 = −1 + C,

C = 2, y = −(x+ 1) + 2ex

xn

y(xn)

abs. error

perc. error

0

1

0

0 3

0.2

1.24

0.04

6 9 11 13

0.4

1.58

0.10

0.6

2.04

0.19

0.8

2.65

0.30

1.0

3.44

0.46

(c)

0.2 0.4 0.6 0.8 1

3

x

y

20. h = 0.1, yn+1 = (xn + 11yn)/10

nxn

yn

0

0

1.00

1

0.1

1.10

2 3 4 5

0.2

1.22

0.3

1.36

0.4

1.53

0.5

1.72

6

0.6

1.94

7

0.7

2.20

8

0.8

2.49

9

0.9

2.82

10

1.0

3.19

With ∆x = 0.2, Euler’s method gives y(1) ≈ 2.98; with ∆x = 0.1, it gives y(1) ≈ 3.19. The truevalue is y(1) = 2e− 2 ≈ 3.44; so the absolute errors are approximately 0.46 and 0.25 respectively.

21.dy

dt= rate in − rate out, where y is the amount of salt at time t,

dy

dt= (4)(2)−

( y50

)(2) = 8− 1

25y, so

dy

dt+

125y = 8 and y(0) = 25.

µ = e∫(1/25)dt = et/25, et/25y =

∫8et/25dt = 200et/25 + C,

y = 200 + Ce−t/25, 25 = 200 + C, C = −175,

(a) y = 200− 175e−t/25 oz (b) when t = 25, y = 200− 175e−1 ≈ 136 oz

22.dy

dt= (5)(20)− y

200(20) = 100− 1

10y, so

dy

dt+

110y = 100 and y(0) = 0.


422 Chapter 8

µ = e∫(1/10)dt = et/10, et/10y =

∫100et/10dt = 1000et/10 + C,

y = 1000 + Ce−t/10, 0 = 1000 + C, C = −1000;

(a) y = 1000− 1000e−t/10 lb (b) when t = 30, y = 1000− 1000e−3 ≈ 950 lb

23. The volume V of the (polluted) water is V (t) = 500 + (20− 10)t = 500 + 10t; if y(t) is the number

of pounds of particulate matter in the water, then y(0) = 50 anddy

dt= 0−10

y

V= − y

50 + t. Using

the method of integrating factors, we havedy

dt+

150 + t

y = 0; µ = e∫

dt50+t = 50+t;

d

dt[(50+t)y] = 0,

(50 + t)y = C, 2500 = 50y(0) = C, y(t) = 2500/(50 + t). (The differential equation may also besolved by separation of variables.)

The tank reaches the point of overflowing when V = 500 + 10t = 1000, t = 50 min, soy = 2500/(50 + 50) = 25 lb.

24. The volume of the lake (in gallons) is V = 264πr2h = 264π(15)23 = 178,200π gals. Let y(t)

denote the number of pounds of mercury salts at time t; thendy

dt= 0 − 103 y

V= − y

178.2πlb/h

and y0 = 10−5V = 1.782π lb;dy

y= − dt

178.2π, ln y = − t

178.2π + C1, y = Ce−t/(178.2π), and

C = y(0) = y0 = 1.782π, y = 1.782πe−t/(178.2π) lb of mercury salts.

t 1 2 3 4 5 6 7 8 9 10 11 12y(t) 5.588 5.578 5.568 5.558 5.549 5.539 5.529 5.519 5.509 5.499 5.489 5.480

We assumed that the mercury is always distributed uniformly throughout the lake, and doesn’tsettle to the bottom.

25. (a)dv

dt+

c

mv = −g, µ = e(c/m)

∫dt = ect/m,

d

dt

[vect/m

]= −gect/m, vect/m = −gm

cect/m + C,

v = −gmc

+Ce−ct/m, but v0 = v(0) = −gmc

+C,C = v0+gm

c, v = −gm

c+(v0 +

gm

c

)e−ct/m

(b) Replacemg

cwith vτ and −ct/m with −gt/vτ in (16).

