EXAMPLE 1 Verification of Solution

Verify that y = c/x (c an arbitrary constant) is a solution of the ODE xy′ = −y for all x ≠ 0. Indeed, differentiate y = c/x to get y′ = −c/x². Multiply this by x, obtaining xy′ = −c/x; thus, xy′ = −y, the given ODE.

EXAMPLE 2 Solution by Calculus. Solution Curves

The ODE y′ = dy/dx = cos x can be solved directly by integration on both sides. Indeed, using calculus, we obtain y = ∫ cos x dx = sin x + c, where c is an arbitrary constant. This is a family of solutions. Each value of c, for instance, 2.75 or 0 or −8, gives one of these curves. Figure 3 shows some of them, for c = −3, −2, −1, 0, 1, 2, 3, 4.

EXAMPLE 3 (A) Exponential Growth. (B) Exponential Decay

From calculus we know that y = ce^0.2t has the derivative

Hence y is a solution of y′ = 0.2y (Fig. 4A). This ODE is of the form y′ = ky. With positive-constant k it can model exponential growth, for instance, of colonies of bacteria or populations of animals. It also applies to humans for small populations in a large country (e.g., the United States in early times) and is then known as Malthus's law.¹ We shall say more about this topic in Sec. 1.5.

(B) Similarly, y′ = −0.2 (with a minus on the right) has the solution y = ce^−0.2t, (Fig. 4B) modeling exponential decay, as, for instance, of a radioactive substance (see Example 5).

EXAMPLE 4 Initial Value Problem

Solve the initial value problem

Solution. The general solution is y(x) = ce^3x; see Example 3. From this solution and the initial condition we obtain y(0) = ce⁰ = c = 5.7. Hence the initial value problem has the solution y(x) = 5.7e^3x. This is a particular solution.

EXAMPLE 5 Radioactivity. Exponential Decay

Given an amount of a radioactive substance, say, 0.5 g (gram), find the amount present at any later time.

Physical Information. Experiments show that at each instant a radioactive substance decomposes—and is thus decaying in time—proportional to the amount of substance present.

Step 1. Setting up a mathematical model of the physical process. Denote by y(t) the amount of substance still present at any time t. By the physical law, the time rate of change y′(t) = dy/dt is proportional to y(t). This gives the first-order ODE

where the constant k is positive, so that, because of the minus, we do get decay (as in [B] of Example 3). The value of k is known from experiments for various radioactive substances (e.g., k = 1.4 · 10⁻¹¹ sec⁻¹, approximately, for radium ).

Now the given initial amount is 0.5 g, and we can call the corresponding instant t = 0. Then we have the initial condition y(0) = 0.5. This is the instant at which our observation of the process begins. It motivates the term initial condition (which, however, is also used when the independent variable is not time or when we choose a t other than t = 0). Hence the mathematical model of the physical process is the initial value problem

Step 2. Mathematical solution. As in (B) of Example 3 we conclude that the ODE (6) models exponential decay and has the general solution (with arbitrary constant c but definite given k)

We now determine c by using the initial condition. Since y(0) = c from (8), this gives y(0) = c = 0.5. Hence the particular solution governing our process is (cf. Fig. 5)

Always check your result—it may involve human or computer errors! Verify by differentiation (chain rule!) that your solution (9) satisfies (7) as well as y(0) = 0.5:

Step 3. Interpretation of result. Formula (9) gives the amount of radioactive substance at time t. It starts from the correct initial amount and decreases with time because k is positive. The limit of y as t → ∞ is zero.

PROBLEM SET 1.1

1–8 CALCULUS

Solve the ODE by integration or by remembering a differentiation formula.

1. y′ + 2 sin 2πx = 0
2. 
3. y′ = y
4. y′ = −1.5y
5. y′ = 4e^−x cos x
6. y″ = −y
7. y′ = cosh 5.13x
8. y″′ = e^−0.2x

9–15 VERIFICATION. INITIAL VALUE PROBLEM (IVP)

(a) Verify that y is a solution of the ODE. (b) Determine from y the particular solution of the IVP. (c) Graph the solution of the IVP.

• 9. y′ + 4y = 1.4, y = ce^−4x + 0.35, y(0) = 2
• 10.
• 11.
• 12. yy′ = 4x, y² − 4x² = c(y > 0), y(1) = 4
• 13.
• 14. y′ tan x = 2y − 8, y = c sin² x + 4,
• 15. Find two constant solutions of the ODE in Prob. 13 by inspection.
• 16. Singular solution. An ODE may sometimes have an additional solution that cannot be obtained from the general solution and is then called a singular solution. The ODE y^′2 − xy′ + y = 0 is of this kind. Show by differentiation and substitution that it has the general solution y = cx − c² and the singular solution y = x²/4. Explain Fig. 6.

  

  Fig. 6. Particular solutions and singular solution in Problem 16

17–20 MODELING, APPLICATIONS

These problems will give you a first impression of modeling. Many more problems on modeling follow throughout this chapter.

• 17. Half-life. The half-life measures exponential decay. It is the time in which half of the given amount of radioactive substance will disappear. What is the half-life of (in years) in Example 5?
• 18. Half-life. Radium has a half-life of about 3.6 days.

  (a) Given 1 gram, how much will still be present after 1 day?

  (b) After 1 year?

• 19. Free fall. In dropping a stone or an iron ball, air resistance is practically negligible. Experiments show that the acceleration of the motion is constant (equal to g = 9.80 m/sec² = 32 ft/sec², called the acceleration of gravity). Model this as an ODE for y(t), the distance fallen as a function of time t. If the motion starts at time t = 0 from rest (i.e., with velocity ν = y′ = 0), show that you obtain the familiar law of free fall

  

• 20. Exponential decay. Subsonic flight. The efficiency of the engines of subsonic airplanes depends on air pressure and is usually maximum near 35,000 ft. Find the air pressure y(x) at this height. Physical information. The rate of change y′(x) is proportional to the pressure. At 18,000 ft it is half its value y₀ = y(0) at sea level. Hint. Remember from calculus that if y = e^kx, then y′ = ke^kx = ky. Can you see without calculation that the answer should be close to y₀/4?

PROBLEM SET 1.2

1–8 DIRECTION FIELDS, SOLUTION CURVES

Graph a direction field (by a CAS or by hand). In the field graph several solution curves by hand, particularly those passing through the given points (x, y).

1. y′ = 1 + y²,
2. yy′ + 4x = 0, (1, 1), (0, 2)
3. y′ = 1 − y², (0, 0), (2, )
4. y′ = 2y − y², (0, 0), (0, 1), (0, 2), (0, 3)
5. y′ = x − 1/y, (1, )
6. y′ = sin²y, (0, −0.4), (0, 1)
7. y′ = e^y/x, (2, 2), (3, 3)
8. y′ = −2xy, (0, ), (0, 1), (0, 2)

9–10 ACCURACY OF DIRECTION FIELDS

Direction fields are very useful because they can give you an impression of all solutions without solving the ODE, which may be difficult or even impossible. To get a feel for the accuracy of the method, graph a field, sketch solution curves in it, and compare them with the exact solutions.

