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4.2

5 5

4.3

y = 2 x 2 + 3 x + 2 y = 2 x 2 + 3 x + 2

4.5

y = 1 3 x 3 2 x 2 + 3 x 6 e x + 14 y = 1 3 x 3 2 x 2 + 3 x 6 e x + 14

4.6

v ( t ) = −9.8 t v ( t ) = −9.8 t

4.8


A direction field with arrows pointing to the right at y = -4 and y = 4. The arrows point up for y > -4 and down for y < -4. Close to y = 4, the arrows are more horizontal, but the further away, the more vertical they become.


The equilibrium solutions are y=−2y=−2 and y=2.y=2. For this equation, y=−2y=−2 is an unstable equilibrium solution, and y=2y=2 is a semi-stable equilibrium solution.

4.9
nn xnxn yn=yn1+hf(xn1,yn1)yn=yn1+hf(xn1,yn1)
00 11 −2−2
11 1.11.1 y1=y0+hf(x0,y0)=−1.5y1=y0+hf(x0,y0)=−1.5
22 1.21.2 y2=y1+hf(x1,y1)=−1.1419y2=y1+hf(x1,y1)=−1.1419
33 1.31.3 y3=y2+hf(x2,y2)=−0.8387y3=y2+hf(x2,y2)=−0.8387
44 1.41.4 y4=y3+hf(x3,y3)=−0.5487y4=y3+hf(x3,y3)=−0.5487
55 1.51.5 y5=y4+hf(x4,y4)=−0.2442y5=y4+hf(x4,y4)=−0.2442
66 1.61.6 y6=y5+hf(x5,y5)=0.0993y6=y5+hf(x5,y5)=0.0993
77 1.71.7 y7=y6+hf(x6,y6)=0.5099y7=y6+hf(x6,y6)=0.5099
88 1.81.8 y8=y7+hf(x7,y7)=1.0272y8=y7+hf(x7,y7)=1.0272
99 1.91.9 y9=y8+hf(x8,y8)=1.7159y9=y8+hf(x8,y8)=1.7159
1010 22 y10=y9+hf(x9,y9)=2.6962y10=y9+hf(x9,y9)=2.6962
4.10

y = 2 + C e x 2 + 3 x y = 2 + C e x 2 + 3 x

4.11

y = 4 + 14 e x 2 + x 1 7 e x 2 + x y = 4 + 14 e x 2 + x 1 7 e x 2 + x

4.12

Initial value problem:

d u d t = 2.4 2 u 25 , u ( 0 ) = 3 d u d t = 2.4 2 u 25 , u ( 0 ) = 3

Solution:u(t)=3027e2t/25Solution:u(t)=3027e2t/25
Concentration: 30-27e-2t25Concentration: 30-27e-2t25

4.13
  1. Initial value problem
    dTdt=k(T70),T(0)=450dTdt=k(T70),T(0)=450
  2. T(t)=70+380ektT(t)=70+380ekt
  3. Approximately 114114 minutes.
4.14
  1. dPdt=0.04(1P750),P(0)=200dPdt=0.04(1P750),P(0)=200


  2. A direction field with horizontal lines on the x axis and y = 15. The other lines are vertical, except for those curving into the x axis and y = 15. A solution is drawn that crosses the y axis at about (0, 4) and asymptotically approaches y = 15.
  3. P(t)=3000e.04t11+4e.04tP(t)=3000e.04t11+4e.04t

  4. After 1212 months, the population will be P(12)278P(12)278 rabbits.

4.15

y+15x+3y=10x20x+3;p(x)=15x+3y+15x+3y=10x20x+3;p(x)=15x+3 and q(x)=10x20x+3q(x)=10x20x+3

4.16

y = x 3 + x 2 + C x 2 y = x 3 + x 2 + C x 2

4.17

y = - 2 x - 5 2 + 1 2 e 2 x y = - 2 x - 5 2 + 1 2 e 2 x

4.18
  1. dvdt=v9.8v(0)=0dvdt=v9.8v(0)=0
  2. v(t)=9.8(et1)v(t)=9.8(et1)
  3. limtv(t)=limt(9.8(et1))=−9.8m/s21.922mphlimtv(t)=limt(9.8(et1))=−9.8m/s21.922mph
4.19

Initial-value problem:

8 q + 1 0.02 q = 20 sin 5 t , q ( 0 ) = 4 8 q + 1 0.02 q = 20 sin 5 t , q ( 0 ) = 4

q ( t ) = 10 sin 5 t 8 cos 5 t + 172 e −6.25 t 41 q ( t ) = 10 sin 5 t 8 cos 5 t + 172 e −6.25 t 41

Section 4.1 Exercises

1.

1 1

3.

3 3

5.

1 1

7.

