1 a 10 zill

-4 -2 2 4 t

-4

-2

2

4

X

1 Introduction to Differential EquationsExercises 1.1

1. Second-order; linear.

2. Third-order; nonlinear because of (dy/dx)4.

3. The dierential equation is rst-order. Writing it in the form x(dy/dx) + y2 = 1, we see that it is nonlinear iny because of y2. However, writing it in the form (y2 1)(dx/dy) + x = 0, we see that it is linear in x.

4. The dierential equation is rst-order. Writing it in the form u(dv/du) + (1 + u)v = ueu we see that it is linearin v. However, writing it in the form (v + uv ueu)(du/dv) + u = 0, we see that it is nonlinear in u.

5. Fourth-order; linear

6. Second-order; nonlinear because of cos(r + u)

7. Second-order; nonlinear because of

1 + (dy/dx)2

8. Second-order; nonlinear because of 1/R2

9. Third-order; linear

10. Second-order; nonlinear because of x2

11. From y = ex/2 we obtain y = 12ex/2. Then 2y + y = ex/2 + ex/2 = 0.12. From y = 65 65e20t we obtain dy/dt = 24e20t, so that

dy

dt+ 20y = 24e20t + 20

(65 6

5e20t

)= 24.

13. From y = e3x cos 2x we obtain y = 3e3x cos 2x 2e3x sin 2x and y = 5e3x cos 2x 12e3x sin 2x, so thaty 6y + 13y = 0.

14. From y = cosx ln(secx+ tanx) we obtain y = 1 + sinx ln(secx+ tanx) andy = tanx+ cosx ln(secx+ tanx). Then y + y = tanx.

15. Writing ln(2X 1) ln(X 1) = t and dierentiating implicitly we obtain2

2X 1dX

dt 1X 1

dX

dt= 1

(2

2X 1 1

X 1)dX

dt= 1

2X 2 2X + 1(2X 1)(X 1)

dX

dt= 1

dX

dt= (2X 1)(X 1) = (X 1)(1 2X).

Exponentiating both sides of the implicit solution we obtain

2X 1X 1 = e

t = 2X 1 = Xet et = (et 1) = (et 2)X = X = et 1et 2 .

Solving et 2 = 0 we get t = ln 2. Thus, the solution is dened on (, ln 2) or on (ln 2,). The graph of thesolution dened on (, ln 2) is dashed, and the graph of the solution dened on (ln 2,) is solid.

1

-4 -2 2 4 x

-4

-2

2

4

y

Exercises 1.1

16. Implicitly dierentiating the solution we obtain

2x2 dydx 4xy + 2y dy

dx= 0 = x2 dy 2xy dx+ y dy = 0

= 2xy dx+ (x2 y)dy = 0.Using the quadratic formula to solve y2 2x2y 1 = 0 for y, we gety =

(2x2 4x4 + 4 )/2 = x2 x4 + 1 . Thus, two explicit solutions are

y1 = x2 +x4 + 1 and y2 = x2

x4 + 1 . Both solutions are dened on

(,). The graph of y1(x) is solid and the graph of y2 is dashed.17. Dierentiating P = c1et/ (1 + c1et) we obtain

dP

dt=

(1 + c1et) c1et c1et c1et(1 + c1et)

2

=c1e

t

1 + c1et[(1 + c1et) c1et]

1 + c1et= P (1 P ).

18. Dierentiating y = ex2 x

0

et2dt+ c1ex

2we obtain

y = ex2ex

2 2xex2 x

0

et2dt 2c1xex2 = 1 2xex2

x0

et2dt 2c1xex2 .

Substituting into the dierential equation, we have

y + 2xy = 1 2xex2 x

0

et2dt 2c1xex2 + 2xex2

x0

et2dt+ 2c1xex

2= 1.

19. From y = c1e2x + c2xe2x we obtaindy

dx= (2c1 + c2)e2x + 2c2xe2x and

d2y

dx2= (4c1 + 4c2)e2x + 4c2xe2x, so that

d2y

dx2 4dy

dx+ 4y = (4c1 + 4c2 8c1 4c2 + 4c1)e2x + (4c2 8c2 + 4c2)xe2x = 0.

20. From y = c1x1 + c2x+ c3x lnx+ 4x2 we obtain

dy

dx= c1x2 + c2 + c3 + c3 lnx+ 8x,

d2y

dx2= 2c1x3 + c3x1 + 8,

and

d3y

dx3= 6c1x4 c3x2,

so that

x3d3y

dx3+ 2x2

d2y

dx2 x dy

dx+ y

= (6c1 + 4c1 + c1 + c1)x1 + (c3 + 2c3 c2 c3 + c2)x+ (c3 + c3)x lnx+ (16 8 + 4)x2

= 12x2.

21. From y ={x2, x < 0x2, x 0 we obtain y

={2x, x < 0

2x, x 0 so that xy 2y = 0.

2

Exercises 1.1

22. The function y(x) is not continuous at x = 0 since limx0

y(x) = 5 and limx0+

y(x) = 5. Thus, y(x) does notexist at x = 0.

23. From x = e2t + 3e6t and y = e2t + 5e6t we obtaindx

dt= 2e2t + 18e6t and dy

dt= 2e2t + 30e6t.

Then

x+ 3y = (e2t + 3e6t) + 3(e2t + 5e6t)= 2e2t + 18e6t = dx

dtand

5x+ 3y = 5(e2t + 3e6t) + 3(e2t + 5e6t)= 2e2t + 30e6t =

dy

dt.

24. From x = cos 2t+ sin 2t+ 15et and y = cos 2t sin 2t 15et we obtain

dx

dt= 2 sin 2t+ 2 cos 2t+ 1

5et and

dy

dt= 2 sin 2t 2 cos 2t 1

5et

andd2x

dt2= 4 cos 2t 4 sin 2t+ 1

5et and

d2y

dt2= 4 cos 2t+ 4 sin 2t 1

5et.

Then

4y + et = 4( cos 2t sin 2t 15et) + et

= 4 cos 2t 4 sin 2t+ 15et =

d2x

dt2and

4x et = 4(cos 2t+ sin 2t+ 15et) et

= 4 cos 2t+ 4 sin 2t 15et =

d2y

dt2.

25. An interval on which tan 5t is continuous is /2 < 5t < /2, so 5 tan 5t will be a solution on (/10, /10).26. For (1 sin t)1/2 to be continuous we must have 1 sin t > 0 or sin t < 1. Thus, (1 sin t)1/2 will be a

solution on (/2, 5/2).

27. (y)2 + 1 = 0 has no real solution.

28. The only solution of (y)2 + y2 = 0 is y = 0, since if y = 0, y2 > 0 and (y)2 + y2 y2 > 0.29. The rst derivative of f(t) = et is et. The rst derivative of f(t) = ekt is kekt. The dierential equations are

y = y and y = ky, respectively.

30. Any function of the form y = cet or y = cet is its own second derivative. The corresponding dierentialequation is y y = 0. Functions of the form y = c sin t or y = c cos t have second derivatives that are thenegatives of themselves. The dierential equation is y + y = 0.

31. Since the nth derivative of (x) must exist if (x) is a solution of the nth order dierential equation, all lower-order derivatives of (x) must exist and be continuous. [Recall that a dierentiable function is continuous.]

32. Solving the systemc1y1(0) + c2y2(0) = 2

c1y1(0) + c2y

2(0) = 0

3

Exercises 1.1

for c1 and c2 we get

c1 =2y2(0)

y1(0)y2(0) y1(0)y2(0)and c2 = 2y

1(0)

y1(0)y2(0) y1(0)y2(0).

Thus, a particular solution is

y =2y2(0)

y1(0)y2(0) y1(0)y2(0)y1 2y

1(0)

y1(0)y2(0) y1(0)y2(0)y2,

where we assume that y1(0)y2(0) y1(0)y2(0) = 0.33. For the rst-order dierential equation integrate f(x). For the second-order dierential equation integrate twice.

In the latter case we get y =

(f(t)dt)dt+ c1t+ c2.

34. Solving for y using the quadratic formula we obtain the two dierential equations

y =1t

(2 + 2

1 + 3t6

)and y =

1t

(2 2

1 + 3t6

),

so the dierential equation cannot be put in the form dy/dt = f(t, y).

35. The dierential equation yy ty = 0 has normal form dy/dt = t. These are not equivalent because y = 0 is asolution of the rst dierential equation but not a solution of the second.

36. Dierentiating we get y = c1 + 3c2t2 and y = 6c2t. Then c2 = y/6t and c1 = y ty/2, so

y =(y ty

2

)t+

(y

6t

)t3 = ty 1

3t2y

and the dierential equation is t2y 3ty + 3y = 0.37. (a) From y = emt we obtain y = memt. Then y + 2y = 0 implies

memt + 2emt = (m+ 2)emt = 0.

Since emt > 0 for all t, m = 2. Thus y = e2t is a solution.(b) From y = emt we obtain y = memt and y = m2emt. Then y 5y + 6y = 0 implies

m2emt 5memt + 6emt = (m 2)(m 3)emt = 0.Since emt > 0 for all t, m = 2 and m = 3. Thus y = e2t and y = e3t are solutions.

(c) From y = tm we obtain y = mtm1 and y = m(m 1)tm2. Then ty + 2y = 0 impliestm(m 1)tm2 + 2mtm1 = [m(m 1) + 2m]tm1 = (m2 +m)tm1

= m(m+ 1)tm1 = 0.

Since tm1 > 0 for t > 0, m = 0 and m = 1. Thus y = 1 and y = t1 are solutions.(d) From y = tm we obtain y = mtm1 and y = m(m 1)tm2. Then t2y 7ty + 15y = 0 implies

t2m(m 1)tm2 7tmtm1 + 15tm = [m(m 1) 7m+ 15]tm

= (m2 8m+ 15)tm = (m 3)(m 5)tm = 0.Since tm > 0 for t > 0, m = 3 and m = 5. Thus y = t3 and y = t5 are solutions.

38. When g(t) = 0, y = 0 is a solution of a linear equation.

39. (a) Solving (10 5y)/3x = 0 we see that y = 2 is a constant solution.(b) Solving y2 + 2y 3 = (y + 3)(y 1) = 0 we see that y = 3 and y = 1 are constant solutions.(c) Since 1/(y 1) = 0 has no solutions, the dierential equation has no constant solutions.

4

ty

5 5

5

10

Exercises 1.1

(d) Setting y = 0 we have y = 0 and 6y = 10. Thus y = 5/3 is a constant solution.

40. From y = (1 y)/(x 2) we see that a tangent line to the graph of y(x) is possibly vertical at x = 2. Intervalsof existence could be (, 2) and (2,).

41. One solution is given by the upper portion of the graph with domain approximately (0, 2.6). The other solutionis given by the lower portion of the graph, also with domain approximately (0, 2.6).

42. One solution, with domain approximately (, 1.6) is the portion of the graph in the second quadrant togetherwith the lower part of the graph in the rst quadrant. A second solution, with domain approximately (0, 1.6)is the upper part of the graph in the rst quadrant. The third solution, with domain (0,), is the part of thegraph in the fourth quadrant.

43. Dierentiating (x3 + y3)/xy = 3c we obtain

xy(3x2 + 3y2y) (x3 + y3)(xy + y)x2y2

= 0

3x3 + 3xy3y x4y x3y xy3y y4 = 0(3xy3 x4 xy3)y = 3x3y + x3y + y4

y =y4 2x3y2xy3 x4 =

y(y3 2x3)x(2y3 x3) .

44. A tangent line will be vertical where y is undened, or in this case, where x(2y3 x3) = 0. This gives x = 0and 2y3 = x3. Substituting y3 = x3/2 into x3 + y3 = 3xy we get

x3 +12x3 = 3x

(1

21/3x

)

32x3 =

321/3

x2

x3 = 22/3x2

x2(x 22/3) = 0.Thus, there are vertical tangent lines at x = 0 and x = 22/3, or at (0, 0) and (22/3, 21/3). Since 22/3 1.59, theestimates of the domains in Problem 42 were close.

45. Since (x) > 0 for all x in I, (x) is an increasing function on I. Hence, it can have no relative extrema on I.

46. (a) When y = 5, y = 0, so y = 5 is a solution of y = 5 y.(b) When y > 5, y < 0, and the solution must be decreasing. When y < 5, y > 0, and the solution must be

increasing. Thus, none of the curves in color can be solutions.

(c)

47. (a) y = 0 and y = a/b.

5

y=a b

y=0x

y

Exercises 1.1

(b) Since dy/dx = y(a by) > 0 for 0 < y < a/b, y = (x) is increasing on this interval. Since dy/dx < 0 fory < 0 or y > a/b, y = (x) is decreasing on these intervals.

(c) Using implicit dierentiation we compute

d2y

dx2= y(by) + y(a by) = y(a 2by).

Solving d2y/dx2 = 0 we obtain y = a/2b. Since d2y/dx2 > 0 for 0 < y < a/2b and d2y/dx2 < 0 fora/2b < y < a/b, the graph of y = (x) has a point of inection at y = a/2b.

