1

CHA P T E R

1

Introduction

1.1

1.

For y > 3/2, the slopes are negative, therefore the solutions are decreasing. Fory < 3/2, the slopes are positive, hence the solutions are increasing. The equilibriumsolution appears to be y(t) = 3/2, to which all other solutions converge.

2 Chapter 1. Introduction

2.

For y > 3/2, the slopes are positive, therefore the solutions increase. For y < 3/2,the slopes are negative, therefore, the solutions decrease. As a result, y divergesfrom 3/2 as t→∞ if y(0) 6= 3/2.

3.

For y > −1/2, the slopes are negative, therefore the solutions decrease. For y <−1/2, the slopes are positive, therefore, the solutions increase. As a result, y →−1/2 as t→∞.

4.

1.1 3

For y > −1/2, the slopes are positive, and hence the solutions increase. For y <−1/2, the slopes are negative, and hence the solutions decrease. All solutions divergeaway from the equilibrium solution y(t) = −1/2.

5. For all solutions to approach the equilibrium solution y(t) = 2/3, we must havey ′ < 0 for y > 2/3, and y ′ > 0 for y < 2/3. The required rates are satisfied by thedifferential equation y ′ = 2− 3y.

6. For solutions other than y(t) = 2 to diverge from y = 2, y(t) must be an increas-ing function for y > 2, and a decreasing function for y < 2. The simplest differentialequation whose solutions satisfy these criteria is y ′ = y − 2.

7.

For y = 0 and y = 4 we have y′ = 0 and thus y = 0 and y = 4 are equilibriumsolutions. For y > 4, y′ < 0 so if y(0) > 4 the solution approaches y = 4 fromabove. If 0 < y(0) < 4, then y′ > 0 and the solutions ”grow” to y = 4 as t→∞.For y(0) < 0 we see that y′ < 0 and the solutions diverge from 0.

8.

Note that y ′ = 0 for y = 0 and y = 5. The two equilibrium solutions are y(t) = 0and y(t) = 5. Based on the direction field, y ′ > 0 for y > 5; thus solutions withinitial values greater than 5 diverge from the solution y(t) = 5. For 0 < y < 5, theslopes are negative, and hence solutions with initial values between 0 and 5 alldecrease toward the solution y(t) = 0. For y < 0, the slopes are all positive; thussolutions with initial values less than 0 approach the solution y(t) = 0.

4 Chapter 1. Introduction

9.

Since y′ = y2, y = 0 is the only equilibrium solution and y′ > 0 for all y. Thus y → 0if the initial value is negative; y diverges from 0 if the initial value is positive.

10.

Observe that y ′ = 0 for y = 0 and y = 2. The two equilibrium solutions are y(t) = 0and y(t) = 2. Based on the direction field, y ′ > 0 for y > 2; thus solutions withinitial values greater than 2 diverge from y(t) = 2. For 0 < y < 2, the slopes arealso positive, and hence solutions with initial values between 0 and 2 all increasetoward the solution y(t) = 2. For y < 0, the slopes are all negative; thus solutionswith initial values less than 0 diverge from the solution y(t) = 0.

11. -(j) y ′ = 2− y.

12. From Figure 1.1.6 we can see that y = 2 is an equilibrium solution and thus(c) and (j) are the only possible differential equations to consider. Since dy/dt > 0for y > 2, and dy/dt < 0 for y < 2 we conclude that (c) is the correct answer:y ′ = y − 2.

13. -(g) y ′ = −2− y.

14. -(b) y ′ = 2 + y.

15. From Figure 1.1.9 we can see that y = 0 and y = 3 are equilibrium solutions,so (e) and (h) are the only possible differential equations. Furthermore, we have

1.1 5

dy/dt < 0 for y > 3 and for y < 0, and dy/dt > 0 for 0 < y < 3. This tells us that(h) is the desired differential equation: y ′ = y (3− y).

16. -(e) y ′ = y (y − 3).

