《微分方程》课程教学资源（书籍资料）DENNIS G. ZILL&MICHAEL R. CULLEN《Differential Equations with Boundary-Value Problems》（SEVENTH EDITION）

1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS 。21 total population at time t,then this assumption can be expressed as dp dp P or =kP. (1) dt dt where k is a constant of proportionality.This simple model,which fails to take into account many factors that can influence human populations to either grow or decline (immigration and emigration,for example),nevertheless turned out to be fairly accu- rate in predicting the population of the United States during the years 1790-1860. Populations that grow at a rate described by(1)are rare;nevertheless,(1)is still used to model growth of small populations over short intervals of time (bacteria growing in a petri dish,for example). RADIOACTIVE DECAY The nucleus of an atom consists of combinations of pro- tons and neutrons.Many of these combinations of protons and neutrons are unstable- that is,the atoms decay or transmute into atoms of another substance.Such nuclei are said to be radioactive.For example,over time the highly radioactive radium,Ra-226, transmutes into the radioactive gas radon,Rn-222.To model the phenomenon of radioactive decay,it is assumed that the rate dA/dt at which the nuclei of a sub- stance decay is proportional to the amount(more precisely,the number of nuclei) A(t)of the substance remaining at time t: dA xA or dA =kA. (2) dt dt Of course,equations(1)and(2)are exactly the same;the difference is only in the in- terpretation of the symbols and the constants of proportionality.For growth,as we expect in (1),k>0,and for decay,as in (2),k<0. The model (1)for growth can also be seen as the equation ds/dt rS,which describes the growth of capital S when an annual rate of interest r is compounded continuously.The model(2)for decay also occurs in biological applications such as determining the half-life of a drug-the time that it takes for 50%of a drug to be eliminated from a body by excretion or metabolism.In chemistry the decay model (2)appears in the mathematical description of a first-order chemical reaction.The point is this: A single differential equation can serve as a mathematical model for many different phenomena. Mathematical models are often accompanied by certain side conditions.For ex- ample,in (1)and (2)we would expect to know,in turn,the initial population Po and the initial amount of radioactive substance Ao on hand.If the initial point in time is taken to be t=0,then we know that P(0)=Po and A(0)=A0.In other words,a mathematical model can consist of either an initial-value problem or,as we shall see later on in Section 5.2,a boundary-value problem. NEWTON'S LAW OF COOLING/WARMING According to Newton's empiri- cal law of cooling/warming,the rate at which the temperature of a body changes is proportional to the difference between the temperature of the body and the temper- ature of the surrounding medium,the so-called ambient temperature.If T(t)repre- sents the temperature of a body at time t,Tm the temperature of the surrounding medium,and d7/dt the rate at which the temperature of the body changes,then Newton's law of cooling/warming translates into the mathematical statement dT dT x T-T'm or =k(T-Tm), (3) dt dt where k is a constant of proportionality.In either case,cooling or warming,if Tm is a constant,it stands to reason that k<0

total population at time t, then this assumption can be expressed as , (1) where k is a constant of proportionality. This simple model, which fails to take into account many factors that can influence human populations to either grow or decline (immigration and emigration, for example), nevertheless turned out to be fairly accu￾rate in predicting the population of the United States during the years 1790–1860. Populations that grow at a rate described by (1) are rare; nevertheless, (1) is still used to model growth of small populations over short intervals of time (bacteria growing in a petri dish, for example). RADIOACTIVE DECAY The nucleus of an atom consists of combinations of pro￾tons and neutrons. Many of these combinations of protons and neutrons are unstable— that is, the atoms decay or transmute into atoms of another substance. Such nuclei are said to be radioactive. For example, over time the highly radioactive radium, Ra-226, transmutes into the radioactive gas radon, Rn-222. To model the phenomenon of radioactive decay, it is assumed that the rate dA
dt at which the nuclei of a sub￾stance decay is proportional to the amount (more precisely, the number of nuclei) A(t) of the substance remaining at time t: . (2) Of course, equations (1) and (2) are exactly the same; the difference is only in the in￾terpretation of the symbols and the constants of proportionality. For growth, as we expect in (1), k  0, and for decay, as in (2), k  0. The model (1) for growth can also be seen as the equation dS