• 9. y′ = cos πx
• 10.
• 11. Autonomous ODE. This means an ODE not showing x (the independent variable) explicitly. (The ODEs in Probs. 6 and 10 are autonomous.) What will the level curves f(x, y) = const (also called isoclines curves of equal inclination) of an autonomous ODE look like? Give reason.

12–15 MOTIONS

Model the motion of a body B on a straight line with velocity as given, y(t) being the distance of B from a point y = 0 at time t. Graph a direction field of the model (the ODE). In the field sketch the solution curve satisfying the given initial condition.

• 12. Product of velocity times distance constant, equal to 2, y(0) = 2.
• 13. Distance = Velocity × Time, y(1) = 1
• 14. Square of the distance plus square of the velocity equal to 1, initial distance
• 15. Parachutist. Two forces act on a parachutist, the attraction by the earth mg (m = mass of person plus equipment, g = 9.8 m/sec² the acceleration of gravity) and the air resistance, assumed to be proportional to the square of the velocity ν(t). Using Newton's second law of motion (mass × acceleration = resultant of the forces), set up a model (an ODE for ν(t)). Graph a direction field (choosing m and the constant of proportionality equal to 1). Assume that the parachute opens when ν = 10 m/sec. Graph the corresponding solution in the field. What is the limiting velocity? Would the parachute still be sufficient if the air resistance were only proportional to ν(t)?
• 16. CAS PROJECT. Direction Fields. Discuss direction fields as follows.

  (a) Graph portions of the direction field of the ODE (2) (see Fig. 7), for instance, −5 x 2, −1 y 5. Explain what you have gained by this enlargement of the portion of the field.

  (b) Using implicit differentiation, find an ODE with the general solution x² + 9y² = c(y > 0). Graph its direction field. Does the field give the impression that the solution curves may be semi-ellipses? Can you do similar work for circles? Hyperbolas? Parabolas? Other curves?

  (c) Make a conjecture about the solutions of y′ = −x/y from the direction field.

  (d) Graph the direction field of and some solutions of your choice. How do they behave? Why do they decrease for? y > 0?

17–20 EULER'S METHOD

This is the simplest method to explain numerically solving an ODE, more precisely, an initial value problem (IVP). (More accurate methods based on the same principle are explained in Sec. 21.1.) Using the method, to get a feel for numerics as well as for the nature of IVPs, solve the IVP numerically with a PC or a calculator, 10 steps. Graph the computed values and the solution curve on the same coordinate axes.

• 17. y′ = y, y(0) = 1, h = 0.1
• 18. y′ = y, y(0) = 1, h = 0.01
• 19. y′ = (y − x)², y(0) = 0, h = 0.1 Sol. y = x − tanh x
• 20. y′ = −5x⁴y², y(0) = 1, h = 0.2 Sol. y = 1/(1 + x)⁵

EXAMPLE 1 Separable ODE

The ODE y′ = 1 + y² is separable because it can be written

It is very important to introduce the constant of integration immediately when the integration is performed. If we wrote arctan y = x, then y = tan x, and then introduced c, we would have obtained y = tan x + c, which is not a solution (when c ≠ 0). Verify this.

EXAMPLE 2 Separable ODE

The ODE y′ = (x + 1)e^−xy² is separable; we obtain y⁻² dy = (x + 1)e^−x dx.

By integration, .

EXAMPLE 3 Initial Value Problem (IVP). Bell-Shaped Curve

Solve y′ = −2xy, y(0) = 1.8.

Solution. By separation and integration,

This is the general solution. From it and the initial condition, y(0) = ce⁰ = c = 1.8. Hence the IVP has the solution . This is a particular solution, representing a bell-shaped curve (Fig. 10).

EXAMPLE 4 Radiocarbon Dating²

In September 1991 the famous Iceman (Oetzi), a mummy from the Neolithic period of the Stone Age found in the ice of the Oetztal Alps (hence the name “Oetzi”) in Southern Tyrolia near the Austrian–Italian border, caused a scientific sensation. When did Oetzi approximately live and die if the ratio of carbon to carbon in this mummy is 52.5% of that of a living organism?

Physical Information. In the atmosphere and in living organisms, the ratio of radioactive carbon (made radioactive by cosmic rays) to ordinary carbon is constant. When an organism dies, its absorption of by breathing and eating terminates. Hence one can estimate the age of a fossil by comparing the radioactive carbon ratio in the fossil with that in the atmosphere. To do this, one needs to know the half-life of , which is 5715 years (CRC Handbook of Chemistry and Physics, 83rd ed., Boca Raton: CRC Press, 2002, page 11–52, line 9).

Solution. Modeling. Radioactive decay is governed by the ODE y′ = ky (see Sec. 1.1, Example 5). By separation and integration (where t is time and y₀ is the initial ratio of to )

Next we use the half-life H = 5715 to determine k. When t = H, half of the original substance is still present. Thus,

Finally, we use the ratio 52.5% for determining the time t when Oetzi died (actually, was killed),

Other methods show that radiocarbon dating values are usually too small. According to recent research, this is due to a variation in that carbon ratio because of industrial pollution and other factors, such as nuclear testing.

EXAMPLE 5 Mixing Problem

Mixing problems occur quite frequently in chemical industry. We explain here how to solve the basic model involving a single tank. The tank in Fig. 11 contains 1000 gal of water in which initially 100 lb of salt is dissolved. Brine runs in at a rate of 10 gal min, and each gallon contains 5 lb of dissoved salt. The mixture in the tank is kept uniform by stirring. Brine runs out at 10 gal min. Find the amount of salt in the tank at any time t.

Solution. Step 1. Setting up a model. Let y(t) denote the amount of salt in the tank at time t. Its time rate of change is

5 lb times 10 gal gives an inflow of 50 lb of salt. Now, the outflow is 10 gal of brine. This is 10/1000 = 0.01 (= 1%) of the total brine content in the tank, hence 0.01 of the salt content y(t), that is, 0.01 y(t). Thus the model is the ODE

Step 2. Solution of the model. The ODE (4) is separable. Separation, integration, and taking exponents on both sides gives

Initially the tank contains 100 lb of salt. Hence y(0) = 100 is the initial condition that will give the unique solution. Substituting y = 100 and t = 0 in the last equation gives 100 − 5000 = ce⁰ = c. Hence c = −4900. Hence the amount of salt in the tank at time t is

This function shows an exponential approach to the limit 5000 lb; see Fig. 11. Can you explain physically that should increase with time? That its limit is 5000 lb? Can you see the limit directly from the ODE?