17.(a) Let a(t) denote the amount of chemical in the pond at time t. The amounta will be measured in grams and the time t will be measured in hours. The rateat which the chemical is entering the pond is given by 300 gal/h · .01 g/gal =3 g/h. The rate at which the chemical leaves the pond is given by 300 gal/h ·a/106 g/gal = (3× 10−4)a g/h. Thus the differential equation is given by da/dt =3− (3× 10−4)a.

(b) The equilibrium solution occurs when a′ = 0, or a = 104 grams. Since a′ > 0 fora < 104 g and a′ < 0 for a > 104 g, all solutions approach the equilibrium solutionindependent of the amount present at t = 0.

(c) Let a(t) denote the amount of chemical in the pond at time t. From part (a)the function a(t) satisfies the differential equation da/dt = 3− (3× 10−4)a. Thus interms of the concentration c(t) = a(t)/106, dc/dt = (1/106)(da/dt) = (1/106)(3−(3× 10−4)a) = (3× 10−6)− (10−6)(3× 10−4)a = (3× 10−6)− (3× 10−4)c.

18. The surface area of a spherical raindrop of radius r is given by S = 4πr2. The

volume of a spherical raindrop is given by V = 4πr3/3. Therefore, we see that thesurface area S = cV 2/3 for some constant c. If the raindrop evaporates at a rateproportional to its surface area, then dV/dt = −kV 2/3 for some k > 0.

19. The difference between the temperature of the object and the ambient tem-perature is u− 70 (u in ◦F). Since the object is cooling when u > 70, and the rateconstant is k = 0.05 min−1, the governing differential equation for the temperatureof the object is du/dt = −.05 (u− 70).

20.(a) Let M(t) be the total amount of the drug (in milligrams) in the patient’sbody at any given time t (hr). The drug enters the body at a constant rate of 500mg/hr. The rate at which the drug leaves the bloodstream is given by 0.4M(t).Hence the accumulation rate of the drug is described by the differential equationdM/dt = 500− 0.4M (mg/hr).

(b)

6 Chapter 1. Introduction

Based on the direction field, the amount of drug in the bloodstream approaches theequilibrium level of 1250 mg (within a few hours).

21.(a) Following the discussion in the text, the differential equation is m(dv/dt) =

mg − γ v2, or equivalently, dv/dt = g − γv2/m.

(b) After a long time, dv/dt ≈ 0. Hence the object attains a terminal velocity givenby v∞ =

√mg/γ .

(c) Using the relation γ v 2∞ = mg, the required drag coefficient is γ = 2/49 kg/s.

(d)

22.

All solutions become asymptotic to the line y = t− 3 as t→∞.

1.1 7

23.

The solutions approach −∞, 0, or ∞ depending upon the value of y(0).

24.

The solutions approach −∞, ∞, or oscillate depending upon the value of y(0).

25.

For all y(0), there is a number tf that depends on the value of y(0) such thaty(tf ) = 0 and the solution does not exist for t > tf .

8 Chapter 1. Introduction

1.2

1.(a) The differential equation can be rewritten as

dy

5− y= dt .

Integrating both sides of this equation results in − ln |5− y| = t+ c1, or equiva-lently, 5− y = c e−t. Applying the initial condition y(0) = y0 results in the specifi-cation of the constant as c = 5− y0. Hence the solution is y(t) = 5 + (y0 − 5)e−t.

All solutions appear to converge to the equilibrium solution y(t) = 5.

(b) The differential equation can be rewritten as

dy

5− 2y= dt .

Integrating both sides of this equation results in −(1/2) ln |5− 2y| = t+ c1, orequivalently, 5− 2y = c e−2t. Applying the initial condition y(0) = y0 results in thespecification of the constant as c = 5− 2y0. Hence y(t) = 5/2 + (y0 − 5/2)e−2t.

All solutions appear to converge to the equilibrium solution y(t) = 5/2, at a fasterrate than in part (a).

(c) Rewrite the differential equation as

dy

10− 2y= dt .

1.2 9

Integrating both sides of this equation results in −(1/2) ln |10− 2y| = t+ c1, orequivalently, 5− y = c e−2t. Applying the initial condition y(0) = y0 results in thespecification of the constant as c = 5− y0. Hence y(t) = 5 + (y0 − 5)e−2t.