dt  rS, which describes the growth of capital S when an annual rate of interest r is compounded continuously. The model (2) for decay also occurs in biological applications such as determining the half-life of a drug—the time that it takes for 50% of a drug to be eliminated from a body by excretion or metabolism. In chemistry the decay model (2) appears in the mathematical description of a first-order chemical reaction. The point is this: A single differential equation can serve as a mathematical model for many different phenomena. Mathematical models are often accompanied by certain side conditions. For ex￾ample, in (1) and (2) we would expect to know, in turn, the initial population P0 and the initial amount of radioactive substance A0 on hand. If the initial point in time is taken to be t  0, then we know that P(0)  P0 and A(0)  A0. In other words, a mathematical model can consist of either an initial-value problem or, as we shall see later on in Section 5.2, a boundary-value problem. NEWTON’S LAW OF COOLING/WARMING According to Newton’s empiri￾cal law of cooling/warming, the rate at which the temperature of a body changes is proportional to the difference between the temperature of the body and the temper￾ature of the surrounding medium, the so-called ambient temperature. If T(t) repre￾sents the temperature of a body at time t, Tm the temperature of the surrounding medium, and dT
dt the rate at which the temperature of the body changes, then Newton’s law of cooling/warming translates into the mathematical statement , (3) where k is a constant of proportionality. In either case, cooling or warming, if Tm is a constant, it stands to reason that k  0. dT dt  T
Tm or dT dt  k(T
Tm) dA dt  A or dA dt  kA dP dt  P or dP dt  kP 1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS ● 21

22.CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS SPREAD OF A DISEASE A contagious disease-for example,a flu virus-is spread throughout a community by people coming into contact with other people.Let x(r)denote the number of people who have contracted the disease and y(t)denote the number of people who have not yet been exposed.It seems reasonable to assume that the rate dx/dt at which the disease spreads is proportional to the number of encoun- ters,or interactions,between these two groups of people.If we assume that the num- ber of interactions is jointly proportional tox(t)and y(r)-that is,proportional to the product xy-then dx =kxy. (4) dt where k is the usual constant of proportionality.Suppose a small community has a fixed population of n people.If one infected person is introduced into this commu- nity,then it could be argued that x(t)and y(t)are related by x+y=n+1.Using this last equation to eliminate y in (4)gives us the model dx =kx(n+1-x. (5) dt An obvious initial condition accompanying equation(5)is x(0)=1. CHEMICAL REACTIONS The disintegration of a radioactive substance,governed by the differential equation (1),is said to be a first-order reaction.In chemistry a few reactions follow this same empirical law:If the molecules of substance A decompose into smaller molecules,it is a natural assumption that the rate at which this decomposition takes place is proportional to the amount of the first substance that has not undergone conversion;that is,if X(t)is the amount of substance A remaining at any time,then dX/dt =kX,where k is a negative constant since X is decreasing.An example of a first-order chemical reaction is the conversion of t-butyl chloride,(CH3)3CCI,into t-butyl alcohol,(CH3)3COH: (CH3 )3CCI NaOH-(CH3)COH NaCI. Only the concentration of the t-butyl chloride controls the rate of reaction.But in the reaction CH,CI NaOH-CH,OH NaCI one molecule of sodium hydroxide,NaOH,is consumed for every molecule of methyl chloride,CH3Cl,thus forming one molecule of methyl alcohol,CH3OH,and one molecule of sodium chloride,NaCl.In this case the rate at which the reaction proceeds is proportional to the product of the remaining concentrations of CH3Cl and NaOH.To describe this second reaction in general,let us suppose one molecule of a substance A combines with one molecule of a substance B to form one molecule of a substance C.If X denotes the amount of chemical C formed at time t and if a and B are,in turn,the amounts of the two chemicals A and B at t =0(the initial amounts). then the instantaneous amounts of A and B not converted to chemical C are a-X and B-X.respectively.Hence the rate of formation of Cis given by dX =k(a-X)(B-X), (6 dr where k is a constant of proportionality.A reaction whose model is equation (6)is said to be a second-order reaction. MIXTURES The mixing of two salt solutions of differing concentrations gives rise to a first-order differential equation for the amount of salt contained in the mix- ture.Let us suppose that a large mixing tank initially holds 300 gallons of brine (that is,water in which a certain number of pounds of salt has been dissolved).Another