(c) From part (b), s(t) = C − vτ t− (v0 + vτ )vτge−gt/vτ ; s0 = s(0) = C − (v0 + vτ )

vτg

,

C = s0 + (v0 + vτ )vτg

, s(t) = s0 − vτ t+vτg

(v0 + vτ )(

1− e−gt/vτ)

26. (a) Let t denote time elapsed in seconds after the moment of the drop. From Exercise 25(b),while the parachute is closedv(t) = e−gt/vτ (v0 + vτ ) − vτ = e−32t/120 (0 + 120) − 120 = 120

(e−4t/15 − 1

)and thus

v(25) = 120(e−20/3 − 1

)≈ −119.85, so the skydiver is falling at a speed of 119.85 ft/s

when the parachute opens. From Exercise 25(c), s(t) = s0 − 120t +12032

120(

1− e−4t/15),

s(25) = 10000− 120 · 25 + 450(

1− e−20/3)≈ 7449.43 ft.

(b) If t denotes time elapsed after the parachute opens, then, by Exercise 25(c),

s(t) ≈ 7449.43−24t+2432

(−119.85+24)(1−e−32t/24). A calculating utility finds that s(t) = 0

for t ≈ 307.4 s, so the skydiver is in the air for about 25 + 307.4 = 332.4 s.


Review Exercises, Chapter 8 423

27.dI

dt+R

LI =

V (t)L

, µ = e(R/L)∫dt = eRt/L,

d

dt(eRt/LI) =

V (t)L

eRt/L,

IeRt/L = I(0) +1L

∫ t

0

V (u)eRu/Ldu, I(t) = I(0)e−Rt/L +1Le−Rt/L

∫ t

0

V (u)eRu/Ldu.

(a) I(t) =15e−2t

∫ t

0

20e2udu = 2e−2te2u]t0

= 2(1− e−2t

)A.

(b) limt→+∞

I(t) = 2 A

28. From Exercise 27 and Endpaper Table #42,

I(t) = 15e−2t +13e−2t

∫ t

0

3e2u sinu du = 15e−2t + e−2t e2u

5(2 sinu− cosu)

]t0

= 15e−2t +15

(2 sin t− cos t) +15e−2t.

29. (a) Let y =1µ

[H(x) + C] where µ = eP (x),dP

dx= p(x),

d

dxH(x) = µq, and C is an arbitrary

constant. Thendy

dx+ p(x)y =

1µH ′(x)− µ′

µ2[H(x) + C] + p(x)y = q − p

µ[H(x) + C] + p(x)y = q

(b) Given the initial value problem, let C = µ(x0)y0−H(x0). Then y =1µ

[H(x)+C] is a solution

of the initial value problem with y(x0) = y0. This shows that the initial value problem hasa solution.

To show uniqueness, suppose u(x) also satisfies (3) together with u(x0) = y0. Following the

arguments in the text we arrive at u(x) =1µ

[H(x) + C] for some constant C. The initial

condition requires C = µ(x0)y0 −H(x0), and thus u(x) is identical with y(x).

30. (a) y = x and y = −x are both solutions of the given initial value problem.

(b)∫y dy = −

∫x dx, y2 = −x2 + C; but y(0) = 0, so C = 0. Thus y2 = −x2, which is

impossible.

REVIEW EXERCISES, CHAPTER 8

2. (a) yes (b) yes (c) no (d) yes

3.dy

1 + y2= x2 dx, tan−1 y =

13x3 + C, y = tan

(13x3 + C

)

4.1

tan ydy =

3secx

dx,cos ysin y

dy = 3 cosx dx, ln | sin y| = 3 sinx+ C1,

sin y = ±e3 sin x+C1 = ±eC1e3 sin x = Ce3 sin x, C 6= 0,

y = sin−1(Ce3 sin x

), as is y = 0 by inspection

5.(

1y

+ y

)dy = exdx, ln |y|+ y2/2 = ex + C; by inspection, y = 0 is also a solution


424 Chapter 8

6.dy

y2 + 1= dx, tan−1 y = x+ C, π/4 = C; y = tan(x+ π/4)

7.(

1y5

+1y

)dy =

dx

x, −1

4y−4 + ln |y| = ln |x|+ C; −1

4= C, y−4 + 4 ln(x/y) = 1

8.dy

y2= 4 sec2 2x dx, −1

y= 2 tan 2x+ C, −1 = 2 tan

(2π

8

)+ C = 2 tan

π

4+ C = 2 + C, C = −3,

y =1

3− 2 tan 2x

9.dy

y2= −2x dx, −1

y= −x2 + C, −1 = C, y = 1/(x2 + 1)

–1 1

1

x

y

10. 2y dy = dx, y2 = x + C; if y(0) = 1 then C = 1, y2 = x + 1, y =√x+ 1; if y(0) = −1 then

C = 1, y2 = x+ 1, y = −√x+ 1.