All solutions appear to converge to the equilibrium solution y(t) = 5, at a fasterrate than in part (a), and at the same rate as in part (b).

2.(a) The differential equation can be rewritten as

dy

y − 5= dt .

Integrating both sides of this equation results in ln |y − 5| = t+ c1, or equivalently,y − 5 = c et. Applying the initial condition y(0) = y0 results in the specification ofthe constant as c = y0 − 5. Hence the solution is y(t) = 5 + (y0 − 5)et.

All solutions appear to diverge from the equilibrium solution y(t) = 5.

(b) Rewrite the differential equation as

dy

2y − 5= dt .

Integrating both sides of this equation results in (1/2) ln |2y − 5| = t+ c1, or equiv-alently, 2y − 5 = c e2t. Applying the initial condition y(0) = y0 results in the speci-fication of the constant as c = 2y0 − 5. So the solution is y(t) = (y0 − 5/2)e2t + 5/2.

10 Chapter 1. Introduction

All solutions appear to diverge from the equilibrium solution y(t) = 5/2.

(c) The differential equation can be rewritten as

dy

2y − 10= dt .

Integrating both sides of this equation results in (1/2) ln |2y − 10| = t+ c1, or equiv-alently, y − 5 = c e2t. Applying the initial condition y(0) = y0 results in the speci-fication of the constant as c = y0 − 5. Hence the solution is y(t) = 5 + (y0 − 5)e2t.

All solutions appear to diverge from the equilibrium solution y(t) = 5.

3.(a) Rewrite the differential equation as

dy

b− ay= dt ,

which is valid for y 6= b /a. Integrating both sides results in −(1/a) ln |b− ay| =t+ c1, or equivalently, b− ay = c e−at. Hence the general solution is y(t) = (b−c e−at)/a. Note that if y = b/a, then dy/dt = 0, and y(t) = b/a is an equilibriumsolution.

1.2 11

(b)

(c) (i) As a increases, the equilibrium solution gets closer to y(t) = 0, from above.The convergence rate of all solutions is a. As a increases, the solutions converge tothe equilibrium solution quicker.

(ii) As b increases, then the equilibrium solution y(t) = b/a also becomes larger.In this case, the convergence rate remains the same.

(iii) If a and b both increase but b/a =constant, then the equilibrium solutiony(t) = b/a remains the same, but the convergence rate of all solutions increases.

4.(a) The equilibrium solution satisfies the differential equation dye/dt = 0. Settingaye − b = 0, we obtain ye(t) = b/a.

(b) Since dY/dt = dy/dt, it follows that dY/dt = a(Y + ye)− b = a Y .

5.(a) Rewrite Eq.(ii) as (1/y)dy/dt = a and thus ln |y| = at+ C, or y1 = ceat.

(b) If y = y1(t) + k, then dy/dt = dy1/dt. Substituting both these into Eq.(i) we getdy1/dt = a(y1 + k)− b. Since dy1/dt = ay1, this leaves ak − b = 0 and thus k = b/a.Hence y = y1(t) + b/a is the solution to Eq(i).

(c) Substitution of y1 = ceat shows this is the same as that given in Eq.(17).

6.(a) Consider the simpler equation dy1/dt = −ay1. As in the previous solutions,rewrite the equation as (1/y1)dy1 = −a dt. Integrating both sides results in y1(t) =c e−at.

(b) Now set y(t) = y1(t) + k, and substitute into the original differential equation.We find that −ay1 + 0 = −a(y1 + k) + b. That is, −ak + b = 0, and hence k = b/a.

(c) The general solution of the differential equation is y(t) = c e−at + b/a. Thisis exactly the form given by Eq.(17) in the text. Invoking an initial conditiony(0) = y0, the solution may also be expressed as y(t) = b/a+ (y0 − b/a)e−at.

