Fundamentals of Differential Equations 8th Nagel

35

Newton’s second law states that force is equal to mass times acceleration. We can express thisby the equation

where F represents the total force on the object, m is the mass of the object, and is theacceleration, expressed as the derivative of velocity with respect to time. It will be convenientin the future to define y as positive when it is directed downward (as opposed to the analysis inSection 1.1).

Near Earth’s surface, the force due to gravity is just the weight of the objects and is alsodirected downward. This force can be expressed by mg, where g is the acceleration due to grav-ity. No general law precisely models the air resistance acting on the object, since this forceseems to depend on the velocity of the object, the density of the air, and the shape of the object,among other things. However, in some instances air resistance can be reasonably representedby �by, where b is a positive constant depending on the density of the air and the shape of theobject. We use the negative sign because air resistance is a force that opposes the motion. Theforces acting on the object are depicted in Figure 2.1 on page 36. (Note that we have general-ized the free-fall model in Section 1.1 by including air resistance.)

Applying Newton’s law, we obtain the first-order differential equation

(1)

To solve this equation, we exploit a technique called separation of variables, which was usedto analyze the radioactive decay model in Section 1.1 and will be developed in full detail inSection 2.2. Treating dy and dt as differentials, we rewrite equation (1) so as to isolate thevariables y and t on opposite sides of the equation:

(Hence, the nomenclature “separation of variables.”)

dymg � by

�dtm

.

m dYdt

� mg � bY .

dy/dt

m dydt

� F ,

An object falls through the air toward Earth. Assuming that the only forces actingon the object are gravity and air resistance, determine the velocity of the object asa function of time.

First-Order Differential Equations

CHAPTER 2

INTRODUCTION: MOTION OF A FALLING BODY2.1

Next we integrate the separated equation

(2)

and derive

(3)

Therefore,

or

where the new constant A has magnitude e�bc and the same sign (�) as (mg � by). Solving fory, we obtain

(4)

which is called a general solution to the differential equation because, as we will see inSection 2.3, every solution to (1) can be expressed in the form given in (4).

In a specific case, we would be given the values of m, g, and b. To determine the constantA in the general solution, we can use the initial velocity of the object y0. That is, we solve theinitial value problem

Substituting and into the general solution to the differential equation, we cansolve for A. With this value for A, the solution to the initial value problem is

(5) Y �mgb

� aY0 �mgbb e �bt/m .

t � 0y � y0

m dydt

� mg � by , y(0) � y0 .

y �mgb

�Ab

e�bt/m ,

mg � by � Ae �bt/m ,

0mg � by 0 � e �bce �bt/m

�1b

ln 0mg � by 0 � tm

� c .

� dymg � by

� � dtm

36 Chapter 2 First-Order Differential Equations

m

mg

–b

Gravity

Velocity

Air resistance

Figure 2.1 Forces on falling object

The preceding formula gives the velocity of the object falling through the air as a func-tion of time if the initial velocity of the object is y0. In Figure 2.2 we have sketched the graphof for various values of y0. It appears from Figure 2.2 that the velocity approaches

regardless of the initial velocity y0. This is easy to see from formula (5) by lettingThe constant is referred to as the limiting or terminal velocity of the

object.From this model for a falling body, we can make certain observations. Because

rapidly tends to zero, the velocity is approximately the weight, mg, divided by the coeffi-cient of air resistance, b. Thus, in the presence of air resistance, the heavier the object, thefaster it will fall, assuming shapes and sizes are the same. Also, when air resistance is less-ened (b is made smaller), the object will fall faster. These observations certainly agree withour experience.

Many other physical problems,† when formulated mathematically, lead to first-order dif-ferential equations or initial value problems. Several of these are discussed in Chapter 3. In thischapter we learn how to recognize and obtain solutions for some special types of first-orderequations. We begin by studying separable equations, then linear equations, and then exactequations. The methods for solving these are the most basic. In the last two sections, we illus-trate how devices such as integrating factors, substitutions, and transformations can be used totransform certain equations into either separable, exact, or linear equations that we can solve.Through our discussion of these special types of equations, you will gain insight into thebehavior of solutions to more general equations and the possible difficulties in finding thesesolutions.

A word of warning is in order: In solving differential equations, integration plays an essen-tial role. In particular, the separable equations in Section 2.2 always entail integration, asdemonstrated in equations (2) and (3) above. For your convenience, Appendix A reviews threestandard techniques for integrating the functions encountered in this text.

e�bt/m

mg/bt S�q. 4 3mg/by AtBy AtB

Section 2.1 Introduction: Motion of a Falling Body 37

m g b

0

(m/sec)

t (sec)

object slows down

object speeds up

0 >m g /b, so

0 <m g /b, so

Figure 2.2 Graph of for six different initial velocities y0. (g � 9.8 m/sec2, 5 sec)m/b �y AtB

†The physical problem just discussed has other mathematical models. For example, one could take into account thevariations in the gravitational field of Earth and the more general equations for air resistance.


Separable Equation

Definition 1. If the right-hand side of the equation

can be expressed as a function that depends only on x times a function that depends only on y, then the differential equation is called separable.†

p AyBg AxBdydx

� f Ax, yB

†Historical Footnote: A procedure for solving separable equations was discovered implicitly by Gottfried Leibniz in1691. The explicit technique called separation of variables was formalized by John Bernoulli in 1694.

In other words, a first-order equation is separable if it can be written in the form

For example, the equation

is separable, since (if one is sufficiently alert to detect the factorization)

However, the equation

admits no such factorization of the right-hand side and so is not separable.Informally speaking, one solves separable equations by performing the separation and

then integrating each side.

dydx

� 1 � xy

2x � xy

y2 � 1� x

2 � y

y2 � 1� g AxB p A yB .

dy

dx�

2x � xy

y2 � 1

dydx

� g AxB p A yB .

A simple class of first-order differential equations that can be solved using integration is theclass of separable equations. These are equations

(1)

that can be rewritten to isolate the variables x and y (together with their differentials dx and dy)on opposite sides of the equation, as in

So the original right-hand side must have the factored form

More formally, we write and present the following definition.p A yB � 1/h A yBf Ax, yB � g AxB # 1

h A yB .f Ax, yBh A yB dy � g AxB dx .

dydx

� f Ax, yB ,

2.2 SEPARABLE EQUATIONS

Caution: Constant functions y c such that p(c) � 0 are also solutions to (2), which mayor may not be included in (3) (as we shall see in Example 3).

We will look at the mathematical justification of this “streamlined” procedure shortly, butfirst we study some examples.

Solve the nonlinear equation

Following the streamlined approach, we separate the variables and rewrite the equation in the form

Integrating, we have

and solving this last equation for y gives

Since C is a constant of integration that can be any real number, 3C can also be any real num-ber. Replacing 3C by the single symbol K, we then have

If we wish to abide by the custom of letting C represent an arbitrary constant, we can go onestep further and use C instead of K in the final answer. This solution family is graphed inFigure 2.3 on page 40. ◆

As Example 1 attests, separable equations are among the easiest to solve. However, theprocedure does require a facility for computing integrals. Many of the procedures to be dis-cussed in the text also require a familiarity with the techniques of integration. For this reason

y � a3x2

2� 15x � Kb 1/3

.

y � a3x2

2� 15x � 3Cb 1/3

.

y3

3�

x2

2� 5x � C ,

� y2 dy � � Ax � 5B dx

y2 dy � Ax � 5B dx .

dy

dx�

x � 5

y2 .

�

Section 2.2 Separable Equations 39

Method for Solving Separable Equations

To solve the equation

(2)

multiply by dx and by to obtain

Then integrate both sides:

(3)

where we have merged the two constants of integration into a single symbol C. The lastequation gives an implicit solution to the differential equation.

H A yB � G AxB � C ,

�h A yB dy � �g AxB dx ,

h A yB dy � g AxB dx .

h A yB J 1/p A yBdydx

� g AxB p A yB

Solution

Example 1

we have provided a review of integration methods in Appendix A and a brief table of integralson the inside front cover.

Solve the initial value problem

(4)

Separating the variables and integrating gives

(5)

At this point, we can either solve for y explicitly (retaining the constant C ) or use the initialcondition to determine C and then solve explicitly for y. Let’s try the first approach.

Exponentiating equation (5), we have

(6)

where †† Now, depending on the values of y, we have ; andsimilarly, . Thus, (6) can be written as

or y � 1 � C1 Ax � 3B ,y � 1 � �C1 Ax � 3B0 x � 3 0 � � Ax � 3B 0 y � 1 0 � � A y � 1BC1 J eC.

0 y � 1 0 � eC 0 x � 3 0 � C1 0 x � 3 0 , e ln 0 y�1 0 � e ln 0 x�3 0�C � eCe ln 0 x�3 0 ,

ln 0 y � 1 0 � ln 0 x � 3 0 � C .