(d)

48. (a) In Mathematica use

Clear[y]

y[x ]:= x Exp[5x] Cos[2x]

y[x]

y''''[x] 20 y'''[x] + 158 y''[x] 580 y'[x] + 841 y[x] // Simplify(b) In Mathematica use

Clear[y]

y[x ]:= 20 Cos[5 Log[x]]/x 3 Sin[5 Log[x]]/xy[x]

x3 y'''[x] + 2x2 y''[x] + 20 x y'[x] 78 y[x] // Simplify

Exercises 1.2

1. Solving 13

=1

1 + c1we get c1 = 4. The solution is y = 11 4et .

2. Solving 2 =1

1 + c1ewe get c1 = 12e

1. The solution is y =2

2 e(t+1) .3. Using x = c1 sin t+ c2 cos t we obtain c1 = 1 and c2 = 8. The solution is x = cos t+ 8 sin t.4. Using x = c1 sin t+ c2 cos t we obtain c2 = 0 and c1 = 1. The solution is x = cos t.5. Using x = c1 sin t+ c2 cos t we obtain

32c1 +

12c2 =

12

12c1 +

3

2c2 = 0.

Solving we nd c1 =

34

and c2 =14

. The solution is x =

34

cos t+14

sin t.

6

Exercises 1.2

6. Using x = c1 sin t+ c2 cos t we obtain

22

c1 +

22

c2 =

2

22

c1 +

22

c2 = 2

2 .

Solving we nd c1 = 1 and c2 = 3. The solution is x = cos t+ 3 sin t.7. From the initial conditions we obtain the system

c1 + c2 = 1

c1 c2 = 2.Solving we get c1 = 32 and c2 = 12 . A solution of the initial-value problem is y = 32ex 12ex.

8. From the initial conditions we obtain the system

c1e+ c2e1 = 0

c1e c2e1 = e.Solving we get c1 = 12 and c2 = 12e2. A solution of the initial-value problem is y = 12ex 12e2x.

9. From the initial conditions we obtainc1e

1 + c2e = 5

c1e1 c2e = 5.

Solving we get c1 = 0 and c2 = 5e1. A solution of the initial-value problem is y = 5ex1.

10. From the initial conditions we obtainc1 + c2 = 0

c1 c2 = 0.Solving we get c1 = c2 = 0. A solution of the initial-value problem is y = 0.

11. Two solutions are y = 0 and y = x3.

12. Two solutions are y = 0 and y = x2. (Also, any constant multiple of x2 is a solution.)

13. For f(x, y) = y2/3 we havef

y=

23y1/3. Thus the dierential equation will have a unique solution in any

rectangular region of the plane where y = 0.

14. For f(x, y) =xy we have

f

y=

12

x

y. Thus the dierential equation will have a unique solution in any

region where x > 0 and y > 0 or where x < 0 and y < 0.

15. For f(x, y) =y

xwe have

f

y=

1x

. Thus the dierential equation will have a unique solution in any region

where x = 0.16. For f(x, y) = x + y we have

f

y= 1. Thus the dierential equation will have a unique solution in the entire

plane.

17. For f(x, y) =x2

4 y2 we havef

y=

2x2y(4 y2)2 . Thus the dierential equation will have a unique solution in

any region where y < 2, 2 < y < 2, or y > 2.

7

Exercises 1.2

18. For f(x, y) =x2

1 + y3we have

f

y=

3x2y2(1 + y3)2

. Thus the dierential equation will have a unique solution in

any region where y = 1.

19. For f(x, y) =y2

x2 + y2we have

f

y=

2x2y(x2 + y2)2

. Thus the dierential equation will have a unique solution in

any region not containing (0, 0).

20. For f(x, y) =y + xy x we have

f

y=

2x(y x)2 . Thus the dierential equation will have a unique solution in any

region where y < x or where y > x.

21. The dierential equation has a unique solution at (1, 4).

22. The dierential equation is not guaranteed to have a unique solution at (5, 3).

23. The dierential equation is not guaranteed to have a unique solution at (2,3).24. The dierential equation is not guaranteed to have a unique solution at (1, 1).25. (a) A one-parameter family of solutions is y = cx. Since y = c, xy = xc = y and y(0) = c 0 = 0.

(b) Writing the equation in the form y = y/x we see that R cannot contain any point on the y-axis. Thus,any rectangular region disjoint from the y-axis and containing (x0, y0) will determine an interval around x0and a unique solution through (x0, y0). Since x0 = 0 in part (a) we are not guaranteed a unique solutionthrough (0, 0).

(c) The piecewise-dened function which satises y(0) = 0 is not a solution since it is not dierentiable atx = 0.

26. (a) Sinced

dxtan(x + c) = sec2(x + c) = 1 + tan2(x + c), we see that y = tan(x + c) satises the dierential

equation.

(b) Solving y(0) = tan c = 0 we obtain c = 0 and y = tanx. Since tanx is discontinuous at x = /2, thesolution is not dened on (2, 2) because it contains /2.

(c) The largest interval on which the solution can exist is (/2, /2).

27. (a) Sinced

dt

( 1t+ c

)=

1(t+ c)2

= y2, we see that y = 1t+ c

is a solution of the dierential equation.

(b) Solving y(0) = 1/c = 1 we obtain c = 1 and y = 1/(1 t). Solving y(0) = 1/c = 1 we obtain c = 1and y = 1/(1 + t). Being sure to include t = 0, we see that the interval of existence of y = 1/(1 t) is(, 1), while the interval of existence of y = 1/(1 + t) is (1,).

(c) Solving y(0) = 1/c = y0 we obtain c = 1/y0 and

y = 11/y0 + t =y0

1 y0t , y0 = 0.

Since we must have 1/y0 + t = 0, the largest interval of existence (which must contain 0) is either(, 1/y0) when y0 > 0 or (1/y0,) when y0 < 0.

(d) By inspection we see that y = 0 is a solution on (,).28. (a) Dierentiating 3x2 y2 = c we get 6x 2yy = 0 or yy = 3x.

8

-4 -2 2 4 x

-4

-2

2

4

y

-4 -2 2 4 x

-4

-2

2

4

y

-4 -2 2 4 x12345y

-4 -2 2 4 x12345y

Exercises 1.2

(b) Solving 3x2 y2 = 3 for y we gety = 1(x) =

3(x2 1) , 1 < x

Exercises 1.3

Exercises 1.3

1.dP

dt= kP + r;

dP

dt= kP r

2. Let b be the rate of births and d the rate of deaths. Then b = k1P and d = k2P . Since dP/dt = b d, thedierential equation is dP/dt = k1P k2P .

3. Let b be the rate of births and d the rate of deaths. Then b = k1P and d = k2P 2. Since dP/dt = b d, thedierential equation is dP/dt = k1P k2P 2.

4. Let P (t) be the number of owls present at time t. Then dP/dt = k(P 200 + 10t).5. From the graph we estimate T0 = 180 and Tm = 75. We observe that when T = 85, dT/dt 1. From the

dierential equation we then have

k =dT/dt

T Tm =1

85 75 = 0.1.

6. By inspecting the graph we take Tm to be Tm(t) = 8030 cost/12. Then the temperature of the body at timet is determined by the dierential equation

dT

dt= k

[T

(80 30 cos

12t)]

, t > 0.

7. The number of students with the u is x and the number not infected is 1000 x, so dx/dt = kx(1000 x).8. By analogy with dierential equation modeling the spread of a disease we assume that the rate at which the

technological innovation is adopted is proportional to the number of people who have adopted the innovationand also to the number of people, y(t), who have not yet adopted it. If one person who has adopted theinnovation is introduced into the population then x+ y = n+ 1 and

dx

dt= kx(n+ 1 x), x(0) = 1.

9. The rate at which salt is leaving the tank is

(3 gal/min) (

A

300lb/gal

)=

A

100lb/min.

Thus dA/dt = A/100.

10. The rate at which salt is entering the tank is

R1 = (3 gal/min) (2 lb/gal) = 6 lb/min.

Since the solution is pumped out at a slower rate, it is accumulating at the rate of (3 2)gal/min = 1 gal/min.After t minutes there are 300 + t gallons of brine in the tank. The rate at which salt is leaving is

R2 = (2 gal/min) (

A

300 + tlb/gal

)=

2A300 + t

lb/min.

The dierential equation isdA

dt= 6 2A

300 + t.

11. The volume of water in the tank at time t is V = Awh. The dierential equation is then

dh

dt=

1Aw

dV

dt=

1Aw

(cA0

2gh

)= cA0

Aw

2gh .

10

Exercises 1.3

Using A0 = (

212

)2=

36, Aw = 102 = 100, and g = 32, this becomes

dh

dt= c/36

100

64h = c

450

h .

12. The volume of water in the tank at time t is V = 13r2h = 13Awh. Using the formula from Problem 11 for the

volume of water leaving the tank we see that the dierential equation is

dh

dt=

3Aw

dV

dt=

3Aw

(cAh

2gh ) = 3cAhAw

2gh .

Using Ah = (2/12)2 = /36, g = 32, and c = 0.6, this becomes

dh

dt= 3(0.6)/36

Aw

64h = 0.4

Awh1/2 .

To nd Aw we let r be the radius of the top of the water. Then r/h = 8/20, so r = 2h/5 and Aw = (2h/5)2 =4h2/25. Thus

dh

dt= 0.4

4h2/25h1/2 = 2.5h3/2.

13. Since i =dq

dtand L

d2q

dt2+R

dq

dt= E(t) we obtain L

di

dt+Ri = E(t).

14. By Kirchos second law we obtain Rdq

dt+

1Cq = E(t).

15. From Newtons second law we obtain mdv

dt= kv2 +mg.

16. We have from Archimedes principle

upward force of water on barrel = weight of water displaced

= (62.4) (volume of water displaced)= (62.4)(s/2)2y = 15.6s2y.

It then follows from Newtons second law thatw

g

d2y

dt2= 15.6s2y or d

2y

dt2+

15.6s2gw

y = 0, where g = 32

and w is the weight of the barrel in pounds.

17. The net force acting on the mass is

F = ma = md2x

dt2= k(s+ x) +mg = kx+mg ks.

Since the condition of equilibrium is mg = ks, the dierential equation is

md2x

dt2= ks.

18. From Problem 17, without a damping force, the dierential equation is md2x/dt2 = kx. With a dampingforce proportional to velocity the dierential equation becomes

md2x

dt2= kx dx

dtor m

d2x

dt2+

dx

dt+ kx = 0.

19. Let x(t) denote the height of the top of the chain at time t with the positive direction upward. The weight ofthe portion of chain o the ground is W = (x ft) (1 lb/ft) = x. The mass of the chain is m = W/g = x/32.The net force is F = 5W = 5 x. By Newtons second law,

d

dt

( x32v)

= 5 x or xdvdt

+ vdx

dt= 160 32x.

11

(x,y)

x

y

x

y

Exercises 1.3

Thus, the dierential equation is

xd2x

dt+

(dxdt

)2+ 32x = 160.

20. The force is the weight of the chain, 2L, so by Newtons second law,d

dt[mv] = 2L. Since the mass of the portion

of chain o the ground is m = 2(L x)/g, we haved

dt

[2(L x)g

v]

= 2L or (L x)dvdt

+ v(dxdt

)= Lg.

Thus, the dierential equation is

(L x)d2x

dt2

(dxdt

)2= Lg.

21. From g = k/R2 we nd k = gR2. Using a = d2r/dt2 and the fact that the positive direction is upward we get

d2r

dt2= a = k

r2= gR

2

r2or

d2r

dt2+gR2

r2= 0.

22. The gravitational force onm is F = kMrm/r2. SinceMr = 4r3/3 andM = 4R3/3 we haveMr = r3M/R3and

F = k Mrmr2

= k r3Mm/R3

r2= k mM

R3r.

Now from F = ma = d2r/dt2 we have

md2r

dt2= k mM

R3r or

d2r

dt2= kM

R3r.

23. The dierential equation isdA

dt= k(M A).

24. The dierential equation isdA

dt= k1(M A) k2A.

25. The dierential equation is x(t) = r kx(t) where k > 0.26. By the Pythagorean Theorem the slope of the tangent line is y =

ys2 y2 .

27. We see from the gure that 2 + = . Thus

y

x = tan = tan( 2) = tan 2 = 2 tan

1 tan2 .

Since the slope of the tangent line is y = tan we have y/x = 2y[1 (y)2] ory y(y)2 = 2xy, which is the quadratic equation y(y)2 + 2xy y = 0 in y.Using the quadratic formula we get

y =2x

4x2 + 4y2

2y=x

x2 + y2

y.

Since dy/dx > 0, the dierential equation is

dy

dx=x+

x2 + y2

yor y

dy

dx

x2 + y2 + x = 0.

28. The dierential equation is dP/dt = kP , so from Problem 29 in Exercises 1.1, P = ekt, and a one-parameterfamily of solutions is P = cekt.

12

Exercises 1.3

29. The dierential equation in (3) is dT/dt = k(T Tm). When the body is cooling, T > Tm, so T Tm > 0.Since T is decreasing, dT/dt < 0 and k < 0. When the body is warming, T < Tm, so T Tm < 0. Since T isincreasing, dT/dt > 0 and k < 0.

30. The dierential equation in (8) is dA/dt = 6A/100. If A(t) attains a maximum, then dA/dt = 0 at this timeand A = 600. If A(t) continues to increase without reaching a maximum then A(t) > 0 for t > 0 and A cannotexceed 600. In this case, if A(t) approaches 0 as t increases to innity, we see that A(t) approaches 600 as tincreases to innity.