1 1

19.

y = 4 + 3 x 4 4 y = 4 + 3 x 4 4

21.

y = 1 2 e x 2 y = 1 2 e x 2

23.

y = 2 e 1 / x y = 2 e 1 / x

25.

u = sin −1 ( e −1 + t ) u = sin −1 ( e −1 + t )

27.

y = x + 1 1 x 1 y = x + 1 1 x 1

29.

y = C x + x ln x ln ( cos x ) y = C x + x ln x ln ( cos x )

31.

y = C + 4 x ln ( 4 ) y = C + 4 x ln ( 4 )

33.

y = 2 3 t 2 + 16 ( t 2 + 16 ) + C y = 2 3 t 2 + 16 ( t 2 + 16 ) + C

35.

x = 2 15 4 + t ( 3 t 2 + 4 t 32 ) + C x = 2 15 4 + t ( 3 t 2 + 4 t 32 ) + C

37.

y = C x y = C x

39.

y = 1 t 2 2 , y = t 2 2 1 y = 1 t 2 2 , y = t 2 2 1

41.

y = e t , y = e t y = e t , y = e t

43.

y = 2 ( t 2 + 5 ) , t = 3 5 y = 2 ( t 2 + 5 ) , t = 3 5

45.

y = 10 e −2 t , t = 1 2 ln ( 1 10 ) y = 10 e −2 t , t = 1 2 ln ( 1 10 )

47.

y=14(41e−4t),y=14(41e−4t), never

49.

Solution changes from increasing to decreasing at y(0)=0y(0)=0

51.

Solution changes from increasing to decreasing at y(0)=0y(0)=0

53.

v ( t ) = −32 t + a v ( t ) = −32 t + a

55.

00 ft/s

57.

52.35452.354 meters

59.

x=50t15π2cos(πt)+3π2,2x=50t15π2cos(πt)+3π2,2 hours 11 minute

61.

y = 4 e 3 t y = 4 e 3 t

63.

y = 3 2 t + t 2 y = 3 2 t + t 2

65.

y=1k(ekt1)y=1k(ekt1) and y=xy=x

Section 4.2 Exercises

67.


A graph of the given direction field with a flat line drawn on the axis. The arrows point up for y < 0 and down for y > 0. The closer they are to the x axis, the more horizontal the arrows are, and the further away they are, the more vertical they become.
69.

y=0y=0 is a stable equilibrium

71.


A direction field with horizontal arrows at y = 0 and y = 2. The arrows point up for y > 2 and for y < 0. The arrows point down for 0 < y < 2. The closer the arrows are to these lines, the more horizontal they are, and the further away, the more vertical the arrows are. A solution is sketched that follows y = 2 in quadrant two, goes through (0, 1), and then follows the x axis.
73.

y=0y=0 is a stable equilibrium and y=2y=2 is unstable

75.

General solution is y=et+Cy=et+C.

A direction field over the four quadrants. As t goes from 0 to infinity, the arrows become more and more vertical after being horizontal closer to x = 0.
77.

General solution is y=et(t-1)+Cy=et(t-1)+C.

A direction field over [-2, 2] in the x and y axes. The arrows point slightly down and to the right over [-2, 0] and gradually become vertical over [0, 2].
79.


A direction field with horizontal arrows pointing to the right at y = 1 and y = -1. The arrows point up for y < -1 and y > 1. The arrows point down for -1 < y < 1. The closer the arrows are to these lines, the more horizontal they are, and the further away they are, the more vertical they are.
81.


A direction field with arrows pointing down and to the right for nearly all points in [-2, 2] on the x and y axes. Close to the origin, the arrows become more horizontal, point to the upper right, become more horizontal, and then point down to the right again.
83.


A direction field with horizontal arrows pointing to the right on the x axis and x = -3. Above the x axis and for x < -3, the arrows point down. For x > -3, the arrows point up. Below the x axis and for x < -3, the arrows point up. For x > -3, the arrows point down. The further away from the x axis and x = -3, the arrows become more vertical, and the closer they become, the more horizontal they become.
85.

E

87.

A

89.

B

91.

A

93.

C

95.

2.24,2.24, exact: 33

97.

7.739364,7.739364, exact: 5(e1)5(e1)

99.

−0.2535−0.2535 exact: 00

101.

1.345,1.345, exact: 1ln(2)1ln(2)

103.

−4,−4, exact: 1/21/2

105.


A direction field with horizontal arrows pointing to the right on the x axis and at y = 4. The arrows below the x axis and above y = 4 point down and to the right. The arrows between the x axis and y = 4 point up and to the right.
107.

y = 2 e t 2 / 2 y = 2 e t 2 / 2

109.

2 2

111.

3.2756 3.2756

113.

