-4 -2 2 4 t-4-224X1 Introduction to Differential EquationsExercises 1.11. 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 (y21)(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 = ueuwe 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; linear6. Second-order; nonlinear because of cos(r +u)7. Second-order; nonlinear because of 1 + (dy/dx)28. Second-order; nonlinear because of 1/R29. Third-order; linear10. Second-order; nonlinear because of x211. From y = ex/2we obtain y

= 12ex/2. Then 2y

+y = ex/2+ex/2= 0.12. From y = 65 65e20twe obtain dy/dt = 24e20t, so thatdydt + 20y = 24e20t+ 20

65 65e20t

= 24.13. From y = e3xcos 2x we obtain y

= 3e3xcos 2x 2e3xsin2x and y

= 5e3xcos 2x 12e3xsin2x, so thaty

6y

+ 13y = 0.14. From y = cos xln(sec x + tanx) we obtain y

= 1 + sinxln(sec x + tanx) andy

= tanx + cos xln(sec x + tanx). Then y

+y = tanx.15. Writing ln(2X 1) ln(X 1) = t and dierentiating implicitly we obtain22X 1dXdt 1X 1dXdt = 1

22X 1 1X 1

dXdt = 12X 2 2X + 1(2X 1)(X 1)dXdt = 1dXdt = (2X 1)(X 1) = (X 1)(1 2X).Exponentiating both sides of the implicit solution we obtain2X 1X 1 = et= 2X 1 = Xetet= (et1) = (et2)X = X = et1et2 .Solving et2 = 0 we get t = ln2. Thus, the solution is dened on (, ln2) or on (ln2, ). The graph of thesolution dened on (, ln2) is dashed, and the graph of the solution dened on (ln2, ) is solid.1-4 -2 2 4 x-4-224yExercises 1.116. Implicitly dierentiating the solution we obtain2x2 dydx 4xy + 2y dydx = 0 = x2dy 2xy dx +y dy = 0= 2xy dx + (x2y)dy = 0.Using the quadratic formula to solve y2 2x2y 1 = 0 for y, we gety = 2x24x4+ 4

/2 = x2x4+ 1 . Thus, two explicit solutions arey1 = x2+x4+ 1 and y2 = x2x4+ 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 obtaindPdt = (1 +c1et) c1etc1et c1et(1 +c1et)2= c1et1 +c1et[(1 +c1et) c1et]1 +c1et = P(1 P).18. Dierentiating y = ex2

x0et2dt +c1ex2we obtainy

= ex2ex22xex2

x0et2dt 2c1xex2= 1 2xex2

x0et2dt 2c1xex2.Substituting into the dierential equation, we havey

+ 2xy = 1 2xex2

x0et2dt 2c1xex2+ 2xex2

x0et2dt + 2c1xex2= 1.19. From y = c1e2x+c2xe2xwe obtain dydx = (2c1 +c2)e2x+ 2c2xe2xand d2ydx2 = (4c1 + 4c2)e2x+ 4c2xe2x, so thatd2ydx2 4dydx + 4y = (4c1 + 4c28c14c2 + 4c1)e2x+ (4c28c2 + 4c2)xe2x= 0.20. From y = c1x1+c2x +c3xlnx + 4x2we obtaindydx = c1x2+c2 +c3 +c3 lnx + 8x,d2ydx2 = 2c1x3+c3x1+ 8,andd3ydx3 = 6c1x4c3x2,so thatx3 d3ydx3 + 2x2 d2ydx2 x dydx +y= (6c1 + 4c1 +c1 +c1)x1+ (c3 + 2c3c2c3 +c2)x+ (c3 +c3)xlnx + (16 8 + 4)x2= 12x2.21. From y =

x2, x < 0x2, x 0 we obtain y

=

2x, x < 02x, x 0 so that xy

2y = 0.2Exercises 1.122. The function y(x) is not continuous at x = 0 since limx0y(x) = 5 and limx0+y(x) = 5. Thus, y

(x) does notexist at x = 0.23. From x = e2t+ 3e6tand y = e2t+ 5e6twe obtaindxdt = 2e2t+ 18e6tand dydt = 2e2t+ 30e6t.Thenx + 3y = (e2t+ 3e6t) + 3(e2t+ 5e6t)= 2e2t+ 18e6t= dxdtand5x + 3y = 5(e2t+ 3e6t) + 3(e2t+ 5e6t)= 2e2t+ 30e6t= dydt .24. From x = cos 2t + sin2t + 15etand y = cos 2t sin2t 15etwe obtaindxdt = 2 sin2t + 2 cos 2t + 15etand dydt = 2 sin2t 2 cos 2t 15etandd2xdt2 = 4 cos 2t 4 sin2t + 15etand d2ydt2 = 4 cos 2t + 4 sin2t 15et.Then4y +et= 4(cos 2t sin2t 15et) +et= 4 cos 2t 4 sin2t + 15et= d2xdt2and4x et= 4(cos 2t + sin2t + 15et) et= 4 cos 2t + 4 sin2t 15et= d2ydt2 .25. An interval on which tan5t is continuous is /2 < 5t < /2, so 5 tan5t will be a solution on (/10, /10).26. For (1 sint)1/2to be continuous we must have 1 sint > 0 or sint < 1. Thus, (1 sint)1/2will be asolution 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) = etis et. The rst derivative of f(t) = ektis kekt. The dierential equations arey

= y and y

= ky, respectively.30. Any function of the form y = cetor y = cetis its own second derivative. The corresponding dierentialequation is y

y = 0. Functions of the form y = c sint 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) = 2c1y

1(0) +c2y

2(0) = 03Exercises 1.1for c1 and c2 we getc1 = 2y

2(0)y1(0)y

2(0) y

1(0)y2(0) and c2 = 2y

1(0)y1(0)y

2(0) y

1(0)y2(0) .Thus, a particular solution isy = 2y

2(0)y1(0)y

2(0) y

1(0)y2(0) y1 2y

1(0)y1(0)y

2(0) y

1(0)y2(0) y2,where we assume that y1(0)y

2(0) y

1(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 equationsy

= 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 + 3c2t2and y

= 6c2t. Then c2 = y

/6t and c1 = y

ty

/2, soy =

y

ty

2

t +

y

6t

t3= ty

13t2y

and the dierential equation is t2y

3ty

+ 3y = 0.37. (a) From y = emtwe obtain y

= memt. Then y

+ 2y = 0 impliesmemt+ 2emt= (m+ 2)emt= 0.Since emt> 0 for all t, m = 2. Thus y = e2tis a solution.(b) From y = emtwe obtain y

= memtand y

= m2emt. Then y

5y

+ 6y = 0 impliesm2emt5memt+ 6emt= (m2)(m3)emt= 0.Since emt> 0 for all t, m = 2 and m = 3. Thus y = e2tand y = e3tare solutions.(c) From y = tmwe obtain y

= mtm1and y

= m(m1)tm2. Then ty

+ 2y

= 0 impliestm(m1)tm2+ 2mtm1= [m(m1) + 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 = t1are solutions.(d) From y = tmwe obtain y

= mtm1and y

= m(m1)tm2. Then t2y

7ty

+ 15y = 0 impliest2m(m1)tm27tmtm1+ 15tm= [m(m1) 7m+ 15]tm= (m28m+ 15)tm= (m3)(m5)tm= 0.Since tm> 0 for t > 0, m = 3 and m = 5. Thus y = t3and y = t5are 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.4ty5 5510Exercises 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 obtainxy(3x2+ 3y2y

) (x3+y3)(xy

+y)x2y2 = 03x3+ 3xy3y

x4y

x3y xy3y

y4= 0(3xy3x4xy3)y

= 3x3y +x3y +y4y

= y42x3y2xy3x4 = y(y32x3)x(2y3x3) .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 getx3+ 12x3= 3x

121/3 x

32x3= 321/3x2x3= 22/3x2x2(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 beincreasing. Thus, none of the curves in color can be solutions.(c)47. (a) y = 0 and y = a/b.5y=a by=0xyExercises 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 computed2ydx2 = 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 useClear[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 useClear[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] // SimplifyExercises 1.21. Solving 13 = 11 +c1we get c1 = 4. The solution is y = 11 4et .2. Solving 2 = 11 +c1e we get c1 = 12e1. The solution is y = 22 e(t+1) .3. Using x

= c1 sint +c2 cos t we obtain c1 = 1 and c2 = 8. The solution is x = cos t + 8 sint.4. Using x