–1 1

–1

1

x

y

–1 1

–1

1

x

y

11.

x

y

1 2 3 4

1

2

3

4

12.dy

y=

18x dx, ln |y| = 1

16x2 + C, y = C1e

x2/16

13. y0 = 1, yn+1 = yn +√yn/2

n 0 1 2 3 4 5 6 7 8xn 0 0.5 1 1.5 2 2.5 3 3.5 4yn 1 1.50 2.11 2.84 3.68 4.64 5.72 6.91 8.23

2 41 3

9

x

y


Review Exercises, Chapter 8 425

14. y0 = 1, yn+1 = yn +12

sin yn

n 0 1 2 3 4tn 0 0.5 1 1.5 2yn 1 1.42 1.92 2.39 2.73

3

3

t

y

15. h = 1/5, y0 = 1, yn+1 = yn +15

cos(2πn/5)

n 0 1 2 3 4 5tn 0 0.2 0.4 0.6 0.8 1.0yn 1.00 1.20 1.26 1.10 0.94 1.00

16. yn+1 = yn + 0.1(1 + 5tn − yn), y0 = 5

n 0 1 2 3 4 5 6 7 8 9 10tn 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2yn 5.00 5.10 5.24 5.42 5.62 5.86 6.13 6.41 6.72 7.05 7.39

17. From formula (19) of Section 8.2, y(t) = y0e−0.000121t, so 0.785y0 = y0e

−0.000121t, t = − ln 0.785/0.000121 ≈2000.6 yr

18. (a)d

dty(t) = 0.01y, y(0) = 5000

(b) y(t) = 5000e0.01t

(c) 2 = e0.01t, t = 100 ln 2 ≈ 69.31 h(d) 30,000 = 5000e0.01t, t = 100 ln 6 ≈ 179.18 h

19. µ = e∫3 dx = e3x, e3xy =

∫ex dx = ex + C, y = e−2x + Ce−3x

20.dy

dx+ y =

11 + ex

, µ = e∫dx = ex, exy =

∫ex

1 + exdx = ln(1 + ex) + C, y = e−x ln(1 + ex) + Ce−x

21. µ = e−∫x dx = e−x

2/2, e−x2/2y =

∫xe−x

2/2dx = −e−x2/2 + C,

y = −1 + Cex2/2, 3 = −1 + C, C = 4, y = −1 + 4ex

2/2

22.dy

dx+

2xy = 4x, µ = e

∫(2/x)dx = x2,

d

dx

[yx2]

= 4x3, yx2 = x4 + C, y = x2 + Cx−2,

2 = y(1) = 1 + C, C = 1, y = x2 + 1/x2

23. By inspection, the left side of the equation isd

dx(y coshx), so

d

dx(y coshx) = cosh2 x =

12

(1 + cosh 2x) and y coshx =12x+

14

sinh 2x+ C =12

(x+ sinhx coshx) + C.

When x = 0, y = 2 so 2 = C, and y = 2 sech x+12

(x sech x+ sinhx).


426 Chapter 8

24. (a) µ = e−∫dx = e−x,

d

dx

[ye−x

]= xe−x sin 3x,

ye−x =∫xe−x sin 3x dx =

(− 3

10x− 3

50

)e−x cos 3x+

(− 1

10x+

225

)e−x sin 3x+ C;

1 = y(0) = − 350

+ C, C =5350, y =

(− 3

10x− 3

50

)cos 3x+

(− 1

10x+

225

)sin 3x+

5350ex

(c)

–10 –2

–2

4

x

y

25. (a) linear (b) both (c) separable (d) neither

26.dy

dx− 4xy = x

(a) IF: e∫(−4x)dx = e−2x2

,d

dx[ye−2x2

] = xe−2x2, ye−2x2

=∫xe−2x2

dx = −14e−2x2

+ C,

y = −14

+ Ce2x2

(b)dy

dx= 4xy + x,

dy

4y + 1= x dx,

14

ln(4y + 1) =12x2 + C, ln(4y + 1) = 2x2 + C1,

4y + 1 = C2e2x2, y = 1

4 (C2e2x2 − 1) = C3e

2x2 − 14 , same as in part (a)