� dyy � 1

� � dxx � 3

,

dy

y � 1�

dxx � 3

,

dydx

�y � 1x � 3

, y A�1B � 0 .


y

x

11

K = 24

K = 0K = −12

K = −24K = 12

Figure 2.3 Family of solutions for Example 1†

Example 2

Solution

†The gaps in the curves reflect the fact that in the original differential equation, y appears in the denominator, so thaty � 0 must be excluded.††Recall that the symbol means “is defined to be.”J

where the choice of sign depends (as we said) on the values of x and y. Because C1 is a positiveconstant (recall that C1 � eC � 0), we can replace �C1 by K, where K now represents anarbitrary nonzero constant. We then obtain

(7)

Finally, we determine K such that the initial condition is satisfied. Putting and in equation (7) gives

and so . Thus, the solution to the initial value problem is

(8)

Alternative Approach. The second approach is to first set x � �1 and y � 0 in equation (5)and solve for C. In this case, we obtain

and so C � �ln 2. Thus, from (5), the solution y is given implicitly by

Here we have replaced y � 1 by 1 � y and x � 3 by x � 3, since we are interested in x and y nearthe initial values x � �1, y � 0 (for such values, y � 1 � 0 and x � 3 � 0). Solving for y, we find

which agrees with the solution (8) found by the first method. ◆

Solve the nonlinear equation

(9)

Separating variables and integrating, we find

At this point, we reach an impasse. We would like to solve for y explicitly, but we cannot. This isoften the case in solving nonlinear first-order equations. Consequently, when we say “solve the equa-tion,” we must on occasion be content if only an implicit form of the solution has been found. ◆

The separation of variables technique, as well as several other techniques discussed in thisbook, entails rewriting a differential equation by performing certain algebraic operations on it.

sin y � ey � x6 � x2 � x � C .

� Acos y � eyB dy � � A6x5 � 2x � 1B dx ,

Acos y � eyB dy � A6x5 � 2x � 1B dx ,

dy

dx�

6x5 � 2x � 1

cos y � ey .

y � 1 �12

Ax � 3B � �12

Ax � 1B , 1 � y �

x � 32

,

ln A1 � yB � ln Ax � 3B � ln 2 � lnax � 32b ,

0000ln A1 � yB � ln Ax � 3B � ln 2 .

0 � ln 1 � ln 2 � C ,

ln 00 � 1 0 � ln 0�1 � 3 0 � C ,

y � 1 �12

Ax � 3B � � 12

Ax � 1B .K � �1/2

0 � 1 � K A�1 � 3B � 1 � 2K ,

y � 0x � �1y A�1B � 0

y � 1 � K Ax � 3B .


Example 3

Solution

“Rewriting as ” amounts to dividing both sides by .You may recall from your algebra days that doing this can be treacherous. For example, theequation has two solutions: x � 2 and x � 4. But if we “rewrite” theequation as x � 4 by dividing both sides by , we lose track of the root x � 2. Thus, weshould record the zeros of itself before dividing by this factor.

By the same token we must take note of the zeros of in the separable equationprior to dividing. After all, if (say) , then

observe that the constant function solves the differential equation :

Indeed, in solving the equation of Example 2,

we obtained as the set of solutions, where K was a nonzero constant (sinceK replaced �eC). But notice that the constant function (which in this case correspondsto K � 0) is also a solution to the differential equation. The reason we lost this solution can betraced back to a division by y � 1 in the separation process. (See Problem 30 for an example ofwhere a solution is lost and cannot be retrieved by setting the constant K � 0.)

Formal Justification of Method

We close this section by reviewing the separation of variables procedure in a more rigorousframework. The original differential equation (2) is rewritten in the form

(10)

where Letting and denote antiderivatives (indefinite integrals) ofand , respectively—that is,

we recast equation (10) as

By the chain rule for differentiation, the left-hand side is the derivative of the composite

function :

Thus, if is a solution to equation (2), then and are two functions of x thathave the same derivative. Therefore, they differ by a constant:

(11)

Equation (11) agrees with equation (3), which was derived informally, and we have thusverified that the latter can be used to construct implicit solutions.

H Ay AxBB � G AxB � C .

G AxBH Ay AxBBy AxBddx

H Ay AxBB � H¿ Ay AxBB dydx

.

H Ay AxBBH¿ A yB dy

dx� G¿ AxB .

H¿ A yB � h A yB , G¿ AxB � g AxB ,g AxBh A yB G AxBH A yBh A yB J 1/p A yB.h A yB dy

dx� g AxB ,

y � 1y � 1 � K Ax � 3B

dydx

�y � 1x � 3

,

g AxBp A yB � Ax � 2B2 A13 � 13B � 0 .

dydx

�d A13B

dx� 0 ,

dy/dx � g AxBp A yBy(x) � 13g AxBp A yB � Ax � 2B2 A y � 13Bg AxBp A yBdy/dx �

p A yBAx � 2B Ax � 2Bx Ax � 2B � 4 Ax � 2Bp A yBh A yB dy � g AxB dxdy/dx � g AxBp A yB


In Problems 1–6, determine whether the given differen-tial equation is separable.

1. 2.

3. 4.

5.

6.

In Problems 7–16, solve the equation.

7. 8.

9. 10.

11. 12.

13. 14.

15.16.

In Problems 17–26, solve the initial value problem.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26. 2y dx � A1 � xB dy � 0 , y A0B � 1

dydx

� x2 A1 � yB , y A0B � 3

dydx

� 8x3e�2y , y A1B � 0

t�1 dydt

� 2 cos 2y , y (0) � p/4

x2 dx � 2y dy � 0 , y A0B � 2

dydu

�y sin u

y2 � 1 , y(p) � 1

1u

x2 dydx

�4x2 � x � 2

(x � 1)(y � 1) , y(1) � 1

12

dydx

� 2y � 1 cos x , y(p) � 0

dydx

� A1 � y2Btan x , y A0B � 23

y¿ � x3 A1 � yB , y A0B � 3

Ax � xy2B dx � ex2y dy � 0

y�1 dy � yecos x sin x dx � 0

dydx

� 3x2(1 � y2)3/2dxdt

� x3 � x

x dydx

�1 � 4y2

3ydy

dx�

sec2y

1 � x2

dxdt

�t

xet�2x

dydx

�x

y221 � x

xdydx

�1y3

dxdt

� 3xt2

s2 �dsdt

�s � 1

st

Axy2 � 3y2B dy � 2x dx � 0

dy

dx�

yex�y

x2 � 2dsdt

� t ln As2tB � 8t2

dydx

� 4y2 � 3y � 1dydx

� sin Ax � yB � 0


27. Solutions Not Expressible in Terms of Elemen-tary Functions. As discussed in calculus, certainindefinite integrals (antiderivatives) such as cannot be expressed in finite terms using elementaryfunctions. When such an integral is encountered whilesolving a differential equation, it is often helpful to usedefinite integration (integrals with variable upperlimit). For example, consider the initial value problem

The differential equation separates if we divide by y2

and multiply by dx. We integrate the separated equa-tion from x � 2 to x � x1 and find

If we let t be the variable of integration and replacex1 by x and by 1, then we can express the solu-tion to the initial value problem by

Use definite integration to find an explicit solution tothe initial value problems in parts (a)–(c).

(a)(b)

(c)(d) Use a numerical integration algorithm (such as

Simpson’s rule, described in Appendix C) toapproximate the solution to part (b) at x � 0.5 tothree decimal places.

28. Sketch the solution to the initial value problem

and determine its maximum value.

29. Uniqueness Questions. In Chapter 1 we indicatedthat in applications most initial value problems willhave a unique solution. In fact, the existence of unique

dydt

� 2y � 2yt , y A0B � 3

dy/dx � 21 � sin x A1 � y2B , y A0B � 1

dy/dx � ex 2y�2 , y A0B � 1

dy/dx � ex 2 , y A0B � 0

y AxB � a1 � �x

2 et 2

dtb�1

.

y A2B � �

1y Ax1B �

1y A2B .

� �1y

` x�x1

x�2

�x�x1

x�2 ex 2

dx � �x�x1

x�2 dy

y2

dydx

� ex 2y2 , y A2B � 1 .

� ex2dx

2.2 EXERCISES

(c) Determine the domains of the solutions in part (b).(d) As found in part (c), the domains of the solu-

tions depend on the initial conditions. For the ini-tial value problem with a,a � 0, show that as a approaches zero from theright the domain approaches the whole real line

and as a approaches the domainshrinks to a single point.

(e) Sketch the solutions to the initial value problemwith a for , �1,

and �2.

32. Analyze the solution to the initial valueproblem

using approximation methods and then compare withits exact form as follows.

(a) Sketch the direction field of the differentialequation and use it to guess the value oflim .

(b) Use Euler’s method with a step size of 0.1 tofind an approximation of .

(c) Find a formula for and graph on thedirection field from part (a).

(d) What is the exact value of ? Compare withyour approximation in part (b).

(e) Using the exact solution obtained in part (c),determine lim and compare with yourguess in part (a).

33. Mixing. Suppose a brine containing 0.3 kilogram(kg) of salt per liter (L) runs into a tank initiallyfilled with 400 L of water containing 2 kg of salt. Ifthe brine enters at 10 L/min, the mixture is kept uni-form by stirring, and the mixture flows out at thesame rate. Find the mass of salt in the tank after 10min (see Figure 2.4). [Hint: Let A denote the numberof kilograms of salt in the tank at t min after theprocess begins and use the fact that

rate of increase in A � rate of input � rate of exit.

A further discussion of mixing problems is given inSection 3.2.]

xSq f AxBf(1)

f(x)f(x)f(1)

xSq f AxB

dydx

� y2 � 3y � 2 , y A0B � 1.5

y � f(x)

a � �1/2y A0B �dy/dx � xy3

�qA�q, q By A0B �dy/dx � xy3

solutions was so important that we stated an exis-tence and uniqueness theorem, Theorem 1, page 11.The method for separable equations can give us asolution, but it may not give us all the solutions (alsosee Problem 30). To illustrate this, consider theequation .

(a) Use the method of separation of variables toshow that

is a solution.(b) Show that the initial value problem

with is satisfied for by

for x � 0.(c) Now show that the constant function also

satisfies the initial value problem given in part(b). Hence, this initial value problem does nothave a unique solution.

(d) Finally, show that the conditions of Theorem 1on page 11 are not satisfied.

(The solution was lost because of the divisionby zero in the separation process.)