7.(a) The general solution is p(t) = 900 + c et/2, that is, p(t) = 900 + (p0 − 900)et/2.With p0 = 850, the specific solution becomes p(t) = 900− 50et/2. This solution is a

12 Chapter 1. Introduction

decreasing exponential, and hence the time of extinction is equal to the number ofmonths it takes, say tf , for the population to reach zero. Solving 900− 50etf/2 = 0,we find that tf = 2 ln(900/50) ≈ 5.78 months.

(b) The solution, p(t) = 900 + (p0 − 900)et/2, is a decreasing exponential as long asp0 < 900. Hence 900 + (p0 − 900)etf/2 = 0 has only one root, given by

tf = 2 ln

(900

900− p0

)months.

(c) The answer in part (b) is a general equation relating time of extinction tothe value of the initial population. Setting tf = 12 months, the equation may bewritten as 900/(900− p0) = e6, which has solution p0 ≈ 897.8. Since p0 is the initialpopulation, the appropriate answer is p0 = 898 mice.

8.(a) The solution to this problem is given by Eq.(26), which has a limiting velocity

of 49 m/s. Substituting v = 48.02 (which is 98% of 49) into Eq.(26) yields et/5 =0.02. Solving for t we have t = −5 ln(0.02) = 19.56 s.

(b) Use Eq.(29) with t = 19.56.

9.(a) If the drag force is proportional to v2 then F = 98− kv2 is the net forceacting on the falling mass (m = 10 kg). Thus 10dv/dt = 98− kv2, which has alimiting velocity of v2 = 98/k. Setting v2 = 492 gives k = 98/492 and hence dv/dt =(492 − v2)/(10/k) = (492 − v2)/245.

(b) From part (a) we have dv/(492 − v2) = (1/245)dt, which after integration yields(1/49) tanh−1(v/49) = t/245 + C0. Setting t = 0 and v = 0, we have 0 = 0 + C0,or C0 = 0. Thus tanh

−1(v/49) = t/5, or v(t) = 49 tanh(t/5) m/s.

(c)

(d) In the graph of part (c), the solution for the linear drag force lies below thesolution for the quadratic drag force. This latter solution approaches equilibriumfaster since, as the velocity increases, there is a larger drag force.

1.2 13

(e) Note that∫

tanh(x) dx = ln(coshx) + C. The distance the object falls is givenby x = 245 ln(cosh(t/5)).

(f) Solving 300 = 245 ln(cosh(T/5)) gives T ≈ 9.48 s.

10.(a,b) The general solution of the differential equation isQ(t) = c e−rt. Given thatQ(0) = 100 mg, the value of the constant is given by c = 100. Hence the amount ofthorium-234 present at any time is given by Q(t) = 100 e−rt. Furthermore, basedon the hypothesis, setting t = 1 results in 82.04 = 100 e−r. Solving for the rateconstant, we find that r = − ln(82.04/100) ≈ .19796/week or r ≈ .02828/day.

(c) Let T be the time that it takes the isotope to decay to one-half of its originalamount. From part (a), it follows that 50 = 100 e−rT , in which r = .19796/week.Taking the natural logarithm of both sides, we find that T ≈ 3.5014 weeks or T ≈24.51 days.

11. The general solution of the differential equation dQ/dt = −r Q isQ(t) = Q0e−rt,in which Q0 = Q(0) is the initial amount of the substance. Let τ be the time thatit takes the substance to decay to one-half of its original amount, Q0. Setting t = τin the solution, we have 0.5Q0 = Q0e

−rτ . Taking the natural logarithm of bothsides, it follows that −rτ = ln(0.5) or rτ = ln 2.

12.(a) Rewrite the differential equation as du/(u− T ) = −kdt and then integrate tofind ln |u− T | = −kt+ c. Thus u− T = ±Ce−kt. For t = 0 we have u0 − T = ±Cand thus u(t) = T + (u0 − T )e−kt.

(b) Set u(τ)− T = (u0 − T )/2 in the solution in part (a). We obtain 1/2 = e−kτ ,or kτ = ln 2.

13.(a) Rewrite the differential equation as dQ/(Q− CV ) = −(1/CR)dt, thus, uponintegrating and simplifying, we get Q = De−t/CR + CV . Q(0) = 0 implies that thesolution of the differential equation is Q(t) = CV (1− e−t/CR).