The model discussed becomes more realistic in problems on pollutants in lakes (see Problem Set 1.5, Prob. 35) or drugs in organs. These types of problems are more difficult because the mixing may be imperfect and the flow rates (in and out) may be different and known only very roughly.

EXAMPLE 6 Heating an Office Building (Newton's Law of Cooling³)

Suppose that in winter the daytime temperature in a certain office building is maintained at 70°F. The heating is shut off at 10 P.M. and turned on again at 6 A.M. On a certain day the temperature inside the building at 2 A.M. was found to be 65°F. The outside temperature was 50°F at 10 P.M. and had dropped to 40°F by 6 A.M. What was the temperature inside the building when the heat was turned on at 6 A.M.?

Physical information. Experiments show that the time rate of change of the temperature T of a body B (which conducts heat well, for example, as a copper ball does) is proportional to the difference between T and the temperature of the surrounding medium (Newton's law of cooling).

Solution. Step 1. Setting up a model. Let T(t) be the temperature inside the building and T_A the outside temperature (assumed to be constant in Newton's law). Then by Newton's law,

Such experimental laws are derived under idealized assumptions that rarely hold exactly. However, even if a model seems to fit the reality only poorly (as in the present case), it may still give valuable qualitative information. To see how good a model is, the engineer will collect experimental data and compare them with calculations from the model.

Step 2. General solution. We cannot solve (6) because we do not know T_A, just that it varied between 50°F and 40°F, so we follow the Golden Rule: If you cannot solve your problem, try to solve a simpler one. We solve (6) with the unknown function T_A replaced with the average of the two known values, or 45°F. For physical reasons we may expect that this will give us a reasonable approximate value of T in the building at 6 A.M.

For constant T_A = 45 (or any other constant value) the ODE (6) is separable. Separation, integration, and taking exponents gives the general solution

Step 3. Particular solution. We choose 10 P.M. to be t = 0. Then the given initial condition is T(0) = 70 and yields a particular solution, call it T_p. By substitution,

Step 4. Determination of k. We use T(4) = 65, where t = 4 is 2 A.M. Solving algebraically for k and inserting k into T_p(t) gives (Fig. 12)

Fig. 12. Particular solution (temperature) in Example 6

Step 5. Answer and interpretation. 6 A.M. is t = 8 (namely, 8 hours after 10 P.M.), and

Hence the temperature in the building dropped 9°F, a result that looks reasonable.

EXAMPLE 7 Leaking Tank. Outflow of Water Through a Hole (Torricelli's Law)

This is another prototype engineering problem that leads to an ODE. It concerns the outflow of water from a cylindrical tank with a hole at the bottom (Fig. 13). You are asked to find the height of the water in the tank at any time if the tank has diameter 2 m, the hole has diameter 1 cm, and the initial height of the water when the hole is opened is 2.25 m. When will the tank be empty?

Physical information. Under the influence of gravity the outflowing water has velocity

where h(t) is the height of the water above the hole at time t, and g = 980 cm/sec² = 32.17 ft/sec² is the acceleration of gravity at the surface of the earth.

Solution. Step 1. Setting up the model. To get an equation, we relate the decrease in water level h(t) to the outflow. The volume ΔV of the outflow during a short time Δt is

ΔV must equal the change ΔV* of the volume of the water in the tank. Now

where Δh (> 0) is the decrease of the height h(t) of the water. The minus sign appears because the volume of the water in the tank decreases. Equating ΔV and ΔV* gives

We now express ν according to Torricelli's law and then let Δt (the length of the time interval considered) approach 0—this is a standard way of obtaining an ODE as a model. That is, we have

and by letting Δt → 0 we obtain the ODE

where . This is our model, a first-order ODE.

Step 2. General solution. Our ODE is separable. A/B is constant. Separation and integration gives

Dividing by 2 and squaring gives h = (c − 13.28At/B)². Inserting 13.28A/B = 13.28 · 0.5²π/100²π = 0.000332 yields the general solution

Step 3. Particular solution. The initial height (the initial condition) is h(0) = 225 cm. Substitution of t = 0 and h = 225 gives from the general solution c² = 225, c = 15.00 and thus the particular solution (Fig. 13)

Step 4. Tank empty. h_p(t) = 0 if t = 15.00/0.000332 = 45,181 [sec] = 12.6 [hours].

Here you see distinctly the importance of the choice of units—we have been working with the cgs system, in which time is measured in seconds! We used g = 980 cm/sec².

Step 5. Checking. Check the result.

EXAMPLE 8 Reduction to Separable Form

Solve

Solution. To get the usual explicit form, divide the given equation by 2xy,

Now substitute y and y′ from (9) and then simplify by subtracting u on both sides,

You see that in the last equation you can now separate the variables,

Take exponents on both sides to get 1 + u² = c/x or 1 + (y/x)² = c/x. Multiply the last equation by x² to obtain (Fig. 14)

This general solution represents a family of circles passing through the origin with centers on the x-axis.

PROBLEM SET 1.3

1. CAUTION! Constant of integration. Why is it important to introduce the constant of integration immediately when you integrate?

2–10 GENERAL SOLUTION

Find a general solution. Show the steps of derivation. Check your answer by substitution.

• 2. y³y′ + x³ = 0
• 3. y′ = sec²y
• 4. y′ sin 2πx = πy cos 2πx
• 5. yy′ + 36x = 0
• 6. y′ = e^2x−1y²
• 7.
• 8. y′ = (y + 4x)² (Set y + 4x = ν)
• 9. xy′ = y² + y (Set y/x = u)
• 10. xy′ = x + y (Set y/x = u)

11–17 INITIAL VALUE PROBLEMS (IVPS)

Solve the IVP. Show the steps of derivation, beginning with the general solution.

• 11. xy′ + y = 0, y(4) = 6
• 12. y′ = 1 + 4y², y(1) = 0
• 13. y′ cosh²x = sin²y,
• 14. dr/dt = −2tr, r(0) = r₀
• 15. y′ = −4x/y, y(2) = 3
• 16. y′ (x + y − 2)², y(0) = 2 (Set ν = x + y − 2)
• 17. xy′ = y + 3x⁴ cos²(y/x), y(1) = 0
• 18. Particular solution. Introduce limits of integration in (3) such that y obtained from (3) satisfies the initial condition y(x₀) = y₀.

19–36 MODELING, APPLICATIONS

• 19. Exponential growth. If the growth rate of the number of bacteria at any time t is proportional to the number present at t and doubles in 1 week, how many bacteria can be expected after 2 weeks? After 4 weeks?
• 20. Another population model.

  (a) If the birth rate and death rate of the number of bacteria are proportional to the number of bacteria present, what is the population as a function of time.

  (b) What is the limiting situation for increasing time? Interpret it.

