1
1
Introduction
1.1
1.
For y >3/2, the slopes are negative, therefore the solutions are decreasing. For y <3/2, the slopes are positive, hence the solutions are increasing. The equilibrium solution appears to be y(t) = 3/2, to which all other solutions converge.
2.
Fory >3/2, the slopes are positive, therefore the solutions increase. Fory <3/2, the slopes are negative, therefore, the solutions decrease. As a result, y diverges from 3/2 ast→ ∞ify(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 ast→ ∞.
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 diverge away from the equilibrium solution y(t) =−1/2.
5. For all solutions to approach the equilibrium solutiony(t) = 2/3, we must have y0 <0 for y >2/3, andy0>0 fory <2/3. The required rates are satisfied by the differential equationy0= 2−3y.
6. For solutions other thany(t) = 2 to diverge fromy= 2,y(t) must be an increas-ing function fory >2, and a decreasing function fory <2. The simplest differential equation whose solutions satisfy these criteria isy0 =y−2.
7.
For y= 0 and y= 4 we have y0= 0 and thus y= 0 and y= 4 are equilibrium
solutions. For y >4, y0<0 so if y(0)>4 the solution approaches y= 4 from above. If 0< y(0)<4, then y0>0 and the solutions ”grow” to y= 4 as t→ ∞. Fory(0)<0 we see thaty0<0 and the solutions diverge from 0.
8.
Note that y0= 0 fory= 0 andy= 5. The two equilibrium solutions are y(t) = 0
and y(t) = 5. Based on the direction field, y0 >0 for y >5; thus solutions with initial values greater than 5 diverge from the solution y(t) = 5. For 0< y <5, the slopes are negative, and hence solutions with initial values between 0 and 5 all decrease toward the solution y(t) = 0. For y <0, the slopes are all positive; thus solutions with initial values less than 0 approach the solutiony(t) = 0.
9.
Sincey0 =y2,y= 0 is the only equilibrium solution andy0>0 for ally. Thusy→0
if the initial value is negative;y diverges from 0 if the initial value is positive. 10.
Observe thaty0= 0 fory= 0 andy= 2. The two equilibrium solutions arey(t) = 0 and y(t) = 2. Based on the direction field, y0>0 for y >2; thus solutions with initial values greater than 2 diverge from y(t) = 2. For 0< y <2, the slopes are also positive, and hence solutions with initial values between 0 and 2 all increase toward the solutiony(t) = 2. Fory <0, the slopes are all negative; thus solutions with initial values less than 0 diverge from the solutiony(t) = 0.
11. -(j) y0 = 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. Sincedy/dt >0 for y >2, and dy/dt <0 for y <2 we conclude that (c) is the correct answer: y0=y−2.
13. -(g) y0=−2−y. 14. -(b) y0= 2 +y.
15. From Figure 1.1.9 we can see that y= 0 andy= 3 are equilibrium solutions, so (e) and (h) are the only possible differential equations. Furthermore, we have
dy/dt <0 fory >3 and fory <0, anddy/dt >0 for 0< y <3. This tells us that (h) is the desired differential equation: y0 =y(3−y).
16. -(e) y0 =y(y−3).
17.(a) Let a(t) denote the amount of chemical in the pond at timet. The amount a will be measured in grams and the timet will be measured in hours. The rate at 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/106g/gal = (3×10−4)ag/h. Thus the differential equation is given byda/dt=
3−(3×10−4)a.
(b) The equilibrium solution occurs whena0= 0, ora= 104grams. Sincea0>0 for
a <104 g anda0<0 fora >104 g, all solutions approach the equilibrium solution
independent of the amount present att= 0.
(c) Let a(t) denote the amount of chemical in the pond at timet. From part (a) the functiona(t) satisfies the differential equationda/dt= 3−(3×10−4)a. Thus in terms 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 radiusr is given byS = 4πr2. The volume of a spherical raindrop is given byV = 4πr3/3. Therefore, we see that the
surface area S=cV2/3 for some constant c. If the raindrop evaporates at a rate
proportional to its surface area, thendV /dt=−kV2/3 for somek >0.
19. The difference between the temperature of the object and the ambient tem-perature is u−70 (uin◦F). Since the object is cooling whenu >70, and the rate constant isk = 0.05 min−1, the governing differential equation for the temperature of the object isdu/dt=−.05 (u−70).