(b) As t → ∞, the exponential term vanishes, and the limiting value is QL = CV .

14 Chapter 1. Introduction

(c) In this case RdQ/dt+Q/C = 0. Q(t1) = CV . The solution of this differentialequation is Q(t) = Ee−t/CR, so Q(t1) = Ee

−t1/CR = CV , or E = CV et1/CR. ThusQ(t) = CV et1/CRe−t/CR = CV e−(t−t1)/CR for t ≥ t1.

14.(a) The accumulation rate of the chemical is (0.01)(300) grams per hour. Atany given time t, the concentration of the chemical in the pond is Q(t)/106 gramsper gallon. Consequently, the chemical leaves the pond at a rate of (3× 10−4)Q(t)grams per hour. Hence, the rate of change of the chemical is given by

dQ

dt= 3− 0.0003Q(t) g/hr.

Since the pond is initially free of the chemical, Q(0) = 0.

(b) The differential equation can be rewritten as dQ/(10000−Q) = 0.0003 dt. In-tegrating both sides of the equation results in − ln |10000−Q| = 0.0003t+ C. Tak-ing the exponential of both sides gives 10000−Q = c e−0.0003t. Since Q(0) = 0, thevalue of the constant is c = 10000. Hence the amount of chemical in the pond atany time is Q(t) = 10000(1− e−0.0003t) grams. Note that 1 year= 8760 hours. Set-ting t = 8760, the amount of chemical present after one year is Q(8760) ≈ 9277.77grams, that is, 9.27777 kilograms.

(c) With the accumulation rate now equal to zero, the governing equation becomesdQ/dt = −0.0003Q(t) g/hr. Resetting the time variable, we now assign the newinitial value as Q(0) = 9277.77 grams.

(d) The solution of the differential equation in part (c) is Q(t) = 9277.77 e−0.0003t.Hence, one year after the source is removed, the amount of chemical in the pond isQ(8760) ≈ 670.1 grams.

(e) Letting t be the amount of time after the source is removed, we obtain the equa-tion 10 = 9277.77 e−0.0003t. Taking the natural logarithm of both sides, −0.0003 t =ln(10/9277.77) or t ≈ 22, 776 hours≈ 2.6 years.

1.3 15

(f)

1.3

1. The differential equation is second order, since the highest derivative in the equa-tion is of order two. The equation is linear, since the left hand side is a linearfunction of y and its derivatives.

2. The differential equation is second order since the highest derivative in the equa-

tion is of order two. The equation is nonlinear due to the y2 term (as well as dueto the y2 term multiplying the y′′ term).

3. The differential equation is fourth order, since the highest derivative of the func-tion y is of order four. The equation is also linear, since the terms containing thedependent variable is linear in y and its derivatives.

4. The differential equation is second order. Furthermore, the equation is nonlinear,since the dependent variable y is an argument of the sine function, which is not alinear function.

5. y1(t) = et, so y ′1(t) = y

′′1 (t) = e

t. Hence y ′′1 − y1 = 0. Also, y2(t) = cosh t, soy ′1(t) = sinh t and y

′′2 (t) = cosh t. Thus y

′′2 − y2 = 0.

6. y1 = e−3t, so y′1 = −3e−3t and y′′1 = 9e−3t. This implies that y′′1 + 2y′1 − 3y1 =

(9− 6− 3)e−3t = 0. Also, y2 = et, so y′2 = y′′2 = et. This gives y′′2 + 2y′2 − 3y2 =(1 + 2− 3)et = 0.

7. y(t) = 3t+ t2, so y ′(t) = 3 + 2t. Substituting into the differential equation, wehave t(3 + 2t)− (3t+ t2) = 3t+ 2t2 − 3t− t2 = t2. Hence the given function is asolution.