30. As stated in this section, the separation of equation(2) on page 39 requires division by , and thismay disguise the fact that the roots of the equation

�0 are actually constant solutions to the differ-ential equation.

(a) To explore this further, separate the equation

to derive the solution,

(b) Show that satisfies the original equation.

(c) Show that there is no choice of the constant Cthat will make the solution in part (a) yield thesolution Thus, we lost the solution

when we divided by .

31. Interval of Definition. By looking at an initialvalue problem with , it isnot always possible to determine the domain of thesolution or the interval over which the function

satisfies the differential equation.

(a) Solve the equation .(b) Give explicitly the solutions to the initial

value problem with 1; ;2.y A0B �

1/2y A0B �y A0B �

dy/dx � xy3

y AxB y AxBy Ax0B � y0dy/dx � f Ax, yB

A y � 1B2/3y � �1y � �1.

dy/dx � Ax � 3B A y � 1B2/3y � �1

y � �1 � Ax2/6 � x � CB3 .dydx

� Ax � 3B A y � 1B2/3

p(y)

p A yBy � 0

y � 0y � A2x/3B3/2

C � 0y A0B � 0y1/3dy/dx �

y � a2x3

� Cb 3/2

dy/dx � y1/3


A(t)

400 L

A( 0) = 2 kg

10 L/min 0.3 kg/L

10 L/min

Figure 2.4 Schematic representation of a mixing problem


34. Newton’s Law of Cooling. According to New-ton’s law of cooling, if an object at temperature T isimmersed in a medium having the constant tempera-ture M, then the rate of change of T is proportional tothe difference of temperature M � T. This gives thedifferential equation

(a) Solve the differential equation for T.(b) A thermometer reading 100F is placed in a

medium having a constant temperature of 70F.After 6 min, the thermometer reads 80F. Whatis the reading after 20 min?

(Further applications of Newton’s law of coolingappear in Section 3.3.)

35. Blood plasma is stored at 40F. Before the plasmacan be used, it must be at 90F. When the plasma isplaced in an oven at 120F, it takes 45 min for theplasma to warm to 90F. Assume Newton’s law ofcooling (Problem 34) applies. How long will it takefor the plasma to warm to 90F if the oven tempera-ture is set at (a) 100F, (b) 140F, and (c) 80F?

36. A pot of boiling water at 100C is removed from astove at time t � 0 and left to cool in the kitchen. After5 min, the water temperature has decreased to 80C,and another 5 min later it has dropped to 65C. Assum-ing Newton’s law of cooling (Problem 34) applies,determine the (constant) temperature of the kitchen.

37. Compound Interest. If is the amount of dol-lars in a savings bank account that pays a yearlyinterest rate of r% compounded continuously, then

t in years.

Assume the interest is 5% annually, $1000,and no monies are withdrawn.

(a) How much will be in the account after 2 yr?(b) When will the account reach $4000?(c) If $1000 is added to the account every 12

months, how much will be in the account afteryr?

38. Free Fall. In Section 2.1, we discussed a model foran object falling toward Earth. Assuming that only airresistance and gravity are acting on the object, wefound that the velocity y must satisfy the equation

where m is the mass, g is the acceleration due togravity, and b � 0 is a constant (see Figure 2.1).

m

dYdt

� mg � bY ,

312

P A0B �

dPdt

�r

100 P ,

P AtB

dT/dt � k AM � TB .

If m � 100 kg, g � 9.8 m sec2, b � 5 kg sec, and10 m sec, solve for . What is the limit-

ing (i.e., terminal) velocity of the object?

39. Grand Prix Race. Driver A had been leadingarchrival B for a while by a steady 3 miles. Only 2miles from the finish, driver A ran out of gas anddecelerated thereafter at a rate proportional to thesquare of his remaining speed. One mile later, driverA’s speed was exactly halved. If driver B’s speedremained constant, who won the race?

40. The atmospheric pressure (force per unit area) on asurface at an altitude z is due to the weight of thecolumn of air situated above the surface. Therefore,the difference in air pressure p between the top andbottom of a cylindrical volume element of height zand cross-section area A equals the weight of the airenclosed (density times volume timesgravity g), per unit area:

.

Let to derive the differential equation. To analyze this further we must pos-

tulate a formula that relates pressure and density. Theperfect gas law relates pressure, volume, mass m, andabsolute temperature T according to ,where R is the universal gas constant and M is themolar mass of the air. Therefore, density and pressureare related by .

(a) Derive the equation and solve it

for the “isothermal” case where T is constant to obtain the barometric pressure equationp z � p z0 exp[�Mg z�z0 RT].

(b) If the temperature also varies with altitude, derive the solution

.

(c) Suppose an engineer measures the barometricpressure at the top of a building to be 99,000 Pa(pascals), and 101,000 Pa at the base z � z0 . If the absolute temperature varies as

, determine theheight of the building. Take R � 8.31 N-m mol-K,M � 0.029 kg mol, and g � 9.8 m sec2. (Anamusing story concerning this problem can befound at http://www.snopes.com/college/exam/barometer.asp)

///

T AzB � 288 � 0.0065 Az � z0BBA

p AzB � p Az0B exp e�MgR �

z

z0

dz

T AzB fT � T AzB /BABABA

dpdz

� �MgRT

p

r :� m/V � Mp/RT

pV � mRT/M

dp/dz � �rg¢zS 0

p Az � ¢zB � p AzB � �r AzB AA¢zBg

A� �r AzBg¢z

V � A¢zr

y AtB/y A0B �//

A type of first-order differential equation that occurs frequently in applications is the linearequation. Recall from Section 1.1 that a linear first-order equation is an equation that can beexpressed in the form

(1)

where , , and depend only on the independent variable x, not on y.For example, the equation

is linear, because it can be rewritten in the form

However, the equation

is not linear; it cannot be put in the form of equation (1) due to the presence of the andterms.

There are two situations for which the solution of a linear differential equation is quiteimmediate. The first arises if the coefficient is identically zero, for then equation (1) reduces to

(2)

which is equivalent to

as long as is not zero .The second is less trivial. Note that if happens to equal the derivative of —that

is, —then the two terms on the left-hand side of equation (1) simply comprise thederivative of the product :

Therefore equation (1) becomes

(3)ddx

3a1 AxBy 4 � b AxBa1 AxBy¿ � a0 AxBy � a1 AxBy¿ � a¿1 AxBy �

ddx

3a1 AxBy 4 .a1 AxBya0 AxB � a¿1 AxB a1 AxBa0 AxB4a1 AxB3

y AxB � � b AxBa1 AxB dx � C

a1 AxB dydx

� b AxB ,a0 AxB

y dy/dxy3

y

dydx

� Asin xBy3 � ex � 1

Asin xB dydx

� Acos xBy � x2sin x .

x2sin x � Acos xBy � Asin xB dydx

b AxBa0 AxBa1 AxBa1 AxB dy

dx� a0 AxBy � b AxB ,


LINEAR EQUATIONS2.3

and the solution is again elementary:

One can seldom rewrite a linear differential equation so that it reduces to a form as simpleas (2). However, the form (3) can be achieved through multiplication of the original equation(1) by a well-chosen function . Such a function is then called an “integrating factor”for equation (1). The easiest way to see this is first to divide the original equation (1) by and put it into standard form

(4)

where and Next we wish to determine so that the left-hand side of the multiplied equation

(5)

is just the derivative of the product :

Clearly, this requires that m satisfy

(6)

To find such a function, we recognize that equation (6) is a separable differential equation,which we can write as Integrating both sides gives

(7)

With this choice† for , equation (5) becomes

which has the solution

(8)

Here C is an arbitrary constant, so (8) gives a one-parameter family of solutions to (4). Thisform is known as the general solution to (4).

y AxB �1m AxB c �m AxBQ AxB dx � C d .

ddx

3m AxBy 4 � m AxBQ AxB ,m AxB

M AxB � e� PAxB dx .

A1/mB dm � P AxBdx.

m¿ � mP .

m AxB dydx

� M AxBP AxBy �ddx

3m AxBy 4 � m AxB dydx

� M� AxBy .

m AxBym AxB dy

dx� m AxBP AxBy � m AxBQ AxB

m AxBQ AxB � b AxB /a1 AxB.P AxB � a0 AxB /a1 AxBdydx

� P AxBy � Q AxB ,a1 AxBm AxBm AxB

y AxB �1

a1 AxB c �b AxB dx � C d .a1 AxBy � �b AxB dx � C ,

Section 2.3 Linear Equations 47

†Any choice of the integration constant in will produce a suitable .m AxB� P AxB dx

Find the general solution to

(9)

To put this linear equation in standard form, we multiply by x to obtain

(10)

Here , so

Thus, an integrating factor is

Multiplying equation (10) by yields

ddx

Ax�2yB � cos x .

x�2

dydx

� 2x�3y � cos x ,

m AxBm AxB � e�2 ln 0 x 0 � e lnAx�2B � x�2 .

�P AxB dx � ��2x

dx � �2 ln 0 x 0 .P AxB � �2/x

dydx

�2x

y � x2cos x .

1

x dy

dx�

2y

x2 � x cos x , x 7 0 .


Example 1

Solution

μMethod for Solving Linear Equations

(a) Write the equation in the standard form

(b) Calculate the integrating factor by the formula

(c) Multiply the equation in standard form by and, recalling that the left-hand

side is just obtain

(d) Integrate the last equation and solve for y by dividing by to obtain (8).m AxB ddx

3m AxBy 4 � m AxBQ AxB . m AxB dy

dx� P AxBm AxBy � m AxBQ AxB ,

ddx 3m AxBy 4 , m AxB

m(x) � exp c�P(x)dx d . m AxBdydx

� P AxBy � Q AxB .μ

We can summarize the method for solving linear equations as follows.