• 21. Radiocarbon dating. What should be the content (in percent of y₀) of a fossilized tree that is claimed to be 3000 years old? (See Example 4.)
• 22. Linear accelerators are used in physics for accelerating charged particles. Suppose that an alpha particle enters an accelerator and undergoes a constant acceleration that increases the speed of the particle from 10³ m/sec to 10⁴ m/sec in 10⁻³ sec. Find the acceleration a and the distance traveled during that period of 10⁻³ sec.
• 23. Boyle–Mariotte's law for ideal gases.⁵ Experiments show for a gas at low pressure p (and constant temperature) the rate of change of the volume V(p) equals −V/p. Solve the model.
• 24. Mixing problem. A tank contains 400 gal of brine in which 100 lb of salt are dissolved. Fresh water runs into the tank at a rate of 2 gals/min. The mixture, kept practically uniform by stirring, runs out at the same rate. How much salt will there be in the tank at the end of 1 hour?
• 25. Newton's law of cooling. A thermometer, reading 5°C, is brought into a room whose temperature is 22°C. One minute later the thermometer reading is 12°C. How long does it take until the reading is practically 22°C, say, 21.9°C?
• 26. Gompertz growth in tumors. The Gompertz model is y′ = −Ay ln y (A > 0), where y(t) is the mass of tumor cells at time t. The model agrees well with clinical observations. The declining growth rate with increasing y > 1 corresponds to the fact that cells in the interior of a tumor may die because of insufficient oxygen and nutrients. Use the ODE to discuss the growth and decline of solutions (tumors) and to find constant solutions. Then solve the ODE.
• 27. Dryer. If a wet sheet in a dryer loses its moisture at a rate proportional to its moisture content, and if it loses half of its moisture during the first 10 min of drying, when will it be practically dry, say, when will it have lost 99% of its moisture? First guess, then calculate.
• 28. Estimation. Could you see, practically without calculation, that the answer in Prob. 27 must lie between 60 and 70 min? Explain.
• 29. Alibi? Jack, arrested when leaving a bar, claims that he has been inside for at least half an hour (which would provide him with an alibi). The police check the water temperature of his car (parked near the entrance of the bar) at the instant of arrest and again 30 min later, obtaining the values 190°F and 110°F, respectively. Do these results give Jack an alibi? (Solve by inspection.)
• 30. Rocket. A rocket is shot straight up from the earth, with a net acceleration (= acceleration by the rocket engine minus gravitational pullback) of 7t m/sec² during the initial stage of flight until the engine cut out at t = 10 sec. How high will it go, air resistance neglected?
• 31. Solution curves of y′ = g(y/x). Show that any (nonvertical) straight line through the origin of the xy-plane intersects all these curves of a given ODE at the same angle.
• 32. Friction. If a body slides on a surface, it experiences friction F (a force against the direction of motion). Experiments show that |F| = μ|N| (Coulomb's⁶ law of kinetic friction without lubrication), where N is the normal force (force that holds the two surfaces together; see Fig. 15) and the constant of proportionality μ is called the coefficient of kinetic friction. In Fig. 15 assume that the body weighs 45 nt (about 10 lb; see front cover for conversion). μ = 0.20 (corresponding to steel on steel), a = 30°, the slide is 10 m long, the initial velocity is zero, and air resistance is negligible. Find the velocity of the body at the end of the slide.

  

  Fig. 15. Problem 32

• 33. Rope. To tie a boat in a harbor, how many times must a rope be wound around a bollard (a vertical rough cylindrical post fixed on the ground) so that a man holding one end of the rope can resist a force exerted by the boat 1000 times greater than the man can exert? First guess. Experiments show that the change ΔS of the force S in a small portion of the rope is proportional to S and to the small angle Δφ in Fig. 16. Take the proportionality constant 0.15. The result should surprise you!

  

  Fig. 16. Problem 33

• 34. TEAM PROJECT. Family of Curves. A family of curves can often be characterized as the general solution of y′ = f(x, y).

  (a) Show that for the circles with center at the origin we get y′ = −x/y.

  (b) Graph some of the hyperbolas xy = c. Find an ODE for them.

  (c) Find an ODE for the straight lines through the origin.

  (d) You will see that the product of the right sides of the ODEs in (a) and (c) equals −1. Do you recognize this as the condition for the two families to be orthogonal (i.e., to intersect at right angles)? Do your graphs confirm this?

  (e) Sketch families of curves of your own choice and find their ODEs. Can every family of curves be given by an ODE?

• 35. CAS PROJECT. Graphing Solutions. A CAS can usually graph solutions, even if they are integrals that cannot be evaluated by the usual analytical methods of calculus.

  (a) Show this for the five initial value problems, , y(0) = 0, ±1, ±2, graphing all five curves on the same axes.

  (b) Graph approximate solution curves, using the first few terms of the Maclaurin series (obtained by termwise integration of that of y′) and compare with the exact curves.

  (c) Repeat the work in (a) for another ODE and initial conditions of your own choice, leading to an integral that cannot be evaluated as indicated.

• 36. TEAM PROJECT. Torricelli's Law. Suppose that the tank in Example 7 is hemispherical, of radius R, initially full of water, and has an outlet of 5 cm² cross-sectional area at the bottom. (Make a sketch.) Set up the model for outflow. Indicate what portion of your work in Example 7 you can use (so that it can become part of the general method independent of the shape of the tank). Find the time t to empty the tank (a) for any R, (b) for R = 1 m. Plot t as function of R. Find the time when h = R/2 (a) for any R, (b) for R = 1 m.

EXAMPLE 1 An Exact ODE

Solve

Solution. Step 1. Test for exactness. Our equation is of the form (1) with

Thus

From this and (5) we see that (7) is exact.

Step 2. Implicit general solution. From (6) we obtain by integration

To find k(y), we differentiate this formula with respect to y and use formula (4b), obtaining

Hence dk/dy = 3y² + 2y. By integration, k = y³ + y² + c*. Inserting this result into (8) and observing (3), we obtain the answer

Step 3. Checking an implicit solution. We can check by differentiating the implicit solution u(x, y) = c implicitly and see whether this leads to the given ODE (7):

This completes the check.

EXAMPLE 2 An Initial Value Problem

Solve the initial value problem

Solution. You may verify that the given ODE is exact. We find u. For a change, let us use (6*),

From this, ∂u/∂x = cos y sinh x + dl/dx = M = cos y sinh x + 1. Hence dl/dx = 1. By integration, l(x) = x + c*. This gives the general solution u(x, y) = cos y cosh x + x = c. From the initial condition, cos 2 cosh 1 + 1 = 0.358 = c. Hence the answer is cos y cosh x + x = 0.358. Figure 17 shows the particular solutions for c = 0, 0.358 (thicker curve), 1, 2, 3. Check that the answer satisfies the ODE. (Proceed as in Example 1.) Also check that the initial condition is satisfied.