20.(a) Let M(t) be the total amount of the drug (in milligrams) in the patient’s body at any given time t(hr). The drug enters the body at a constant rate of 500 mg/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 equation dM/dt= 500−0.4M (mg/hr).
Based on the direction field, the amount of drug in the bloodstream approaches the equilibrium level of 1250 mg (within a few hours).
21.(a) Following the discussion in the text, the differential equation ism(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 given byv∞ =pmg/γ.
(c) Using the relationγ v2
∞=mg, the required drag coefficient isγ= 2/49 kg/s.
(d)
22.
23.
The solutions approach−∞, 0, or∞depending upon the value ofy(0). 24.
The solutions approach−∞, ∞, or oscillate depending upon the value ofy(0). 25.
For all y(0), there is a number tf that depends on the value of y(0) such that
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 conditiony(0) =y
0 results in the
specifi-cation of the constant asc= 5−y0. Hence the solution isy(t) = 5 + (y0−5)e−t.
All solutions appear to converge to the equilibrium solutiony(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, or
equivalently, 5−2y=c e−2t. Applying the initial conditiony(0) =y
0results in the
specification of the constant asc= 5−2y0. Hencey(t) = 5/2 + (y0−5/2)e−2t.
All solutions appear to converge to the equilibrium solutiony(t) = 5/2, at a faster rate than in part (a).
(c) Rewrite the differential equation as dy
Integrating both sides of this equation results in −(1/2) ln|10−2y|=t+c1, or
equivalently, 5−y=c e−2t. Applying the initial condition y(0) =y
0results in the
specification of the constant asc= 5−y0. Hencey(t) = 5 + (y0−5)e−2t.
All solutions appear to converge to the equilibrium solution y(t) = 5, at a faster rate 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 conditiony(0) =y
0 results in the specification of
the constant as c=y0−5. Hence the solution isy(t) = 5 + (y0−5)et.
All solutions appear to diverge from the equilibrium solutiony(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 conditiony(0) =y
0results in the
All solutions appear to diverge from the equilibrium solutiony(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 conditiony(0) =y0 results in the
speci-fication of the constant asc=y0−5. Hence the solution isy(t) = 5 + (y0−5)e2t.
All solutions appear to diverge from the equilibrium solutiony(t) = 5. 3.(a) Rewrite the differential equation as
dy
b−ay =dt,
which is valid for y6=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/ais an equilibrium
(b)
(c) (i) As aincreases, the equilibrium solution gets closer toy(t) = 0, from above. The convergence rate of all solutions isa. Asaincreases, the solutions converge to the equilibrium solution quicker.
(ii) Asb increases, then the equilibrium solutiony(t) =b/aalso 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 solution y(t) =b/aremains the same, but the convergence rate of all solutions increases. 4.(a) The equilibrium solution satisfies the differential equationdye/dt= 0. Setting
aye−b= 0, we obtainye(t) =b/a.
(b) SincedY /dt=dy/dt, it follows thatdY /dt=a(Y +ye)−b=a Y.
5.(a) Rewrite Eq.(ii) as (1/y)dy/dt=aand thus ln|y|=at+C, or y1=ceat.
(b) Ify=y1(t) +k, thendy/dt=dy1/dt. Substituting both these into Eq.(i) we get
dy1/dt=a(y1+k)−b. Sincedy1/dt=ay1, this leavesak−b= 0 and thusk=b/a.
Hence y=y1(t) +b/ais the solution to Eq(i).
(c) Substitution ofy1=ceatshows 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 iny1(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 hencek=b/a.
(c) The general solution of the differential equation is y(t) =c e−at+b/a. This
is exactly the form given by Eq.(17) in the text. Invoking an initial condition y(0) =y0, the solution may also be expressed as y(t) =b/a+ (y0−b/a)e−at.
7.(a) The general solution isp(t) = 900 +c et/2, that is,p(t) = 900 + (p
0−900)et/2.
decreasing exponential, and hence the time of extinction is equal to the number of months it takes, saytf, for the population to reach zero. Solving 900−50etf/2= 0,
we find thattf = 2 ln(900/50)≈5.78 months.
(b) The solution,p(t) = 900 + (p0−900)et/2, is a decreasing exponential as long as
p0<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 to the value of the initial population. Setting tf = 12 months, the equation may be
written as 900/(900−p0) =e6, which has solutionp0≈897.8. Sincep0is the initial
population, the appropriate answer isp0= 898 mice.