We now integrate both sides and solve for y to find

(11)

It is easily checked that this solution is valid for all x � 0. In Figure 2.5 we have sketched solu-tions for various values of the constant C in (11). ◆

y � x2 sin x � Cx2 .

x�2y � � cos x dx � sin x � C


Example 2

Solution

C = − 1 / 2

100

0

−100

−200

2 4 6 8 10 12

C = 1

C = 1 / 2

C = 0

C = −1

y

x

Figure 2.5 Graph of for five values of the constant Cy � x2 sin x � Cx2

In the next example, we encounter a linear equation that arises in the study of the radioac-tive decay of an isotope.

A rock contains two radioactive isotopes, RA1 and RA2, that belong to the same radioactiveseries; that is, RA1 decays into RA2, which then decays into stable atoms. Assume that the rateat which RA1 decays into RA2 is kg/sec. Because the rate of decay of RA2 is propor-tional to the mass of RA2 present, the rate of change in RA2 is

(12)

where k � 0 is the decay constant. If k � and initially 40 kg, find the mass ofRA2 for

Equation (12) is linear, so we begin by writing it in standard form

(13)

where we have substituted k � 2 and displayed the initial condition. We now see that ,so Thus, an integrating factor is Multiplying equation (13)by yields

ddt

Ae2tyB � 50e�8t .

e2t

dydt

� 2e2ty � 50e�10t�2t � 50e�8t ,

m AtB m AtB � e2t.� P AtBdt � � 2 dt � 2t.P AtB � 2

dydt

� 2y � 50e�10t , y A0B � 40 ,

t � 0.y AtBy A0B �2/sec

dydt

� 50e�10t � ky ,

dydt

� rate of creation � rate of decay ,

y AtB 50e�10t

μ

Integrating both sides and solving for y, we find

Substituting t � 0 and y(0) � 40 gives

so . Thus, the mass of RA2 at time t is given by

(14) ◆

For the initial value problem

find the value of .

The integrating factor for the differential equation is, from equation (7),

The general solution form (8) thus reads

However, this indefinite integral cannot be expressed in finite terms with elementary functions(recall a similar situation in Problem 27 of Exercises 2.2). Because we can use numerical algo-rithms such as Simpson’s rule (Appendix C) to perform definite integration, we revert to theform (5), which in this case reads

and take the definite integral from the initial value x � 1 to the desired value x � 2:

Inserting the given value of and solving, we express

Using Simpson’s rule, we find that the definite integral is approximately 4.841, so

◆y A2B � 4e�1 � 4.841e�2 � 2.127 .

y A2B � e�2�1 A4B � e�2 �2

1 ex21 � cos2x dx .

y A1Bexy ` x�2

x�1� e2y A2B � e1y A1B � �

x�2

x�1 ex21 � cos2x dx .

ddx

AexyB � ex21 � cos2x ,

y AxB � e�x a� ex21 � cos2x dx � Cb .

m AxB � e� 1 dx � ex .

y A2By¿ � y � 21 � cos2x , y A1B � 4 ,

y AtB � a1854b e�2t � a25

4b e�10t , t � 0 .

y AtBC � 40 � 25/4 � 185/4

40 � � 254

e0 � Ce0 � � 254

� C ,

y � � 254

e�10t � Ce�2t .

e2ty � � 254

e�8t � C ,


Example 3

Solution

In Example 3 we had no difficulty expressing the integral for the integrating factorClearly, situations will arise where this integral, too, cannot be expressed

with elementary functions. In such cases we must again resort to a numerical procedure such asEuler’s method (Section 1.4) or to a “nested loop” implementation of Simpson’s rule. You areinvited to explore such a possibility in Problem 27.

Because we have established explicit formulas for the solutions to linear first-order differ-ential equations, we get as a dividend a direct proof of the following theorem.

m AxB � e� 1 dx � ex.


Existence and Uniqueness of Solution

Theorem 1. Suppose and are continuous on an interval that contains the point x0. Then for any choice of initial value y0, there exists a unique solution on to the initial value problem

(15)

In fact, the solution is given by (8) for a suitable value of C.

dydx

� P AxBy � Q AxB , y Ax0B � y0 .

Aa, bB y AxBAa, bBQ AxBP AxB

The essentials of the proof of Theorem 1 are contained in the deliberations leading to equa-tion (8); Problem 34 provides the details. This theorem differs from Theorem 1 on page 11 inthat for the linear initial value problem (15), we have the existence and uniqueness of thesolution on the whole interval , rather than on some smaller unspecified interval about x0.

The theory of linear differential equations is an important branch of mathematics not onlybecause these equations occur in applications but also because of the elegant structure associ-ated with them. For example, first-order linear equations always have a general solution givenby equation (8). Some further properties of first-order linear equations are described in Prob-lems 28 and 36. Higher-order linear equations are treated in Chapters 4, 6, and 8.

Aa, bB

In Problems 1–6, determine whether the given equationis separable, linear, neither, or both.

1. 2.

3. 4.

5. 6.

In Problems 7–16, obtain the general solution to theequation.

7. 8.dydx

� y � e3x � 0dydx

�yx

� 2x � 1

x dxdt

� t2x � sin t3r �drdu

� u3

At2 � 1B dydt

� yt � y3t � et

dydt

� y ln t

x2 dydx

� sin x � y � 0dxdt

� xt � ex

9. 10.

11. 12.

13.

14.

15.

16. A1 � x2Bdydx

� x2y � A1 � xB21 � x2

Ax2 � 1B dydx

� xy � x � 0

xdydx

� 3 Ay � x2B �sin x

x

y dxdy

� 2x � 5y3

dydx

� x2e�4x � 4yAt � y � 1B dt � dy � 0

drdu

� r tan u � sec uxdydx

� 2y � x�3

2.3 EXERCISES


17.

18.

19.

20.

21.

22.

23. Radioactive Decay. In Example 2 assume that therate at which RA1 decays into RA2 is kg/secand the decay constant for RA2 is k � 5/sec. Find themass of RA2 for if initially 10 kg.

24. In Example 2 the decay constant for isotope RA1 was10/sec, which expresses itself in the exponent of therate term kg/sec. When the decay constantfor RA2 is k � 2/sec, we see that in formula (14) fory the term eventually dominates (hasgreater magnitude for t large).

(a) Redo Example 2 taking k � 20/sec. Now whichterm in the solution eventually dominates?

(b) Redo Example 2 taking k � 10/sec.

25. (a) Using definite integration, show that the solutionto the initial value problem

can be expressed as

(b) Use numerical integration (such as Simpson’srule, Appendix C) to approximate the solution atx � 3.

26. Use numerical integration (such as Simpson’s rule,Appendix C) to approximate the solution, at x � 1,to the initial value problem

Ensure your approximation is accurate to three deci-mal places.

dy

dx�

sin 2x

2 A1 � sin2xB y � 1 , y A0B � 0 .

y AxB � e�x 2 ae4 � �x

2 et 2

dtb .

dydx

� 2xy � 1 , y A2B � 1 ,

A185/4Be�2t

50e�10t

y A0B �t � 0y AtB40e�20t

sin x

dydx

� y cos x � x sin x , y ap2b � 2

y ap4b �

�1522p2

32

cos x

dydx

� y sin x � 2x cos2x ,

dydx

�3yx

� 2 � 3x , y A1B � 1

t 2

dxdt

� 3tx � t 4 ln t � 1, x(1) � 0

dydx

� 4y � e�x � 0 , y A0B �43

dydx

�yx

� xex , y A1B � e � 1


27. Consider the initial value problem

(a) Using definite integration, show that the inte-grating factor for the differential equation can bewritten as

and that the solution to the initial value problem is

(b) Obtain an approximation to the solution at x � 1by using numerical integration (such as Simp-son’s rule, Appendix C) in a nested loop to esti-mate values of and, thereby, the value of

[Hint: First, use Simpson’s rule to approximateat x � 0.1, 0.2, . . . , 1. Then use these

values and apply Simpson’s rule again toapproximate .]

(c) Use Euler’s method (Section 1.4) to approxi-mate the solution at x � 1, with step sizes h �0.1 and 0.05.

[A direct comparison of the merits of the two numer-ical schemes in parts (b) and (c) is very complicated,since it should take into account the number of func-tional evaluations in each algorithm as well as theinherent accuracies.]

28. Constant Multiples of Solutions.(a) Show that is a solution of the linear

equation

(16)

and is a solution of the nonlinear equation

(17)

(b) Show that for any constant the function is a solution of equation (16), while is asolution of equation (17) only when C � 0 or 1.

(c) Show that for any linear equation of the form

if is a solution, then for any constant C thefunction is also a solution.C y AxBy AxBdydx

� P AxBy � 0 ,

Cx�1Ce�xC,

dydx

� y2 � 0 .

y � x�1

dydx

� y � 0 ,

y � e�x

�10 m AsB s ds

m AxB�

1

0 m AsB s ds .

m AxB

y AxB �1m AxB � x

0 m AsB s ds �

2m AxB .

m AxB � expa� x

0 21 � sin2t dtb

dydx

� 21 � sin2x y � x , y A0B � 2 .

29. Use your ingenuity to solve the equation

[Hint: The roles of the independent and dependentvariables may be reversed.]

30. Bernoulli Equations. The equation

(18)

is an example of a Bernoulli equation. (Further dis-cussion of Bernoulli equations is in Section 2.6.)

(a) Show that the substitution reduces equa-tion (18) to the equation

(19)

(b) Solve equation (19) for y. Then make the substi-tution to obtain the solution to equation(18).