EXAMPLE 3 WARNING! Breakdown in the Case of Nonexactness

The equation −y dx + x dy = 0 is not exact because M = −y and N = x, so that in (5), but ∂M/∂y = −1 but ∂N/∂x = 1. Let us show that in such a case the present method does not work. From (6),

Now, ∂u/∂y should equal N = x, by (4b). However, this is impossible because k(y) can depend only on y. Try (6*); it will also fail. Solve the equation by another method that we have discussed.

EXAMPLE 4 Integrating Factor

The integrating factor in (11) is F = 1/x². Hence in this case the exact equation (13) is

These are straight lines y = cx through the origin. (Note that x = 0 is also a solution of −y dx + x dy = 0.)

It is remarkable that we can readily find other integrating factors for the equation −y dx + x dy = 0, namely, 1/y², 1/(xy), and 1/(x² + y²), because

THEOREM 1 Integrating Factor F(x)

If (12) is such that the right side R of (16) depends only on x, then (12) has an integrating factor F = F(x), which is obtained by integrating (16) and taking exponents on both sides.

THEOREM 2 Integrating Factor F*(y)

If (12) is such that the right side R* of (18) depends only on y, then (12) has an integrating factor F* = F*(y), which is obtained from (18) in the form

EXAMPLE 5 Application of Theorems 1 and 2. Initial Value Problem

Using Theorem 1 or 2, find an integrating factor and solve the initial value problem

Solution. Step 1. Nonexactness. The exactness check fails:

Step 2. Integrating factor. General solution. Theorem 1 fails because R [the right side of (16)] depends on both x and y.

Try Theorem 2. The right side of (18) is

Hence (19) gives the integrating factor F*(y) = e^−y. From this result and (20) you get the exact equation

Test for exactness; you will get 1 on both sides of the exactness condition. By integration, using (4a),

Differentiate this with respect to y and use (4b) to get

Hence the general solution is

Setp 3. Particular solution. The initial condition y(0) = −1 gives u(0, −1) = 1 + 0 + e = 3.72. Hence the answer is e^x + xy + e^−y = 1 + e = 3.72. Figure 18 shows several particular solutions obtained as level curves of u(x, y) = c, obtained by a CAS, a convenient way in cases in which it is impossible or difficult to cast a solution into explicit form. Note the curve that (nearly) satisfies the initial condition.

Step 4. Checking. Check by substitution that the answer satisfies the given equation as well as the initial condition.

PROBLEM SET 1.4

1–14 ODEs. INTEGRATING FACTORS

Test for exactness. If exact, solve. If not, use an integrating factor as given or obtained by inspection or by the theorems in the text. Also, if an initial condition is given, find the corresponding particular solution.

1. 2xy dx + x² dy = 0
2. x³dx + y³dy = 0
3. sin x cos y dx + cos x sin y dy = 0
4. e^3θ(dr + 3r dθ) = 0
5. (x² + y²)dx − 2xy dy = 0
6. 3(y + 1) dx = 2x dy, (y + 1)x⁻⁴
7. 2x tan y dx + sec² y dy = 0
8. e^x(cos y dx − sin y dy) = 0
9. e^2x(2 cos y dx − sin y dy) = 0, y(0) = 0
10. y dx + [y + tan (x + y)] dy = 0, cos (x + y)
11. 2 cosh x cos y dx = sinh x sin y dy
12. 
13. e^−y dx + e^−x(−e^−y + 1)dy = 0, F = e^x+y
14. (x + 1)y dx + (b + 1)x dy = 0, y(1) = 1, F = x^ay^b
15. Exactness. Under what conditions for the constants a, b, k, l is (ax + by) dx + (kx + ly) dy = 0 exact? Solve the exact ODE.
16. TEAM PROJECT. Solution by Several Methods. Show this as indicated. Compare the amount of work.

    (a) e^y(sinh x dx + cosh x dy) = 0 as an exact ODE and by separation.

    (b) (1 + 2x)cos y dx + dy/cos y = 0 by Theorem 2 and by separation.

    (c) (x² + y²)dx − 2xy dy = 0 by Theorem 1 or 2 and by separation with ν = y/x.

    (d) 3x²y dx + 4x³ dy = 0 by Theorems 1 and 2 and by separation.

    (e) Search the text and the problems for further ODEs that can be solved by more than one of the methods discussed so far. Make a list of these ODEs. Find further cases of your own.

17. WRITING PROJECT. Working Backward. Working backward from the solution to the problem is useful in many areas. Euler, Lagrange, and other great masters did it. To get additional insight into the idea of integrating factors, start from a u(x, y) of your choice, find du = 0, destroy exactness by division by some F(x, y), and see what ODE's solvable by integrating factors you can get. Can you proceed systematically, beginning with the simplest F(x, y)?
18. CAS PROJECT. Graphing Particular Solutions. Graph particular solutions of the following ODE, proceeding as explained.

    

    (a) Show that (21) is not exact. Find an integrating factor using either Theorem 1 or 2. Solve (21).

    (b) Solve (21) by separating variables. Is this simpler than (a)?

    (c) Graph the seven particular solutions satisfying the following initial conditions y(0) = 1, , , ±1 (see figure below).

    (d) Which solution of (21) do we not get in (a) or (b)?

    

    Particular solutions in CAS Project 18

EXAMPLE 1 First-Order ODE, General Solution, Initial Value Problem

Solve the initial value problem

Solution. Here p = tan x, r = sin 2x = 2 sin x cos x, and

From this we see that in (4),

and the general solution of our equation is

From this and the initial condition, 1 = c · 1 − 2 · 1²; thus c = 3 and the solution of our initial value problem is y = 3 cos x − 2 cos²x. Here 3 cos x is the response to the initial data, and −2 cos² x is the response to the input sin 2x.

EXAMPLE 2 Electric Circuit

Model the RL-circuit in Fig. 19 and solve the resulting ODE for the current I(t) A (amperes), where t is time. Assume that the circuit contains as an EMF E(t) (electromotive force) a battery of E = 48 V (volts), which is constant, a resistor of R = 11 Ω (ohms), and an inductor of L = 0.1 H (henrys), and that the current is initially zero.

Physical Laws. A current I in the circuit causes a voltage drop RI across the resistor (Ohm's law) and a voltage drop LI′ = L dI/dt across the conductor, and the sum of these two voltage drops equals the EMF (Kirchhoff's Voltage Law, KVL).

Solution. According to these laws the model of the RL-circuit is LI′ + RI = E(t), in standard form

We can solve this linear ODE by (4) with x = t, y = I, p = R/L, h = (R/L)t, obtaining the general solution

By integration,

In our case, R/L = 11/0.1 = 110 and E(t) = 48/0.1 = 480 = const; thus,

In modeling, one often gets better insight into the nature of a solution (and smaller roundoff errors) by inserting given numeric data only near the end. Here, the general solution (7) shows that the current approaches the limit E/R = 48/11 faster the larger R/L is, in our case, R/L = 11/0.1 = 110, and the approach is very fast, from below if I(0) < 48/11 or from above if I(0) > 48/11. If I(0) = 48/11, the solution is constant (48/11 A). See Fig. 19.