= c1 sint +c2 cos t we obtain c2 = 0 and c1 = 1. The solution is x = cos t.5. Using x

= c1 sint +c2 cos t we obtain32 c1 + 12c2 = 1212c1 +32 c2 = 0.Solving we nd c1 =34 and c2 = 14 . The solution is x =34 cos t + 14 sint.6Exercises 1.26. Using x

= c1 sint +c2 cos t we obtain22 c1 +22 c2 =222 c1 +22 c2 = 22 .Solving we nd c1 = 1 and c2 = 3. The solution is x = cos t + 3 sint.7. From the initial conditions we obtain the systemc1 +c2 = 1c1c2 = 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 systemc1e +c2e1= 0c1e 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 obtainc1e1+c2e = 5c1e1c2e = 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 = 0c1c2 = 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 x2is a solution.)13. For f(x, y) = y2/3we have fy = 23y1/3. Thus the dierential equation will have a unique solution in anyrectangular region of the plane where y = 0.14. For f(x, y) = xy we have fy = 12

xy . Thus the dierential equation will have a unique solution in anyregion where x > 0 and y > 0 or where x < 0 and y < 0.15. For f(x, y) = yx we have fy = 1x . Thus the dierential equation will have a unique solution in any regionwhere x = 0.16. For f(x, y) = x + y we have fy = 1. Thus the dierential equation will have a unique solution in the entireplane.17. For f(x, y) = x24 y2 we have fy = 2x2y(4 y2)2 . Thus the dierential equation will have a unique solution inany region where y < 2, 2 < y < 2, or y > 2.7Exercises 1.218. For f(x, y) = x21 +y3 we have fy = 3x2y2(1 +y3)2 . Thus the dierential equation will have a unique solution inany region where y = 1.19. For f(x, y) = y2x2+y2 we have fy = 2x2y(x2+y2)2 . Thus the dierential equation will have a unique solution inany region not containing (0, 0).20. For f(x, y) = y +xy x we have fy = 2x(y x)2 . Thus the dierential equation will have a unique solution in anyregion 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) Since ddx tan(x + c) = sec2(x + c) = 1 + tan2(x + c), we see that y = tan(x + c) satises the dierentialequation.(b) Solving y(0) = tanc = 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) Since ddt

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 andy = 11/y0 +t = y01 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 3x2y2= c we get 6x 2yy

= 0 or yy

= 3x.8-4 -2 2 4 x-4-224y-4 -2 2 4 x-4-224y-4 -2 2 4 x12345y-4 -2 2 4 x12345yExercises 1.2(b) Solving 3x2y2= 3 for y we gety = 1(x) = 3(x21) , 1 < x < ,y = 2(x) =

3(x21) , 1 < x < ,y = 3(x) = 3(x21) , < x < 1,y = 4(x) =

3(x21) , < x < 1.Only y = 3(x) satises y(2) = 3.(c) Setting x = 2 and y = 4 in 3x2y2= c we get 12 16 = 4 = c, so theexplicit solution isy =

3x2+ 4 , < x < .(d) Setting c = 0 we have y =3x and y = 3x, both dened on (, ).29. When x = 0 and y = 12 , y

= 1, so the only plausible solution curve is the one with negative slope at (0, 12 ),or the black curve.30. If the solution is tangent to the x-axis at (x0, 0), then y

= 0 when x = x0 and y = 0. Substituting these valuesinto y

+ 2y = 3x 6 we get 0 + 0 = 3x06 or x0 = 2.31. The theorem guarantees a unique (meaning single) solution through any point. Thus, there cannot be twodistinct solutions through any point.32. Dierentiating y = x4/16 we obtain y

= x3/4 = x(x4/16)1/2, so the rstfunction is a solution of the dierential equation. Since y = 0 satises thedierential equation and limx0+ x3/4 = 0, the second function is also a solutionof the dierential equation. Both functions satisfy the condition y(2) = 1.Theorem 1.1 simply guarantees the existence of a unique solution on someinterval containing (2, 1). In this, such an interval could be (0, 4), where the twofunctions are identical.33. The antiderivative of y

= 8e2x+ 6x is y = 4e2x+ 3x2+ c. Setting x = 0 and y = 9 we get 9 = 4 + c, so c = 5and y = 4e2x+ 3x2+ 5.34. The antiderivative of y

= 12x 2 is y

= 6x2 2x + c1. From the equation of the tangent line we see thatwhen x = 1, y = 4 and y

= 1 (the slope of the tangent line). Solving 1 = 6(1)22(1) +c1 we get c1 = 5.The antiderivative of y

= 6x22x 5 is y = 2x3x25x +c2. Setting x = 1 and y = 4 we get 4 = 4 +c2,so c2 = 8 and y = 2x3x25x + 8.9Exercises 1.3Exercises 1.31. dPdt = kP +r; dPdt = kP r2. 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 = k2P2. Since dP/dt = b d, thedierential equation is dP/dt = k1P k2P2.4. 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 thedierential equation we then havek = dT/dtT Tm= 185 75 = 0.1.6. By inspecting the graph we take Tm to be Tm(t) = 8030 cos t/12. Then the temperature of the body at timet is determined by the dierential equationdTdt = 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 thetechnological 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 anddxdt = kx(n + 1 x), x(0) = 1.9. The rate at which salt is leaving the tank is(3 gal/min)

A300 lb/gal

= A100 lb/min.Thus dA/dt = A/100.10. The rate at which salt is entering the tank isR1 = (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 isR2 = (2 gal/min)

A300 +t lb/gal

= 2A300 +t lb/min.The dierential equation isdAdt = 6 2A300 +t .11. The volume of water in the tank at time t is V = Awh. The dierential equation is thendhdt = 1AwdVdt = 1Aw

cA0

2gh

= cA0Aw

2gh.10Exercises 1.3Using A0 =

212

2= 36 , Aw = 102= 100, and g = 32, this becomesdhdt = c/3610064h = c450h.12. The volume of water in the tank at time t is V = 13r2h = 13Awh. Using the formula from Problem 11 for thevolume of water leaving the tank we see that the dierential equation isdhdt = 3AwdVdt = 3Aw(cAh

2gh) = 3cAhAw

2gh.Using Ah = (2/12)2= /36, g = 32, and c = 0.6, this becomesdhdt = 3(0.6)/36Aw64h = 0.4Awh1/2.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. Thusdhdt = 0.44h2/25h1/2= 2.5h3/2.13. Since i = dqdt and Ld2qdt2 +Rdqdt = E(t) we obtain Ldidt +Ri = E(t).14. By Kirchos second law we obtain Rdqdt + 1Cq = E(t).15. From Newtons second law we obtain mdvdt = kv2+mg.16. We have from Archimedes principleupward 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 that wgd2ydt2 = 15.6s2y or d2ydt2 + 15.6s2gw y = 0, where g = 32and w is the weight of the barrel in pounds.17. The net force acting on the mass isF = ma = m d2xdt2 = k(s +x) +mg = kx +mg ks.Since the condition of equilibrium is mg = ks, the dierential equation ism d2xdt2 = 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 becomesm d2xdt2 = kx dxdt or md2xdt2 +dxdt +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 = 5 W = 5 x. By Newtons second law,ddt

x32v

= 5 x or xdvdt +vdxdt = 160 32x.11(x,y)xyxyExercises 1.3Thus, the dierential equation isxd2xdt +

dxdt

2+ 32x = 160.20. The force is the weight of the chain, 2L, so by Newtons second law, ddt[mv] = 2L. Since the mass of the portionof chain o the ground is m = 2(L x)/g, we haveddt

2(L x)g v

= 2L or (L x)dvdt +v

dxdt

= Lg.Thus, the dierential equation is(L x)d2xdt2

dxdt

2= Lg.21. From g = k/R2we nd k = gR2. Using a = d2r/dt2and the fact that the positive direction is upward we getd2rdt2 = a = kr2 = gR2r2 or d2rdt2 + gR2r2 = 0.22. The gravitational force on mis F = kMrm/r2. Since Mr = 4r3/3 and M = 4R3/3 we have Mr = r3M/R3andF = k Mrmr2 = k r3Mm/R3r2 = k mMR3 r.Now from F = ma = d2r/dt2we havem d2rdt2 = k mMR3 r or d2rdt2 = kMR3 r.23. The dierential equation is dAdt = k(M A).24. The dierential equation is dAdt = 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

= y

s2y2.27. We see from the gure that 2 + = . Thusyx = tan = tan( 2) = tan2 = 2 tan1 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 gety