SPREAD OF A DISEASE A contagious disease—for example, a flu virus—is spread throughout a community by people coming into contact with other people. Let x(t) denote the number of people who have contracted the disease and y(t) denote the number of people who have not yet been exposed. It seems reasonable to assume that the rate dx
dt at which the disease spreads is proportional to the number of encoun￾ters, or interactions, between these two groups of people. If we assume that the num￾ber of interactions is jointly proportional to x(t) and y(t)—that is, proportional to the product xy—then , (4) where k is the usual constant of proportionality. Suppose a small community has a fixed population of n people. If one infected person is introduced into this commu￾nity, then it could be argued that x(t) and y(t) are related by x y  n 1. Using this last equation to eliminate y in (4) gives us the model . (5) An obvious initial condition accompanying equation (5) is x(0)  1. CHEMICAL REACTIONS The disintegration of a radioactive substance, governed by the differential equation (1), is said to be a first-order reaction. In chemistry a few reactions follow this same empirical law: If the molecules of substance A decompose into smaller molecules, it is a natural assumption that the rate at which this decomposition takes place is proportional to the amount of the first substance that has not undergone conversion; that is, if X(t) is the amount of substance A remaining at any time, then dX
dt  kX, where k is a negative constant since X is decreasing. An example of a first-order chemical reaction is the conversion of t-butyl chloride, (CH3)3CCl, into t-butyl alcohol, (CH3)3COH: Only the concentration of the t-butyl chloride controls the rate of reaction. But in the reaction one molecule of sodium hydroxide, NaOH, is consumed for every molecule of methyl chloride, CH3Cl, thus forming one molecule of methyl alcohol, CH3OH, and one molecule of sodium chloride, NaCl. In this case the rate at which the reaction proceeds is proportional to the product of the remaining concentrations of CH3Cl and NaOH. To describe this second reaction in general, let us suppose one molecule of a substance A combines with one molecule of a substance B to form one molecule of a substance C. If X denotes the amount of chemical C formed at time t and if  and  are, in turn, the amounts of the two chemicals A and B at t  0 (the initial amounts), then the instantaneous amounts of A and B not converted to chemical C are 
X and 
X, respectively. Hence the rate of formation of C is given by , (6) where k is a constant of proportionality. A reaction whose model is equation (6) is said to be a second-order reaction. MIXTURES The mixing of two salt solutions of differing concentrations gives rise to a first-order differential equation for the amount of salt contained in the mix￾ture. Let us suppose that a large mixing tank initially holds 300 gallons of brine (that is, water in which a certain number of pounds of salt has been dissolved). Another dX dt  k(
X)(
X) CH3Cl NaOH : CH3OH NaCl (CH3)3CCl NaOH : (CH3)3COH NaCl. dx dt  kx(n 1
x) dx dt  kxy 22 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS

1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS.23 input rate of brine brine solution is pumped into the large tank at a rate of 3 gallons per minute;the 3 gal/min concentration of the salt in this inflow is 2 pounds per gallon.When the solution in the tank is well stirred,it is pumped out at the same rate as the entering solution.See Figure 1.3.1.If A(t)denotes the amount of salt (measured in pounds)in the tank at time t,then the rate at which A(t)changes is a net rate: input rate oufput rate constant =Rm一Ramr (7) 300 gal dt of salt of salt The input rate Ri at which salt enters the tank is the product of the inflow concentra- tion of salt and the inflow rate of fluid.Note that Rin is measured in pounds per minute: concentration output rate of brine of salt input rate input rate 3 gal/min in inflow of brine of salt FIGURE 1.3.1 Mixing tank Rin =(2 Ib/gal)(3 gal/min)=(6 Ib/min). Now,since the solution is being pumped out of the tank at the same rate that it is pumped in,the number of