The initial value I(0) = 0 gives I(0) = E/R + c = 0, c = −E/R and the particular solution

EXAMPLE 3 Hormone Level

Assume that the level of a certain hormone in the blood of a patient varies with time. Suppose that the time rate of change is the difference between a sinusoidal input of a 24-hour period from the thyroid gland and a continuous removal rate proportional to the level present. Set up a model for the hormone level in the blood and find its general solution. Find the particular solution satisfying a suitable initial condition.

Solution. Step 1. Setting up a model. Let y(t) be the hormone level at time t. Then the removal rate is Ky(t). The input rate is A + B cos ωt, where ω = 2π/24 = π/12 and A is the average input rate; here A B to make the input rate nonnegative. The constants A, B, K can be determined from measurements. Hence the model is the linear ODE

The initial condition for a particular solution y_part is y_part(0) with t = 0 suitably chosen, for example, 6:00 A.M.

Step 2. General solution. In (4) we have p = K = const, h = Kt, and r = A + B cos ωt. Hence (4) gives the general solution (evaluate cos ωt dt by integration by parts)

The last term decreases to 0 as t increases, practically after a short time and regardless of c (that is, of the initial condition). The other part of y(t) is called the steady-state solution because it consists of constant and periodic terms. The entire solution is called the transient-state solution because it models the transition from rest to the steady state. These terms are used quite generally for physical and other systems whose behavior depends on time.

Step 3. Particular solution. Setting t = 0 in y(t) and choosing y₀ = 0, we have

Inserting this result into y(t), we obtain the particular solution

with the steady-state part as before. To plot y_part we must specify values for the constants, say, A = B = 1 and K = 0.05. Figure 20 shows this solution. Notice that the transition period is relatively short (although K is small), and the curve soon looks sinusoidal; this is the response to the input .

EXAMPLE 4 Logistic Equation

Solve the following Bernoulli equation, known as the logistic equation (or Verhulst equation⁸):

Solution. Write (11) in the form (9), that is,

to see that a = 2, so that u = y^1−a = y⁻¹. Differentiate this u and substitute y′ from (11),

The last term is −Ay⁻¹ = Hence we have obtained the linear ODE

The general solution is [by (4)]

Since u = 1/y, this gives the general solution of (11),

Directly from (11) we see that y = ≡ 0 (y(t) = 0 for all t) is also a solution.

EXAMPLE 5 Stable and Unstable Equilibrium Solutions. “Phase Line Plot”

The ODE y′ = (y − 1)(y − 2) has the stable equilibrium solution y₁ = 1 and the unstable y₂ = 2, as the direction field in Fig. 22 suggests. The values y₁ and y₂ are the zeros of the parabola f(y) = (y − 1)(y − 2) in the figure. Now, since the ODE is autonomous, we can “condense” the direction field to a “phase line plot” giving y₁ and y₂, and the direction (upward or downward) of the arrows in the field, and thus giving information about the stability or instability of the equilibrium solutions.

PROBLEM SET 1.5

1. CAUTION! Show that e^−lnx = 1/x(not −x) and e^{−ln(sec x)} = cos x.
2. Integration constant. Give a reason why in (4) you may choose the constant of integration in ∫p dx to be zero.

3–13 GENERAL SOLUTION. INITIAL VALUE PROBLEMS

Find the general solution. If an initial condition is given, find also the corresponding particular solution and graph or sketch it. (Show the details of your work.)

• 3. y′ − y = 5.2
• 4. y′ = 2y − 4x
• 5. y′ + ky = e^−kx
• 6.
• 7. xy′ = 2y + x³e^x
• 8. y′ + y tan x = e^−0.01x cos x, y(0) = 0
• 9. y′ + y sin x = e^cosx, y(0) = −2.5
• 10.
• 11. y′ = (y − 2) cot x
• 12. xy′ +4y = 8x⁴, y(1) = 2
• 13. y′ = 6(y − 2.5) tanh 1.5x
• 14. CAS EXPERIMENT. (a) Solve the ODE y′ − y/x = −x⁻¹ cos (1/x). Find an initial condition for which the arbitrary constant becomes zero. Graph the resulting particular solution, experimenting to obtain a good figure near x = 0.

  (b) Generalizing (a) from n = 1 to arbitrary n, solve the ODE y′ − ny/x = −xⁿ⁻² cos (1/x). Find an initial condition as in (a) and experiment with the graph.

15–20 GENERAL PROPERTIES OF LINEAR ODEs

These properties are of practical and theoretical importance because they enable us to obtain new solutions from given ones. Thus in modeling, whenever possible, we prefer linear ODEs over nonlinear ones, which have no similar properties.

Show that nonhomogeneous linear ODEs (1) and homogeneous linear ODEs (2) have the following properties. Illustrate each property by a calculation for two or three equations of your choice. Give proofs.

• 15. The sum y₁ + y₂ of two solutions y₁ and y₂ of the homogeneous equation (2) is a solution of (2), and so is a scalar multiple ay₁ for any constant a. These properties are not true for (1)!
• 16. y = 0 (that is, y(x = 0 for all x, also written y(x) ≡ 0) is a solution of (2) [not of (1) if r(x) ≠ 0!], called the trivial solution.
• 17. The sum of a solution of (1) and a solution of (2) is a solution of (1).
• 18. The difference of two solutions of (1) is a solution of (2).
• 19. If y₁ is a solution of (1), what can you say about cy₁?
• 20. If y₁ and y₂ are solutions of and respectively (with the same p!), what can you say about the sum y₁ + y₂?
• 21. Variation of parameter. Another method of obtaining (4) results from the following idea. Write (3) as cy* where y* is the exponential function, which is a solution of the homogeneous linear ODE . Replace the arbitrary constant c in (3) with a function u to be determined so that the resulting function y = uy* is a solution of the nonhomogeneous linear ODE y′ + py = r.

22–28 NONLINEAR ODEs

Using a method of this section or separating variables, find the general solution. If an initial condition is given, find also the particular solution and sketch or graph it.

• 22. y′ + y = y²,
• 23. y′ + xy = xy⁻¹, y(0) = 3
• 24. y′ + y = −x/y
• 25. y′ = 3.2y −10y²
• 26. y′ = (tan y)/(x − 1),
• 27. y′ = 1/(6e^y − 2x)
• 28. 2xyy′ + (x − 1)y² = x²e^x (Set y² = z)
• 29. REPORT PROJECT. Transformation of ODEs. We have transformed ODEs to separable form, to exact form, and to linear form. The purpose of such transformations is an extension of solution methods to larger classes of ODEs. Describe the key idea of each of these transformations and give three typical examples of your choice for each transformation. Show each step (not just the transformed ODE).
• 30. TEAM PROJECT. Riccati Equation. Clairaut Equation. Singular Solution.