= 2x

4x2+ 4y22y = x

x2+y2y .Since dy/dx > 0, the dierential equation isdydx = x + x2+y2y or y dydx

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.12Exercises 1.329. 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 = 6 A/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): dPdt = kP is linear (2): dAdt = kA is linear(3): dTdt = k(T Tm) is linear (5): dxdt = kx(n + 1 x) is nonlinear(6): dXdt = k( X)( X) is nonlinear (8): dAdt = 6 A100 is linear(10): dhdt = AhAw

2gh is nonlinear (11): Ld2qdt2 +Rdqdt + 1Cq = E(t) is linear(12): d2sdt2 = g is linear (14): mdvdt = mg kv is linear(15): md2sdt2 +kdsdt = mg is linear (16): d2xdt2 64L x = 0 is linear33. From Problem 21, d2r/dt2= gR2/r2. Since R is a constant, if r = R +s, then d2r/dt2= d2s/dt2and, usinga Taylor series, we getd2sdt2 = g R2(R +s)2 = gR2(R +s)2 gR2[R22sR3+ ] = g + 2gsR3 + .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 anddmdt = dVdt = ddt

43r3

=

4r2drdt

= S drdt .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, ddt[mv] = mg, where v is the velocity of the raindrop. Thenm dvdt +v dmdt = mg or

43r3

dvdt +v(k4r2) =

43r3

g.Dividing by 4r3/3 we getdvdt + 3krv = g or dvdt + 3k/kt/ +r0v = 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 giveswxdydt = k13Exercises 1.3where 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 becomesdydt = kws1t +t0.Integrating we obtainy = kws [ ln(t +t0) +c ]where c is a constant. Now when t = 0, y = 0 so c = lnt0 andy = kws ln

1 + tt0

.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 +t01 = 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 Exercises1. ddxc1x = c1x2 = c1/xx ; dydx = yx2. ddx(5 +c1e2x) = 2c1e2x= 2(5 +c1e2x5); dydx = 2(y 5) or dydx = 2y + 103. ddx(c1 cos kx +c2 sinkx) = kc1 sinkx +kc2 cos kx;d2dx2(c1 cos kx +c2 sinkx) = k2c1 cos kx k2c2 sinkx = k2(c1 cos kx +c2 sinkx);d2ydx2 = k2y or d2ydx2 +k2y = 04. ddx(c1 cosh kx +c2 sinhkx) = kc1 sinhkx +kc2 cosh kx;d2dx2(c1 cosh kx +c2 sinhkx) = k2c1 cosh kx +k2c2 sinhkx = k2(c1 cosh kx +c2 sinhkx);d2ydx2 = k2y or d2ydx2 k2y = 05. y

= c1ex+c2xex+c2ex; y

= c1ex+c2xex+ 2c2ex;y

+y = 2(c1ex+c2xex) + 2c2ex= 2(c1ex+c2xex+c2ex) = 2y

; y

2y

+y = 06. y

= c1exsinx +c1excos x +c2excos x +c2exsinx;y

= c1excos x c1exsinx c1exsinx +c1excos x c2exsinx +c2excos x +c2excos x +c2exsinx= 2c1exsinx + 2c2excos x;y

2y

= 2c1excos x 2c2exsinx = 2y; y

2y

+ 2y = 07. a,d 8. c 9. b 10. a,c 11. b 12. a,b,d13. A few solutions are y = 0, y = c, and y = ex.14-3 -2 -1 1 2 3 x-3-2-1123y-3 -2 -1 1 2 3 x-3-2-1123yChapter 1 Review Exercises14. 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 = x2/3is undened at x = 0. This function is a solution of the dierentialequation on (, 0) and also on (0, ).18. (a) Dierentiating y22y = x2x +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 theinitial-value problem is y22y = x2x 1.(c) Using the quadratic formula to solve y2 2y (x2 x 1) = 0 we get y = (2

4 + 4(x2x 1) )/2= 1x2x = 1

x(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)= 64 + 5(1)3= 7.21. Dierentiating y = sin(lnx) we obtain y

= cos(lnx)/x and y

= [sin(lnx) + cos(lnx)]/x2. Thenx2y

+xy

+y = x2

sin(lnx) + cos(lnx)x2

+xcos(ln x)x + sin(lnx) = 0.22. Dierentiating y = cos(lnx) ln(cos(lnx)) + (lnx) sin(lnx) we obtainy

= cos(lnx) 1cos(ln x)

sin(lnx)x

+ ln(cos(lnx))

sin(lnx)x

+ lnxcos(lnx)x + sin(lnx)x= ln(cos(lnx)) sin(lnx)x + (lnx) cos(lnx)xandy

= xln(cos(lnx))cos(ln x)x + sin(lnx) 1cos(ln x)

sin(lnx)x

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

sin(lnx)x

+ cos(ln x)x

1x2 (lnx) cos(lnx) 1x2= 1x2ln(cos(lnx) cos(lnx) + sin2(lnx)cos(ln x) + ln(cos(lnx) sin(lnx)(lnx) sin(lnx) + cos(lnx) (lnx) cos(lnx)

.15Chapter 1 Review ExercisesThenx2y

+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(ln x) + cos(lnx) = sin2(lnx) + cos2(lnx)cos(ln x) = 1cos(lnx) = sec(ln x).(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 isdhdt = cA0Aw

2gh.Using A0 = (1/24)2= /576, Aw = (2)2= 4, and g = 32, this becomesdhdt = c/576464h = c288h.25. From Newtons second law we obtainmdvdt = 12mg 32 mg or dvdt = 16

1 3

.16yttyxyttyxxy yttyxxy yxyx xy2 First-Order Differential EquationsExercises 2.11. 2.3. 4.5. 6.7. 8.17yx xyyx xyyx xy-1 0 11 2 x54321y1 2 -1 -2 x1y1 2 -1 -2 x-1y1 2 x-5-4-3-2-1yExercises 2.19. 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 11 2 x54321y1 2 -1 -2 x1y1 2 -1 -2 x-1y-1 -2 x-5-4-3-2-1y0 31 02-2 5-2 0 20 2 4Exercises 2.116. 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 y23y = 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 y2y3= 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 00 ln 9mg/kmg/k Exercises 2.123. 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 yey9y = y(ey9) = 0 we obtain the critical points 0 and ln9.From the phase portrait we see that 0 is asymptotically stable and ln9 is unstable.25. (a) Writing the dierential equation in the formdvdt = km

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,limtv = mg/k.(b) Writing the dierential equation in the formdvdt = km

mgk v2

= km

mgk v

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,limtv = 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 .20tX2//2 tX1/2xtyxc 0 c> < >3 - 33- 3yx-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 3-3-2-10123xyxy-5 5-55Exercises 2.130. When y < 12x2, y

= x2 2y is positive and the portions of solution curvesinside the nullcline parabola are increasing. When y > 12x2, y

= x2 2y isnegative 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, limxy(x) = L,where L c2. We want to show that L = c2. Since L is a horizontal asymptote of y(x), limxy

(x) = 0.Using the fact that f(y) is continuous we havef(L) = f( limxy(x)) = limxf(y(x)) = limxy

(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 wherey

(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.22Exercises 2.2Exercises 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 = ecwe obtain g(y) = c1ef(x).1. From dy = sin5xdx we obtain y = 15 cos 5x +c.2. From dy = (x + 1)2dx we obtain y = 13(x + 1)3+c.3. From dy = e3xdx we obtain y = 13e3x+c.4. From 1(y 1)2 dy = dx we obtain 1y 1 = x +c or y = 1 1x +c .5. From 1y dy = 4x dx we obtain ln|y| = 4 ln|x| +c or y = c1x4.6. From 1y dy = 2xdx 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 yeyey+ex+ 13e3x= c.9. From

y + 2 + 1y

dy = x2lnxdx we obtain y22 + 2y + ln|y| = x33 ln|x| 19x3+c.10. From 1(2y + 3)2 dy = 1(4x + 5)2 dx we obtain 22y + 3 = 14x + 5 +c.11. From 1csc y dy = 1sec2x dx or siny dy = cos2xdx = 12(1 + cos 2x) dx we obtaincos y = 12x 14 sin2x +c or 4 cos y = 2x + sin2x +c1.12. From 2y dy = sin3xcos33x dx = tan3xsec23xdx we obtain y2= 16 sec23x +c.13. From ey(ey+ 1)2 dy = ex(ex+ 1)3 dx we obtain (ey+ 1)1= 12 (ex+ 1)2+c.14. From y(1 +y2)1/2 dy = x(1 +x2)1/2 dx we obtain 1 +y2