31. This dierential equation could describe a population that undergoes periodic uctuations.

32. (1):dP

dt= kP is linear (2):

dA

dt= kA is linear

(3):dT

dt= k(T Tm) is linear (5): dx

dt= kx(n+ 1 x) is nonlinear

(6):dX

dt= k(X)( X) is nonlinear (8): dA

dt= 6 A

100is linear

(10):dh

dt= Ah

Aw

2gh is nonlinear (11): L

d2q

dt2+R

dq

dt+

1Cq = E(t) is linear

(12):d2s

dt2= g is linear (14): mdv

dt= mg kv is linear

(15): md2s

dt2+ k

ds

dt= mg is linear (16):

d2x

dt2 64

Lx = 0 is linear

33. From Problem 21, d2r/dt2 = gR2/r2. Since R is a constant, if r = R + s, then d2r/dt2 = d2s/dt2 and, usinga Taylor series, we get

d2s

dt2= g R

2

(R + s)2= gR2(R + s)2 gR2[R2 2sR3 + ] = g + 2gs

R3+ .

Thus, for R much larger than s, the dierential equation is approximated by d2s/dt2 = g.34. If is the mass density of the raindrop, then m = V and

dm

dt=

dV

dt=

d

dt

[43r3

]=

(4r2

dr

dt

)= S

dr

dt.

If dr/dt is a constant, then dm/dt = kS where dr/dt = k or dr/dt = k/. Since the radius is decreasing,k < 0. Solving dr/dt = k/ we get r = (k/)t+ c0. Since r(0) = r0, c0 = r0 and r = kt/+ r0.

From Newtons second law,d

dt[mv] = mg, where v is the velocity of the raindrop. Then

mdv

dt+ v

dm

dt= mg or

(43r3

)dvdt

+ v(k4r2) = (4

3r3

)g.

Dividing by 4r3/3 we get

dv

dt+

3krv = g or

dv

dt+

3k/kt/+ r0

v = g, k < 0.

35. We assume that the plow clears snow at a constant rate of k cubic miles per hour. Let t be the time in hoursafter noon, x(t) the depth in miles of the snow at time t, and y(t) the distance the plow has moved in t hours.Then dy/dt is the velocity of the plow and the assumption gives

wxdy

dt= k

13

Exercises 1.3

where w is the width of the plow. Each side of this equation simply represents the volume of snow plowed inone hour. Now let t0 be the number of hours before noon when it started snowing and let s be the constant ratein miles per hour at which x increases. Then for t > t0, x = s(t+ t0). The dierential equation then becomes

dy

dt=

k

ws

1t+ t0

.

Integrating we obtain

y =k

ws[ ln(t+ t0) + c ]

where c is a constant. Now when t = 0, y = 0 so c = ln t0 and

y =k

wsln

(1 +

t

t0

).

Finally, from the fact that when t = 1, y = 2 and when t = 2, y = 3, we obtain(

1 +2t0

)2=

(1 +

1t0

)3.

Expanding and simplifying gives t20 + t0 1 = 0. Since t0 > 0, we nd t0 0.618 hours 37 minutes. Thus itstarted snowing at about 11:23 in the morning.

Chapter 1 Review Exercises

1.d

dx

c1x

= c1x2

= c1/xx

;dy

dx= y

x

2.d

dx(5 + c1e2x) = 2c1e2x = 2(5 + c1e2x 5); dy

dx= 2(y 5) or dy

dx= 2y + 10

3.d

dx(c1 cos kx+ c2 sin kx) = kc1 sin kx+ kc2 cos kx;

d2

dx2(c1 cos kx+ c2 sin kx) = k2c1 cos kx k2c2 sin kx = k2(c1 cos kx+ c2 sin kx);

d2y

dx2= k2y or d

2y

dx2+ k2y = 0

4.d

dx(c1 cosh kx+ c2 sinh kx) = kc1 sinh kx+ kc2 cosh kx;

d2

dx2(c1 cosh kx+ c2 sinh kx) = k2c1 cosh kx+ k2c2 sinh kx = k2(c1 cosh kx+ c2 sinh kx);

d2y

dx2= k2y or

d2y

dx2 k2y = 0

5. y = c1ex + c2xex + c2ex; y = c1ex + c2xex + 2c2ex;y + y = 2(c1ex + c2xex) + 2c2ex = 2(c1ex + c2xex + c2ex) = 2y; y 2y + y = 0

6. y = c1ex sinx+ c1ex cosx+ c2ex cosx+ c2ex sinx;y = c1ex cosx c1ex sinx c1ex sinx+ c1ex cosx c2ex sinx+ c2ex cosx+ c2ex cosx+ c2ex sinx

= 2c1ex sinx+ 2c2ex cosx;y 2y = 2c1ex cosx 2c2ex sinx = 2y; y 2y + 2y = 0

7. a,d 8. c 9. b 10. a,c 11. b 12. a,b,d

13. A few solutions are y = 0, y = c, and y = ex.

14

-3 -2 -1 1 2 3 x

-3

-2

-1

1

2

3y

-3 -2 -1 1 2 3 x

-3

-2

-1

1

2

3y


14. Easy solutions to see are y = 0 and y = 3.

15. The slope of the tangent line at (x, y) is y, so the dierential equation is y = x2 + y2.

16. The rate at which the slope changes is dy/dx = y, so the dierential equation is y = y or y + y = 0.17. (a) The domain is all real numbers.

(b) Since y = 2/3x1/3, the solution y = x2/3 is undened at x = 0. This function is a solution of the dierentialequation on (, 0) and also on (0,).

18. (a) Dierentiating y2 2y = x2 x+ c we obtain 2yy 2y = 2x 1 or (2y 2)y = 2x 1.(b) Setting x = 0 and y = 1 in the solution we have 1 2 = 0 0 + c or c = 1. Thus, a solution of the

initial-value problem is y2 2y = x2 x 1.(c) Using the quadratic formula to solve y2 2y (x2 x 1) = 0 we get y = (2 4 + 4(x2 x 1) )/2

= 1x2 x = 1x(x 1) . Since x(x1) 0 for x 0 or x 1, we see that neither y = 1+x(x 1)nor y = 1 x(x 1) is dierentiable at x = 0. Thus, both functions are solutions of the dierentialequation, but neither is a solution of the initial-value problem.

19. (a)

y = x2 + c1 y = x2 + c2(b) When y = x2 + c1, y = 2x and (y)2 = 4x2. When y = x2 + c2, y = 2x and (y)2 = 4x2.

(c) Pasting together x2, x 0, and x2, x 0, we get y ={x2, x 0x2, x > 0

.

20. The slope of the tangent line is y(1,4)= 6

4 + 5(1)3 = 7.

21. Dierentiating y = sin(lnx) we obtain y = cos(lnx)/x and y = [sin(lnx) + cos(lnx)]/x2. Then

x2y + xy + y = x2( sin(lnx) + cos(lnx)

x2

)+ x

cos(lnx)x

+ sin(lnx) = 0.

22. Dierentiating y = cos(lnx) ln(cos(lnx)) + (lnx) sin(lnx) we obtain

y = cos(lnx)1

cos(lnx)

( sin(lnx)

x

)+ ln(cos(lnx))

( sin(lnx)

x

)+ lnx

cos(lnx)x

+sin(lnx)

x

= ln(cos(lnx)) sin(lnx)x

+(lnx) cos(lnx)

xand

y = x[ln(cos(lnx))

cos(lnx)x

+ sin(lnx)1

cos(lnx)

( sin(lnx)

x

)] 1x2

+ ln(cos(lnx)) sin(lnx)1x2

+ x[(lnx)

( sin(lnx)

x

)+

cos(lnx)x

]1x2 (lnx) cos(lnx) 1

x2

=1x2

[ ln(cos(lnx) cos(lnx) + sin

2(lnx)cos(lnx)

+ ln(cos(lnx) sin(lnx)

(lnx) sin(lnx) + cos(lnx) (lnx) cos(lnx)].

15


Then

x2y + xy + y = ln(cos(lnx)) cos(lnx) + sin2(lnx)

cos(lnx)+ ln(cos(lnx)) sin(lnx) (lnx) sin(lnx)

+ cos(lnx) (lnx) cos(lnx) ln(cos(lnx)) sin(lnx)+ (lnx) cos(lnx) + cos(lnx) ln(cos(lnx)) + (lnx) sin(lnx)

=sin2(lnx)cos(lnx)

+ cos(lnx) =sin2(lnx) + cos2(lnx)

cos(lnx)=

1cos(lnx)

= sec(lnx).

(This problem is easily done using Mathematica.)

23. From the graph we see that estimates for y0 and y1 are y0 = 3 and y1 = 0.24. The dierential equation is

dh

dt= cA0

Aw

2gh .

Using A0 = (1/24)2 = /576, Aw = (2)2 = 4, and g = 32, this becomes

dh

dt= c/576

4

64h =

c

288

h .

25. From Newtons second law we obtain

mdv

dt=

12mg

3

2mg or

dv

dt= 16

(1

3

).

16

ytt

y

x

y

tt

y

x

x

y y

tt

y

x

x

y y

x

y

x x

y

2 First-Order Differential Equations

Exercises 2.1

1. 2.

3. 4.

5. 6.

7. 8.

17

yx x

y

y

x x

y

y

x x

y

-1 0 1

1 2 x

5

4

3

2

1

y

1 2-1-2 x

1

y

1 2-1-2 x

-1

y

1 2 x

-5

-4

-3

-2

-1

y

Exercises 2.1

9. 10.

11. 12.

13. 14.

15. Writing the dierential equation in the form dy/dx = y(1 y)(1 + y) we see that critical points are located aty = 1, y = 0, and y = 1. The phase portrait is shown below.

(a) (b)

(c) (d)

18

-1 0 1

1 2 x

5

4

3

2

1

y

1 2-1-2 x

1

y

1 2-1-2 x

-1

y

-1-2 x

-5

-4

-3

-2

-1

y

0 3

10

2

-2 5

-2 0 2

0 2 4

Exercises 2.1

16. Writing the dierential equation in the form dy/dx = y2(1 y)(1 + y) we see that critical points are located aty = 1, y = 0, and y = 1. The phase portrait is shown below.

(a) (b)

(c) (d)

17. Solving y2 3y = y(y 3) = 0 we obtain the critical points 0 and 3.

From the phase portrait we see that 0 is asymptotically stable and 3 is unstable.

18. Solving y2 y3 = y2(1 y) = 0 we obtain the critical points 0 and 1.

From the phase portrait we see that 1 is asymptotically stable and 0 is semi-stable.

19. Solving (y 2)2 = 0 we obtain the critical point 2.

From the phase portrait we see that 2 is semi-stable.

20. Solving 10 + 3y y2 = (5 y)(2 + y) = 0 we obtain the critical points 2 and 5.

From the phase portrait we see that 5 is asymptotically stable and 2 is unstable.21. Solving y2(4 y2) = y2(2 y)(2 + y) = 0 we obtain the critical points 2, 0, and 2.

From the phase portrait we see that 2 is asymptotically stable, 0 is semi-stable, and 2 is unstable.22. Solving y(2 y)(4 y) = 0 we obtain the critical points 0, 2, and 4.

From the phase portrait we see that 2 is asymptotically stable and 0 and 4 are unstable.

19

-2 -1 0

0 ln 9

mg/k

mg/k

Exercises 2.1

23. Solving y ln(y + 2) = 0 we obtain the critical points 1 and 0.

From the phase portrait we see that 1 is asymptotically stable and 0 is unstable.24. Solving yey 9y = y(ey 9) = 0 we obtain the critical points 0 and ln 9.

From the phase portrait we see that 0 is asymptotically stable and ln 9 is unstable.

25. (a) Writing the dierential equation in the form

dv

dt=

k

m

(mgk v

)we see that a critical point is mg/k.

From the phase portrait we see that mg/k is an asymptotically stable critical point. Thus,

limt v = mg/k.

(b) Writing the dierential equation in the form

dv

dt=

k

m

(mgk v2

)=

k

m

(mg

k v

) (mg

k+ v

)

we see that a critical point ismg/k.

From the phase portrait we see thatmg/k is an asymptotically stable critical point. Thus,

limt v =mg/k.

26. (a) From the phase portrait we see that critical points are and . Let X(0) = X0.

If X0 < , we see that X as t. If < X0 < , we see that X as t.If X0 > , we see that X(t) increases in an unbounded manner, but more specic behavior of X(t) ast is not known.

(b) When = the phase portrait is as shown.

If X0 < , then X(t) as t . If X0 > , then X(t) increases in an unbounded manner. Thiscould happen in a nite amount of time. That is, the phase portrait does not indicate that X becomesunbounded as t.

20

tX

2/

/2 t

X

1/

2

x

t

y

x

c0 c

> < >

3- 3

3

- 3

y

x

-2.2 0.5 1.7

> < > , X(t) increases without bound up to t = 1/. For t > 1/, X(t) increases but X ast.

27. Critical points are y = 0 and y = c.

28. Critical points are y = 2.2, y = 0.5, and y = 1.7.

29. At each point on the circle of radius c the lineal element has slope c2.

21

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

x

y

x

y

-5 5

-5

5

Exercises 2.1

30. When y < 12x2, y = x2 2y is positive and the portions of solution curves

inside the nullcline parabola are increasing. When y > 12x2, y = x2 2y is

negative and the portions of the solution curves outside the nullcline parabolaare decreasing.