31. Discontinuous Coefficients. As we will see inChapter 3, occasions arise when the coefficient in a linear equation fails to be continuous because ofjump discontinuities. Fortunately, we may still obtaina “reasonable” solution. For example, consider theinitial value problem

where

(a) Find the general solution for .(b) Choose the constant in the solution of part (a) so

that the initial condition is satisfied.(c) Find the general solution for x > 2.(d) Now choose the constant in the general solution

from part (c) so that the solution from part (b)and the solution from part (c) agree at x � 2. Bypatching the two solutions together, we canobtain a continuous function that satisfies thedifferential equation except at x � 2, where itsderivative is undefined.

(e) Sketch the graph of the solution from x � 0 to x � 5.

32. Discontinuous Forcing Terms. There are occa-sions when the forcing term in a linear equationfails to be continuous because of jump discontinuities.Fortunately, we may still obtain a reasonable solution

Q AxB

0 � x � 2

P AxB J e1 , 0 � x � 2 ,3 , x 7 2 .

dydx

� P AxBy � x , y A0B � 1 ,

P AxBy � y3

dydx

� 6y � 3x .

y � y3

dydx

� 2y � xy�2

dy

dx�

1

e4y � 2x .


imitating the procedure discussed in Problem 31. Usethis procedure to find the continuous solution to theinitial value problem.

where

Sketch the graph of the solution from x � 0 to x � 7.

33. Singular Points. Those values of x for which in equation (4) is not defined are called singularpoints of the equation. For example, x � 0 is a singu-lar point of the equation since whenthe equation is written in the standard form,

we see that is notdefined at x � 0. On an interval containing a singularpoint, the questions of the existence and uniqueness ofa solution are left unanswered, since Theorem 1 doesnot apply. To show the possible behavior of solutionsnear a singular point, consider the following equations.

(a) Show that has only one solutiondefined at x � 0. Then show that the initial valueproblem for this equation with initial condition

has a unique solution when and no solution when

(b) Show that has an infinite num-ber of solutions defined at x � 0. Then show thatthe initial value problem for this equation withinitial condition has an infinite numberof solutions.

34. Existence and Uniqueness. Under the assump-tions of Theorem 1, we will prove that equation (8)gives a solution to equation (4) on . We canthen choose the constant C in equation (8) so that theinitial value problem (15) is solved.

(a) Show that since is continuous on ,then defined in (7) is a positive, continuousfunction satisfying on .

(b) Since

verify that y given in equation (8) satisfies equation(4) by differentiating both sides of equation (8).

(c) Show that when we let be theantiderivative whose value at x0 is 0 (i.e.,

) and choose C to be theinitial condition is satisfied.y Ax0B � y0

y0 m Ax0B,�xx0

m AtBQ AtB dt

� m AxBQ AxB dx

ddx

�m AxBQ AxB dx � m AxBQ AxB ,Aa, bBdm/dx � P AxBm(x)

m AxB Aa, bBP AxBAa, bB

y A0B � 0

xy¿ � 2y � 3xy0 � 0.

y0 � 0y A0B � y0

xy¿ � 2y � 3x

P AxB � 2/xy¿ � A2/xBy � 3,

xy¿ � 2y � 3x,

P AxBQ AxB J e 2 , 0 � x � 3 ,

�2 , x 7 3 .

dydx

� 2y � Q AxB , y A0B � 0 ,

(d) Start with the assumption that is a solutionto the initial value problem (15) and argue thatthe discussion leading to equation (8) impliesthat must obey equation (8). Then arguethat the initial condition in (15) determines theconstant C uniquely.

35. Mixing. Suppose a brine containing 0.2 kg of saltper liter runs into a tank initially filled with 500 L ofwater containing 5 kg of salt. The brine enters thetank at a rate of 5 L/min. The mixture, kept uniformby stirring, is flowing out at the rate of 5 L/min (seeFigure 2.6).

y AxBy AxB


idea that just by knowing the form of the solution,we can substitute into the given equation and solvefor any unknowns. Here we illustrate the method forfirst-order equations (see Sections 4.6 and 6.4 for thegeneralization to higher-order equations).

(a) Show that the general solution to

(20)

has the form

where yh ( ) is a solution to equation (20)when C is a constant, and �

for a suitable function . [Hint:Show that we can take and then useequation (8).]

We can in fact determine the unknown func-tion yh by solving a separable equation. Thendirect substitution of yyh in the original equationwill give a simple equation that can be solvedfor y.

Use this procedure to find the general solution to

(21)

by completing the following steps:(b) Find a nontrivial solution yh to the separable

equation

(22)

(c) Assuming (21) has a solution of the formsubstitute this into equation

(21), and simplify to obtain (d) Now integrate to get .(e) Verify that is a gen-

eral solution to (21).

37. Secretion of Hormones. The secretion of hor-mones into the blood is often a periodic activity. If ahormone is secreted on a 24-h cycle, then the rate ofchange of the level of the hormone in the blood maybe represented by the initial value problem

where is the amount of the hormone in the bloodat time t, is the average secretion rate, is theamount of daily variation in the secretion, and k is apositive constant reflecting the rate at which thebody removes the hormone from the blood. If

and x0 � 10, solve for .x AtBa � b � 1, k � 2,

ba

x AtBdxdt

� a � b cos pt12

� kx , x A0B � x0 ,

y AxB � Cyh AxB � y AxByh AxBy AxB y¿ AxB � x2/yh AxB.yp AxB � y AxByh AxB,dydx

�3x

y � 0 , x 7 0 .

dydx

�3x

y � x2 , x 7 0 ,

yh � m�1 AxB y AxBy AxByh AxB yp AxBQ AxB � 0,[ 0

y AxB � Cyh AxB � yp AxB ,dydx

� P AxBy � Q AxB

A(t)

? L

A (10) = ? kg

5 L/min 0.2 kg/L

5 L/min

1 L/min

Figure 2.7 Mixing problem with unequal flow rates

A(t)

500 L

A (0) = 5 kg

5 L/min 0.2 kg/L

5 L/min

Figure 2.6 Mixing problem with equal flow rates

(a) Find the concentration, in kilograms per liter, of saltin the tank after 10 min. [Hint: Let A denote thenumber of kilograms of salt in the tank at t minutesafter the process begins and use the fact that

rate of increase in A � rate of input � rate of exit.A further discussion of mixing problems isgiven in Section 3.2.]

(b) After 10 min, a leak develops in the tank andan additional liter per minute of mixture flowsout of the tank (see Figure 2.7). What will bethe concentration, in kilograms per liter, of saltin the tank 20 min after the leak develops?[Hint: Use the method discussed in Problems31 and 32.]

36. Variation of Parameters. Here is another proce-dure for solving linear equations that is particularlyuseful for higher-order linear equations. This methodis called variation of parameters. It is based on the

38. Use the separation of variables technique to derivethe solution (7) to the differential equation (6).

39. The temperature T (in units of 100F) of a universityclassroom on a cold winter day varies with time t (inhours) as

dTdt

� e1 � T , if heating unit is ON.�T , if heating unit is OFF.

Section 2.4 Exact Equations 55

Suppose T � 0 at 9:00 A.M., the heating unit is ONfrom 9–10 A.M., OFF from 10–11 A.M., ON againfrom 11 A.M.–noon, and so on for the rest of the day.How warm will the classroom be at noon? At 5:00 P.M.?

Suppose the mathematical function F(x,y) represents some physical quantity, such as tempera-ture, in a region of the xy-plane. Then the level curves of F, where F(x,y) � constant, could beinterpreted as isotherms on a weather map, as depicted in Figure 2.8.

EXACT EQUATIONS2.4

50°60°

70°

80°

90°

Figure 2.8 Level curves of F Ax, yBHow does one calculate the slope of the tangent to a level curve? It is accomplished by

implicit differentiation: One takes the derivative, with respect to x, of both sides of the equationF(x,y) � C, taking into account that y depends on x along the curve:

or

(1) ,

and solves for the slope:

(2) .

The expression obtained by formally multiplying the left-hand member of (1) by dx is knownas the total differential of F, written dF:

,

and our procedure for obtaining the equation for the slope f(x,y) of the level curve F(x,y) � Ccan be expressed as setting the total differential dF � 0 and solving.

Because equation (2) has the form of a differential equation, we should be able to reverse thislogic and come up with a very easy technique for solving some differential equations. After all,any first-order differential equation can be rewritten in the (differential) form

(3) M Ax, yB dx � N Ax, yB dy � 0

dy/dx � f Ax, yB

dF :� 0F0x dx �

0F0y dy

dydx

� f Ax,yB � �

0F/0x0F/0y

0F0x �

0F0y

dy dx

� 0

ddx

F Ax,yB �ddx

ACB ˛

(in a variety of ways). Now, if the left-hand side of equation (3) can be identified as a totaldifferential,

then its solutions are given (implicitly) by the level curves

for an arbitrary constant C.

Solve the differential equation

Some of the choices of differential forms corresponding to this equation are

However, the first form is best for our purposes because it is a total differential of the function:

Thus, the solutions are given implicitly by the formula See Figure 2.9. ◆x2y2 � x � C.

�00x Ax2y2 � xB dx �

00y Ax2y2 � xB dy .

A2xy2 � 1B dx � 2x2y dy � d 3 x2y2 � x 4F Ax, yB � x2y2 � x

dx �2x2y

2xy2 � 1 dy � 0 , etc.

2xy2 � 1

2x2y dx � dy � 0 ,

A2xy2 � 1B dx � 2x2y dy � 0 ,

dy

dx� �

2xy2 � 1

2x2y .