8.(a) The solution to this problem is given by Eq.(26), which has a limiting velocity of 49 m/s. Substitutingv= 48.02 (which is 98% of 49) into Eq.(26) yieldset/5= 0.02. Solving fortwe have t=−5 ln(0.02) = 19.56 s.
(b) Use Eq.(29) witht= 19.56.
9.(a) If the drag force is proportional to v2 then F = 98−kv2 is the net force
acting on the falling mass (m= 10 kg). Thus 10dv/dt= 98−kv2, which has a
limiting velocity ofv2= 98/k. Settingv2= 492givesk= 98/492and hencedv/dt=
(492−v2)/(10/k) = (492−v2)/245.
(b) From part (a) we havedv/(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,
orC0= 0. Thus tanh−1(v/49) =t/5, orv(t) = 49 tanh(t/5) m/s.
(c)
(d) In the graph of part (c), the solution for the linear drag force lies below the solution for the quadratic drag force. This latter solution approaches equilibrium faster since, as the velocity increases, there is a larger drag force.
(e) Note thatR
tanh(x)dx= ln(coshx) +C. The distance the object falls is given byx= 245 ln(cosh(t/5)).
(f) Solving 300 = 245 ln(cosh(T /5)) givesT ≈9.48 s.
10.(a,b) The general solution of the differential equation isQ(t) =c e−rt. Given that Q(0) = 100 mg, the value of the constant is given byc= 100. Hence the amount of thorium-234 present at any time is given by Q(t) = 100e−rt. Furthermore, based
on the hypothesis, setting t= 1 results in 82.04 = 100e−r. Solving for the rate
constant, we find thatr=−ln(82.04/100)≈.19796/week orr≈.02828/day. (c) Let T be the time that it takes the isotope to decay to one-half of its original amount. From part (a), it follows that 50 = 100e−rT, in which r=.19796/week.
Taking the natural logarithm of both sides, we find that T ≈3.5014 weeks orT ≈
24.51 days.
11. The general solution of the differential equationdQ/dt=−r QisQ(t) =Q0e−rt,
in which Q0=Q(0) is the initial amount of the substance. Letτ be the time that
it takes the substance to decay to one-half of its original amount,Q0. Settingt=τ
in the solution, we have 0.5Q0=Q0e−rτ. Taking the natural logarithm of both
sides, it follows that −rτ = ln(0.5) orrτ = ln 2.
12.(a) Rewrite the differential equation asdu/(u−T) =−kdtand then integrate to find ln|u−T|=−kt+c. Thusu−T =±Ce−kt. Fort= 0 we haveu0−T =±C
and thusu(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 asdQ/(Q−CV) =−(1/CR)dt, thus, upon integrating and simplifying, we getQ=De−t/CR+CV. Q(0) = 0 implies that the
solution of the differential equation isQ(t) =CV(1−e−t/CR).
(c) In this case RdQ/dt+Q/C = 0. Q(t1) =CV. The solution of this differential
equation isQ(t) =Ee−t/CR, soQ(t
1) =Ee−t1/CR =CV, orE=CV et1/CR. Thus
Q(t) =CV et1/CRe−t/CR=CV e−(t−t1)/CR fort≥t
1.
14.(a) The accumulation rate of the chemical is (0.01)(300) grams per hour. At any given time t, the concentration of the chemical in the pond isQ(t)/106 grams
per 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 asdQ/(10000−Q) = 0.0003dt. 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. SinceQ(0) = 0, the
value of the constant is c= 10000. Hence the amount of chemical in the pond at any time isQ(t) = 10000(1−e−0.0003t) grams. Note that 1 year= 8760 hours.
Set-tingt= 8760, the amount of chemical present after one year is Q(8760)≈9277.77 grams, that is, 9.27777 kilograms.
(c) With the accumulation rate now equal to zero, the governing equation becomes dQ/dt=−0.0003Q(t) g/hr. Resetting the time variable, we now assign the new initial value asQ(0) = 9277.77 grams.
(d) The solution of the differential equation in part (c) isQ(t) = 9277.77e−0.0003t.
Hence, one year after the source is removed, the amount of chemical in the pond is Q(8760)≈670.1 grams.
(e) Lettingtbe the amount of time after the source is removed, we obtain the equa-tion 10 = 9277.77e−0.0003t. Taking the natural logarithm of both sides,−0.0003t=
(f)
1.3
1. The differential equation is second order, since the highest derivative in the equa-tion is of order two. The equaequa-tion is linear, since the left hand side is a linear function ofy and its derivatives.