31. Recall that for dy/dx = f(y) we are assuming that f and f are continuous functions of y on some intervalI. Now suppose that the graph of a nonconstant solution of the dierential equation crosses the line y = c.If the point of intersection is taken as an initial condition we have two distinct solutions of the initial-valueproblem. This violates uniqueness, so the graph of any nonconstant solution must lie entirely on one side of anyequilibrium solution. Since f is continuous it can only change signs around a point where it is 0. But this is acritical point. Thus, f(y) is completely positive or completely negative in each region Ri. If y(x) is oscillatoryor has a relative extremum, then it must have a horizontal tangent line at some point (x0, y0). In this case y0would be a critical point of the dierential equation, but we saw above that the graph of a nonconstant solutioncannot intersect the graph of the equilibrium solution y = y0.

32. By Problem 31, a solution y(x) of dy/dx = f(y) cannot have relative extrema and hence must be monotone.Since y(x) = f(y) > 0, y(x) is monotone increasing, and since y(x) is bounded above by c2, limx y(x) = L,where L c2. We want to show that L = c2. Since L is a horizontal asymptote of y(x), limx y(x) = 0.Using the fact that f(y) is continuous we have

f(L) = f( limx y(x)) = limx f(y(x)) = limx y

(x) = 0.

But then L is a critical point of f . Since c1 < L c2, and f has no critical points between c1 and c2, L = c2.33. (a) Assuming the existence of the second derivative, points of inection of y(x) occur where

y(x) = 0. From dy/dx = g(y) we have d2y/dx2 = g(y) dy/dx. Thus, the y-coordinate of a point ofinection can be located by solving g(y) = 0. (Points where dy/dx = 0 correspond to constant solutionsof the dierential equation.)

(b) Solving y2 y 6 = (y 3)(y + 2) = 0 we see that 3 and 2 are criticalpoints. Now d2y/dx2 = (2y 1) dy/dx = (2y 1)(y 3)(y+ 2), so the onlypossible point of inection is at y = 12 , although the concavity of solutionscan be dierent on either side of y = 2 and y = 3. Since y(x) < 0 fory < 2 and 12 < y < 3, and y(x) > 0 for 2 < y < 12 and y > 3, wesee that solution curves are concave down for y < 2 and 12 < y < 3 andconcave up for 2 < y < 12 and y > 3. Points of inection of solutions ofautonomous dierential equations will have the same y-coordinates becausebetween critical points they are horizontal translates of each other.

22

Exercises 2.2

Exercises 2.2In many of the following problems we will encounter an expression of the form ln |g(y)| = f(x)+ c. To solve for g(y)we exponentiate both sides of the equation. This yields |g(y)| = ef(x)+c = ecef(x) which implies g(y) = ecef(x).Letting c1 = ec we obtain g(y) = c1ef(x).

1. From dy = sin 5x dx we obtain y = 15

cos 5x+ c.

2. From dy = (x+ 1)2 dx we obtain y =13(x+ 1)3 + c.

3. From dy = e3x dx we obtain y = 13e3x + c.

4. From1

(y 1)2 dy = dx we obtain 1

y 1 = x+ c or y = 11

x+ c.

5. From1ydy =

4xdx we obtain ln |y| = 4 ln |x|+ c or y = c1x4.

6. From1ydy = 2x dx we obtain ln |y| = x2 + c or y = c1ex2 .

7. From e2ydy = e3xdx we obtain 3e2y + 2e3x = c.

8. From yeydy =(ex + e3x

)dx we obtain yey ey + ex + 1

3e3x = c.

9. From(y + 2 +

1y

)dy = x2 lnx dx we obtain

y2

2+ 2y + ln |y| = x

3

3ln |x| 1

9x3 + c.

10. From1

(2y + 3)2dy =

1(4x+ 5)2

dx we obtain2

2y + 3=

14x+ 5

+ c.

11. From1

csc ydy = 1

sec2 xdx or sin y dy = cos2 x dx = 1

2(1 + cos 2x) dx we obtain

cos y = 12x 1

4sin 2x+ c or 4 cos y = 2x+ sin 2x+ c1.

12. From 2y dy = sin 3xcos3 3x

dx = tan 3x sec2 3x dx we obtain y2 = 16

sec2 3x+ c.

13. Fromey

(ey + 1)2dy =

ex(ex + 1)3

dx we obtain (ey + 1)1 = 12

(ex + 1)2 + c.

14. Fromy

(1 + y2)1/2dy =

x

(1 + x2)1/2dx we obtain

(1 + y2

)1/2=

(1 + x2

)1/2+ c.

15. From1SdS = k dr we obtain S = cekr.

16. From1

Q 70 dQ = k dt we obtain ln |Q 70| = kt+ c or Q 70 = c1ekt.

17. From1

P P 2 dP =(

1P

+1

1 P)dP = dt we obtain ln |P | ln |1 P | = t + c so that ln P

1 P = t + c orP

1 P = c1et. Solving for P we have P =

c1et

1 + c1et.

18. From1NdN =

(tet+2 1) dt we obtain ln |N | = tet+2 et+2 t+ c.

19. Fromy 2y + 3

dy =x 1x+ 4

dx or(

1 5y + 3

)dy =

(1 5

x+ 4

)dx we obtain

y 5 ln |y + 3| = x 5 ln |x+ 4|+ c or(x+ 4y + 3

)5= c1exy.

23

Exercises 2.2

20. Fromy + 1y 1 dy =

x+ 2x 3 dx or

(1 +

2y 1

)dy =

(1 +

5x 3

)dx we obtain

y + 2 ln |y 1| = x+ 5 ln |x 3|+ c or (y 1)2

(x 3)5 = c1exy.

21. From x dx =1

1 y2 dy we obtain12x2 = sin1 y + c or y = sin

(x2

2+ c1

).

22. From1y2

dy =1

ex + exdx =

ex

(ex)2 + 1dx we obtain 1

y= tan1 ex + c or y = 1

tan1 ex + c.

23. From1

x2 + 1dx = 4 dt we obtain tan1 x = 4t + c. Using x(/4) = 1 we nd c = 3/4. The solution of the

initial-value problem is tan1 x = 4t 34

or x = tan(

4t 34

).

24. From1

y2 1 dy =1

x2 1 dx or12

(1

y 1 1

y + 1

)dy =

12

(1

x 1 1

x+ 1

)dx we obtain

ln |y 1| ln |y + 1| = ln |x 1| ln |x + 1| + ln c or y 1y + 1

=c(x 1)x+ 1

. Using y(2) = 2 we nd

c = 1. The solution of the initial-value problem isy 1y + 1

=x 1x+ 1

or y = x.

25. From1ydy =

1 xx2

dx =(

1x2 1x

)dx we obtain ln |y| = 1

x ln |x| = c or xy = c1e1/x. Using y(1) = 1

we nd c1 = e1. The solution of the initial-value problem is xy = e11/x.

26. From1

1 2y dy = dt we obtain 12

ln |1 2y| = t + c or 1 2y = c1e2t. Using y(0) = 5/2 we nd c1 = 4.

The solution of the initial-value problem is 1 2y = 4e2t or y = 2e2t + 12

.

27. Separating variables and integrating we obtain

dx1 x2

dy1 y2 = 0 and sin

1 x sin1 y = c.

Setting x = 0 and y =

3/2 we obtain c = /3. Thus, an implicit solution of the initial-value problem issin1 x sin1 y = /3. Solving for y and using a trigonometric identity we get

y = sin(sin1 x+

3

)= x cos

3+

1 x2 sin

3=x

2+

3

1 x22

.

28. From1

1 + (2y)2dy =

x1 + (x2)2

dx we obtain

12

tan1 2y = 12

tan1 x2 + c or tan1 2y + tan1 x2 = c1.

Using y(1) = 0 we nd c1 = /4. The solution of the initial-value problem is

tan1 2y + tan1 x2 =

4.

29. (a) The equilibrium solutions y(x) = 2 and y(x) = 2 satisfy the initial conditions y(0) = 2 and y(0) = 2,respectively. Setting x = 14 and y = 1 in y = 2(1 + ce

4x)/(1 ce4x) we obtain

1 = 21 + ce1 ce , 1 ce = 2 + 2ce, 1 = 3ce, and c =

13e

.

24

Exercises 2.2

The solution of the corresponding initial-value problem is

y = 21 13e4x11 + 13e

4x1 = 23 e4x13 + e4x1

.

(b) Separating variables and integrating yields14

ln |y 2| 14

ln |y + 2|+ ln c1 = xln |y 2| ln |y + 2|+ ln c = 4x

ln c(y 2)

y + 2

= 4xcy 2y + 2

= e4x .

Solving for y we get y = 2(c + e4x)/(c e4x). The initial condition y(0) = 2 can be solved for, yieldingc = 0 and y(x) = 2. The initial condition y(0) = 2 does not correspond to a value of c, and it mustsimply be recognized that y(x) = 2 is a solution of the initial-value problem. Setting x = 14 and y = 1 iny = 2(c+ e4x)/(c e4x) leads to c = 3e. Thus, a solution of the initial-value problem is

y = 23e+ e4x3e e4x = 2

3 e4x13 + e4x1

.

30. From(

1y 1 +

1y

)dy =

1xdx we obtain ln |y 1| ln |y| = ln |x| + c or y = 1

1 c1x . Another solution isy = 0.

(a) If y(0) = 1 then y = 1.

(b) If y(0) = 0 then y = 0.

(c) If y(1/2) = 1/2 then y =1

1 + 2x.

(d) Setting x = 2 and y = 14 we obtain

14

=1

1 c1(2) , 1 2c1 = 4, and c1 = 32.

Thus, y =1

1 + 32x=

22 + 3x

.

31. Singular solutions of dy/dx = x

1 y2 are y = 1 and y = 1. A singular solution of (ex + ex)dy/dx = y2 isy = 0.

32. Dierentiating ln(x2 + 10) + csc y = c we get

2xx2 + 10

cot y csc y dydx

= 0,

2xx2 + 10

cos ysin y

1sin y

dy

dx= 0,

or

2x sin2 y dx (x2 + 10) cos y dy = 0.Writing the dierential equation in the form

dy

dx=

2x sin2 y(x2 + 10) cos y

we see that singular solutions occur when sin2 y = 0, or y = k, where k is an integer.

25

-0.004-0.002 0.002 0.004 x

0.97

0.98

1

1.01y

-0.004-0.002 0.002 0.004 x

0.98

0.99

1.01

1.02y

-0.004-0.002 0.002 0.004 x

0.9996

0.9998

1.0002

1.0004

y

-0.004-0.002 0.002 0.004 x

0.9996

0.9998

1.0002

1.0004

y

Exercises 2.2

33. The singular solution y = 1 satises the initial-value problem.

34. Separating variables we obtaindy

(y 1)2 = dx. Then

1y 1 = x+ c and y =

x+ c 1x+ c

.

Setting x = 0 and y = 1.01 we obtain c = 100. The solution is

y =x 101x 100 .


(y 1)2 + 0.01 = dx. Then

10 tan1 10(y 1) = x+ c and y = 1 + 110

tanx+ c10

.

Setting x = 0 and y = 1 we obtain c = 0. The solution is

y = 1 +110

tanx

10.


(y 1)2 0.01 = dx. Then

5 ln10y 1110y 9

= x+ c.Setting x = 0 and y = 1 we obtain c = 5 ln 1 = 0. The solution is

5 ln10y 1110y 9

= x.

37. Separating variables, we have

dy

y y3 =dy

y(1 y)(1 + y) =(1y

+1/2

1 y 1/2

1 + y

)dy = dx.

Integrating, we get

ln |y| 12

ln |1 y| 12

ln |1 + y| = x+ c.

26

1 2 3 4 5 x

-4

-2

2

4

y

-4 -2 2 4 x

-4

-2

2

4

y

-4 -2 2 4 x

-4

-2

2

4

y

1 2 3 4 5 x

-4

-2

2

4

y

-4 -2 2 4 x

-2

2

4

6

8y

3< >

Exercises 2.2

When y > 1, this becomes

ln y 12

ln(y 1) 12

ln(y + 1) = lnyy2 1 = x+ c.

Letting x = 0 and y = 2 we nd c = ln(2/

3 ). Solving for y we get y1(x) = 2ex/

4e2x 3 , where x > ln(3/2).When 0 < y < 1 we have

ln y 12

ln(1 y) 12

ln(1 + y) = lny

1 y2 = x+ c.

Letting x = 0 and y = 12 we nd c = ln(1/

3 ). Solving for y we get y2(x) = ex/e2x + 3 , where < x 0 or y > 3, and concavedown when d2y/dx2 < 0 or y < 3. From the phase portrait we see that thesolution curve is decreasing when y < 3 and increasing when y > 3.

27

-1 1 2 3 4 5 x

-2

2

4

6

8y

-5 -4 -3 -2 -1 1 2 x

-5

-4

-3

-2

-1

1

2y

-2 -1.5 -1 -0.5 x

-2

-1

1

2y

Exercises 2.2

(b) Separating variables and integrating we obtain

(y 3) dy = dx12y2 3y = x+ c

y2 6y + 9 = 2x+ c1(y 3)2 = 2x+ c1

y = 32x+ c1 .The initial condition dictates whether to use the plus or minus sign.When y1(0) = 4 we have c1 = 1 and y1(x) = 3 +

2x+ 1 .