gallons of brine in the tank at time t is a constant 300 gal- lons.Hence the concentration of the salt in the tank as well as in the outflow is c(t)=A(t)/300 Ib/gal,so the output rate Rour of salt is concentration of salt output rate output rate in outflow of brine of salt A(1 A(1) 300 lb/gal·(3 gal/min)= Ib/min. 100 The net rate(7)then becomes dr =6-0 dA or =6. +100 (8) dt If rim and rom denote general input and output rates of the brine solutions,then there are three possibilities:rin=rou,rin>rout,and rin<rou.In the analysis lead- ing to (8)we have assumed that rin=romt.In the latter two cases the number of gal- lons of brine in the tank is either increasing (rin>ro)or decreasing (rin rou)at the net rate rin-rou See Problems 10-12 in Exercises 1.3. DRAINING A TANK In hydrodynamics Torricelli's law states that the speed v of efflux of water though a sharp-edged hole at the bottom of a tank filled to a depth h is the same as the speed that a body (in this case a drop of water)would acquire in falling freely from a height h-that is,v=V2gh,where g is the acceleration due to gravity.This last expression comes from equating the kinetic energy mv2 with the potential energy mgh and solving for v.Suppose a tank filled with water is allowed to drain through a hole under the influence of gravity.We would like to find the depth h of water remaining in the tank at time t.Consider the tank shown in Figure 1.3.2.If the area of the hole is A(in ft2)and the speed of the water leaving the tank is FIGURE 1.3.2 Draining tank v=V2gh (in ft/s),then the volume of water leaving the tank per second is AV2gh (in ft3/s).Thus if V(t)denotes the volume of water in the tank at time t,then =-Anv2gh, (9) dt Don't confuse these symbols with Rand Rwhich are input and output rates of salt

brine solution is pumped into the large tank at a rate of 3 gallons per minute; the concentration of the salt in this inflow is 2 pounds per gallon. When the solution in the tank is well stirred, it is pumped out at the same rate as the entering solution. See Figure 1.3.1. If A(t) denotes the amount of salt (measured in pounds) in the tank at time t, then the rate at which A(t) changes is a net rate: . (7) The input rate Rin at which salt enters the tank is the product of the inflow concentra￾tion of salt and the inflow rate of fluid. Note that Rin is measured in pounds per minute: Now, since the solution is being pumped out of the tank at the same rate that it is pumped in, the number of gallons of brine in the tank at time t is a constant 300 gal￾lons. Hence the concentration of the salt in the tank as well as in the outflow is c(t)  A(t)
300 lb/gal, so the output rate Rout of salt is The net rate (7) then becomes (8) If rin and rout denote general input and output rates of the brine solutions,* then there are three possibilities: rin  rout, rin  rout, and rin  rout. In the analysis lead￾ing to (8) we have assumed that rin  rout. In the latter two cases the number of gal￾lons of brine in the tank is either increasing (rin  rout) or decreasing (rin  rout) at the net rate rin
rout. See Problems 10–12 in Exercises 1.3. DRAINING A TANK In hydrodynamics Torricelli’s law states that the speed v of efflux of water though a sharp-edged hole at the bottom of a tank filled to a depth h is the same as the speed that a body (in this case a drop of water) would acquire in falling freely from a height h—that is, , where g is the acceleration due to gravity. This last expression comes from equating the kinetic energy with the potential energy mgh and solving for v. Suppose a tank filled with water is allowed to drain through a hole under the influence of gravity. We would like to find the depth h of water remaining in the tank at time t. Consider the tank shown in Figure 1.3.2. If the area of the hole is Ah (in ft2 ) and the speed of the water leaving the tank is (in ft/s), then the volume of water leaving the tank per second is (in ft3 /s). Thus if V(t) denotes the volume of water in the tank at time t, then , (9) dV dt 
Ah12gh v  12gh Ah12gh 1 2mv2 v  12gh dA dt  6
A 100 or dA dt 1 100 A  6. Rout  ( lb/gal) (3 gal/min)  lb/min. A(t) –––– 300 A(t) –––– 100 concentration of salt in outflow output rate of brine output rate of salt concentration of salt in inflow input rate of brine input rate of salt Rin  (2 lb/gal) (3 gal/min)  (6 lb/min). dA dt  input rate of salt
output rate of salt  Rin
Rout 1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS ● 23 input rate of brine 3 gal/min output rate of brine 3 gal/min constant 300 gal FIGURE 1.3.1 Mixing tank h Aw Ah FIGURE 1.3.2 Draining tank * Don’t confuse these symbols with Rin and Rout, which are input and output rates of salt.