2e2e

Step Size Relative Error
h=0.1h=0.1 0.39350.3935
h=0.01h=0.01 0.061630.06163
h=0.001h=0.001 0.0066120.006612
h=0.0001h=0.0001 0.00066610.0006661
115.


A direction field with horizontal arrows pointing to the right on the x axis. Above the x axis, the arrows point down and to the right. Below the x axis, the arrows point up and to the right. The closer the arrows are to the x axis, the more horizontal the arrows are, and the further away they are from the x axis, the more vertical the arrows are.
117.

4.0741 e −10 4.0741 e −10

Section 4.3 Exercises

119.

y = e t 1 y = e t 1

121.

y = 1 + C e t y = 1 + C e t

123.

y = C x e −1 / x y = C x e −1 / x

125.

y = 1 C x 2 y = 1 C x 2

127.

y = 2 C + ln x y = 2 C + ln x

129.

y = C e x ( x + 1 ) + 1 y = C e x ( x + 1 ) + 1

131.

y = sin ( ln t + C ) y = sin ( ln t + C )

133.

y = ln ( e x ) y = ln ( e x )

135.

y = 1 2 e x 2 y = 1 2 e x 2

137.

y = tanh −1 ( x 2 2 ) y = tanh −1 ( x 2 2 )

139.

x = sin ( 1 - t + t ln t ) x = sin ( 1 - t + t ln t )

141.

y = ln ( ln ( 5 ) ) ln ( 2 5 x ) y = ln ( ln ( 5 ) ) ln ( 2 5 x )

143.

y=Ce−2x+12y=Ce−2x+12

A direction field with horizontal arrows pointing to the right at y = 0.5. Above 0.5, the arrows slope down and to the right and are increasingly vertical the further they are from y = 0.5 Below0.5, the arrows slope up and to the right and are increasingly vertical the further they are from y = 0.5.
145.

y=12Cexy=12Cex

A direction field with arrows pointing to the right. They are horizontal at the y axis. The further the arrows are from the axis, the more vertical they become. They point up above the x axis and down below the x axis.
147.

y=Cexxxy=Cexxx

A direction field with arrows pointing to the right. The arrows are flat on y = 1. The further the arrows are from that, the steeper they become. They point up above that line and down below that line.
149.

y = r d ( 1 e d t ) y = r d ( 1 e d t )

151.

y ( t ) = 10 9 e x / 50 y ( t ) = 10 9 e x / 50

153.

134.3134.3 kilograms

155.

720720 seconds

157.

2424 hours 5757 minutes

159.

T ( t ) = 20 + 50 e −0.125 t T ( t ) = 20 + 50 e −0.125 t

161.

T ( t ) = 20 + 38.5 e −0.125 t T ( t ) = 20 + 38.5 e −0.125 t

163.

y = ( c + b a ) e a x b a y = ( c + b a ) e a x b a

165.

y ( t ) = c L + ( I c L ) e r t / L y ( t ) = c L + ( I c L ) e r t / L

167.

y=40(1e−0.1t),40y=40(1e−0.1t),40 g/cm2

Section 4.4 Exercises

169.


A direction field with arrows pointing to the right. The arrows are horizontal along the x axis. The arrows point down above the x axis and below the x axis. The further away the arrows are from the x axis, the more vertical the lines become.


P=0P=0 semi-stable

171.

P = 10 e 10 x e 10 x + 4 P = 10 e 10 x e 10 x + 4

173.

P ( t ) = 10000 e 0.02 t 150 + 50 e 0.02 t P ( t ) = 10000 e 0.02 t 150 + 50 e 0.02 t

175.

6969 hours 55 minutes

177.

88 years 1111 months

179.


A direction field with arrows down for P < 1,000, pointing up for 1,000 < P < 8,500, and pointing down for P > 8,500. Right above P = 8,500, the arrows point down and to the right.
181.


A direction field with arrows pointing down and to the right. Around y = 4,000, the arrows are more horizontal. The further the arrows are from this line, the more vertical the arrows become.


P1P1 semi-stable

183.


A direction field with arrows pointing up for P < 10,000 and arrows pointing down for P > 10,000.


P2>0P2>0 stable

185.


A direction field with arrows pointing to the right at P = 0. Below 0, the arrows point down and to the right. Above 0, the arrows point down and to the right. The further away from 0, the more vertical the arrows become.


P1=0P1=0 is semi-stable

187.

y = −20 4 × 10 −6 0.002 e 0.01 t y = −20 4 × 10 −6 0.002 e 0.01 t

189.


A direction field with arrows pointing horizontally to the right along y = 2 and y = 10. For P < 2, the arrows point down and to the right. For 2 < P < 10, the arrows point up and to the right. For P > 10, the arrows point down and to the right. The further the arrows are from 2 and 10, the steeper they become, and the closer they are from 2 and 10, the more horizontal the arrows become.
191.