1/2= 1 +x2

1/2+c.15. From 1S dS = k dr we obtain S = cekr.16. From 1Q70 dQ = k dt we obtain ln|Q70| = kt +c or Q70 = c1ekt.17. From 1P P2dP =

1P + 11 P

dP = dt we obtain ln|P| ln|1 P| = t + c so that ln P1 P = t + c orP1 P = c1et. Solving for P we have P = c1et1 +c1et .18. From 1N dN = tet+21

dt we obtain ln|N| = tet+2et+2t +c.19. From y 2y + 3 dy = x 1x + 4 dx or

1 5y + 3

dy =

1 5x + 4

dx we obtainy 5 ln|y + 3| = x 5 ln|x + 4| +c or

x + 4y + 3

5= c1exy.23Exercises 2.220. From y + 1y 1 dy = x + 2x 3 dx or

1 + 2y 1

dy =

1 + 5x 3

dx we obtainy + 2 ln|y 1| = x + 5 ln|x 3| +c or (y 1)2(x 3)5 = c1exy.21. From xdx = 1

1 y2dy we obtain 12x2= sin1y +c or y = sin

x22 +c1

.22. From 1y2 dy = 1ex+ex dx = ex(ex)2+ 1 dx we obtain 1y = tan1ex+c or y = 1tan1ex+c .23. From 1x2+ 1 dx = 4 dt we obtain tan1x = 4t + c. Using x(/4) = 1 we nd c = 3/4. The solution of theinitial-value problem is tan1x = 4t 34 or x = tan

4t 34

.24. From 1y21 dy = 1x21 dx or 12

1y 1 1y + 1

dy = 12

1x 1 1x + 1

dx we obtainln|y 1| ln|y + 1| = ln|x 1| ln|x + 1| + lnc or y 1y + 1 = c(x 1)x + 1 . Using y(2) = 2 we ndc = 1. The solution of the initial-value problem is y 1y + 1 = x 1x + 1 or y = x.25. From 1y dy = 1 xx2 dx =

1x2 1x

dx we obtain ln|y| = 1x ln|x| = c or xy = c1e1/x. Using y(1) = 1we nd c1 = e1. The solution of the initial-value problem is xy = e11/x.26. From 11 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 = 4e2tor y = 2e2t+ 12 .27. Separating variables and integrating we obtaindx1 x2 dy

1 y2= 0 and sin1x sin1y = c.Setting x = 0 and y = 3/2 we obtain c = /3. Thus, an implicit solution of the initial-value problem issin1x sin1y = /3. Solving for y and using a trigonometric identity we gety = sin

sin1x + 3

= xcos 3 +

1 x2sin 3 = x2 +31 x22 .28. From 11 + (2y)2 dy = x1 + (x2)2 dx we obtain12 tan12y = 12 tan1x2+c or tan12y + tan1x2= c1.Using y(1) = 0 we nd c1 = /4. The solution of the initial-value problem istan12y + tan1x2= 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 +ce4x)/(1 ce4x) we obtain1 = 21 +ce1 ce , 1 ce = 2 + 2ce, 1 = 3ce, and c = 13e .24Exercises 2.2The solution of the corresponding initial-value problem isy = 21 13e4x11 + 13e4x1 = 23 e4x13 +e4x1 .(b) Separating variables and integrating yields14 ln|y 2| 14 ln|y + 2| + lnc1 = xln|y 2| ln|y + 2| + lnc = 4xln

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 isy = 23e +e4x3e e4x = 23 e4x13 +e4x1 .30. From

1y 1 + 1y

dy = 1x dx we obtain ln|y 1| ln|y| = ln|x| + c or y = 11 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 = 11 + 2x .(d) Setting x = 2 and y = 14 we obtain14 = 11 c1(2) , 1 2c1 = 4, and c1 = 32 .Thus, y = 11 + 32x= 22 + 3x .31. Singular solutions of dy/dx = x

1 y2are y = 1 and y = 1. A singular solution of (ex+ex)dy/dx = y2isy = 0.32. Dierentiating ln(x2+ 10) + csc y = c we get2xx2+ 10 cot y csc y dydx = 0,2xx2+ 10 cos ysiny1sinydydx = 0,or2xsin2y dx (x2+ 10) cos y dy = 0.Writing the dierential equation in the formdydx = 2xsin2y(x2+ 10) cos ywe see that singular solutions occur when sin2y = 0, or y = k, where k is an integer.25-0.004-0.002 0.002 0.004 x0.970.9811.01y-0.004-0.002 0.002 0.004 x0.980.991.011.02y-0.004-0.002 0.002 0.004 x0.99960.99981.00021.0004y-0.004-0.002 0.002 0.004 x0.99960.99981.00021.0004yExercises 2.233. The singular solution y = 1 satises the initial-value problem.34. Separating variables we obtain dy(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 isy = x 101x 100 .35. Separating variables we obtain dy(y 1)2+ 0.01 = dx. Then10 tan110(y 1) = x +c and y = 1 + 110 tan x +c10 .Setting x = 0 and y = 1 we obtain c = 0. The solution isy = 1 + 110 tan x10 .36. Separating variables we obtain dy(y 1)20.01 = dx. Then5 ln

10y 1110y 9

= x +c.Setting x = 0 and y = 1 we obtain c = 5 ln1 = 0. The solution is5 ln

10y 1110y 9

= x.37. Separating variables, we havedyy y3 = dyy(1 y)(1 +y) =

1y + 1/21 y 1/21 +y

dy = dx.Integrating, we getln|y| 12 ln|1 y| 12 ln|1 +y| = x +c.261 2 3 4 5 x-4-224y-4 -2 2 4 x-4-224y-4 -2 2 4 x-4-224y1 2 3 4 5 x-4-224y-4 -2 2 4 x-22468y3< >Exercises 2.2When y > 1, this becomeslny 12 ln(y 1) 12 ln(y + 1) = ln y

y21 = x +c.Letting x = 0 and y = 2 we nd c = ln(2/3 ). Solving for y we get y1(x) = 2ex/4e2x3 , where x > ln(3/2).When 0 < y < 1 we havelny 12 ln(1 y) 12 ln(1 +y) = ln y

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 < .When 1 < y < 0 we haveln(y) 12 ln(1 y) 12 ln(1 +y) = ln y

1 y2= x +c.Letting x = 0 and y = 12 we nd c = ln(1/3 ). Solving for y we get y3(x) = ex/e2x+ 3 , where< x < .When y < 1 we haveln(y) 12 ln(1 y) 12 ln(1 y) = ln y

y21 = x +c.Letting x = 0 and y = 2 we nd c = ln(2/3 ). Solving for y we get y4(x) = 2ex/4e2x3 , wherex > ln(3/2).38. (a) The second derivative of y isd2ydx2 = dy/dx(y 1)2 = 1/(y 3)(y 3)2 = 1(y 3)3 .The solution curve is concave up when d2y/dx2> 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-22468y-5 -4 -3 -2 -1 1 2 x-5-4-3-2-112y-2 -1.5 -1 -0.5 x-2-112yExercises 2.2(b) Separating variables and integrating we obtain(y 3) dy = dx12y23y = x +cy26y + 9 = 2x +c1(y 3)2= 2x +c1y = 3 2x +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 = 12 125 , so the exact interval of denition is (, 12 125 ).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 x2is dened at x = 5 and x = 5, y

1(x) is not dened at these values, and so theinterval 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 hyperbolacentered at the origin. All four initial conditions imply c = 0 and y = x. Since the dierential equation is28-6 -4 -2 2 4 6 8 x0.511.522.533.5yxy3 333Exercises 2.2not 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 +y2sin2y

= 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 andget

2uu +u2 du =

21 +u du = 2 ln|1 +u| +c = 2 ln(1 +x) +c.Integrating the separated dierential equation we have2 ln(1 +y ) = 2 ln(1 +x) +c or ln(1 +y ) = ln(1 +x) + lnc1.Solving for y we get y = [c1(1 +x) 1]2.45. We are looking for a function y(x) such thaty2+

dydx

2= 1.Using the positive square root givesdydx = 1 y2= dy

1 y2= dx = sin1y = x +c.Thus a solution is y = sin(x +c). If we use the negative square root we obtainy = 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 = y21 /x21 . Separating variables andintegrating, we obtaindy

y21 = dxx21 and cosh1y = cosh1x +c.29-4 -2 0 2 4-4-2024xy-4 -2 0 2 4-4-2024xyc=-2c=10c=67c=-31-6 -4 -2 0 2 4 6-4-2024xy-2 0 2 4 6-4-2024xy-4 -2 0 2 4 6 8 10-8-6-4-2024xyExercises 2.2Setting x = 2 and y = 2 we nd c = cosh12 cosh12 = 0 and cosh1y = cosh1x. An explicit solutionis y = x.47. (a) Separating variables and integrating, we have(3y2+ 1)dy = (8x + 5)dx and y3+y = 4x25x +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 31.48. (a) Separating variables and integrating, we have(2y +y2)dy = (x x2)dxandy2+ 13y3= 12x2 13x3+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 . Theportion 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 2y36y2+ 2x33x2= 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, ).30Exercises 2.3Exercises 2.31. For y