24 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS where the minus sign indicates that V is decreasing.Note here that we are ignoring the possibility of friction at the hole that might cause a reduction of the rate of flow there.Now if the tank is such that the volume of water in it at time t can be written V(t)=Awh,where Aw(in ft2)is the constant area of the upper surface of the water (see Figure 1.3.2),then dv/dt=Awdh/dt.Substituting this last expression into (9) gives us the desired differential equation for the height of the water at time t: (a)LRC-series circuit =_An v2gh. dh dt A (10) Inductor inductance L:henries (h) It is interesting to note that(10)remains valid even when Aw is not constant.In this voltage drop across:L case we must express the upper surface area of the water as a function of h-that is, Aw=A(h).See Problem 14 in Exercises 1.3. i 000 SERIES CIRCUITS Consider the single-loop series circuit shown in Figure 1.3.3(a), containing an inductor,resistor,and capacitor.The current in a circuit after a switch is closed is denoted by i(t);the charge on a capacitor at time t is denoted by g(t).The Resistor letters L.R,and C are known as inductance,resistance,and capacitance,respectively, resistance R:ohms (2) and are generally constants.Now according to Kirchhoff's second law,the im- voltage drop across:iR pressed voltage E(t)on a closed loop must equal the sum of the voltage drops in the loop.Figure 1.3.3(b)shows the symbols and the formulas for the respective voltage drops across an inductor,a capacitor,and a resistor.Since current i(r)is related to i- R charge g(t)on the capacitor by i=dq/dt,adding the three voltages inductor resistor capacitor di d'g iR=Rdg 1 Capacitor =L capacitance C:farads (f) dp and dt Cq voltage drop across:q and equating the sum to the impressed voltage yields a second-order differential equation 'drCq=E(0). (11) (b) FIGURE 1.3.3 Symbols,units,and We will examine a differential equation analogous to (11)in great detail in voltages.Current i(t)and charge g(t)are Section 5.1. measured in amperes (A)and coulombs (C),respectively FALLING BODIES To construct a mathematical model of the motion of a body moving in a force field.one often starts with Newton's second law of motion.Recall from elementary physics that Newton's first law of motion states that a body either will remain at rest or will continue to move with a constant velocity unless acted on by an external force.In each case this is equivalent to saying that when the sum of rock the forces F=F-that is,the net or resultant force-acting on the body is zero, then the acceleration a of the body is zero.Newton's second law of motion indicates that when the net force acting on a body is not zero,then the net force is proportional to its acceleration a or,more precisely,F=ma,where m is the mass of s(t) the body. Now suppose a rock is tossed upward from the roof of a building as illustrated in Figure 1.3.4.What is the position s(t)of the rock relative to the ground at time t?The building acceleration of the rock is the second derivative d2s/dt2.If we assume that the up- ground ward direction is positive and that no force acts on the rock other than the force of FIGURE 1.3.4 Position of rock gravity,then Newton's second law gives measured from ground level d2s dp =-mg Or d (12) dr =一8 In other words,the net force is simply the weight F=F=-W of the rock near the surface of the Earth.Recall that the magnitude of the weight is W=mg,where m is

where the minus sign indicates that V is decreasing. Note here that we are ignoring the possibility of friction at the hole that might cause a reduction of the rate of flow there. Now if the tank is such that the volume of water in it at time t can be written V(t)  Awh, where Aw (in ft2 ) is the constant area of the upper surface of the water (see Figure 1.3.2), then dV
dt  Aw dh
dt. Substituting this last expression into (9) gives us the desired differential equation for the height of the water at time t: . (10) It is interesting to note that (10) remains valid even when Aw is not constant. In this case we must express the upper surface area of the water as a function of h—that is, Aw  A(h). See Problem 14 in Exercises 1.3. SERIES CIRCUITS Consider the single-loop series circuit shown in Figure 1.3.3(a), containing an inductor, resistor, and capacitor. The current in a circuit after a switch is closed is denoted by i(t); the charge on a capacitor at time t is denoted by q(t). The letters L, R, and C are known as inductance, resistance, and capacitance, respectively, and are generally constants. Now according to Kirchhoff’s second law, the im￾pressed voltage E(t) on a closed loop must equal the sum of the voltage drops in the loop. Figure 1.3.3(b) shows the symbols and the formulas for the respective voltage drops across an inductor, a capacitor, and a resistor. Since current i(t) is related to charge q(t) on the capacitor by i  dq