16 Chapter 1. Introduction

8. y1(t) = t/3, so y′1(t) = 1/3 and y

′′1 (t) = y

′′′1 (t) = y

′′′′1 (t) = 0. Clearly, y1(t)

is a solution. Likewise, y2(t) = e−t + t/3, so y ′2(t) = −e−t + 1/3, y ′′2 (t) = e−t,

y ′′′2 (t) = −e−t, y ′′′′2 (t) = e−t. Substituting into the left hand side of the equation,we find that e−t + 4(−e−t) + 3(e−t + t/3) = e−t − 4e−t + 3e−t + t = t. Hence bothfunctions are solutions of the differential equation.

9. y1(t) = t−2, so y ′1(t) = −2t−3 and y ′′1 (t) = 6 t−4. Substituting into the left hand

side of the differential equation, we have t2(6 t−4) + 5t(−2t−3) + 4 t−2 = 6 t−2 −10 t−2 + 4 t−2 = 0. Likewise, y2(t) = t

−2 ln t, so y ′2(t) = t−3 − 2t−3 ln t and y ′′2 (t) =

−5 t−4 + 6 t−4 ln t. Substituting into the left hand side of the equation, we have

t2(−5 t−4 + 6 t−4 ln t) + 5t(t−3 − 2t−3 ln t) + 4(t−2 ln t) =

= −5 t−2 + 6 t−2 ln t+ 5 t−2 − 10 t−2 ln t+ 4 t−2 ln t = 0.

Hence both functions are solutions of the differential equation.

10. Recall that if u(t) =∫ t0f(s) ds, then u′(t) = f(t). Now y = et

2 ∫ t0e−s

2

ds+

et2

, so y′ = 2tet2 ∫ t

0e−s

2

ds+ 1 + 2tet2

. Therefore, y′ − 2ty = 2tet2∫ t0e−s

2

ds+ 1 +

2tet2 − 2t(et2

∫ t0e−s

2

ds+ et2

) = 1.

11. Let y(t) = ert. Then y ′(t) = rert, and substitution into the differential equationresults in rert + 2ert = 0. Since ert 6= 0, we obtain the algebraic equation r + 2 = 0.The root of this equation is r = −2.

12. y(t) = ert, so y ′(t) = r ert and y ′′(t) = r2ert. Substituting into the differen-tial equation, we have r2ert + rert − 6 ert = 0. Since ert 6= 0, we obtain the alge-braic equation r2 + r − 6 = 0, that is, (r − 2)(r + 3) = 0. The roots are r1 = 2,r2 = −3.

13. Let y(t) = ert. Then y ′(t) = rert, y ′′(t) = r2ert and y ′′′(t) = r3ert. Substitut-ing the derivatives into the differential equation, we have r3ert − 3r2ert + 2rert = 0.Since ert 6= 0, we obtain the algebraic equation r3 − 3r2 + 2r = 0 . By inspection,it follows that r(r − 1)(r − 2) = 0. Clearly, the roots are r1 = 0, r2 = 1 and r3 = 2.

14. Let y = tr. Then y′ = rtr−1 and y′′ = r(r − 1)tr−2. Substituting these termsinto the differential equation, t2y′′ + 4ty′ + 2y = t2(r(r − 1)tr−2) + 4t(rtr−1) + 2tr= (r(r − 1) + 4r + 2)tr = 0. If this is to hold for all t, then we need r(r − 1) + 4r +2 = 0. Simplifying this expression, we need r2 + 3r + 2 = 0. The solutions of thisequation are r = −1,−2.

15. y(t) = tr, so y ′(t) = r tr−1 and y ′′(t) = r(r − 1)tr−2. Substituting the deriva-tives into the differential equation, we have t2

[r(r − 1)tr−2

]− 4t(r tr−1) + 4 tr = 0.

After some algebra, it follows that r(r − 1)tr − 4r tr + 4 tr = 0. For t 6= 0, we obtainthe algebraic equation r2 − 5r + 4 = 0 . The roots of this equation are r1 = 1 andr2 = 4.

1.3 17

16. The order of the partial differential equation is two, since the highest derivative,in fact each one of the derivatives, is of second order. The equation is linear, sincethe left hand side is a linear function of the partial derivatives.