When y2(0) = 2 we have c1 = 1 and y2(x) = 3

2x+ 1 .When y3(1) = 2 we have c1 = 1 and y3(x) = 3

2x 1 .

When y4(1) = 4 we have c1 = 3 and y4(x) = 3 +

2x+ 3 .

39. (a) Separating variables we have 2y dy = (2x+ 1)dx. Integrating gives y2 = x2 + x+ c. When y(2) = 1 wend c = 1, so y2 = x2 + x 1 and y = x2 + x 1 . The negative square root is chosen because of theinitial condition.

(b) The interval of denition appears to be approximately (,1.65).

(c) Solving x2 + x 1 = 0 we get x = 12 12

5 , so the exact interval of denition is (, 12 12

5 ).

40. (a) From Problem 7 the general solution is 3e2y+2e3x = c. When y(0) = 0 we nd c = 5, so 3e2y+2e3x = 5.Solving for y we get y = 12 ln 13 (5 2e3x).

(b) The interval of denition appears to be approximately (, 0.3).

(c) Solving 13 (5 2e3x) = 0 we get x = 13 ln(52 ), so the exact interval of denition is (, 13 ln(52 )).41. (a) While y2(x) =

25 x2 is dened at x = 5 and x = 5, y1(x) is not dened at these values, and so the

interval of denition is the open interval (5, 5).(b) At any point on the x-axis the derivative of y(x) is undened, so no solution curve can cross the x-axis.

Since x/y is not dened when y = 0, the initial-value problem has no solution.42. (a) Separating variables and integrating we obtain x2 y2 = c. For c = 0 the graph is a square hyperbola

centered at the origin. All four initial conditions imply c = 0 and y = x. Since the dierential equation is

28

-6 -4 -2 2 4 6 8 x0.51

1.52

2.53

3.5y

x

y

3 3

3

3

Exercises 2.2

not dened for y = 0, solutions are y = x, x < 0 and y = x, x > 0. The solution for y(a) = a is y = x,x > 0; for y(a) = a is y = x; for y(a) = a is y = x, x < 0; and for y(a) = a is y = x, x < 0.

(b) Since x/y is not dened when y = 0, the initial-value problem has no solution.

(c) Setting x = 1 and y = 2 in x2 y2 = c we get c = 3, so y2 = x2 + 3 and y(x) = x2 + 3 , wherethe positive square root is chosen because of the initial condition. The domain is all real numbers sincex2 + 3 > 0 for all x.

43. Separating variables we have dy/(

1 + y2 sin2 y)

= dx which isnot readily integrated (even by a CAS). We note that dy/dx 0for all values of x and y and that dy/dx = 0 when y = 0 and y = ,which are equilibrium solutions.

44. Separating variables we have dy/(y + y) = dx/(

x+ x). To integrate

dt/(

t+ t) we substitute u2 = t and

get 2u

u+ u2du =

2

1 + udu = 2 ln |1 + u|+ c = 2 ln(1 +x ) + c.

Integrating the separated dierential equation we have

2 ln(1 +y ) = 2 ln(1 +

x ) + c or ln(1 +

y ) = ln(1 +

x ) + ln c1.

Solving for y we get y = [c1(1 +x ) 1]2.

45. We are looking for a function y(x) such that

y2 +(dy

dx

)2= 1.

Using the positive square root gives

dy

dx=

1 y2 = dy

1 y2 = dx = sin1 y = x+ c.

Thus a solution is y = sin(x+ c). If we use the negative square root we obtain

y = sin(c x) = sin(x c) = sin(x+ c1).Note also that y = 1 and y = 1 are solutions.

46. (a)

(b) For |x| > 1 and |y| > 1 the dierential equation is dy/dx =y2 1 /x2 1 . Separating variables and

integrating, we obtain

dyy2 1 =

dxx2 1 and cosh

1 y = cosh1 x+ c.

29

-4 -2 0 2 4

-4

-2

0

2

4

x

y

-4 -2 0 2 4

-4

-2

0

2

4

x

y

c=-2

c=10

c=67

c=-31

-6 -4 -2 0 2 4 6

-4

-2

0

2

4

x

y

-2 0 2 4 6

-4

-2

0

2

4

x

y

-4 -2 0 2 4 6 8 10-8

-6

-4

-2

0

2

4

x

y

Exercises 2.2

Setting x = 2 and y = 2 we nd c = cosh1 2 cosh1 2 = 0 and cosh1 y = cosh1 x. An explicit solutionis y = x.

47. (a) Separating variables and integrating, we have

(3y2 + 1)dy = (8x+ 5)dx and y3 + y = 4x2 5x+ c.Using a CAS we show various contours of f(x, y) = y3 + y + 4x2 + 5x. The plotsshown on [5, 5] [5, 5] correspond to c-values of 0, 5, 20, 40, 80, and125.

(b) The value of c corresponding to y(0) = 1 is f(0,1) = 2; to y(0) = 2 isf(0, 2) = 10; to y(1) = 4 is f(1, 4) = 67; and to y(1) = 3 is 31.


(2y + y2)dy = (x x2)dxand

y2 + 13y3 =

12x2 1

3x3 + c.

Using a CAS we show some contours of f(x, y) = 2y36y2+2x33x2.The plots shown on [7, 7] [5, 5] correspond to c-values of 450,300, 200, 120, 60, 20, 10, 8.1, 5, 0.8, 20, 60, and 120.

(b) The value of c corresponding to y(0) = 32 is f(0, 32

)= 274 . The

portion of the graph between the dots corresponds to the solutioncurve satisfying the intial condition. To determine the interval ofdenition we nd dy/dx for 2y3 6y2 + 2x3 3x2 = 274 . Usingimplicit dierentiation we get y = (xx2)/(y22y), which is innitewhen y = 0 and y = 2. Letting y = 0 in 2y36y2 +2x33x2 = 274and using a CAS to solve for x we get x = 1.13232. Similarly,letting y = 2, we nd x = 1.71299. The largest interval of denitionis approximately (1.13232, 1.71299).

(c) The value of c corresponding to y(0) = 2 is f(0,2) = 40. Theportion of the graph to the right of the dot corresponds to the solu-tion curve satisfying the initial condition. To determine the intervalof denition we nd dy/dx for 2y3 6y2 + 2x3 3x2 = 40. Usingimplicit dierentiation we get y = (xx2)/(y22y), which is innitewhen y = 0 and y = 2. Letting y = 0 in 2y36y2 +2x33x2 = 40and using a CAS to solve for x we get x = 2.29551. The largestinterval of denition is approximately (2.29551,).

30

Exercises 2.3

Exercises 2.3

1. For y 5y = 0 an integrating factor is e

5 dx = e5x so thatd

dx

[e5xy

]= 0 and y = ce5x for < x

Exercises 2.3

16. Fordx

dy+

2yx = ey an integrating factor is e

(2/y)dy = y2 so that

d

dy

[y2x

]= y2ey and x = ey 2

yey +

2y2

+c

y2

for 0 < y

xy

5

1

x

y

5

1

-1

x

y

3

2

x

y

5

-1

1

Exercises 2.3

30. For y + (tanx)y = cos2 x an integrating factor is e

tan x dx = secx so thatd

dx[(secx) y] = cosx and y =

sinx cosx+ c cosx for /2 < x < /2. If y(0) = 1 then c = 1 and y = sinx cosx cosx.31. For y + 2y = f(x) an integrating factor is e2x so that

ye2x ={ 1

2e2x + c1, 0 x 3;

c2, x > 3.

If y(0) = 0 then c1 = 1/2 and for continuity we must have c2 = 12e6 12so that

y =

{12 (1 e2x), 0 x 3;12 (e

6 1)e2x, x > 3.

32. For y + y = f(x) an integrating factor is ex so that

yex ={ex + c1, 0 x 1;ex + c2, x > 1.

If y(0) = 1 then c1 = 0 and for continuity we must have c2 = 2e so that

y ={

1, 0 x 1;2e1x 1, x > 1.

33. For y + 2xy = f(x) an integrating factor is ex2

so that

yex2

=

{12e

x2 + c1, 0 x 1;c2, x > 1.

If y(0) = 2 then c1 = 3/2 and for continuity we must have c2 = 12e +32

so that

y =

{12 +

32ex2 , 0 x 1;(

12e+

32

)ex

2, x > 1.

34. For y +2x

1 + x2y =

x

1 + x2, 0 x 1;

x1 + x2

, x > 1

an integrating factor is 1 + x2 so that

(1 + x2

)y =

{ 12x

2 + c1, 0 x 1; 12x2 + c2, x > 1.

If y(0) = 0 then c1 = 0 and for continuity we must have c2 = 1 so that

y =

12 1

2 (1 + x2), 0 x 1;

32 (1 + x2)

12, x > 1.

33

xy

3

5

10

15

20

0 1 2 3 4 5x

0.2

0.4

0.6

0.8

1y

Exercises 2.3

35. We need P (x)dx =

{2x, 0 x 12 lnx, x > 1 .

An integrating factor is

eP (x)dx =

{e2x, 0 x 11/x2, x > 1

andd

dx

[{ye2x, 0 x 1y/x2, x > 1

]=

{4xe2x, 0 x 14/x, x > 1

.

Integrating we get {ye2x, 0 x 1y/x2, x > 1

={

2xe2x e2x + c1, 0 x 14 lnx+ c2, x > 1

.

Using y(0) = 3 we nd c1 = 4. For continuity we must have c2 = 2 1 + 4e2 = 1 + 4e2. Then

y ={

2x 1 + 4e2x, 0 x 14x2 lnx+ (1 + 4e2)x2, x > 1

.

36. (a) An integrating factor for y 2xy = 1 is ex2 . Thusd

dx[ex

2y] = ex2

ex2y =

x0

et2dt =

2erf(x) + c.

From y(0) =/2, and noting that erf(0) = 0, we get c =

/2. Thus

y = ex2(

2erf(x) +

2

)=

2ex

2(1 erf(x)) =

2ex

2erfc(x).

(b) Using Mathematica we nd y(2) 0.226339.

37. An integrating factor for y 2xy = 1 is ex2 . Thusd

dx[ex

2y ] = ex

2

ex2y =

x0

et2dt = erf(x) + c

and

y = ex2erf(x) + cex

2.

From y(1) = 1 we get 1 = e erf(1) + ce, so that c = e1 erf(1). Thusy = ex

2erf(x) + (e1 erf(1))ex2 = ex21 + ex2(erf(x) erf(1)).

38. (a) An integrating factor for

y +2xy =

10 sinxx3

34

1 2 3 4 5x

-5-4-3-2-1

12y

2 4 6 8 x

2

4

6

8

10

12

14

y

Exercises 2.3

is x2. Thusd

dx[x2y] = 10

sinxx

x2y = 10 x

0

sin tt

dt+ c

y = 10x2Si(x) + cx2.

From y(1) = 0 we get c = 10Si(1). Thusy = 10x2Si(x) 10x2Si(1) = 10x2(Si(x) Si(1)).

(b) Using Mathematica we nd y(2) 1.64832.


dy

y= sinx2 dx and ln |y| =

x0

sin t2 dt+ c.

Now, letting t =/2u we have

x0

sin t2 dt =

2

2/ x0

sin(

2u2

)du,

so

y = c1e x

0sin t2 dt = c1e

/2

2/ x0

sin(u2/2) du = c1e/2S(

2/ x).

Using S(0) = 0 and y(0) = 5 we see that c1 = 5 and y = 5e/2S(

2/ x).

(b) Using Mathematica we nd y(2) 11.181. From the graph wesee that as x, y(x) oscillates with decreasing amplitudeapproaching 9.35672.

40. For y + exy = 1 an integrating factor is eex

. Thus

d

dx

[ee

x

y]

= eex

and eex

y = x

0

eet

dt+ c.

From y(0) = 1 we get c = e, so y = eex x

0ee

t

dt+ e1ex

.

When y + exy = 0 we can separate variables and integrate:

dy

y= ex dx and ln |y| = ex + c.

Thus y = c1eex

. From y(0) = 1 we get c1 = e, so y = e1ex

.

When y + exy = ex we can see by inspection that y = 1 is a solution.

35

Exercises 2.3

41. On the interval (3, 3) the integrating factor is

ex dx/(x29) = e

x dx/(9x2) = e

12 ln(9x2) =

9 x2

and sod

dx

[9 x2 y

]= 0 and y =

c9 x2 .

42. We want 4 to be a critical point, so use y = 4 y.43. (a) All solutions of the form y = x5ex x4ex + cx4 satisfy the initial condition. In this case, since 4/x is

discontinuous at x = 0, the hypotheses of Theorem 1.1 are not satised and the initial-value problem doesnot have a unique solution.

(b) The dierential equation has no solution satisfying y(0) = y0, y0 = 0.(c) In this case, since x0 = 0, Theorem 1.1 applies and the initial-value problem has a unique solution given by

y = x5ex x4ex + cx4 where c = y0/x40 x0ex0 + ex0 .44. We want the general solution to be y = 3x 5 + cex. (Rather than ex, any function that approaches 0 as

x could be used.) Dierentiating we get

y = 3 cex = 3 (y 3x+ 5) = y + 3x 2,

so the dierential equation y + y = 3x 2 has solutions asymptotic to the line y = 3x 5.45. The left-hand derivative of the function at x = 1 is 1/e and the right-hand derivative at x = 1 is 1 1/e. Thus,

y is not dierentiable at x = 1.