P ( t ) = 850 + 500 e 0.009 t 85 + 5 e 0.009 t P ( t ) = 850 + 500 e 0.009 t 85 + 5 e 0.009 t

193.

1313 years months

195.


A direction field with arrows pointing down and to the right for P < 0, up for 0 < P < 1,000, and down for P > 1,000. The further the arrows are from P = 0 and P = 1,000, the more vertical they become, and the closer they are, the more horizontal they are.
197.

31.46531.465 days

199.

September 20082008

201.

K + T 2 K + T 2

203.

r = 0.0405 r = 0.0405

205.

α = 0.0081 α = 0.0081

207.

Logistic: 361,361, Threshold: 436,436, Gompertz: 309.309.

Section 4.5 Exercises

209.

Yes

211.

Yes

213.

y x 3 y = sin x y x 3 y = sin x

215.

y + ( 3 x + 2 ) x y = e x y + ( 3 x + 2 ) x y = e x

217.

d y d t y x ( x + 1 ) = 0 d y d t y x ( x + 1 ) = 0

219.

e x e x

221.

ln ( cosh x ) ln ( cosh x )

223.

y = C e 3 x 2 3 y = C e 3 x 2 3

225.

y = C x 3 + 6 x 2 y = C x 3 + 6 x 2

227.

y = C e x 2 / 2 3 y = C e x 2 / 2 3

229.

y = C tan ( x 2 ) 2 x + 4 tan ( x 2 ) ln ( sin ( x 2 ) ) y = C tan ( x 2 ) 2 x + 4 tan ( x 2 ) ln ( sin ( x 2 ) )

231.

y = C x 3 x 2 y = C x 3 x 2

233.

y = C ( x + 2 ) 2 + 1 2 y = C ( x + 2 ) 2 + 1 2

235.

y = C x + 2 sin ( 3 t ) y = C x + 2 sin ( 3 t )

237.

y = C ( x + 1 ) 3 x 2 2 x 1 y = C ( x + 1 ) 3 x 2 2 x 1

239.

y = C e sinh −1 x 2 y = C e sinh −1 x 2

241.

y = x + 4 e x 1 y = x + 4 e x 1

243.

y = 3 x 2 ( x 2 1 ) y = 3 x 2 ( x 2 1 )

245.

y = 1 e tan −1 x y = 1 e tan −1 x

247.

y = ( x + 2 ) ln ( x + 2 2 ) y = ( x + 2 ) ln ( x + 2 2 )

249.

y = 2 e 2 x 2 x 2 x 1 y = 2 e 2 x 2 x 2 x 1

251.

v ( t ) = g m k ( 1 e k t / m ) v ( t ) = g m k ( 1 e k t / m )

253.

40.45140.451 seconds

255.

g m k g m k

257.

y = C e x a ( x + 1 ) y = C e x a ( x + 1 )

259.

y = C e x 2 / 2 a y = C e x 2 / 2 a

261.

y = e k t e t k 1 y = e k t e t k 1

Review Exercises

263.

F

265.

T

267.

y ( x ) = 2 x ln ( 2 ) + x cos −1 x 1 x 2 + C y ( x ) = 2 x ln ( 2 ) + x cos −1 x 1 x 2 + C

269.

y ( x ) = ln ( C cos x ) y ( x ) = ln ( C cos x )

271.

y ( x ) = e e C + x y ( x ) = e e C + x

273.

y ( x ) = 4 + 3 2 x 2 + 2 x sin x y ( x ) = 4 + 3 2 x 2 + 2 x sin x

275.

y ( x ) = 2 1 + 3 ( x 2 + 2 sin x ) y ( x ) = 2 1 + 3 ( x 2 + 2 sin x )

277.

y ( x ) = −2 x 2 2 x 1 3 2 3 e 3 x y ( x ) = −2 x 2 2 x 1 3 2 3 e 3 x

279.


A direction field with arrows pointing up and to the right along a logarithmic curve that approaches negative infinity as x goes to zero and increases as x goes to infinity.


y(x)=Cex+lnxy(x)=Cex+lnx

281.

Euler: 0.6939,0.6939, exact solution: y(x)=3xe−2x2+ln(3)y(x)=3xe−2x2+ln(3)

283.

40494049 second

285.

x(t)=5000+2459493t2459e5/3t,t=307.8x(t)=5000+2459493t2459e5/3t,t=307.8 seconds

287.

T ( t ) = 200 ( 1 e t / 1000 ) T ( t ) = 200 ( 1 e t / 1000 )

289.

P ( t ) = 1600000 e 0.02 t 9840 + 160 e 0.02 t P ( t ) = 1600000 e 0.02 t 9840 + 160 e 0.02 t

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