F Ax, yB � C

M Ax, yB dx � N Ax, yB dy �0F0x dx �

0F0y dy � dF Ax, yB ,


Example 1

Solution

y

x

1

10

C = 4

C = 2

C = 2

C = 0

C = −2

C = 4

Figure 2.9 Solutions of Example 1

As you might suspect, in applications a differential equation is rarely given to us in exactdifferential form. However, the solution procedure is so quick and simple for such equationsthat we devote this section to it. From Example 1, we see that what is needed is (i) a test todetermine if a differential form is exact and, if so, (ii) a procedure forfinding the function itself.

The test for exactness arises from the following observation. If

then the calculus theorem concerning the equality of continuous mixed partial derivatives

would dictate a “compatibility condition” on the functions M and N:

In fact, Theorem 2 states that the compatibility condition is also sufficient for the differentialform to be exact.

00y M Ax, yB �

00x N Ax, yB .

00y 0F0x �

00x 0F0y

M Ax, yB dx � N Ax, yB dy �0F0x dx �

0F0y dy ,

F Ax, yB M Ax, yB dx � N Ax, yB dy


Next we introduce some terminology.

Exact Differential Form

Definition 2. The differential form is said to be exact in a rectangle R if there is a function such that

(4)

for all in R. That is, the total differential of satisfies

If is an exact differential form, then the equation

is called an exact equation.

M Ax, yB dx � N Ax, yB dy � 0

M Ax, yB dx � N Ax, yB dy

dF Ax, yB � M Ax, yB dx � N Ax, yB dy .

F Ax, yBAx, yB�F�x

Ax, yB � M Ax, yB and �F�y

Ax, yB � N Ax, yBF Ax, yBM Ax, yB dx � N Ax, yB dy

Test for Exactness

Theorem 2. Suppose the first partial derivatives of are continuous in a rectangle R. Then

is an exact equation in R if and only if the compatibility condition

(5)

holds for all in R.†Ax, yB�M�y

Ax, yB ��N�x

Ax, yBM Ax, yB dx � N Ax, yB dy � 0

M Ax, yB and N Ax, yB

Before we address the proof of Theorem 2, note that in Example 1 the differential formthat led to the total differential wasA2xy2 � 1B dx � A2x2yB dy � 0 .

†Historical Footnote: This theorem was proven by Leonhard Euler in 1734.

The compatibility conditions are easily confirmed:

Also clear is the fact that the other differential forms considered,

do not meet the compatibility conditions.

Proof of Theorem 2. There are two parts to the theorem: Exactness implies compatibil-ity, and compatibility implies exactness. First, we have seen that if the differential equation isexact, then the two members of equation (5) are simply the mixed second partials of a function

. As such, their equality is ensured by the theorem of calculus that states that mixed sec-ond partials are equal if they are continuous. Because the hypothesis of Theorem 2 guaranteesthe latter condition, equation (5) is validated.

Rather than proceed directly with the proof of the second part of the theorem, let’s derive aformula for a function that satisfies Integrating the firstequation with respect to x yields

(6)

Notice that instead of using C to represent the constant of integration, we have written .This is because y is held fixed while integrating with respect to x, and so our “constant” maywell depend on y. To determine , we differentiate both sides of (6) with respect to y to obtain

(7)

As g is a function of y alone, we can write and solving (7) for gives

Since this last equation becomes

(8)

Notice that although the right-hand side of (8) indicates a possible dependence on x, theappearances of this variable must cancel because the left-hand side, depends only on y. Byintegrating (8), we can determine up to a numerical constant, and therefore we can deter-mine the function up to a numerical constant from the functions and

To finish the proof of Theorem 2, we need to show that the condition (5) implies thatis an exact equation. This we do by actually exhibiting a function that

satisfies Fortunately, we needn’t look too far for such a function.0F/ 0x � M and 0F/ 0y � N.F Ax, yBM dx � N dy � 0

N Ax, yB.M Ax, yBF Ax, yB g A yB g¿ A yB,g¿ A yB � N Ax, yB �

00y �M Ax, yB dx .

0F/ 0y � N,

g¿ A yB �0F0y Ax, yB �

00y �M Ax, yB dx .

g¿ A yB0g/ 0y � g¿ A yB,0F0y Ax, yB �

00y �M Ax, yB dx �

00y g A yB .

g A yBg A yBF Ax, yB � �M Ax, yB dx � g A yB .

0F/ 0x � M and 0F/ 0y � N.F Ax, yBF Ax, yB

2xy2 � 1

2x2y dx � dy � 0 , dx �

2x2y

2xy2 � 1 dy � 0 ,

0N0x �

00x A2x2yB � 4xy .

0M0y �

00y A2xy2 � 1B � 4xy ,


The discussion in the first part of the proof suggests (6) as a candidate, where is given by(8). Namely, we by

(9)

where is a fixed point in the rectangle R and is determined, up to a numerical con-stant, by the equation

(10)

Before proceeding we must address an extremely important question concerning the defin-ition of . That is, how can we be sure (in this portion of the proof) that as given inequation (10), is really a function of just y alone? To show that the right-hand side of (10) isindependent of x (that is, that the appearances of the variable x cancel), all we need to do isshow that its partial derivative with respect to x is zero. This is where condition (5) is utilized.We leave to the reader this computation and the verification that satisfies conditions (4)(see Problems 35 and 36). ◆

The construction in the proof of Theorem 2 actually provides an explicit procedure forsolving exact equations. Let’s recap and look at some examples.

F Ax, yBg¿ A yB,F Ax, yB

g¿ AyB J N Ax, yB �00y �

x

x0

M At, yB dt .

g A yBAx0, y0BF Ax, yB J �

x

x0

M At, yB dt � g A yB ,define F Ax, yB g¿ AyB


Method for Solving Exact Equations

(a) If is exact, then Integrate this last equation withrespect to x to get

(11)

(b) To determine , take the partial derivative with respect to y of both sides of equa-tion (11) and substitute N for We can now solve for

(c) Integrate to obtain up to a numerical constant. Substituting intoequation (11) gives .

(d) The solution to is given implicitly by

(Alternatively, starting with the implicit solution can be found by firstintegrating with respect to y; see Example 3.)

0F/ 0y � N,

F Ax, yB � C .

M dx � N dy � 0F Ax, yB g A yBg A yBg¿ A yB g¿ A yB.0F/ 0y.

g A yBF Ax, yB � �M Ax, yB dx � g A yB .

0F/ 0x � M.M dx � N dy � 0

Solve

(12)

Here Because

0M0y � 2x �

0N0x ,

M Ax, yB � 2xy � sec2x and N Ax, yB � x2 � 2y.

A2xy � sec2xB dx � Ax2 � 2yB dy � 0 .

Example 2

Solution

equation (12) is exact. To find , we begin by integrating M with respect to x:

(13)

Next we take the partial derivative of (13) with respect to y and substitute for N:

Thus, and since the choice of the constant of integration is not important, we cantake Hence, from (13), we have and the solution toequation (12) is given implicitly by ◆

Remark. The procedure for solving exact equations requires several steps. As a check on ourwork, we observe that when we solve for we must obtain a function that is independentof x. If this is not the case, then we have erred either in our computation of or incomputing

In the construction of we can first integrate with respect to y to get

(14)

and then proceed to find . We illustrate this alternative method in the next example.

Solve

(15)

Here and . Because

equation (15) is exact. If we now integrate with respect to y, we obtain

When we take the partial derivative with respect to x and substitute for M, we get

Thus, so we take Hence, and the solution toequation (15) is given implicitly by In this case we can solve explicitlyfor y to obtain ◆y � AC � xB / A2 � xexB.xexy � 2y � x � C.

F Ax, yB � xexy � 2y � x,h AxB � x.h¿ AxB � 1,

exy � xexy � h¿ AxB � 1 � exy � xexy .

0F0x Ax, yB � M Ax, yB

F Ax, yB � � Axex � 2B dy � h AxB � xexy � 2y � h AxB .N Ax, yB

0M0y � ex � xex �

0N0x ,

N � xex � 2M � 1 � exy � xexy

A1 � exy � xexyB dx � Axex � 2B dy � 0 .

h AxBF Ax, yB � �N Ax, yB dy � h AxB

N Ax, yBF Ax, yB,0M/ 0y or 0N/ 0x.

F Ax, yBg¿ A yB,x2y � tan x � y2 � C.

F Ax, yB � x2y � tan x � y2,g A yB � y2.g¿ A yB � 2y,

x2 � g¿ A yB � x2 � 2y .

0F0y Ax, yB � N Ax, yB ,

x2 � 2y

� x2y � tan x � g A yB . F Ax, yB � � A2xy � sec2xB dx � g A yB

F Ax, yB60 Chapter 2 First-Order Differential Equations

Example 3

Solution

Remark. Since we can use either procedure for finding it may be worthwhile to con-sider each of the integrals and If one is easier to evaluate than theother, this would be sufficient reason for us to use one method over the other. [The skepticalreader should try solving equation (15) by first integrating .]

Show that

(16)

is not exact but that multiplying this equation by the factor yields an exact equation. Usethis fact to solve (16).

In equation (16), and Because

equation (16) is not exact. When we multiply (16) by the factor we obtain

(17)

For this new equation, If we test for exactness, we nowfind that

and hence (17) is exact. Upon solving (17), we find that the solution is given implicitly bySince equations (16) and (17) differ only by a factor of x, then any solution

to one will be a solution for the other whenever Hence the solution to equation (16) isgiven implicitly by ◆

In Section 2.5 we discuss methods for finding factors that, like in Example 4, changeinexact equations into exact equations.

x�1

x � x3sin y � C.x � 0.

x � x3sin y � C.

0M0y � 3x2cos y �

0N0x ,

M � 1 � 3x2sin y and N � x3cos y.