46. Dierentiating yc = c/x3 we get

yc = 3cx4

= 3x

c

x3= 3

xyc

so a dierential equation with general solution yc = c/x3 is xy + 3y = 0. Now

xyp + 3yp = x(3x2) + 3(x3) = 6x3

so a dierential equation with general solution y = c/x3 + x3 is xy + 3y = 6x3.

47. Since eP (x)dx+c = ece

P (x)dx = c1e

P (x)dx, we would have

c1eP (x)dxy = c2 +

c1e

P (x)dxf(x) dx and e

P (x)dxy = c3 +

eP (x)dxf(x) dx,

which is the same as (6) in the text.

48. We see by inspection that y = 0 is a solution.

36

Exercises 2.4

Exercises 2.4

1. Let M = 2x 1 and N = 3y + 7 so that My = 0 = Nx. From fx = 2x 1 we obtain f = x2 x + h(y),h(y) = 3y + 7, and h(y) =

32y2 + 7y. The solution is x2 x+ 32y2 + 7y = c.

2. Let M = 2x+ y and N = x 6y. Then My = 1 and Nx = 1, so the equation is not exact.3. Let M = 5x+ 4y and N = 4x 8y3 so that My = 4 = Nx. From fx = 5x+ 4y we obtain f = 52x2 + 4xy+h(y),

h(y) = 8y3, and h(y) = 2y4. The solution is 52x2 + 4xy 2y4 = c.4. Let M = sin y y sinx and N = cosx+ x cos y y so that My = cos y sinx = Nx. From fx = sin y y sinx

we obtain f = x sin y + y cosx+ h(y), h(y) = y, and h(y) = 12y2. The solution is x sin y + y cosx 12y2 = c.5. Let M = 2y2x3 and N = 2yx2+4 so that My = 4xy = Nx. From fx = 2y2x3 we obtain f = x2y23x+h(y),

h(y) = 4, and h(y) = 4y. The solution is x2y2 3x+ 4y = c.6. Let M = 4x33y sin 3xy/x2 and N = 2y1/x+cos 3x so that My = 3 sin 3x1/x2 and Nx = 1/x23 sin 3x.

The equation is not exact.

7. Let M = x2 y2 and N = x2 2xy so that My = 2y and Nx = 2x 2y. The equation is not exact.8. Let M = 1 + lnx + y/x and N = 1 + lnx so that My = 1/x = Nx. From fy = 1 + lnx we obtain

f = y + y lnx+ h(y), h(x) = 1 + lnx, and h(y) = x lnx. The solution is y + y lnx+ x lnx = c.9. Let M = y3y2 sinxx and N = 3xy2 +2y cosx so that My = 3y22y sinx = Nx. From fx = y3y2 sinxx

we obtain f = xy3 + y2 cosx 12x2 + h(y), h(y) = 0, and h(y) = 0. The solution is xy3 + y2 cosx 12x2 = c.10. Let M = x3 + y3 and N = 3xy2 so that My = 3y2 = Nx. From fx = x3 + y3 we obtain f = 14x

4 + xy3 + h(y),h(y) = 0, and h(y) = 0. The solution is 14x

4 + xy3 = c.

11. Let M = y ln y exy and N = 1/y + x ln y so that My = 1 + ln y + yexy and Nx = ln y. The equation is notexact.

12. Let M = 3x2y + ey and N = x3 + xey 2y so that My = 3x2 + ey = Nx. From fx = 3x2y + ey we obtainf = x3y + xey + h(y), h(y) = 2y, and h(y) = y2. The solution is x3y + xey y2 = c.

13. Let M = y 6x2 2xex and N = x so that My = 1 = Nx. From fx = y 6x2 2xex we obtain f =xy 2x3 2xex + 2ex + h(y), h(y) = 0, and h(y) = 0. The solution is xy 2x3 2xex + 2ex = c.

14. Let M = 1 3/x + y and N = 1 3/y + x so that My = 1 = Nx. From fx = 1 3/x + y we obtainf = x 3 ln |x|+ xy + h(y), h(y) = 1 3

y, and h(y) = y 3 ln |y|. The solution is x+ y + xy 3 ln |xy| = c.

15. Let M = x2y3 1/ (1 + 9x2) and N = x3y2 so that My = 3x2y2 = Nx. From fx = x2y3 1/ (1 + 9x2) weobtain f =

13x3y3 1

3arctan(3x) + h(y), h(y) = 0, and h(y) = 0. The solution is x3y3 arctan(3x) = c.

16. Let M = 2y and N = 5y2x so that My = 2 = Nx. From fx = 2y we obtain f = 2xy+h(y), h(y) = 5y,and h(y) =

52y2. The solution is 2xy + 5

2y2 = c.

17. Let M = tanx sinx sin y and N = cosx cos y so that My = sinx cos y = Nx. From fx = tanx sinx sin ywe obtain f = ln | secx|+ cosx sin y + h(y), h(y) = 0, and h(y) = 0. The solution is ln | secx|+ cosx sin y = c.

18. Let M = 2y sinx cosx y + 2y2exy2 and N = x+ sin2 x+ 4xyexy2 so thatMy = 2 sinx cosx 1 + 4xy3exy2 + 4yexy2 = Nx.

From fx = 2y sinx cosx y+ 2y2exy2 we obtain f = y sin2 x xy+ 2exy2 + h(y), h(y) = 0, and h(y) = 0. Thesolution is y sin2 x xy + 2exy2 = c.

37

Exercises 2.4

19. Let M = 4t3y 15t2 y and N = t4 + 3y2 t so that My = 4t3 1 = Nt. From ft = 4t3y 15t2 y we obtainf = t4y 5t3 ty + h(y), h(y) = 3y2, and h(y) = y3. The solution is t4y 5t3 ty + y3 = c.

20. Let M = 1/t+ 1/t2 y/ (t2 + y2) and N = yey + t/ (t2 + y2) so that My = (y2 t2) / (t2 + y2)2 = Nt. Fromft = 1/t+ 1/t2 y/

(t2 + y2

)we obtain f = ln |t| 1

t arctan

(t

y

)+ h(y), h(y) = yey, and h(y) = yey ey.

The solution is

ln |t| 1t arctan

(t

y

)+ yey ey = c.

21. Let M = x2 + 2xy+ y2 and N = 2xy+x2 1 so that My = 2(x+ y) = Nx. From fx = x2 + 2xy+ y2 we obtainf =

13x3 + x2y + xy2 + h(y), h(y) = 1, and h(y) = y. The general solution is 1

3x3 + x2y + xy2 y = c. If

y(1) = 1 then c = 4/3 and the solution of the initial-value problem is13x3 + x2y + xy2 y = 4

3.

22. Let M = ex + y and N = 2 + x+ yey so that My = 1 = Nx. From fx = ex + y we obtain f = ex + xy + h(y),h(y) = 2 + yey, and h(y) = 2y + yey y. The general solution is ex + xy + 2y + yey ey = c. If y(0) = 1 thenc = 3 and the solution of the initial-value problem is ex + xy + 2y + yey ey = 3.

23. Let M = 4y + 2t 5 and N = 6y + 4t 1 so that My = 4 = Nt. From ft = 4y + 2t 5 we obtainf = 4ty+ t2 5t+ h(y), h(y) = 6y 1, and h(y) = 3y2 y. The general solution is 4ty+ t2 5t+ 3y2 y = c.If y(1) = 2 then c = 8 and the solution of the initial-value problem is 4ty + t2 5t+ 3y2 y = 8.

24. Let M = t/2y4 and N =(3y2 t2) /y5 so that My = 2t/y5 = Nt. From ft = t/2y4 we obtain f = t24y4 +h(y),

h(y) =3y3

, and h(y) = 32y2

. The general solution ist2

4y4 3

2y2= c. If y(1) = 1 then c = 5/4 and the

solution of the initial-value problem ist2

4y4 3

2y2= 5

4.

25. Let M = y2 cosx 3x2y 2x and N = 2y sinx x3 + ln y so that My = 2y cosx 3x2 = Nx. Fromfx = y2 cosx 3x2y 2x we obtain f = y2 sinx x3y x2 + h(y), h(y) = ln y, and h(y) = y ln y y. Thegeneral solution is y2 sinxx3yx2 + y ln y y = c. If y(0) = e then c = 0 and the solution of the initial-valueproblem is y2 sinx x3y x2 + y ln y y = 0.

26. Let M = y2 + y sinx and N = 2xy cosx 1/ (1 + y2) so that My = 2y + sinx = Nx. From fx =y2 + y sinx we obtain f = xy2 y cosx + h(y), h(y) = 1

1 + y2, and h(y) = tan1 y. The general solution

is xy2 y cosx tan1 y = c. If y(0) = 1 then c = 1 /4 and the solution of the initial-value problem isxy2 y cosx tan1 y = 1

4.

27. Equating My = 3y2 + 4kxy3 and Nx = 3y2 + 40xy3 we obtain k = 10.

28. Equating My = 18xy2 sin y and Nx = 4kxy2 sin y we obtain k = 9/2.29. Let M = x2y2 sinx + 2xy2 cosx and N = 2x2y cosx so that My = 2x2y sinx + 4xy cosx = Nx. From

fy = 2x2y cosx we obtain f = x2y2 cosx + h(y), h(y) = 0, and h(y) = 0. The solution of the dierentialequation is x2y2 cosx = c.

30. Let M =(x2 + 2xy y2) / (x2 + 2xy + y2) and N = (y2 + 2xy x2) / (y2 + 2xy + x2) so that My =

4xy/(x+ y)3 = Nx. From fx =(x2 + 2xy + y2 2y2) /(x+ y)2 we obtain f = x+ 2y2

x+ y+ h(y), h(y) = 1,

and h(y) = y. The solution of the dierential equation is x2 + y2 = c(x+ y).

38

1 2 3 4 5 6 7 x0.51

1.52

2.5y

-4 -2 2 4 x

-6

-4

-2

2

4y

y1

y2

Exercises 2.4

31. We note that (My Nx)/N = 1/x, so an integrating factor is edx/x = x. Let M = 2xy2 + 3x2 and N = 2x2y

so that My = 4xy = Nx. From fx = 2xy2 + 3x2 we obtain f = x2y2 + x3 + h(y), h(y) = 0, and h(y) = 0. Thesolution of the dierential equation is x2y2 + x3 = c.

32. We note that (My Nx)/N = 1, so an integrating factor is edx = ex. Let M = xyex + y2ex + yex and

N = xex + 2yex so that My = xex + 2yex + ex = Nx. From fy = xex + 2yex we obtain f = xyex + y2ex + h(x),h(y) = 0, and h(y) = 0. The solution of the dierential equation is xyex + y2ex = c.

33. We note that (NxMy)/M = 2/y, so an integrating factor is e

2dy/y = y2. Let M = 6xy3 and N = 4y3+9x2y2

so that My = 18xy2 = Nx. From fx = 6xy3 we obtain f = 3x2y3 + h(y), h(y) = 4y3, and h(y) = y4. Thesolution of the dierential equation is 3x2y3 + y4 = c.

34. We note that (MyNx)/N = cotx, so an integrating factor is e

cot x dx = cscx. Let M = cosx cscx = cotxand N = (1 + 2/y) sinx cscx = 1 + 2/y, so that My = 0 = Nx. From fx = cotx we obtain f = ln(sinx) + h(y),h(y) = 1 + 2/y, and h(y) = y + ln y2. The solution of the dierential equation is ln(sinx) + y + ln y2 = c.

35. (a) Write the separable equation as g(x) dx+ dy/h(y). Identifying M = g(x) and N = 1/h(y), we see thatMy = 0 = Nx, so the dierential equation is exact.

(b) Separating variables and integrating we have

(sin y) dy = (cosx) dx and cos y = sinx+ c.Using y(7/6) = /2 we get c = 1/2, so the solution of the initial-value problem is cos y = sinx+ 12 .

(c) Solving for y we have y = cos1(sinx+ 12 ). Since the domain of cos1 t

is [1, 1] we see that 1 sinx + 12 1 or 32 sinx 12 , which isequivalent to sinx 12 . We also want the interval to contain 7/6, sothe interval of denition is (5/6, 13/6).

36. (a) Implicitly dierentiating x3 + 2x2y + y2 = c and solving for dy/dx we obtain

3x2 + 2x2dy

dx+ 4xy + 2y

dy

dx= 0 and

dy

dx= 3x

2 + 4xy2x2 + 2y

.

Separating variables we get (4xy + 3x2)dx+ (2y + 2x2)dy = 0.

(b) Setting x = 0 and y = 2 in x3 + 2x2y + y2 = c we nd c = 4, and setting x = y = 1 we also nd c = 4.Thus, both initial conditions determine the same implicit solution.

(c) Solving x3 + 2x2y + y2 = 4 for y we get

y1(x) = x2

4 x3 + x4 and y2(x) = x2 +

4 x3 + x4 .

37. To see that the equations are not equivalent consider dx = (x/y)dy = 0. An integrating factor is (x, y) = yresulting in y dx+ x dy = 0. A solution of the latter equation is y = 0, but this is not a solution of the originalequation.