dt, adding the three voltages inductor resistor capacitor and equating the sum to the impressed voltage yields a second-order differential equation (11) We will examine a differential equation analogous to (11) in great detail in Section 5.1. FALLING BODIES To construct a mathematical model of the motion of a body moving in a force field, one often starts with Newton’s second law of motion. Recall from elementary physics that Newton’s first law of motion states that a body either will remain at rest or will continue to move with a constant velocity unless acted on by an external force. In each case this is equivalent to saying that when the sum of the forces —that is, the net or resultant force—acting on the body is zero, then the acceleration a of the body is zero. Newton’s second law of motion indicates that when the net force acting on a body is not zero, then the net force is proportional to its acceleration a or, more precisely, F  ma, where m is the mass of the body. Now suppose a rock is tossed upward from the roof of a building as illustrated in Figure 1.3.4. What is the position s(t) of the rock relative to the ground at time t? The acceleration of the rock is the second derivative d2 s
dt2 . If we assume that the up￾ward direction is positive and that no force acts on the rock other than the force of gravity, then Newton’s second law gives . (12) In other words, the net force is simply the weight F  F1 
W of the rock near the surface of the Earth. Recall that the magnitude of the weight is W  mg, where m is m d2 s dt2 
mg or d2 s dt2 
g F  Fk L d2 q dt2 R dq dt 1 C q  E(t). L di dt  L d2 q dt2 , iR  R dq dt, and 1 C q dh dt 
Ah Aw 12gh 24 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS (a) (b) E(t) L C R (a) LRC-series circuit (b) L R Inductor inductance L: henries (h) voltage drop across: L di dt i Capacitor capacitance C: farads (f) voltage drop across: 1 C i Resistor resistance R: ohms (Ω) voltage drop across: iR i q C FIGURE 1.3.3 Symbols, units, and voltages. Current i(t) and charge q(t) are measured in amperes (A) and coulombs (C), respectively ground building rock s(t) s0 v0 FIGURE 1.3.4 Position of rock measured from ground level

1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS.25 the mass of the body and g is the acceleration due to gravity.The minus sign in(12)is used because the weight of the rock is a force directed downward,which is opposite to the positive direction.If the height of the building is so and the initial velocity of the rock is vo.then s is determined from the second-order initial-value problem d2s dt2 =-g,s(0)=5o,s'(0)=o (13) Although we have not been stressing solutions of the equations we have con- structed,note that (13)can be solved by integrating the constant-g twice with respect to t.The initial conditions determine the two constants of integration. From elementary physics you might recognize the solution of(13)as the formula s(t)=-78+Vot +5o. FALLING BODIES AND AIR RESISTANCE Before Galileo's famous experiment kv from the leaning tower of Pisa,it was generally believed that heavier objects in free positive air resistance fall,such as a cannonball,fell with a greater acceleration than lighter objects,such as direction a feather.Obviously,a cannonball and a feather when dropped simultaneously from the same height do fall at different rates,but it is not because a cannonball is heavier. The difference in rates is due to air resistance.The resistive force of air was ignored gravity in the model given in (13).Under some circumstances a falling body of mass m,such as a feather with low density and irregular shape,encounters air resistance propor- mg tional to its instantaneous velocity v.If we take,in this circumstance,the positive FIGURE 1.3.5 Falling body of mass m direction to be oriented downward,then the net force acting on the mass is given by F=F1+F2 mg -kv,where the weight F1=mg of the body is force acting in the positive direction and air resistance F2=-kv is a force,called viscous damping, acting in the opposite or upward direction.See Figure 1.3.5.Now since vis related to acceleration a by a=dv/dt,Newton's second law becomes F=ma m dv/dt.By equating the net force to this form of Newton's second law,we obtain a first-order differential equation for the velocity v(t)of the body at time t, dv mg -kv. (14) dt (a)suspension bridge cable Here k is a positive constant of proportionality.If s(t)is the distance the body falls in time t from its initial point of release,then v=ds/dt and a dv/dt =d2s/dr2.In terms ofs,(14)is a second-order differential equation dr2=mg-k ds d's ds or m =mg. (15) dt (b)telephone wires SUSPENDED CABLES Suppose a flexible cable,wire,or heavy rope is suspended FIGURE 1.3.6 Cables suspended between two vertical supports.Physical examples of this could be one of the