A1 � 3x2sin yB dx � Ax3cos yB dy � 0 .

x �1,

0M0y � 3x3cos y [ 4x3cos y �

0N0x ,

N � x4cos y.M � x � 3x3sin y

x�1

Ax � 3x3sin yB dx � Ax4cos yB dy � 0

M Ax, yB�N Ax, yB dy.�M Ax, yB dx

F Ax, yB,Section 2.4 Exact Equations 61

In Problems 1–8, classify the equation as separable, lin-ear, exact, or none of these. Notice that some equationsmay have more than one classification.

1.2.

3.4.5. xy dx � dy � 0

Ayexy � 2xB dx � Axexy � 2yB dy � 0

2�2y � y2 dx � A3 � 2x � x2B dy � 0

Ax2y � x4cos xB dx � x3 dy � 0

Ax10/3 � 2yB dx � x dy � 0

6.

7.

8.

In Problems 9–20, determine whether the equation isexact. If it is, then solve it.

9.

10. A2xy � 3B dx � Ax2 � 1B dy � 0

A2x � yB dx � Ax � 2yB dy � 0

u dr � A3r � u � 1B du � 0

3 2x � y cos AxyB 4 dx � 3 x cos AxyB � 2y 4 dy � 0

y2 dx � A2xy � cos yB dy � 0

Example 4

Solution

2.4 EXERCISES

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.


21.

22.

23.

24.

25.

26.

27. For each of the following equations, find the mostgeneral function so that the equation is exact.

(a)(b)

28. For each of the following equations, find the mostgeneral function so that the equation is exact.

(a)(b)

29. Consider the equationA y2 � 2xyB dx � x2 dy � 0 .

Ayexy � 4x3y � 2B dx � N Ax, yB dy � 03 y cos AxyB � ex 4 dx � N Ax, yB dy � 0

N Ax, yBM Ax, yB dx � Asin x cos y � xy � e�yB dy � 0M Ax, yB dx � Asec2y � x/yB dy � 0

M Ax, yBy A0B � 1

Atan y � 2B dx � Ax sec2y � 1/yB dy � 0 , A y2 sin xB dx � A1/x � y/xB dy � 0 , y ApB � 1

Aetx � 1B dt � Aet � 1B dx � 0 , x A1B � 1

Aety � tetyB dt � Atet � 2B dy � 0 , y A0B � �1

y A1B � 1

A yexy � 1/yB dx � Axexy � x/y2B dy � 0 , y A1B � p

A1/x � 2y2xB dx � A2yx2 � cos yB dy � 0 ,

� 3 x cos AxyB � y�1/3 4 dy � 0

c 2

21 � x2� y cos AxyB d dx

a2x �y

1 � x2y2b dx � a x

1 � x2y2 � 2yb dy � 0

� 3 2xy � cos Ax � yB � ey 4 dy � 0

32x � y2 � cos Ax � yB 4 dx

A1/yB dx � A3y � x/y2B dy � 0

A yexy � 1 /yB dx � Axexy � x/y2B dy � 0

cos u dr � Ar sin u � euB du � 0

et A y � tB dt � A1 � etB dy � 0

At /yB dy � A1 � ln yB dt � 0

Aexsin y � 3x2B dx � Aexcos y � y�2/3/3B dy � 0

Acos x cos y � 2xB dx � Asin x sin y � 2yB dy � 0


(a) Show that this equation is not exact.(b) Show that multiplying both sides of the equation

by yields a new equation that is exact.(c) Use the solution of the resulting exact equation

to solve the original equation.(d) Were any solutions lost in the process?

30. Consider the equation

(a) Show that the equation is not exact.(b) Multiply the equation by and determine

values for n and m that make the resulting equa-tion exact.

(c) Use the solution of the resulting exact equationto solve the original equation.

31. Argue that in the proof of Theorem 2 the function gcan be taken as

which can be expressed as

This leads ultimately to the representation

(18)

Evaluate this formula directly with to rework

(a) Example 1.(b) Example 2.(c) Example 3.

32. Orthogonal Trajectories. A geometric problemoccurring often in engineering is that of finding afamily of curves (orthogonal trajectories) that inter-sects a given family of curves orthogonally at eachpoint. For example, we may be given the lines offorce of an electric field and want to find the equation

x0 � 0, y0 � 0

F Ax, yB � �y

y0

N Ax, tB dt � �x

x0

M As, y0B ds .

� �x

x0

M As, y0B ds .

g AyB � �y

y0

N Ax, tB dt � �x

x0

M As, yB ds

g AyB � �y

y0

N Ax, tB dt � �y

y0

c 00t � x

x0

M As, tB ds ddt ,

xnym

� A2x3 � 3x4y � 3x2yB dy � 0 .

A5x2y � 6x3y2 � 4xy2B dx

y �2

for the equipotential curves. Consider the family ofcurves described by , where k is a param-eter. Recall from the discussion of equation (2) thatfor each curve in the family, the slope is given by

(a) Recall that the slope of a curve that is orthogo-nal (perpendicular) to a given curve is just thenegative reciprocal of the slope of the givencurve. Using this fact, show that the curvesorthogonal to the family satisfy thedifferential equation

(b) Using the preceding differential equation, showthat the orthogonal trajectories to the family ofcircles are just straight linesthrough the origin (see Figure 2.10).

x2 � y2 � k

�F�yAx, yB dx �

�F�x

Ax, yB dy � 0 .

F Ax, yB � k

dydx

� ��F�x~�F

�y .

F Ax, yB � k


33. Use the method in Problem 32 to find the orthogo-nal trajectories for each of the given families ofcurves, where k is a parameter.

(a)(b)(c)(d)[Hint: First express the family in the form .]

34. Use the method described in Problem 32 to showthat the orthogonal trajectories to the family ofcurves a parameter, satisfy

Find the orthogonal trajectories by solving the aboveequation. Sketch the family of curves, along withtheir orthogonal trajectories. [Hint: Try multiplyingthe equation by as in Problem 30.]

35. Using condition (5), show that the right-hand side of(10) is independent of x by showing that its partialderivative with respect to x is zero. [Hint: Since thepartial derivatives of M are continuous, Leibniz’stheorem allows you to interchange the operations ofintegration and differentiation.]

36. Verify that as defined by (9) and (10) satisfiesconditions (4).

F Ax, yB

xmyn

A2yx�1B dx � A y2x�2 � 1B dy � 0 .

x2 � y2 � kx, k

F(x, y) � k

y2 � kxy � ekxy � kx4

2x2 � y2 � ky

x

Figure 2.10 Orthogonal trajectories for concentric circles are lines through the center

y

x

Figure 2.11 Families of orthogonal hyperbolas

(c) Show that the orthogonal trajectories to thefamily of hyperbolas are the hyper-bolas (see Figure 2.11).x2 � y2 � k

xy � k

If we take the standard form for the linear differential equation of Section 2.3,

and rewrite it in differential form by multiplying through by dx, we obtain

This form is certainly not exact, but it becomes exact upon multiplication by the integrating

factor We have

as the form, and the compatibility condition is precisely the identity (seeProblem 20).

This leads us to generalize the notion of an integrating factor.

m AxBP AxB � m¿AxB3m AxBP AxBy � m AxBQ AxB 4 dx � m AxB dy � 0

m AxB � e� P AxB dx.

3P AxBy � Q AxB 4 dx � dy � 0 .

dydx

� P AxBy � Q AxB ,


†Historical Footnote: A general theory of integrating factors was developed by Alexis Clairaut in 1739. LeonhardEuler also studied classes of equations that could be solved using a specific integrating factor.

2.5 SPECIAL INTEGRATING FACTORS

Integrating Factor

Definition 3. If the equation

(1)

is not exact, but the equation

(2)

which results from multiplying equation (1) by the function exact, then is called an integrating factor† of the equation (1).

m Ax, yBm Ax, yB, ism Ax, yBM Ax, yB dx � m Ax, yBN Ax, yB dy � 0 ,


Show that is an integrating factor for

(3)

Use this integrating factor to solve the equation.

We leave it to you to show that (3) is not exact. Multiplying (3) by we obtain

(4)

For this equation we have Because

equation (4) is exact. Hence, is indeed an integrating factor of equation (3).m Ax, yB � xy2

0M0y Ax, yB � 6xy2 � 12x2y �

0N0x Ax, yB ,

M � 2xy3 � 6x2y2 and N � 3x2y2 � 4x3y.

A2xy3 � 6x2y2B dx � A3x2y2 � 4x3yB dy � 0 .

m Ax, yB � xy2,

A2y � 6xB dx � A3x � 4x2y�1B dy � 0 .

m Ax, yB � xy2Example 1

Solution

Let’s now solve equation (4) using the procedure of Section 2.4. To find we beginby integrating M with respect to x:

When we take the partial derivative with respect to y and substitute for N, we find

Thus, so we can take Hence, and the solution toequation (4) is given implicitly by

Although equations (3) and (4) have essentially the same solutions, it is possible to lose orgain solutions when multiplying by In this case is a solution of equation (4) butnot of equation (3). The extraneous solution arises because, when we multiply (3) by to obtain (4), we are actually multiplying both sides of (3) by zero if This gives us

as a solution to (4), but it is not a solution to (3). ◆

Generally speaking, when using integrating factors, you should check whether any solu-tions to are in fact solutions to the original differential equation.

How do we find an integrating factor? If is an integrating factor of (1) with contin-uous first partial derivatives, then testing (2) for exactness, we must have

By use of the product rule, this reduces to the equation

(5)

But solving the partial differential equation (5) for m is usually more difficult than solving theoriginal equation (1). There are, however, two important exceptions.