5y = 0 an integrating factor is e

5 dx= e5xso that ddx

e5xy

= 0 and y = ce5xfor < x < .2. For y

+ 2y = 0 an integrating factor is e

2 dx= e2xso that ddx

e2xy

= 0 and y = ce2xfor < x < .The transient term is ce2x.3. For y

+y = e3xan integrating factor is e

dx= exso that ddx [exy] = e4xand y = 14e3x+cexfor < x < .The transient term is cex.4. For y

+ 4y = 43 an integrating factor is e

4 dx= e4xso that ddx

e4xy

= 43e4xand y = 13 + ce4xfor< x < . The transient term is ce4x.5. For y

+ 3x2y = x2an integrating factor is e

3x2dx= ex3so that ddx

ex3y

= x2ex3and y = 13 + cex3for< x < . The transient term is cex3.6. For y

+ 2xy = x3an integrating factor is e

2x dx= ex2so that ddx

ex2y

= x3ex2and y = 12x2 12 + cex2for < x < . The transient term is cex2.7. For y

+ 1x y = 1x2 an integrating factor is e

(1/x)dx= x so that ddx [xy] = 1x and y = 1x lnx+ cx for 0 < x < .8. For y

2y = x2+ 5 an integrating factor is e

2 dx= e2xso that ddx

e2xy

= x2e2x+ 5e2xandy = 12x2 12x 114 +ce2xfor < x < .9. For y

1x y = xsinx an integrating factor is e

(1/x)dx= 1x so that ddx1x y

= sinx and y = cx xcos x for0 < x < .10. For y

+ 2x y = 3x an integrating factor is e

(2/x)dx= x2so that ddx

x2y

= 3x and y = 32+cx2for 0 < x < .11. For y

+ 4x y = x21 an integrating factor is e

(4/x)dx= x4so that ddx

x4y

= x6x4and y = 17x315x+cx4for 0 < x < .12. For y

x(1 +x) y = x an integrating factor is e

[x/(1+x)]dx= (x+1)exso that ddx

(x + 1)exy

= x(x+1)exand y = x 2x + 3x + 1 + cexx + 1 for 1 < x < .13. For y

+

1 + 2x

y = exx2 an integrating factor is e

[1+(2/x)]dx= x2exso that ddx

x2exy

= e2xand y =12exx2 + cexx2 for 0 < x < . The transient term is cexx2 .14. For y

+

1 + 1x

y = 1xexsin2x an integrating factor is e

[1+(1/x)]dx= xexso that ddx [xexy] = sin2x andy = 12xexcos 2x + cexx for 0 < x < . The entire solution is transient.15. For dxdy 4y x = 4y5an integrating factor is e

(4/y)dy= y4so that ddy

y4x

= 4y and x = 2y6+ cy4for0 < y < .31Exercises 2.316. For dxdy + 2y x = eyan integrating factor is e

(2/y)dy= y2so that ddy

y2x

= y2eyand x = ey2yey+ 2y2 + cy2for 0 < y < . The transient term is 2 +cy2 .17. For y

+ (tanx)y = sec x an integrating factor is e

tan x dx= sec x so that ddx [(sec x) y] = sec2x andy = sinx +c cos x for /2 < x < /2.18. For y

+ (cot x)y = sec2xcsc x an integrating factor is e

cot x dx= sinx so that ddx [(sinx) y] = sec2x andy = sec x +c csc x for 0 < x < /2.19. For y

+ x + 2x + 1 y = 2xexx + 1 an integrating factor is e

[(x+2)/(x+1)]dx= (x + 1)exso that ddx [(x + 1)exy] = 2xand y = x2x + 1 ex+ cx + 1 exfor 1 < x < . The entire solution is transient.20. For y

+ 4x + 2 y = 5(x + 2)2 an integrating factor is e

[4/(x+2)]dx= (x + 2)4so that ddx

(x + 2)4y

= 5(x + 2)2and y = 53(x + 2)1+c(x + 2)4for 2 < x < . The entire solution is transient.21. For drd +r sec = cos an integrating factor is e

sec d= sec + tan so that dd [r(sec + tan)] = 1 + sinand r(sec + tan) = cos +c for /2 < < /2 .22. For dPdt +(2t 1)P = 4t 2 an integrating factor is e

(2t1) dt= et2tso that ddt

Pet2t

= (4t 2)et2tandP = 2 +cett2for < t < . The transient term is cett2.23. For y

+

3 + 1x

y = e3xx an integrating factor is e

[3+(1/x)]dx= xe3xso that ddx

xe3xy

= 1 andy = e3x+ ce3xx for 0 < x < . The transient term is ce3x/x.24. For y

+ 2x21 y = x + 1x 1 an integrating factor is e

[2/(x21)]dx= x 1x + 1 so that ddxx 1x + 1y

= 1 and(x 1)y = x(x + 1) +c(x + 1) for 1 < x < 1.25. For y

+ 1x y = 1x exan integrating factor is e

(1/x)dx= x so that ddx [xy] = exand y = 1xex+ cx for 0 < x < .If y(1) = 2 then c = 2 e and y = 1xex+ 2 ex .26. For dxdy 1y x = 2y an integrating factor is e

(1/y)dy= 1y so that ddy1yx

= 2 and x = 2y2+ cy for< y < . If y(1) = 5 then c = 49/5 and x = 2y2 495 y.27. For didt+RL i = EL an integrating factor is e

(R/L) dt= eRt/Lso that ddt

ieRt/L

= ELeRt/Land i = ER+ceRt/Lfor < t < . If i(0) = i0 then c = i0E/R and i = ER +

i0 ER

eRt/L.28. For dTdt kT = Tmk an integrating factor is e

(k)dt= ektso that ddt [Tekt] = Tmkektand T = Tm+cektfor < t < . If T(0) = T0 then c = T0Tm and T = Tm + (T0Tm)ekt.29. For y

+ 1x + 1 y = lnxx + 1 an integrating factor is e

[1/(x+1)]dx= x + 1 so that ddx[(x + 1)y] = lnx andy = xx + 1 lnx xx + 1 + cx + 1 for 0 < x < . If y(1) = 10 then c = 21 and y = xx + 1 lnx xx + 1 + 21x + 1 .32xy51xy51-1xy32xy5-11Exercises 2.330. For y

+ (tanx)y = cos2x an integrating factor is e

tan x dx= sec x so that ddx [(sec x) y] = cos x and y =sinxcos x +c cos x for /2 < x < /2. If y(0) = 1 then c = 1 and y = sinxcos x cos x.31. For y

+ 2y = f(x) an integrating factor is e2xso thatye2x=

12e2x+c1, 0 x 3;c2, x > 3.If y(0) = 0 then c1 = 1/2 and for continuity we must have c2 = 12e612so thaty =

12(1 e2x), 0 x 3;12(e61)e2x, x > 3.32. For y

+y = f(x) an integrating factor is exso thatyex=

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 thaty =

1, 0 x 1;2e1x1, x > 1.33. For y

+ 2xy = f(x) an integrating factor is ex2so thatyex2=

12ex2+c1, 0 x 1;c2, x > 1.If y(0) = 2 then c1 = 3/2 and for continuity we must have c2 = 12e + 32so thaty =

12 + 32ex2, 0 x 1;

12e + 32

ex2, x > 1.34. For y

+ 2x1 +x2 y =

x1 +x2 , 0 x 1;x1 +x2 , x > 1an integrating factor is 1 +x2so that

1 +x2

y =

12x2+c1, 0 x 1;12x2+c2, x > 1.If y(0) = 0 then c1 = 0 and for continuity we must have c2 = 1 so thaty =

12 12 (1 +x2) , 0 x 1;32 (1 +x2) 12 , x > 1.33xy351015200 1 2 3 4 5 x0.20.40.60.81yExercises 2.335. We need P(x)dx =