two between vertical supports cables supporting the roadbed of a suspension bridge as shown in Figure 1.3.6(a)or a long telephone wire strung between two posts as shown in Figure 1.3.6(b).Our goal is to construct a mathematical model that describes the shape that such a cable assumes. To begin.let's agree to examine only a portion or element of the cable between T? its lowest point P and any arbitrary point P2.As drawn in blue in Figure 1.3.7.this element of the cable is the curve in a rectangular coordinate system with y-axis cho- P2 wire T cos 0 sen to pass through the lowest point PI on the curve and the x-axis chosen a units below P1.Three forces are acting on the cable:the tensions TI and T2 in the cable (0,a that are tangent to the cable at P and P2,respectively,and the portion W of the total (x,0) vertical load between the points P and P2.Let T=Ti,T2=T2,and W=W denote the magnitudes of these vectors.Now the tension T2 resolves FIGURE 1.3.7 Element of cable into horizontal and vertical components (scalar quantities)T2 cos and T2 sin 0

the mass of the body and g is the acceleration due to gravity. The minus sign in (12) is used because the weight of the rock is a force directed downward, which is opposite to the positive direction. If the height of the building is s0 and the initial velocity of the rock is v0, then s is determined from the second-order initial-value problem . (13) Although we have not been stressing solutions of the equations we have con￾structed, note that (13) can be solved by integrating the constant
g twice with respect to t. The initial conditions determine the two constants of integration. From elementary physics you might recognize the solution of (13) as the formula FALLING BODIES AND AIR RESISTANCE Before Galileo’s famous experiment from the leaning tower of Pisa, it was generally believed that heavier objects in free fall, such as a cannonball, fell with a greater acceleration than lighter objects, such as a feather. Obviously, a cannonball and a feather when dropped simultaneously from the same height do fall at different rates, but it is not because a cannonball is heavier. The difference in rates is due to air resistance. The resistive force of air was ignored in the model given in (13). Under some circumstances a falling body of mass m, such as a feather with low density and irregular shape, encounters air resistance propor￾tional to its instantaneous velocity v. If we take, in this circumstance, the positive direction to be oriented downward, then the net force acting on the mass is given by F  F1 F2  mg
kv, where the weight F1  mg of the body is force acting in the positive direction and air resistance F2 
kv is a force, called viscous damping, acting in the opposite or upward direction. See Figure 1.3.5. Now since v is related to acceleration a by a  dv
dt, Newton’s second law becomes F  ma  m dv
dt. By equating the net force to this form of Newton’s second law, we obtain a first-order differential equation for the velocity v(t) of the body at time t, . (14) Here k is a positive constant of proportionality. If s(t) is the distance the body falls in time t from its initial point of release, then v  ds
dt and a  dv
dt  d2 s
dt2 . In terms of s, (14) is a second-order differential equation (15) SUSPENDED CABLES Suppose a flexible cable, wire, or heavy rope is suspended between two vertical supports. Physical examples of this could be one of the two cables supporting the roadbed of a suspension bridge as shown in Figure 1.3.6(a) or a long telephone wire strung between two posts as shown in Figure 1.3.6(b). Our goal is to construct a mathematical model that describes the shape that such a cable assumes. To begin, let’s agree to examine only a portion or element of the cable between its lowest point P1 and any arbitrary point P2. As drawn in blue in Figure 1.3.7, this element of the cable is the curve in a rectangular coordinate system with y-axis cho￾sen to pass through the lowest point P1 on the curve and the x-axis chosen a units below P1. Three forces are acting on the cable: the tensions T1 and T2 in the cable that are tangent to the cable at P1 and P2, respectively, and the portion W of the total vertical load between the points P1 and P2. Let T1  T1, T2  T2, and W  W denote the magnitudes of these vectors. Now the tension T2 resolves into horizontal and vertical components (scalar quantities) T2 cos and T2 sin . m d2 s dt 2  mg
k ds dt or m d2 s dt 2 k ds dt  mg. m dv dt  mg
kv s(t) 
1 2gt2 v0t s0. d2 s dt 2 
g, s(0)  s0, s(0)  v0 1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS ● 25 positive direction air resistance gravity kv mg FIGURE 1.3.5 Falling body of mass m (a) suspension bridge cable (b) telephone wires FIGURE 1.3.6 Cables suspended between vertical supports FIGURE 1.3.7 Element of cable wire T cos 2 θ θ T2 sin T2 P2 T1 W P1 θ y (x, 0) x (0, a)