38. The explicit solution is y =

(3 + cos2 x)/(1 x2) . Since 3 + cos2 x > 0 for all x we must have 1 x2 > 0 or1 < x < 1. Thus, the interval of denition is (1, 1).

39

Exercises 2.4

39. (a) Since fy = N(x, y) = xexy+2xy+1/x we obtain f = exy+xy2+y

x+h(x) so that fx = yexy+y2 y

x2+h(x).

Let M(x, y) = yexy + y2 yx2

.

(b) Since fx = M(x, y) = y1/2x1/2 + x(x2 + y

)1 we obtain f = 2y1/2x1/2 + 12

lnx2 + y + g(y) so that

fy = y1/2x1/2 +12

(x2 + y

)1+ g(x). Let N(x, y) = y1/2x1/2 +

12

(x2 + y

)1.

40. First note thatd(

x2 + y2)

=x

x2 + y2dx+

yx2 + y2

dy.

Then x dx+ y dy =x2 + y2 dx becomes

xx2 + y2

dx+y

x2 + y2dy = d

(x2 + y2

)= dx.

The left side is the total dierential ofx2 + y2 and the right side is the total dierential of x + c. Thus

x2 + y2 = x+ c is a solution of the dierential equation.

Exercises 2.5

1. Letting y = ux we have(x ux) dx+ x(u dx+ x du) = 0

dx+ x du = 0

dx

x+ du = 0

ln |x|+ u = cx ln |x|+ y = cx.

2. Letting y = ux we have(x+ ux) dx+ x(u dx+ x du) = 0

(1 + 2u) dx+ x du = 0

dx

x+

du

1 + 2u= 0

ln |x|+ 12

ln |1 + 2u| = c

x2(1 + 2

y

x

)= c1

x2 + 2xy = c1.

3. Letting x = vy we havevy(v dy + y dv) + (y 2vy) dy = 0

vy dv +(v2 2v + 1) dy = 0

v dv

(v 1)2 +dy

y= 0

ln |v 1| 1v 1 + ln |y| = c

lnxy 1

1x/y 1 + ln y = c(x y) ln |x y| y = c(x y).

40

Exercises 2.5

4. Letting x = vy we havey(v dy + y dv) 2(vy + y) dy = 0

y dv (v + 2) dy = 0dv

v + 2 dy

y= 0

ln |v + 2| ln |y| = c

lnxy + 2

ln |y| = cx+ 2y = c1y2.

5. Letting y = ux we have (u2x2 + ux2

)dx x2(u dx+ x du) = 0

u2 dx x du = 0dx

x duu2

= 0

ln |x|+ 1u

= c

ln |x|+ xy

= c

y ln |x|+ x = cy.6. Letting y = ux we have (

u2x2 + ux2)dx+ x2(u dx+ x du) = 0(u2 + 2u

)dx+ x du = 0

dx

x+

du

u(u+ 2)= 0

ln |x|+ 12

ln |u| 12

ln |u+ 2| = cx2u

u+ 2= c1

x2y

x= c1

(yx

+ 2)

x2y = c1(y + 2x).

7. Letting y = ux we have(ux x) dx (ux+ x)(u dx+ x du) = 0(

u2 + 1)dx+ x(u+ 1) du = 0

dx

x+

u+ 1u2 + 1

du = 0

ln |x|+ 12

ln(u2 + 1

)+ tan1 u = c

lnx2(y2

x2+ 1

)+ 2 tan1

y

x= c1

ln(x2 + y2

)+ 2 tan1

y

x= c1.

41

Exercises 2.5

8. Letting y = ux we have

(x+ 3ux) dx (3x+ ux)(u dx+ x du) = 0(u2 1) dx+ x(u+ 3) du = 0dx

x+

u+ 3(u 1)(u+ 1) du = 0

ln |x|+ 2 ln |u 1| ln |u+ 1| = cx(u 1)2u+ 1

= c1

x(yx 1

)2= c1

(yx

+ 1)

(y x)2 = c1(y + x).

9. Letting y = ux we have

ux dx+ (x+ux)(u dx+ x du) = 0(x+ x

u ) du+ u3/2 dx = 0(

u3/2 +1u

)du+

dx

x= 0

2u1/2 + ln |u|+ ln |x| = cln |y/x|+ ln |x| = 2

x/y + c

y(ln |y| c)2 = 4x.

10. Letting y = ux we have (ux+

x2 + u2x2

)dx x(u dx+ x du) = 0

x

1 + u2 dx x2 du = 0dx

x du

1 + u2= 0

ln |x| lnu+ 1 + u2 = cu+

1 + u2 = c1x

y +y2 + x2 = c1x2.

11. Letting y = ux we have (x3 u3x3) dx+ u2x3(u dx+ x du) = 0

dx+ u2x du = 0

dx

x+ u2 du = 0

ln |x|+ 13u3 = c

3x3 ln |x|+ y3 = c1x3.

Using y(1) = 2 we nd c1 = 8. The solution of the initial-value problem is 3x3 ln |x|+ y3 = 8x3.

42

Exercises 2.5

12. Letting y = ux we have (x2 + 2u2x2

)dx ux2(u dx+ x du) = 0(

1 + u2)dx ux du = 0

dx

x u du

1 + u2= 0

ln |x| 12

ln(1 + u2

)= c

x2

1 + u2= c1

x4 = c1(y2 + x2

).

Using y(1) = 1 we nd c1 = 1/2. The solution of the initial-value problem is 2x4 = y2 + x2.13. Letting y = ux we have

(x+ uxeu) dx xeu(u dx+ x du) = 0dx xeu du = 0dx

x eu du = 0

ln |x| eu = cln |x| ey/x = c.

Using y(1) = 0 we nd c = 1. The solution of the initial-value problem is ln |x| = ey/x 1.14. Letting x = vy we have

y(v dy + y dv) + vy(ln vy ln y 1) dy = 0y dv + v ln v dy = 0

dv

v ln v+dy

y= 0

ln |ln |v||+ ln |y| = c

y lnxy

= c1.Using y(1) = e we nd c1 = e. The solution of the initial-value problem is y ln

xy = e.

15. From y +1xy =

1xy2 and w = y3 we obtain

dw

dx+

3xw =

3x

. An integrating factor is x3 so that x3w = x3 + c

or y3 = 1 + cx3.

16. From yy = exy2 and w = y1 we obtain dwdx

+w = ex. An integrating factor is ex so that exw = 12e2x+ c

or y1 = 12ex + cex.

17. From y + y = xy4 and w = y3 we obtaindw

dx 3w = 3x. An integrating factor is e3x so that e3x =

xe3x +13e3x + c or y3 = x+

13

+ ce3x.

18. From y(

1 +1x

)y = y2 and w = y1 we obtain

dw

dx+

(1 +

1x

)w = 1. An integrating factor is xex so that

xexw = xex + ex + c or y1 = 1 + 1x

+c

xex.

43

Exercises 2.5

19. From y 1ty = 1

t2y2 and w = y1 we obtain

dw

dt+

1tw =

1t2

. An integrating factor is t so that tw = ln t+ c

or y1 =1t

ln t+c

t. Writing this in the form

t

y= lnx+ c, we see that the solution can also be expressed in the

form et/y = c1x.

20. From y +2

3 (1 + t2)y =

2t3 (1 + t2)

y4 and w = y3 we obtaindw

dt 2t

1 + t2w =

2t1 + t2

. An integrating factor is

11 + t2

so thatw

1 + t2=

11 + t2

+ c or y3 = 1 + c(1 + t2

).

21. From y 2xy =

3x2y4 and w = y3 we obtain

dw

dx+

6xw = 9

x2. An integrating factor is x6 so that

x6w = 95x5 + c or y3 = 9

5x1 + cx6. If y(1) =

12

then c =495

and y3 = 95x1 +

495x6.

22. From y + y = y1/2 and w = y3/2 we obtaindw

dx+

32w =

32

. An integrating factor is e3x/2 so that e3x/2w =

e3x/2 + c or y3/2 = 1 + ce3x/2. If y(0) = 4 then c = 7 and y3/2 = 1 + 7e3x/2.

23. Let u = x+ y + 1 so that du/dx = 1 + dy/dx. Thendu

dx 1 = u2 or 1

1 + u2du = dx. Thus tan1 u = x+ c or

u = tan(x+ c), and x+ y + 1 = tan(x+ c) or y = tan(x+ c) x 1.

24. Let u = x+y so that du/dx = 1+dy/dx. Thendu

dx1 = 1 u

uor u du = dx. Thus

12u2 = x+c or u2 = 2x+c1,

and (x+ y)2 = 2x+ c1.

25. Let u = x+ y so that du/dx = 1 + dy/dx. Thendu

dx 1 = tan2 u or cos2 u du = dx. Thus 1

2u+

14

sin 2u = x+ c

or 2u+ sin 2u = 4x+ c1, and 2(x+ y) + sin 2(x+ y) = 4x+ c1 or 2y + sin 2(x+ y) = 2x+ c1.

26. Let u = x + y so that du/dx = 1 + dy/dx. Thendu

dx 1 = sinu or 1

1 + sinudu = dx. Multiplying by

(1 sinu)/(1 sinu) we have 1 sinucos2 u

du = dx or(sec2 u tanu secu) du = dx. Thus tanu secu = x + c

or tan(x+ y) sec(x+ y) = x+ c.

27. Let u = y 2x+ 3 so that du/dx = dy/dx 2. Then dudx

+ 2 = 2 +u or

1udu = dx. Thus 2

u = x+ c and

2y 2x+ 3 = x+ c.

28. Let u = y x+ 5 so that du/dx = dy/dx 1. Then dudx

+ 1 = 1 + eu or eudu = dx. Thus eu = x+ c andeyx+5 = x+ c.

29. Let u = x+ y so that du/dx = 1 + dy/dx. Thendu

dx 1 = cosu and 1

1 + cosudu = dx. Now

11 + cosu

=1 cosu1 cos2 u =

1 cosusin2 u

= csc2 u cscu cotu

so we have

(csc2 u cscu cotu)du = dx and cotu+ cscu = x+ c. Thus cot(x+ y) + csc(x+ y) = x+ c.Setting x = 0 and y = /4 we obtain c =

2 1. The solution is

csc(x+ y) cot(x+ y) = x+

2 1.

30. Let u = 3x+ 2y so that du/dx = 3 + 2 dy/dx. Thendu

dx= 3 +

2uu+ 2

=5u+ 6u+ 2

andu+ 25u+ 6

du = dx. Now

u+ 25u+ 6

=15

+4

25u+ 30

44

Exercises 2.5

so we have (15

+4

25u+ 30

)du = dx

and15u+

425

ln |25u+ 30| = x+ c. Thus15(3x+ 2y) +

425

ln |75x+ 50y + 30| = x+ c.

Setting x = 1 and y = 1 we obtain c = 45 ln 95. The solution is15(3x+ 2y) +

425

ln |75x+ 50y + 30| = x+ 45

ln 95or

5y 5x+ 2 ln |75x+ 50y + 30| = 10 ln 95.

31. We write the dierential equation M(x, y)dx+N(x, y)dy = 0 as dy/dx = f(x, y) where

f(x, y) = M(x, y)N(x, y)

.

The function f(x, y) must necessarily be homogeneous of degree 0 when M and N are homogeneous of degree. Since M is homogeneous of degree , M(tx, ty) = tM(x, y), and letting t = 1/x we have

M(1, y/x) =1x

M(x, y) or M(x, y) = xM(1, y/x).

Thusdy

dx= f(x, y) = x

M(1, y/x)xN(1, y/x)

= M(1, y/x)N(1, y/x)

= F(yx

).

To show that the dierential equation also has the form

dy

dx= G

(x

y

)

we use the fact that M(x, y) = yM(x/y, 1). The forms F (y/x) and G(x/y) suggest, respectively, the substitu-tions u = y/x or y = ux and v = x/y or x = vy.

32. As x , e6x 0 and y 2x + 3. Now write (1 + ce6x)/(1 ce6x) as (e6x + c)/(e6x c). Then, asx, e6x 0 and y 2x 3.

33. (a) The substitutions y = y1 + u anddy

dx=dy1dx

+du

dx

lead tody1dx

+du

dx= P +Q(y1 + u) +R(y1 + u)2

= P +Qy1 +Ry21 +Qu+ 2y1Ru+Ru2

ordu

dx (Q+ 2y1R)u = Ru2.

This is a Bernoulli equation with n = 2 which can be reduced to the linear equation

dw

dx+ (Q+ 2y1R)w = R

by the substitution w = u1.

45

Exercises 2.5

(b) Identify P (x) = 4/x2, Q(x) = 1/x, and R(x) = 1. Then dwdx

+( 1x

+4x

)w = 1. An integrating

factor is x3 so that x3w = 14x4 + c or u =

[1

4x+ cx3

]1. Thus, y =

2x

+ u.

34. Write the dierential equation in the form x(y/y) = lnx + ln y and let u = ln y. Then du/dx = y/y and thedierential equation becomes x(du/dx) = lnx+ u or du/dx u/x = (lnx)/x, which is rst-order, linear. Anintegrating factor is e

dx/x = 1/x, so that (using integration by parts)

d

dx

[ 1xu]

=lnxx2

andu

x= 1

x lnx

x+ c.

The solution is

ln y = 1 lnx+ cx or y = ecx1

x.