2x, 0 x 12 lnx, x > 1 .An integrating factor ise

P(x)dx=

e2x, 0 x 11/x2, x > 1andddxye2x, 0 x 1y/x2, x > 1

=

4xe2x, 0 x 14/x, x > 1 .Integrating we get

ye2x, 0 x 1y/x2, x > 1 =

2xe2xe2x+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. Theny =

2x 1 + 4e2x, 0 x 14x2lnx + (1 + 4e2)x2, x > 1 .36. (a) An integrating factor for y

2xy = 1 is ex2. Thusddx[ex2y] = ex2ex2y =

x0et2dt = 2 erf(x) +c.From y(0) =/2, and noting that erf(0) = 0, we get c =/2. Thusy = ex2

2 erf(x) +2

=2 ex2(1 erf(x)) =2 ex2erfc(x).(b) Using Mathematica we nd y(2) 0.226339.37. An integrating factor for y

2xy = 1 is ex2. Thusddx [ex2y ] = ex2ex2y =

x0et2dt = erf(x) +candy = ex2erf(x) +cex2.From y(1) = 1 we get 1 = e erf(1) +ce, so that c = e1erf(1). Thusy = ex2erf(x) + (e1erf(1))ex2= ex21+ex2(erf(x) erf(1)).38. (a) An integrating factor fory

+ 2x y = 10 sinxx3341 2 3 4 5 x-5-4-3-2-112y2 4 6 8 x2468101214yExercises 2.3is x2. Thusddx[x2y] = 10sinxxx2y = 10

x0sintt dt +cy = 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) 1.64832.39. (a) Separating variables and integrating, we havedyy = sinx2dx and ln|y| =

x0sint2dt +c.Now, letting t = /2 u we have

x0sint2dt =

2

2/ x0sin

2u2

du,soy = c1e

x0 sin t2dt= c1e/2

2/ x0 sin(u2/2) du= c1e/2 S(2/ x).Using S(0) = 0 and y(0) = 5 we see that c1 = 5 and y = 5e/2 S(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. Thusddx

eexy

= eexand eexy =

x0eetdt +c.From y(0) = 1 we get c = e, so y = eex x0 eetdt +e1ex.When y

+exy = 0 we can separate variables and integrate:dyy = exdx and ln|y| = ex+c.Thus y = c1eex. From y(0) = 1 we get c1 = e, so y = e1ex.When y

+exy = exwe can see by inspection that y = 1 is a solution.35Exercises 2.341. On the interval (3, 3) the integrating factor ise

x dx/(x29)= e

x dx/(9x2)= e12 ln(9x2)=

9 x2and soddx

9 x2y

= 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+ cx4satisfy the initial condition. In this case, since 4/x isdiscontinuous 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 byy = x5exx4ex+cx4where c = y0/x40x0ex0+ex0.44. We want the general solution to be y = 3x 5 + cex. (Rather than ex, any function that approaches 0 asx could be used.) Dierentiating we gety

= 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/x3we gety

c = 3cx4 = 3xcx3 = 3x ycso a dierential equation with general solution yc = c/x3is xy

+ 3y = 0. Nowxy

p + 3yp = x(3x2) + 3(x3) = 6x3so a dierential equation with general solution y = c/x3+x3is xy

+ 3y = 6x3.47. Since e

P(x)dx+c= ece

P(x)dx= c1e

P(x)dx, we would havec1e

P(x)dxy = c2 +

c1e

P(x)dxf(x) dx and e

P(x)dxy = c3 +

e

P(x)dxf(x) dx,which is the same as (6) in the text.48. We see by inspection that y = 0 is a solution.36Exercises 2.4Exercises 2.41. 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 x2x + 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 8y3so 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 = siny y sinx and N = cos x +xcos y y so that My = cos y sinx = Nx. From fx = siny y sinxwe obtain f = xsiny +y cos x +h(y), h

(y) = y, and h(y) = 12y2. The solution is xsiny +y cos x 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 x2y23x + 4y = c.6. Let M = 4x33y sin3xy/x2and N = 2y1/x+cos 3x so that My = 3 sin3x1/x2and Nx = 1/x23 sin3x.The equation is not exact.7. Let M = x2y2and N = x22xy 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 obtainf = y +y lnx +h(y), h

(x) = 1 + lnx, and h(y) = xlnx. The solution is y +y lnx +xlnx = c.9. Let M = y3y2sinxx and N = 3xy2+2y cos x so that My = 3y22y sinx = Nx. From fx = y3y2sinxxwe obtain f = xy3+y2cos x 12x2+h(y), h

(y) = 0, and h(y) = 0. The solution is xy3+y2cos x 12x2= c.10. Let M = x3+y3and N = 3xy2so that My = 3y2= Nx. From fx = x3+y3we obtain f = 14x4+xy3+h(y),h

(y) = 0, and h(y) = 0. The solution is 14x4+xy3= c.11. Let M = y lny exyand N = 1/y +xlny so that My = 1 + lny +yexyand Nx = lny. The equation is notexact.12. Let M = 3x2y + eyand N = x3+ xey 2y so that My = 3x2+ ey= Nx. From fx = 3x2y + eywe obtainf = x3y +xey+h(y), h

(y) = 2y, and h(y) = y2. The solution is x3y +xeyy2= c.13. Let M = y 6x2 2xexand N = x so that My = 1 = Nx. From fx = y 6x2 2xexwe obtain f =xy 2x32xex+ 2ex+h(y), h

(y) = 0, and h(y) = 0. The solution is xy 2x32xex+ 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 3y, 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 = x3y2so that My = 3x2y2= Nx. From fx = x2y3 1/

1 + 9x2

weobtain f = 13x3y3 13 arctan(3x) +h(y), h

(y) = 0, and h(y) = 0. The solution is x3y3arctan(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 + 52y2= c.17. Let M = tanx sinxsiny and N = cos xcos y so that My = sinxcos y = Nx. From fx = tanx sinxsinywe obtain f = ln| sec x| + cos xsiny +h(y), h

(y) = 0, and h(y) = 0. The solution is ln| sec x| + cos xsiny = c.18. Let M = 2y sinxcos x y + 2y2exy2and N = x + sin2x + 4xyexy2so thatMy = 2 sinxcos x 1 + 4xy3exy2+ 4yexy2= Nx.From fx = 2y sinxcos x y +2y2exy2we obtain f = y sin2x xy +2exy2+h(y), h

(y) = 0, and h(y) = 0. Thesolution is y sin2x xy + 2exy2= c.37Exercises 2.419. Let M = 4t3y 15t2y and N = t4+3y2t so that My = 4t31 = Nt. From ft = 4t3y 15t2y we obtainf = t4y 5t3ty +h(y), h

(y) = 3y2, and h(y) = y3. The solution is t4y 5t3ty +y3= c.20. Let M = 1/t + 1/t2y/

t2+y2

and N = yey+t/

t2+y2

so that My = y2t2

/

t2+y2

2= Nt. Fromft = 1/t + 1/t2y/

t2+y2

we obtain f = ln|t| 1t arctan

ty

+h(y), h

(y) = yey, and h(y) = yeyey.The solution isln|t| 1t arctan

ty

+yeyey= c.21. Let M = x2+2xy +y2and N = 2xy +x21 so that My = 2(x+y) = Nx. From fx = x2+2xy +y2we obtainf = 13x3+x2y +xy2+h(y), h

(y) = 1, and h(y) = y. The general solution is 13x3+x2y +xy2y = c. Ify(1) = 1 then c = 4/3 and the solution of the initial-value problem is 13x3+x2y +xy2y = 43 .22. Let M = ex+y and N = 2 +x +yeyso that My = 1 = Nx. From fx = ex+y we obtain f = ex+xy +h(y),h

(y) = 2 +yey, and h(y) = 2y +yeyy. The general solution is ex+xy +2y +yeyey= c. If y(0) = 1 thenc = 3 and the solution of the initial-value problem is ex+xy + 2y +yeyey= 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 +t25t +h(y), h

(y) = 6y 1, and h(y) = 3y2y. The general solution is 4ty +t25t +3y2y = c.If y(1) = 2 then c = 8 and the solution of the initial-value problem is 4ty +t25t + 3y2y = 8.24. Let M = t/2y4and N = 3y2t2