17. The partial differential equation is fourth order, since the highest derivative,and in fact each of the derivatives, is of order four. The equation is linear, sincethe left hand side is a linear function of the partial derivatives.

18. The partial differential equation is second order, since the highest derivative ofthe function u(x, y) is of order two. The equation is nonlinear, due to the productu · ux on the left hand side of the equation.

19. If u1(x, y) = cos x cosh y, then ∂2u1/∂x

2 = − cos x cosh y and ∂2u1/∂y2 =cosx cosh y. It is evident that ∂2u1/∂x

2 + ∂2u1/∂y2 = 0. Also, when u2(x, y) =

ln(x2 + y2), the second derivatives are

∂2u2∂x2

=2

x2 + y2− 4x

2

(x2 + y2)2and

∂2u2∂y2

=2

x2 + y2− 4y

2

(x2 + y2)2.

Adding the partial derivatives,

∂2u2∂x2

+∂2u2∂y2

=2

x2 + y2− 4x

2

(x2 + y2)2+

2

x2 + y2− 4y

2

(x2 + y2)2=

=4

x2 + y2− 4(x

2 + y2)

(x2 + y2)2= 0.

Hence u2(x, y) is also a solution of the differential equation.

20. Since ∂u1/∂t = −α2e−α2t sinx and ∂2u1/∂x

2 = −e−α2t sinx we have α2uxx =ut, for all t and x.

21. Let u1(x, t) = sin (λx) sin (λat). Then the second derivatives are

∂2u1∂x2

= −λ2 sin λx sin λat and ∂2u1∂t2

= −λ2a2 sin λx sin λat.

It is easy to see that a2∂2u1/∂x2 = ∂2u1/∂t

2. Likewise, given u2(x, t) = sin(x− at),we have

∂2u2∂x2

= − sin(x− at) and ∂2u2∂t2

= −a2 sin(x− at).

Clearly, u2(x, t) is also a solution of the partial differential equation.

18 Chapter 1. Introduction

22.(a)

W mg=

=

µ

µ

µ

T tensionL

(b) The path of the particle is a circle, therefore polar coordinates are intrinsic tothe problem. The variable r is radial distance and the angle θ is measured fromthe vertical.

Newton’s Second Law states that∑F = ma. In the tangential direction, the

equation of motion may be expressed as∑Fθ = maθ, in which the tangential

acceleration, that is, the linear acceleration along the path is aθ = Ld2θ/dt2. ( aθ is

positive in the direction of increasing θ ). Since the only force acting in the tangen-tial direction is the component of weight, the equation of motion is

−mg sin θ = mLd2θ

dt2.

(c) Rearranging the terms results in the differential equation

d2θ

dt2+g

Lsin θ = 0 .

23.(a) The kinetic energy of a particle of mass m is given by T = mv2/2, in whichv is its speed. A particle in motion on a circle of radius L has speed L (dθ/dt),where θ is its angular position and dθ/dt is its angular speed.

(b) Gravitational potential energy is given by V = mgh, where h is the heightabove a certain datum. Choosing the lowest point of the swing as the datum, itfollows from trigonometry that h = L(1− cos θ).

(c) From parts (a) and (b),

E =1

2mL2(

dθ

dt)2 +mgL(1− cos θ) .

Applying the chain rule for differentiation,

dE

dt= mL2

dθ

dt

d2θ

dt2+mgL sin θ

dθ

dt.

1.3 19

Setting dE/dt = 0 and dividing both sides of the equation by dθ/dt results in

mL2d2θ

dt2+mgL sin θ = 0,

which leads to Equation (12).

24.(a) The angular momentum is the moment of the linear momentum about a givenpoint. The linear momentum is given by mv = mLdθ/dt. Taking the moment aboutthe point of support, the angular momentum is

M = mvL = mL2dθ

dt.

(b) The moment of the gravitational force is −mgL sin θ. The negative sign isincluded since positive moments are counterclockwise. Setting dM/dt equal to themoment of the gravitational force gives

dM

dt= mL2

d2θ

dt2= −mgL sin θ,

which leads to Equation (12).