2. The differential equation is second order since the highest derivative in the equa-tion is of order two. The equaequa-tion is nonlinear due to they2 term (as well as due
to they2 term multiplying they00 term).
3. The differential equation is fourth order, since the highest derivative of the func-tiony is of order four. The equation is also linear, since the terms containing the dependent variable is linear in yand its derivatives.
4. The differential equation is second order. Furthermore, the equation is nonlinear, since the dependent variabley is an argument of the sine function, which is not a linear function.
5. y1(t) =et, so y10(t) =y100(t) =et. Hence y100−y1= 0. Also, y2(t) = cosht, so
y10(t) = sinhtandy200(t) = cosht. Thusy200−y2 = 0.
6. y1=e−3t, soy01=−3e−3tand y100= 9e−3t. This implies thaty001+ 2y10 −3y1=
(9−6−3)e−3t= 0. Also, y
2=et, so y02=y002 =et. This gives y002+ 2y20 −3y2=
(1 + 2−3)et= 0.
7. y(t) = 3t+t2, soy0(t) = 3 + 2t. Substituting into the differential equation, we
have t(3 + 2t)−(3t+t2) = 3t+ 2t2−3t−t2=t2. Hence the given function is a
8. y1(t) =t/3, so y10(t) = 1/3 and y100(t) =y1000(t) =y10000(t) = 0. Clearly, y1(t)
is a solution. Likewise, y2(t) =e−t+t/3, so y20(t) =−e−t+ 1/3, y200(t) =e−t,
y2000(t) =−e−t, y20000(t) =e−t. Substituting into the left hand side of the equation, we find thate−t+ 4(−e−t) + 3(e−t+t/3) =e−t−4e−t+ 3e−t+t=t. Hence both functions are solutions of the differential equation.
9. y1(t) =t−2, soy10(t) =−2t−3 andy100(t) = 6t−4. Substituting into the left hand
side of the differential equation, we have t2(6t−4) + 5t(−2t−3) + 4t−2= 6t−2−
10t−2+ 4t−2= 0. Likewise,y
2(t) =t−2ln t, soy20(t) =t−3−2t−3lntandy200(t) =
−5t−4 + 6t−4lnt. Substituting into the left hand side of the equation, we have
t2(−5t−4 + 6t−4ln t) + 5t(t−3−2t−3lnt) + 4(t−2lnt) = =−5t−2+ 6t−2lnt+ 5t−2−10t−2lnt+ 4t−2lnt= 0. Hence both functions are solutions of the differential equation. 10. Recall that if u(t) =Rt
0f(s)ds, then u 0(t) =f(t). Now y=et2Rt 0e −s2ds+ et2, soy0= 2tet2Rt 0e
−s2ds+ 1 + 2tet2. Therefore,y0−2ty= 2tet2Rt
0e
−s2ds+ 1 +
2tet2−2t(et2R0te−s2ds+et2) = 1.
11. Lety(t) =ert. Theny0(t) =rert, and substitution into the differential equation results inrert+ 2ert= 0. Sinceert 6= 0, we obtain the algebraic equationr+ 2 = 0. The root of this equation isr=−2.
12. y(t) =ert, so y0(t) =r ert and y00(t) =r2ert. Substituting into the
differen-tial equation, we have r2ert+rert−6ert= 0. Since ert6= 0, we obtain the
alge-braic equation r2+r−6 = 0, that is, (r−2)(r+ 3) = 0. The roots are r 1= 2,
r2=−3.
13. Lety(t) =ert. Then y0(t) =rert,y00(t) =r2ert andy000(t) =r3ert.
Substitut-ing the derivatives into the differential equation, we haver3ert−3r2ert+ 2rert = 0.
Sinceert6= 0, we obtain the algebraic equation r3−3r2+ 2r= 0 . By inspection,
it follows that r(r−1)(r−2) = 0. Clearly, the roots arer1= 0,r2= 1 andr3= 2.
14. Let y=tr. Then y0=rtr−1 and y00=r(r−1)tr−2. Substituting these terms into the differential equation,t2y00+ 4ty0+ 2y=t2(r(r−1)tr−2) + 4t(rtr−1) + 2tr
= (r(r−1) + 4r+ 2)tr= 0. If this is to hold for allt, then we needr(r−1) + 4r+
2 = 0. Simplifying this expression, we needr2+ 3r+ 2 = 0. The solutions of this
equation arer=−1,−2.