/y5so that My = 2t/y5= Nt. From ft = t/2y4we obtain f = t24y4 +h(y),h

(y) = 3y3, and h(y) = 32y2 . The general solution is t24y4 32y2 = c. If y(1) = 1 then c = 5/4 and thesolution of the initial-value problem is t24y4 32y2 = 54 .25. Let M = y2cos x 3x2y 2x and N = 2y sinx x3+ lny so that My = 2y cos x 3x2= Nx. Fromfx = y2cos x 3x2y 2x we obtain f = y2sinx x3y x2+ h(y), h

(y) = lny, and h(y) = y lny y. Thegeneral solution is y2sinxx3y x2+y lny y = c. If y(0) = e then c = 0 and the solution of the initial-valueproblem is y2sinx x3y x2+y lny y = 0.26. Let M = y2+ y sinx and N = 2xy cos x 1/

1 +y2

so that My = 2y + sinx = Nx. From fx =y2+ y sinx we obtain f = xy2 y cos x + h(y), h

(y) = 11 +y2 , and h(y) = tan1y. The general solutionis xy2 y cos x tan1y = c. If y(0) = 1 then c = 1 /4 and the solution of the initial-value problem isxy2y cos x tan1y = 1 4 .27. Equating My = 3y2+ 4kxy3and Nx = 3y2+ 40xy3we obtain k = 10.28. Equating My = 18xy2siny and Nx = 4kxy2siny we obtain k = 9/2.29. Let M = x2y2sinx + 2xy2cos x and N = 2x2y cos x so that My = 2x2y sinx + 4xy cos x = Nx. Fromfy = 2x2y cos x we obtain f = x2y2cos x + h(y), h

(y) = 0, and h(y) = 0. The solution of the dierentialequation is x2y2cos x = 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 +y22y2

/(x +y)2we obtain f = x + 2y2x +y +h(y), h

(y) = 1,and h(y) = y. The solution of the dierential equation is x2+y2= c(x +y).381 2 3 4 5 6 7 x0.511.522.5y-4 -2 2 4 x-6-4-224yy1y2Exercises 2.431. We note that (MyNx)/N = 1/x, so an integrating factor is e

dx/x= x. Let M = 2xy2+ 3x2and N = 2x2yso that My = 4xy = Nx. From fx = 2xy2+ 3x2we 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 e

dx= ex. Let M = xyex+ y2ex+ yexandN = xex+2yexso that My = xex+2yex+ex= Nx. From fy = xex+2yexwe 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 = 6xy3and N = 4y3+9x2y2so that My = 18xy2= Nx. From fx = 6xy3we 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 = cot x, so an integrating factor is e

cot x dx= csc x. Let M = cos xcsc x = cot xand N = (1 +2/y) sinxcsc x = 1 +2/y, so that My = 0 = Nx. From fx = cot x we obtain f = ln(sinx) +h(y),h

(y) = 1 + 2/y, and h(y) = y + lny2. The solution of the dierential equation is ln(sinx) +y + lny2= 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(siny) dy = (cos x) 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 cos1tis [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 obtain3x2+ 2x2 dydx + 4xy + 2y dydx = 0 and dydx = 3x2+ 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 gety1(x) = x2

4 x3+x4and 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 +xdy = 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 + cos2x)/(1 x2) . Since 3 + cos2x > 0 for all x we must have 1 x2> 0 or1 < x < 1. Thus, the interval of denition is (1, 1).39Exercises 2.439. (a) Since fy = N(x, y) = xexy+2xy+1/x we obtain f = exy+xy2+yx+h(x) so that fx = yexy+y2 yx2 +h

(x).Let M(x, y) = yexy+y2 yx2 .(b) Since fx = M(x, y) = y1/2x1/2+ x

x2+y

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

x2+y

+ g(y) so thatfy = 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 + y

x2+y2dy.Then xdx +y dy = x2+y2dx becomesx

x2+y2dx + y

x2+y2dy = d

x2+y2

= dx.The left side is the total dierential of x2+y2and the right side is the total dierential of x + c. Thus

x2+y2= x +c is a solution of the dierential equation.Exercises 2.51. Letting y = ux we have(x ux) dx +x(udx +xdu) = 0dx +xdu = 0dxx +du = 0ln|x| +u = cxln|x| +y = cx.2. Letting y = ux we have(x +ux) dx +x(udx +xdu) = 0(1 + 2u) dx +xdu = 0dxx + du1 + 2u = 0ln|x| + 12 ln|1 + 2u| = cx2

1 + 2yx

= c1x2+ 2xy = c1.3. Letting x = vy we havevy(v dy +y dv) + (y 2vy) dy = 0vy dv +

v22v + 1

dy = 0v dv(v 1)2 + dyy = 0ln|v 1| 1v 1 + ln|y| = cln

xy 1

1x/y 1 + lny = c(x y) ln|x y| y = c(x y).40Exercises 2.54. Letting x = vy we havey(v dy +y dv) 2(vy +y) dy = 0y dv (v + 2) dy = 0dvv + 2 dyy = 0ln|v + 2| ln|y| = cln

xy + 2

ln|y| = cx + 2y = c1y2.5. Letting y = ux we have

u2x2+ux2

dx x2(udx +xdu) = 0u2dx xdu = 0dxx duu2 = 0ln|x| + 1u = cln|x| + xy = cy ln|x| +x = cy.6. Letting y = ux we have

u2x2+ux2

dx +x2(udx +xdu) = 0

u2+ 2u

dx +xdu = 0dxx + duu(u + 2) = 0ln|x| + 12 ln|u| 12 ln|u + 2| = cx2uu + 2 = c1x2yx = c1

yx + 2

x2y = c1(y + 2x).7. Letting y = ux we have(ux x) dx (ux +x)(udx +xdu) = 0

u2+ 1

dx +x(u + 1) du = 0dxx + u + 1u2+ 1 du = 0ln|x| + 12 ln

u2+ 1

+ tan1u = clnx2

y2x2 + 1

+ 2 tan1 yx = c1ln

x2+y2

+ 2 tan1 yx = c1.41Exercises 2.58. Letting y = ux we have(x + 3ux) dx (3x +ux)(udx +xdu) = 0

u21

dx +x(u + 3) du = 0dxx + u + 3(u 1)(u + 1) du = 0ln|x| + 2 ln|u 1| ln|u + 1| = cx(u 1)2u + 1 = c1x

yx 1

2= c1

yx + 1

(y x)2= c1(y +x).9. Letting y = ux we haveuxdx + (x +ux)(udx +xdu) = 0(x +xu) du +u3/2dx = 0

u3/2+ 1u

du + dxx = 02u1/2+ ln|u| + ln|x| = cln|y/x| + ln|x| = 2

x/y +cy(ln|y| c)2= 4x.10. Letting y = ux we have

ux +

x2+u2x2

dx x(udx +xdu) = 0x

1 +u2dx x2du = 0dxx du1 +u2= 0ln|x| ln

u +

1 +u2

= cu +

1 +u2= c1xy +

y2+x2= c1x2.11. Letting y = ux we have

x3u3x3

dx +u2x3(udx +xdu) = 0dx +u2xdu = 0dxx +u2du = 0ln|x| + 13u3= c3x3ln|x| +y3= c1x3.Using y(1) = 2 we nd c1 = 8. The solution of the initial-value problem is 3x3ln|x| +y3= 8x3.42Exercises 2.512. Letting y = ux we have

x2+ 2u2x2

dx ux2(udx +xdu) = 0

1 +u2

dx uxdu = 0dxx udu1 +u2 = 0ln|x| 12 ln

1 +u2

= cx21 +u2 = c1x4= 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(udx +xdu) = 0dx xeudu = 0dxx eudu = 0ln|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/x1.14. Letting x = vy we havey(v dy +y dv) +vy(lnvy lny 1) dy = 0y dv +v lnv dy = 0dvv lnv + dyy = 0ln|ln|v|| + ln|y| = cy ln

xy

= 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 = 1xy2and w = y3we obtain dwdx + 3xw = 3x . An integrating factor is x3so that x3w = x3+cor y3= 1 +cx3.16. From y

y = exy2and w = y1we obtain dwdx +w = ex. An integrating factor is exso that exw = 12e2x+cor y1= 12ex+cex.17. From y

+ y = xy4and w = y3we obtain dwdx 3w = 3x. An integrating factor is e3xso that e3x=xe3x+ 13e3x+c or y3= x + 13 +ce3x.18. From y

1 + 1x

y = y2and w = y1we obtain dwdx +

1 + 1x

w = 1. An integrating factor is xexso thatxexw = xex+ex+c or y1= 1 + 1x + cxex.43Exercises 2.519. From y