27. Assume the tank contains y(t) oz of salt at time t. Then y0 = 0 and for 0 < t < 15,dy

dt= 5 · 10− y

100010 = (50− y/100) oz/min, with solution y = 5000 +Ce−t/100. But y(0) = 0 so

C = −5000, y = 5000(1− e−t/100) for 0 ≤ t ≤ 15, and y(15) = 5000(1− e−0.15). For 15 < t < 30,dy

dt= 0− y

10005, y = C1e

−t/200, C1e−0.075 = y(15) = 5000(1−e−0.15), C1 = 5000(e0.075−e−0.075),

y = 5000(e0.075 − e−0.075)e−t/200, y(30) = 5000(e0.075 − e−0.075)e−0.15 ≈ 646.14 oz.

28. (a) Assume the air contains y(t) ft3 of carbon monoxide at time t. Then y0 = 0 and for

t > 0,dy

dt= 0.04(0.1)− y

1200(0.1) = 1/250− y/12000,

d

dt

[yet/12000

]=

1250

et/12000,

yet/12000 = 48et/12000 + C, y(0) = 0, C = −48; y = 48(1− e−t/12000). Thus the percentage

of carbon monoxide is P =y

1200100 = 4(1− e−t/12000) percent.

(b) 0.012 = 4(1− e−t/12000), t = 36.05 min

MAKING CONNECTIONS, CHAPTER 8

1. (a) u(x) = q − p y(x) sodu

dx= −pdy

dx= −p(q − py(x)) = (−p)u(x).

If p < 0 then −p > 0 so u(x) grows exponentially.If p > 0 then −p < 0 so u(x) decays exponentially.


Making Connections, Chapter 8 427

(b) From (a), u(x) = 4 − 2y(x) satisfiesdu

dx= −2u(x), so equation (14) of Section 8.2 gives

u(x) = u0e−2x for some constant u0. Since u(0) = 4 − 2y(0) = 6, we have u(x) = 6e−2x;

hence y(x) = 2− 3e−2x.

2. (a)du

dx=

d

dx(ax+ b y(x)+ c) = a+ b

dy

dx= a+ b f(ax+ by+ c) = a+ b f(u), so

1a+ b f(u)

du

dx= 1.

(b) From (a) with a = b = 1, c = 0, f(t) = 1/t, we have1

1 + 1/udu

dx= 1, where u = x + y. So

u

u+ 1du = dx,

∫u

u+ 1du =

∫dx, u− ln |u+ 1| = x+C, x+ y− ln |x+ y+ 1| = x+C, and

y − ln |x+ y + 1| = C.

3. (a)du

dx=

d

dx

(yx

)=xdy

dx− y

x2=x f(yx

)− y

x2. Since y = ux,

du

dx=x f(u)− ux

x2=f(u)− u

x

and1

f(u)− udu

dx=

1x

.

(b)dy

dx=x− yx+ y

=1− y/x1 + y/x

has the form given in (a), with f(t) =1− t1 + t

. So1

1− u1 + u

− u

du

dx=

1x

,

1 + u

1− 2u− u2du =

dx

x,∫

1 + u

1− 2u− u2du =

∫dx

x, −1

2ln |1 − 2u − u2| = ln |x| + C1, and

|1 − 2u − u2| = e−2C1x−2. Hence 1 − 2u − u2 = Cx−2 where C is either e−2C1 or −e−2C1 .

Substituting u =y

xgives 1− 2y

x− y2

x2= Cx−2, and x2 − 2xy − y2 = C.

4. (a)du

dx= (1 − n)y−n

dy

dx= (1 − n)y−n[q(x)yn − p(x)y] = (1 − n)q(x) − (1 − n)p(x)y1−n =

(1− n)q(x)− (1− n)p(x)u. Hencedu

dx+ (1− n)p(x)u = (1− n)q(x).

(b)dy

dx− 1xy = −2y2, so this has the form given in (a) with p(x) = −1/x, q(x) = −2, and n = 2.

So u = y−1 satisfiesdu

dx+

1xu = 2. An integrating factor is given by µ = e

∫dx/x = eln x = x.

Sod

dx(xu) = x

du

dx+u = 2x, xu = x2+C, u = x+Cx−1, and y = u−1 =

1x+ Cx−1

=x

x2 + C.

Since y(1) =12

, C = 1 and y =x

x2 + 1.

chapter 8 mathematical modeling with di erential...

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