15. y(t) =tr, so y0(t) =r tr−1 and y00(t) =r(r−1)tr−2. Substituting the
deriva-tives into the differential equation, we havet2
r(r−1)tr−2
−4t(r tr−1) + 4tr= 0.
After some algebra, it follows thatr(r−1)tr−4r tr+ 4tr= 0. Fort6= 0, we obtain the algebraic equation r2−5r+ 4 = 0 . The roots of this equation arer1= 1 and
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, since the 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, since the left hand side is a linear function of the partial derivatives.
18. The partial differential equation is second order, since the highest derivative of the function u(x, y) is of order two. The equation is nonlinear, due to the product u·ux on the left hand side of the equation.
19. If u1(x, y) = cosxcoshy, then ∂2u1/∂x2=−cosxcoshy and ∂2u1/∂y2=
cosxcosh y. It is evident that ∂2u
1/∂x2+∂2u1/∂y2= 0. Also, when u2(x, y) =
ln(x2+y2), the second derivatives are
∂2u 2 ∂x2 = 2 x2+y2− 4x2 (x2+y2)2 and ∂2u 2 ∂y2 = 2 x2+y2− 4y2 (x2+y2)2.
Adding the partial derivatives,
∂2u 2 ∂x2 + ∂2u 2 ∂y2 = 2 x2+y2 − 4x2 (x2+y2)2 + 2 x2+y2 − 4y2 (x2+y2)2 = = 4 x2+y2 − 4(x2+y2) (x2+y2)2 = 0.
Hence u2(x, y) is also a solution of the differential equation.
20. Since ∂u1/∂t=−α2e−α
2t
sinxand ∂2u1/∂x2=−e−α
2t
sinxwe haveα2uxx=
ut, for alltandx.
21. Letu1(x, t) = sin (λx) sin (λat). Then the second derivatives are
∂2u1
∂x2 =−λ
2sinλx sinλat and ∂ 2u
1
∂t2 =−λ
2a2sin λxsinλat.
It is easy to see thata2∂2u
1/∂x2=∂2u1/∂t2. Likewise, givenu2(x, t) = sin(x−at),
we have
∂2u 2
∂x2 =−sin(x−at) and
∂2u 2
∂t2 =−a
2sin(x−at).
22.(a)
W mg
= =µ
µ
µ
T
tension
L
(b) The path of the particle is a circle, therefore polar coordinates are intrinsic to the problem. The variable r is radial distance and the angle θ is measured from the vertical.
Newton’s Second Law states that PF =ma. In the tangential direction, the
equation of motion may be expressed as PF
θ=m aθ, in which the tangential
acceleration, that is, the linear acceleration along the path isaθ=L d2θ/dt2. (aθis
positive in the direction of increasingθ). Since the only force acting in the tangen-tial direction is the component of weight, the equation of motion is
−mg sinθ=mLd
2θ
dt2 .
(c) Rearranging the terms results in the differential equation d2θ
dt2 +
g
Lsinθ= 0.
23.(a) The kinetic energy of a particle of massmis given byT =m v2/2, in which
v is its speed. A particle in motion on a circle of radius L has speed L(dθ/dt), whereθ is its angular position anddθ/dtis its angular speed.
(b) Gravitational potential energy is given by V = mgh, where h is the height above a certain datum. Choosing the lowest point of the swing as the datum, it follows from trigonometry thath=L(1−cosθ).
(c) From parts (a) and (b),
E= 1 2mL
2(dθ
dt)
2+mgL(1−cosθ).
Applying the chain rule for differentiation, dE dt =mL 2dθ dt d2θ dt2 +mgLsinθ dθ dt .
Setting dE/dt = 0 and dividing both sides of the equation bydθ/dtresults in mL2d
2θ
dt2 +mgLsin θ= 0,
which leads to Equation (12).
24.(a) The angular momentum is the moment of the linear momentum about a given point. The linear momentum is given bymv=mLdθ/dt. Taking the moment about the point of support, the angular momentum is
M =mvL=mL2dθ dt.
(b) The moment of the gravitational force is −mgLsinθ. The negative sign is included since positive moments are counterclockwise. SettingdM/dtequal to the moment of the gravitational force gives
dM dt =mL
2d2θ
dt2 =−mgLsinθ,