1ty = 1t2y2and w = y1we obtain dwdt + 1tw = 1t2 . An integrating factor is t so that tw = lnt +cor y1= 1t lnt + ct. Writing this in the form ty = lnx +c, we see that the solution can also be expressed in theform et/y= c1x.20. From y

+ 23 (1 +t2)y = 2t3 (1 +t2)y4and w = y3we obtain dwdt 2t1 +t2w = 2t1 +t2 . An integrating factor is11 +t2 so that w1 +t2 = 11 +t2 +c or y3= 1 +c

1 +t2

.21. From y

2xy = 3x2y4and w = y3we obtain dwdx + 6xw = 9x2 . An integrating factor is x6so thatx6w = 95x5+c or y3= 95x1+cx6. If y(1) = 12 then c = 495 and y3= 95x1+ 495 x6.22. From y

+ y = y1/2and w = y3/2we obtain dwdx + 32w = 32 . An integrating factor is e3x/2so 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. Then dudx 1 = u2or 11 +u2 du = dx. Thus tan1u = x +c oru = 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. Then dudx1 = 1 uu or udu = 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. Then dudx1 = tan2u or cos2udu = dx. Thus 12u + 14 sin2u = x +cor 2u + sin2u = 4x +c1, and 2(x +y) + sin2(x +y) = 4x +c1 or 2y + sin2(x +y) = 2x +c1.26. Let u = x + y so that du/dx = 1 + dy/dx. Then dudx 1 = sinu or 11 + sinu du = dx. Multiplying by(1 sinu)/(1 sinu) we have 1 sinucos2u du = dx or sec2u tanusec u

du = dx. Thus tanu sec u = x + cor 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 1u du = dx. Thus 2u = x +c and2y 2x + 3 = x +c.28. Let u = y x + 5 so that du/dx = dy/dx 1. Then dudx + 1 = 1 +euor eudu = dx. Thus eu= x +c andeyx+5= x +c.29. Let u = x +y so that du/dx = 1 +dy/dx. Then dudx 1 = cos u and 11 + cos u du = dx. Now11 + cos u = 1 cos u1 cos2u = 1 cos usin2u= csc2u csc ucot uso we have (csc2u csc ucot u)du = dx and cot u +csc u = x +c. Thus cot(x +y) +csc(x +y) = x +c.Setting x = 0 and y = /4 we obtain c =2 1. The solution iscsc(x +y) cot(x +y) = x +2 1.30. Let u = 3x + 2y so that du/dx = 3 + 2 dy/dx. Then dudx = 3 + 2uu + 2 = 5u + 6u + 2 and u + 25u + 6 du = dx. Nowu + 25u + 6 = 15 + 425u + 3044Exercises 2.5so we have 15 + 425u + 30

du = dxand 15u + 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 ln95. The solution is15(3x + 2y) + 425 ln|75x + 50y + 30| = x + 45 ln95or5y 5x + 2 ln|75x + 50y + 30| = 10 ln95.31. We write the dierential equation M(x, y)dx +N(x, y)dy = 0 as dy/dx = f(x, y) wheref(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 haveM(1, y/x) = 1x M(x, y) or M(x, y) = xM(1, y/x).Thusdydx = f(x, y) = xM(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 formdydx = G

xy

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 , e6x0 and y 2x 3.33. (a) The substitutions y = y1 +u anddydx = dy1dx + dudxlead tody1dx + dudx = P +Q(y1 +u) +R(y1 +u)2= P +Qy1 +Ry21 +Qu + 2y1Ru +Ru2ordudx (Q+ 2y1R)u = Ru2.This is a Bernoulli equation with n = 2 which can be reduced to the linear equationdwdx + (Q+ 2y1R)w = Rby the substitution w = u1.45Exercises 2.5(b) Identify P(x) = 4/x2, Q(x) = 1/x, and R(x) = 1. Then dwdx +

1x + 4x

w = 1. An integratingfactor is x3so that x3w = 14x4+c or u =14x +cx3

1. Thus, y = 2x +u.34. Write the dierential equation in the form x(y

/y) = lnx + lny and let u = lny. 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)ddx

1x u

= lnxx2 and ux = 1x lnxx +c.The solution islny = 1 lnx +cx or y = ecx1x .Exercises 2.61. We identify f(x, y) = 2x 3y + 1. Then, for h = 0.1,yn+1 = yn + 0.1(2xn3yn + 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 = 0.05yn+1 = yn + 0.05(2xn3yn + 1) = 0.1xn + 0.85yn + 0.1,andy(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 = 3.115142. 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 = 0.05yn+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= 0.0150.46h=0.1 h=0.05xn ynTrue ValueAbs. Error% Rel. Error xn ynTrue ValueAbs. Error% Rel. Error0.00 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 1.200.55 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 2.39h=0.1 h=0.05xn ynTrue ValueAbs. Error% Rel. Error xn ynTrue ValueAbs. Error% Rel. Error1.00 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 4.001.30 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 9.08h=0.1 h=0.05xn 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 0.22780.30 0.26760.35 0.30580.40 0.34270.45 0.37820.50 0.4124h=0.1 h=0.05xn 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 1.31440.30 1.40390.35 1.50700.40 1.62670.45 1.76700.50 1.9332Exercises 2.63. Separating variables and integrating, we havedyy = dx and ln|y| = x +c.Thus y = c1exand, using y(0) = 1, we nd c = 1, so y = exis the solution of the initial-value problem.4. Separating variables and integrating, we havedyy = 2xdx and ln|y| = x2+c.Thus y = c1ex2and, using y(1) = 1, we nd c = e1, so y = ex21is the solution of the initial-value problem.5. 6.47h=0.1 h=0.05xn 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 0.54520.30 0.54960.35 0.55270.40 0.55470.45 0.55590.50 0.5565h=0.1 h=0.05xn 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 1.29210.30 1.36510.35 1.44400.40 1.52930.45 1.62170.50 1.7219h=0.1 h=0.05xn 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 1.05061.30 1.07751.35 1.11151.40 1.15381.45 1.20571.50 1.2696h=0.1 h=0.05xn 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 0.56230.30 0.57460.35 0.58680.40 0.59890.45 0.61090.50 0.62285 1 0 - 5 - 1 05xyxEulerRunge-Kuttah=0.255 10 - 5 - 105h=0.1xyRunge-KuttaEuler5 10 - 5 - 105EulerRunge-Kuttaxyh=0.052 4 - 25h=0.25Runge-KuttaEulerxy2 4 - 25yxEulerRunge-Kuttah=0.12 4 - 25h=0.05EulerRunge-KuttaxyExercises 2.67. 8.9. 10.11.12.48h=0.1 Euler h=0.05 Euler h=0.1 R-K h=0.05 R-Kxn 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 1.33330.55 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 84.0132Exercises 2.713. 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 2.71. 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 ln2 and P = P0e(ln 2)t/5. Setting P(t) = 3P0 we have3 = e(ln 2)t/5= ln3 = (ln2)t5 = t = 5 ln3ln2 7.9 years.Setting P(t) = 4P0 we have4 = e(ln 2)t/5= ln4 = (ln2)t5 = t = 10 years.2. Setting P =10,000 and t = 3 in Problem 1 we obtain10,000 = 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 ln1.15. Then P(30) = 500e3 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 obtainN = N0ekt. Using N(3) = 400 and N(10) = 2000 we nd 400 = N0e3kor ek= (400/N0)1/3. From N(10) = 2000we then have2000 = N0e10k= N0

400N0

10/3= 200040010/3 = N7/30 = N0 =

200040010/3

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 ln0.25, and I(15) = 0.00098I0.6. From dS/dt = rS we obtain S = S0ertwhere S(0) = S0.49Exercises 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 $6651.827. 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 ln1/2. When 90% of the lead has decayed, 0.1 grams will remain. SettingN(t) = 0.1 we haveet(1/3.3) ln(1/2)= 0.1 = t3.3 ln 12 = ln0.1 = t = 3.3 ln0.1ln1/2 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 ln0.97. 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 obtain50 = 100ekt= kt = ln 12 = t = ln1/2(1/6) ln0.97 136.5 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 = (ln2)/k.(b) Since k = (ln2)/T we haveA(t) = A0e(ln 2)t/T= A02t/T.11. Assume that A = A0ektand 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 solvingfor 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. If T(t) = 15 then t = 3.06 minutes.14. Assume that dT/dt = k(T 5) so that T = 5 + cekt. If T(1)