Exercises 2.6

1. We identify f(x, y) = 2x 3y + 1. Then, for h = 0.1,yn+1 = yn + 0.1(2xn 3yn + 1) = 0.2xn + 0.7yn + 0.1,

andy(1.1) y1 = 0.2(1) + 0.7(5) + 0.1 = 3.8y(1.2) y2 = 0.2(1.1) + 0.7(3.8) + 0.1 = 2.98.

For h = 0.05

yn+1 = yn + 0.05(2xn 3yn + 1) = 0.1xn + 0.85yn + 0.1,and

y(1.05) y1 = 0.1(1) + 0.85(5) + 0.1 = 4.4y(1.1) y2 = 0.1(1.05) + 0.85(4.4) + 0.1 = 3.895y(1.15) y3 = 0.1(1.1) + 0.85(3.895) + 0.1 = 3.47075y(1.2) y4 = 0.1(1.15) + 0.85(3.47075) + 0.1 = 3.11514

2. We identify f(x, y) = x+ y2. Then, for h = 0.1,

yn+1 = yn + 0.1(xn + y2n) = 0.1xn + yn + 0.1y2n,

andy(0.1) y1 = 0.1(0) + 0 + 0.1(0)2 = 0y(0.2) y2 = 0.1(0.1) + 0 + 0.1(0)2 = 0.01.

For h = 0.05

yn+1 = yn + 0.05(xn + y2n) = 0.05xn + yn + 0.05y2n,

andy(0.05) y1 = 0.05(0) + 0 + 0.05(0)2 = 0y(0.1) y2 = 0.05(0.05) + 0 + 0.05(0)2 = 0.0025y(0.15) y3 = 0.05(0.1) + 0.0025 + 0.05(0.0025)2 = 0.0075y(0.2) y4 = 0.05(0.15) + 0.0075 + 0.05(0.0075)2 = 0.0150.

46

h=0.1 h=0.05

xn yn True ValueAbs. Error

% Rel. Error xn yn

True Value

Abs. Error

% Rel. Error

0.00 1.0000 1.0000 0.0000 0.00 0.00 1.0000 1.0000 0.0000 0.000.10 1.1000 1.1052 0.0052 0.47 0.05 1.0500 1.0513 0.0013 0.120.20 1.2100 1.2214 0.0114 0.93 0.10 1.1025 1.1052 0.0027 0.240.30 1.3310 1.3499 0.0189 1.40 0.15 1.1576 1.1618 0.0042 0.360.40 1.4641 1.4918 0.0277 1.86 0.20 1.2155 1.2214 0.0059 0.480.50 1.6105 1.6487 0.0382 2.32 0.25 1.2763 1.2840 0.0077 0.600.60 1.7716 1.8221 0.0506 2.77 0.30 1.3401 1.3499 0.0098 0.720.70 1.9487 2.0138 0.0650 3.23 0.35 1.4071 1.4191 0.0120 0.840.80 2.1436 2.2255 0.0820 3.68 0.40 1.4775 1.4918 0.0144 0.960.90 2.3579 2.4596 0.1017 4.13 0.45 1.5513 1.5683 0.0170 1.081.00 2.5937 2.7183 0.1245 4.58 0.50 1.6289 1.6487 0.0198 1.20

0.55 1.7103 1.7333 0.0229 1.320.60 1.7959 1.8221 0.0263 1.440.65 1.8856 1.9155 0.0299 1.560.70 1.9799 2.0138 0.0338 1.680.75 2.0789 2.1170 0.0381 1.800.80 2.1829 2.2255 0.0427 1.920.85 2.2920 2.3396 0.0476 2.040.90 2.4066 2.4596 0.0530 2.150.95 2.5270 2.5857 0.0588 2.271.00 2.6533 2.7183 0.0650 2.39

h=0.1 h=0.05

xn yn True ValueAbs. Error

% Rel. Error xn yn

True Value

Abs. Error

% Rel. Error

1.00 1.0000 1.0000 0.0000 0.00 1.00 1.0000 1.0000 0.0000 0.001.10 1.2000 1.2337 0.0337 2.73 1.05 1.1000 1.1079 0.0079 0.721.20 1.4640 1.5527 0.0887 5.71 1.10 1.2155 1.2337 0.0182 1.471.30 1.8154 1.9937 0.1784 8.95 1.15 1.3492 1.3806 0.0314 2.271.40 2.2874 2.6117 0.3243 12.42 1.20 1.5044 1.5527 0.0483 3.111.50 2.9278 3.4903 0.5625 16.12 1.25 1.6849 1.7551 0.0702 4.00

1.30 1.8955 1.9937 0.0982 4.931.35 2.1419 2.2762 0.1343 5.901.40 2.4311 2.6117 0.1806 6.921.45 2.7714 3.0117 0.2403 7.981.50 3.1733 3.4903 0.3171 9.08

h=0.1 h=0.05

xn yn xn yn0.00 0.0000 0.00 0.00000.10 0.1000 0.05 0.05000.20 0.1905 0.10 0.09760.30 0.2731 0.15 0.14290.40 0.3492 0.20 0.18630.50 0.4198 0.25 0.2278

0.30 0.26760.35 0.30580.40 0.34270.45 0.37820.50 0.4124

h=0.1 h=0.05

xn yn xn yn0.00 1.0000 0.00 1.00000.10 1.1000 0.05 1.05000.20 1.2220 0.10 1.10530.30 1.3753 0.15 1.16680.40 1.5735 0.20 1.23600.50 1.8371 0.25 1.3144

0.30 1.40390.35 1.50700.40 1.62670.45 1.76700.50 1.9332

Exercises 2.6

3. Separating variables and integrating, we havedy

y= dx and ln |y| = x+ c.

Thus y = c1ex and, using y(0) = 1, we nd c = 1, so y = ex is the solution of the initial-value problem.

4. Separating variables and integrating, we havedy

y= 2x dx and ln |y| = x2 + c.

Thus y = c1ex2

and, using y(1) = 1, we nd c = e1, so y = ex21 is the solution of the initial-value problem.

5. 6.

47

h=0.1 h=0.05

xn yn xn yn0.00 0.5000 0.00 0.50000.10 0.5250 0.05 0.51250.20 0.5431 0.10 0.52320.30 0.5548 0.15 0.53220.40 0.5613 0.20 0.53950.50 0.5639 0.25 0.5452

0.30 0.54960.35 0.55270.40 0.55470.45 0.55590.50 0.5565

h=0.1 h=0.05

xn yn xn yn0.00 1.0000 0.00 1.00000.10 1.1000 0.05 1.05000.20 1.2159 0.10 1.10390.30 1.3505 0.15 1.16190.40 1.5072 0.20 1.22450.50 1.6902 0.25 1.2921

0.30 1.36510.35 1.44400.40 1.52930.45 1.62170.50 1.7219

h=0.1 h=0.05

xn yn xn yn1.00 1.0000 1.00 1.00001.10 1.0000 1.05 1.00001.20 1.0191 1.10 1.00491.30 1.0588 1.15 1.01471.40 1.1231 1.20 1.02981.50 1.2194 1.25 1.0506

1.30 1.07751.35 1.11151.40 1.15381.45 1.20571.50 1.2696

h=0.1 h=0.05

xn yn xn yn0.00 0.5000 0.00 0.50000.10 0.5250 0.05 0.51250.20 0.5499 0.10 0.52500.30 0.5747 0.15 0.53750.40 0.5991 0.20 0.54990.50 0.6231 0.25 0.5623

0.30 0.57460.35 0.58680.40 0.59890.45 0.61090.50 0.6228

5 1 0- 5- 1 0

5

x

y

x

Euler

Runge-Kutta

h=0.25

5 10- 5-10

5

h=0.1

x

y

Runge-Kutta

Euler

5 10- 5-10

5Euler

Runge-Kutta

x

yh=0.05

2 4- 2

5h=0.25

Runge-Kutta

Euler

x

y

2 4- 2

5

y

x

Euler

Runge-Kutta

h=0.1

2 4- 2

5h=0.05

Euler

Runge-Kutta

x

y

Exercises 2.6

7. 8.

9. 10.

11.

12.

48

h=0.1 Euler h=0.05 Euler h=0.1 R-K h=0.05 R-K

xn yn xn yn xn yn xn yn0.00 1.0000 0.00 1.0000 0.00 1.0000 0.00 1.00000.10 1.0000 0.05 1.0000 0.10 1.0101 0.05 1.00250.20 1.0200 0.10 1.0050 0.20 1.0417 0.10 1.01010.30 1.0616 0.15 1.0151 0.30 1.0989 0.15 1.02300.40 1.1292 0.20 1.0306 0.40 1.1905 0.20 1.04170.50 1.2313 0.25 1.0518 0.50 1.3333 0.25 1.06670.60 1.3829 0.30 1.0795 0.60 1.5625 0.30 1.09890.70 1.6123 0.35 1.1144 0.70 1.9607 0.35 1.13960.80 1.9763 0.40 1.1579 0.80 2.7771 0.40 1.19050.90 2.6012 0.45 1.2115 0.90 5.2388 0.45 1.25391.00 3.8191 0.50 1.2776 1.00 42.9931 0.50 1.3333

0.55 1.3592 0.55 1.43370.60 1.4608 0.60 1.56250.65 1.5888 0.65 1.73160.70 1.7529 0.70 1.96080.75 1.9679 0.75 2.28570.80 2.2584 0.80 2.77770.85 2.6664 0.85 3.60340.90 3.2708 0.90 5.26090.95 4.2336 0.95 10.19731.00 5.9363 1.00 84.0132

Exercises 2.7

13. Using separation of variables we nd that the solution of the dierential equation is y = 1/(1 x2), which isundened at x = 1, where the graph has a vertical asymptote.

Exercises 2.7

1. Let P = P (t) be the population at time t, and P0 the initial population. From dP/dt = kP we obtain P = P0ekt.Using P (5) = 2P0 we nd k = 15 ln 2 and P = P0e

(ln 2)t/5. Setting P (t) = 3P0 we have

3 = e(ln 2)t/5 = ln 3 = (ln 2)t5

= t = 5 ln 3ln 2

7.9 years.

Setting P (t) = 4P0 we have

4 = e(ln 2)t/5 = ln 4 = (ln 2)t5

= t = 10 years.

2. Setting P =10,000 and t = 3 in Problem 1 we obtain

10,000 = P0e(ln 2)3/5 = P0 = 10,000e0.6 ln 2 6597.5.Then P (10) = P0e2 ln 2 = 4P0 26,390.

3. Let P = P (t) be the population at time t. From dP/dt = kt and P (0) = P0 = 500 we obtain P = 500ekt. UsingP (10) = 575 we nd k = 110 ln 1.15. Then P (30) = 500e

3 ln 1.15 760 years.4. Let N = N(t) be the number of bacteria at time t and N0 the initial number. From dN/dt = kN we obtain

N = N0ekt. Using N(3) = 400 and N(10) = 2000 we nd 400 = N0e3k or ek = (400/N0)1/3. From N(10) = 2000we then have

2000 = N0e10k = N0

(400N0

)10/3= 2000

40010/3= N7/30 = N0 =

(2000

40010/3

)3/7 201.

5. Let I = I(t) be the intensity, t the thickness, and I(0) = I0. If dI/dt = kI and I(3) = 0.25I0 then I = I0ekt,k = 13 ln 0.25, and I(15) = 0.00098I0.

6. From dS/dt = rS we obtain S = S0ert where S(0) = S0.

49

Exercises 2.7

(a) If S0 = $5000 and r = 5.75% then S(5) = $6665.45.(b) If S(t) =$10,000 then t = 12 years.(c) S $6651.82

7. Let N = N(t) be the amount of lead at time t. From dN/dt = kN and N(0) = 1 we obtain N = ekt. UsingN(3.3) = 1/2 we nd k = 13.3 ln 1/2. When 90% of the lead has decayed, 0.1 grams will remain. SettingN(t) = 0.1 we have

et(1/3.3) ln(1/2) = 0.1 = t3.3

ln12

= ln 0.1 = t = 3.3 ln 0.1ln 1/2

10.96 hours.

8. Let N = N(t) be the amount at time t. From dN/dt = kt and N(0) = 100 we obtain N = 100ekt. UsingN(6) = 97 we nd k = 16 ln 0.97. Then N(24) = 100e

(1/6)(ln 0.97)24 = 100(0.97)4 88.5 mg.9. Setting N(t) = 50 in Problem 8 we obtain

50 = 100ekt = kt = ln 12

= t = ln 1/2(1/6) ln 0.97

136.5 hours.

10. (a) The solution of dA/dt = kA is A(t) = A0ekt. Letting A = 12A0 and solving for t we obtain the half-lifeT = (ln 2)/k.

(b) Since k = (ln 2)/T we haveA(t) = A0e(ln 2)t/T = A02t/T .

11. Assume that A = A0ekt and k = 0.00012378. If A(t) = 0.145A0 then t 15,600 years.12. From Example 3, the amount of carbon present at time t is A(t) = A0e0.00012378t. Letting t = 660 and solving

for A0 we have A(660) = A0e0.0001237(660) = 0.921553A0. Thus, approximately 92% of the original amount ofC-14 remained in the cloth as of 1988.

13. Assume that dT/dt = k(T 10) so that T = 10 + cekt. If T (0) = 70 and T (1/2) = 50 then c = 60 andk = 2 ln(2/3) so that T (1) = 36.67.

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