  A Riccati equation is of the form

  

  A Clairaut equation is of the form

  

  (a) Apply the transformation y = Y + 1/u to the Riccati equation (14), where Y is a solution of (14), and obtain for u the linear ODE u′ + (2Yg − p)u = −g. Explain the effect of the transformation by writing it as y = Y + ν, ν = 1/ν.

  (b) Show that y = Y = x is a solution of the ODE y′ − (2x³ + 1) y = −x²y² − x⁴ − x + 1 and slove this Riccati equation, showing the details.

  (c) Solve the Clairaut equation y′² − xy′ + y = 0 follows. Differentiate it with respect to x, obtaining y″(2y′ − x) = 0. Then solve (A) y″ = 0 and (B) 2y′ − x = 0 separately and substitute the two solutions (a) and (b) of (A) and (B) into the given ODE. Thus obtain (a) a general solution (straight lines) and (b) a parabola for which those lines (a) are tangents (Fig. 6 in Prob. Set 1.1); so (b) is the envelope of (a). Such a solution (b) that cannot be obtained from a general solution is called a singular solution.

  (d) Show that the Clairaut equation (15) has as solutions a family of straight lines y = cx + g(c) and a singular solution determined by g′ (s) = −x, where s = y′, that forms the envelope of that family.

31–40 MODELING. FURTHER APPLICATIONS

• 31. Newton's law of cooling. If the temperature of a cake is 300°F when it leaves the oven and is 200°F ten minutes later, when will it be practically equal to the room temperature of 60°F, say, when will it be 61°F?
• 32. Heating and cooling of a building. Heating and cooling of a building can be modeled by the ODE

  

  where T = T(t) is the temperature in the building at time t, T_a the outside temperature, T_w the temperature wanted in the building, and P the rate of increase of T due to machines and people in the building, and k₁ and k₂ are (negative) constants. Solve this ODE, assuming andP = const, T_w = const, and T_a varying sinusoidally over 24 hours, say, T_a = A − C cos (2π/24)t. Discuss the effect of each term of the equation on the solution.

• 33. Drug injection. Find and solve the model for drug injection into the bloodstream if, beginning at t = 0, a constant amount A g/min is injected and the drug is simultaneously removed at a rate proportional to the amount of the drug present at time t.
• 34. Epidemics. A model for the spread of contagious diseases is obtained by assuming that the rate of spread is proportional to the number of contacts between infected and noninfected persons, who are assumed to move freely among each other. Set up the model. Find the equilibrium solutions and indicate their stability or instability. Solve the ODE. Find the limit of the proportion of infected persons as t → ∞ and explain what it means.
• 35. Lake Erie. Lake Erie has a water volume of about 450 km³ and a flow rate (in and out) of about 175 km² per year. If at some instant the lake has pollution concentration p = 0.04%, how long, approximately, will it take to decrease it to p/2, assuming that the inflow is much cleaner, say, it has pollution concentration p/4, and the mixture is uniform (an assumption that is only imperfectly true)? First guess.
• 36. Harvesting renewable resources. Fishing. Suppose that the population y(t) of a certain kind of fish is given by the logistic equation (11), and fish are caught at a rate Hy proportional to y. Solve this so-called Schaefer model. Find the equilibrium solutions y₁ and y₂ (> 0) when H < A. The expression Y = Hy₂ is called the equilibrium harvest or sustainable yield corresponding to H. Why?
• 37. Harvesting. In Prob. 36 find and graph the solution satisfying y(0) = 2 when (for simplicity) A = B = 1 and H = 0.2. What is the limit? What does it mean? What if there were no fishing?
• 38. Intermittent harvesting. In Prob. 36 assume that you fish for 3 years, then fishing is banned for the next 3 years. Thereafter you start again. And so on. This is called intermittent harvesting. Describe qualitatively how the population will develop if intermitting is continued periodically. Find and graph the solution for the first 9 years, assuming that A = B = 1, H = 0.2, and y(0) = 2.

  

  Fig. 23. Fish population in Problem 38

• 39. Extinction vs. unlimited growth. If in a population y(t) the death rate is proportional to the population, and the birth rate is proportional to the chance encounters of meeting mates for reproduction, what will the model be? Without solving, find out what will eventually happen to a small initial population. To a large one. Then solve the model.
• 40. Air circulation. In a room containing 20,000 ft³ of air, 600 ft³ of fresh air flows in per minute, and the mixture (made practically uniform by circulating fans) is exhausted at a rate of 600 cubic feet per minute (cfm). What is the amount of fresh air y(t) at any time if y(0) = 0? After what time will 90% of the air be fresh?

PROBLEM SET 1.6

1–3 FAMILIES OF CURVES

Represent the given family of curves in the form G(x, y; c) = 0 and sketch some of the curves.

1. All ellipses with foci −3 and 3 on the x-axis.
2. All circles with centers on the cubic parabola y = x³ and passing through the origin (0, 0).
3. The catenaries obtained by translating the catenary y = cosh x in the direction of the straight line y = x.

4–10 ORTHOGONAL TRAJECTORIES (OTs)

Sketch or graph some of the given curves. Guess what their OTs may look like. Find these OTs.

• 4. y = x² + c
• 5. y = cx
• 6. xy = c
• 7. y = c/x²
• 8.
• 9.
• 10. x² + (y − c)² = c²

11–16 APPLICATIONS, EXTENSIONS

• 11. Electric field. Let the electric equipotential lines (curves of constant potential) between two concentric cylinders with the z-axis in space be given by u(x, y) = x² + y² = c (these are circular cylinders in the xyz-space). Using the method in the text, find their orthogonal trajectories (the curves of electric force).
• 12. Electric field. The lines of electric force of two opposite charges of the same strength at (−1, 0) and (1, 0) are the circles through (−1, 0) and (1, 0). Show that these circles are given by x² + (y − c)² = 1 + c². Show that the equipotential lines (which are orthogonal trajectories of those circles) are the circles given by (dashed in Fig. 25).

  

  Fig. 25. Electric field in Problem 12

• 13. Temperature field. Let the isotherms (curves of constant temperature) in a body in the upper half-plane y > 0 be given by 4x² + 9y² = c. Find the orthogonal trajectories (the curves along which heat will flow in regions filled with heat-conducting material and free of heat sources or heat sinks).
• 14. Conic sections. Find the conditions under which the orthogonal trajectories of families of ellipses x²/a² + y²/b² = c are again conic sections. Illustrate your result graphically by sketches or by using your CAS. What happens if a → 0? If b → 0?
• 15. Cauchy–Riemann equations. Show that for a family u(x, y) = c = const the orthogonal trajectories ν(x, y) = c* = const can be obtained from the following Cauchy–Riemann equations (which are basic in complex analysis in Chap. 13) and use them to find the orthogonal trajectories of e^x sin y = const. (Here, subscripts denote partial derivatives.)

  

• 16. Congruent OTs. If y′ = f(x) with f independent of y, show that the curves of the corresponding family are congruent, and so are their OTs.

THEOREM 1 Existence Theorem

Let the right side f(x, y) of the ODE in the initial value problem

be continuous at all points (x, y) in some rectangle

and bounded in R; that is, there is a number K such that

Then the initial value problem (1) has at least one solution y(x). This solution exists at least for all x in the subinterval |x − x₀| < α of the interval |x − x₀| < a; here, α is the smaller of the two numbers a and b/K.

THEOREM 2 Uniqueness Theorem

Let f and its partial derivative f_y = ∂f/∂y be continuous for all (x, y) in the rectangle R (Fig. 26) and bounded, say,

Then the initial value problem (1) has at most one solution y(x). Thus, by Theorem 1, the problem has precisely one solution. This solution exists at least for all x in that subinterval |x − x₀| < α.

EXAMPLE 1 Choice of a Rectangle

Consider the initial value problem

and take the rectangle R; |x| < 5, |y| < 3. Then a = 5, b = 3, and

Indeed, the solution of the problem is y = tan x (see Sec. 1.3, Example 1). This solution is discontinuous at ±π/2, and there is no continuous solution valid in the entire interval |x| < 5 from which we started.

EXAMPLE 2 Nonuniqueness

The initial value problem

has the two solutions

although is continuous for all y. The Lipschitz condition (4) is violated in any region that includes the line y = 0, because for y₁ = 0 and positive y₂ we have

and this can be made as large as we please by choosing y₂ sufficiently small, whereas (4) requires that the quotient on the left side of (5) should not exceed a fixed constant M.

PROBLEM SET 1.7

1. Linear ODE. If p and r in y′ + p(x)y = r(x are continuous for all x in an interval |x − x₀| ≤ a, show that f(x, y) in this ODE satisfies the conditions of our present theorems, so that a corresponding initial value problem has a unique solution. Do you actually need these theorems for this ODE?
2. Existence? Does the initial value problem (x − 2)y′ = y, y(2) = 1 have a solution? Does your result contradict our present theorems?
3. Vertical strip. If the assumptions of Theorems 1 and 2 are satisfied not merely in a rectangle but in a vertical infinite strip |x − x₀| < a, in what interval will the solution of (1) exist?
4. Change of initial condition. What happens in Prob. 2 if you replace y(2) = 1 with y(2) = k?
5. Length of x-interval. In most cases the solution of an initial value problem (1) exists in an x-interval larger than that guaranteed by the present theorems. Show this fact for y′ = 2y², y(1) = 1 by finding the best possible a (choosing b optimally) and comparing the result with the actual solution.
6. CAS PROJECT. Picard Iteration. (a) Show that by integrating the ODE in (1) and observing the initial condition you obtain

   

   This form (6) of (1) suggests Picard's Iteration Method¹⁰ which is defined by

   

   It gives approximations y₁, y₂, y₃, … of the unknown solution y of (1). Indeed, you obtain y₁ by substituting y = y₀ on the right and integrating—this is the first step—then y₂ by substituting y = y₁ on the right and integrating—this is the second step—and so on. Write a program of the iteration that gives a printout of the first approximations y₀, y₁, …, y_N as well as their graphs on common axes. Try your program on two initial value problems of your own choice.

   (b) Apply the iteration to y′ = x + y, y(0) = 0. Also solve the problem exactly.

   (c) Apply the iteration to y′ = 2y², y(0) = 1. Also solve the problem exactly.

   (d) Find all solutions of , y(1) = 0. Which of them does Picard's iteration approximate?

   (e) Experiment with the conjecture that Picard's iteration converges to the solution of the problem for any initial choice of y in the integrand in (7) (leaving y₀ outside the integral as it is). Begin with a simple ODE and see what happens. When you are reasonably sure, take a slightly more complicated ODE and give it a try.

7. Maximum α. What is the largest possible α in Example 1 in the text?
8. Lipschitz condition. Show that for a linear ODE y′ + p(x)y = r(x) with continuous p and r in |x − x₀| a a Lipschitz condition holds. This is remarkable because it means that for a linear ODE the continuity of f(x, y) guarantees not only the existence but also the uniqueness of the solution of an initial value problem. (Of course, this also follows directly from (4) in Sec. 1.5.)
9. Common points. Can two solution curves of the same ODE have a common point in a rectangle in which the assumptions of the present theorems are satisfied?
10. Three possible cases. Find all initial conditions such that (x² − x)y′ = (2x − 1)y has no solution, precisely one solution, and more than one solution.

This chapter concerns ordinary differential equations (ODEs) of first order and their applications. These are equations of the form

involving the derivative y′ = dy/dx of an unknown function y, given functions of x, and, perhaps, y itself. If the independent variable x is time, we denote it by t.

In Sec. 1.1 we explained the basic concepts and the process of modeling, that is, of expressing a physical or other problem in some mathematical form and solving it. Then we discussed the method of direction fields (Sec. 1.2), solution methods and models (Secs. 1.3–1.6), and, finally, ideas on existence and uniqueness of solutions (Sec. 1.7).

A first-order ODE usually has a general solution, that is, a solution involving an arbitrary constant, which we denote by c. In applications we usually have to find a unique solution by determining a value of c from an initial condition y(x₀) = y₀. Together with the ODE this is called an initial value problem

and its solution is a particular solution of the ODE. Geometrically, a general solution represents a family of curves, which can be graphed by using direction fields (Sec. 1.2). And each particular solution corresponds to one of these curves.

A separable ODE is one that we can put into the form

by algebraic manipulations (possibly combined with transformations, such as y/x = u) and solve by integrating on both sides.

An exact ODE is of the form

where M dx + N dy is the differential

of a function u(x, y), so that from du = 0 we immediately get the implicit general solution u(x, y) = c. This method extends to nonexact ODEs that can be made exact by multiplying them by some function F(x, y,), called an integrating factor (Sec. 1.4).

Linear ODEs

are very important. Their solutions are given by the integral formula (4), Sec. 1.5. Certain nonlinear ODEs can be transformed to linear form in terms of new variables. This holds for the Bernoulli equation

Applications and modeling are discussed throughout the chapter, in particular in Secs. 1.1, 1.3, 1.5 (population dynamics, etc.), and 1.6 (trajectories).

Picard's existence and uniqueness theorems are explained in Sec. 1.7 (and Picard's iteration in Problem Set 1.7).

Numeric methods for first-order ODEs can be studied in Secs. 21.1 and 21.2 immediately after this chapter, as indicated in the chapter opening.