Let’s assume that equation (1) has an integrating factor that depends only on x; that is,In this case equation (5) reduces to the separable equation

(6)

where is (presumably) just a function of x. In a similar fashion, if equa-tion (1) has an integrating factor that depends only on y, then equation (5) reduces to theseparable equation

(7)

where is just a function of y.A0N/ 0x � 0M/ 0yB /Mdmdy

� a0N/ 0x � 0M/ 0yM

bm ,

A0M/ 0y � 0N/ 0xB /Ndmdx

� a0M/ 0y � 0N/ 0xN

bm ,

m � m AxB.

M

�M

�y� N

�M

�x� a�N

�x�

�M�ybM .

00y 3m Ax, yBM Ax, yB 4 �

00x 3m Ax, yBN Ax, yB 4 .

m Ax, yBm Ax, yB � 0

y � 0y � 0.

m � xy2y � 0m Ax, yB.

x2y3 � 2x3y2 � C .

F Ax, yB � x2y3 � 2x3y2,g A yB � 0.g¿ A yB � 0,

3x2y2 � 4x3y � g¿A yB � 3x2y2 � 4x3y .

0F0y Ax, yB � N Ax, yB

F Ax, yB � � A2xy3 � 6x2y2B dx � g A yB � x2y3 � 2x3y2 � g A yB .F Ax, yB,

Section 2.5 Special Integrating Factors 65

We can reverse the above argument. In particular, if is a function thatdepends only on x, then we can solve the separable equation (6) to obtain the integrating factor

for equation (1). We summarize these observations in the following theorem.

m AxB � exp c� a0M/ 0y � 0N/ 0xN

bdx dA0M/ 0y � 0N/ 0xB /N


Special Integrating Factors

Theorem 3. If is continuous and depends only on x, then

(8)

is an integrating factor for equation (1).If is continuous and depends only on y, then

(9)

is an integrating factor for equation (1).

m A yB � exp c� a0N/ 0x � 0M/ 0yM

bdy dA0N/ 0x � 0M/ 0yB /Mm AxB � exp c�a0M/ 0y � 0N/ 0x

Nbdx d

A0M/ 0y � 0N/ 0xB /N

Theorem 3 suggests the following procedure.

Method for Finding Special Integrating Factors

If is neither separable nor linear, compute . Ifthen the equation is exact. If it is not exact, consider

(10)

If (10) is a function of just x, then an integrating factor is given by formula (8). If not,consider

(11)

If (11) is a function of just y, then an integrating factor is given by formula (9).

0N/ 0x � 0M/ 0yM

.

0M/ 0y � 0N/ 0xN

.

0M/ 0y � 0N/ 0x,0M/ 0y and 0N/ 0xM dx � N dy � 0

Solve

(12)

A quick inspection shows that equation (12) is neither separable nor linear. We also note that

0M0y � 1 [ A2xy � 1B �

0N0x .

A2x2 � yB dx � Ax2y � xB dy � 0 .

Example 2

Solution

Because (12) is not exact, we compute

We obtain a function of only x, so an integrating factor for (12) is given by formula (8). That is,

When we multiply (12) by we get the exact equation

Solving this equation, we ultimately derive the implicit solution

(13)

Notice that the solution was lost in multiplying by Hence, (13) and aresolutions to equation (12). ◆

There are many differential equations that are not covered by Theorem 3 but for which anintegrating factor nevertheless exists. The major difficulty, however, is in finding an explicitformula for these integrating factors, which in general will depend on both x and y.

x � 0m � x�2.x � 0

2x � yx�1 �y2

2� C .

A2 � yx�2B dx � A y � x�1B dy � 0 .

m � x�2,

m AxB � expa��2x

dxb � x�2 .

0M/ 0y � 0N/ 0xN

�1 � A2xy � 1B

x2y � x�

2 A1 � xyB�x A1 � xyB �

�2

x .

Section 2.5 Special Integrating Factors 67

In Problems 1–6, identify the equation as separable, lin-ear, exact, or having an integrating factor that is a func-tion of either x alone or y alone.

1.2.3.4.5.6.

In Problems 7–12, solve the equation.7.8.9.

10.11.12.

In Problems 13 and 14, find an integrating factor of theform and solve the equation.

13.14. A12 � 5xyB dx � A6xy�1 � 3x2B dy � 0

A2y2 � 6xyB dx � A3xy � 4x2B dy � 0

xnym

A3x2y2 � y�1B dy � 0A2xy3 � 1B dx �

A y2 � 2xyB dx � x2 dy � 0

A2y2 � 2y � 4x2B dx � A2xy � xB dy � 0

Ax4 � x � yB dx � x dy � 0

A3x2 � yB dx � Ax2y � xB dy � 0

A2xyB dx � A y2 � 3x2B dy � 0

A2y2x � yB dx � x dy � 0

Ax2sin x � 4yB dx � x dy � 0

A y2 � 2xyB dx � x2 dy � 0

A2x � yB dx � Ax � 2yB dy � 0

A2y3 � 2y2B dx � A3y2x � 2xyB dy � 0

A2x � yx�1B dx � Axy � 1B dy � 0

15. (a) Show that if depends only on the product xy, that is,

then the equation has an integrating factor of the form Givethe general formula for

(b) Use your answer to part (a) to find an implicitsolution to

satisfying the initial condition

16. (a) Prove that has an integratingfactor that depends only on the sum if andonly if the expression

depends only on .(b) Use part (a) to solve the equation

(3 � y � xy)dx � (3 � x � xy)dy � 0.

x � y

0N/ 0x � 0M/ 0yM � N

x � yMdx � Ndy � 0

y(1) � 1.

(3y � 2xy2)dx � (x � 2x2y)dy � 0 ,

m AxyB. m AxyB.M Ax, yB dx � N Ax, yB dy � 0

0N/ 0x � 0M/ 0yxM � yN

� H AxyB ,A0N/ 0x � 0M/ 0yB / AxM � yNB

2.5 EXERCISES

17. (a) Find a condition on M and N that is necessaryand sufficient for to have anintegrating factor that depends only on theproduct .

(b) Use part (a) to solve the equation

18. If find the solution to theequation

19. Fluid Flow. The streamlines associated with a cer-tain fluid flow are represented by the family of curves

The velocity potentials of theflow are just the orthogonal trajectories of this family.y � x � 1 � ke�x.

M Ax, yB dx � N Ax, yB dy � 0.xM Ax, yB � yN Ax, yB � 0,

� (2x � x4 � 2x3y) dy � 0 .

(2x � 2y � 2x3y � 4x2y2) dx

x2y

Mdx � Ndy � 0


(a) Use the method described in Problem 32 ofExercises 2.4 to show that the velocity potentialssatisfy

[Hint: First express the family inthe form � k.](b) Find the velocity potentials by solving the equa-

tion obtained in part (a).

20. Verify that when the linear differential equationis multiplied by

the result is exact.e� PAxB dx,

m AxB �3P AxBy � QAxB 4 dx � dy � 0

F Ax, yB y � x � 1 � ke�x

dx � Ax � yB dy � 0.

SUBSTITUTIONS AND TRANSFORMATIONS2.6When the equation

is not a separable, exact, or linear equation, it may still be possible to transform it into one thatwe know how to solve. This was in fact our approach in Section 2.5, where we used anintegrating factor to transform our original equation into an exact equation.

In this section we study four types of equations that can be transformed into either a sepa-rable or linear equation by means of a suitable substitution or transformation.


Substitution Procedure

(a) Identify the type of equation and determine the appropriate substitution ortransformation.

(b) Rewrite the original equation in terms of new variables.(c) Solve the transformed equation.(d) Express the solution in terms of the original variables.

Homogeneous Equations

Homogeneous Equation

Definition 4. If the right-hand side of the equation

(1)

can be expressed as a function of the ratio alone, then we say the equation is homogeneous.

y/x

dydx

� f Ax, yB

For example, the equation

(2)

can be written in the form

Since we have expressed as a function of the ratio that is,where then equation (2) is homogeneous.

The equation

(3)

can be written in the form

Here the right-hand side cannot be expressed as a function of alone because of the term in the numerator. Hence, equation (3) is not homogeneous.

One test for the homogeneity of equation (1) is to replace x by tx and y by ty. Then (1) ishomogeneous if and only if

for all [see Problem 43(a)].To solve a homogeneous equation, we make a rather obvious substitution. Let

Our homogeneous equation now has the form

(4)

and all we need is to express in terms of x and y. Since , then . Keeping inmind that both y and y are functions of x, we use the product rule for differentiation to deducefrom y � yx that

We then substitute the above expression for into equation (4) to obtain

(5)

The new equation (5) is separable, and we can obtain its implicit solution from

All that remains to do is to express the solution in terms of the original variables x and y.

� 1G AyB � y

dy � � 1x

dx .

y � x dydx

� G AyB .dy/dx

dydx

� y � x

dydx

.

y � yxy � y/xdy/dx

dydx

� G AyB ,Y �

yx

.

t � 0

f Atx, tyB � f Ax, yB1/xy/x

dy

dx�

x � 2y � 1

y � x�

1 � 2 A y/xB � A1/xBA y/xB � 1 .

Ax � 2y � 1B dx � Ax � yB dy � 0

G AyB J y � 1 4 , A y � xB /x � G A y/xB,3y/xA y � xB /xdydx

�y � x

x�

yx

� 1 .

Ax � yB dx � x dy � 0

Section 2.6 Substitutions and Transformations 69

Solve

(6)

A check will show that equation (6) is not separable, exact, or linear. If we express (6) in thederivative form

(7)

then we see that the right-hand side of (7) is a function of just . Thus, equation (6) is homo-geneous.

Now let and recall that With these substitutions, equation(7) becomes

The above equation is separable, and, on separating the variables and integrating, we obtain

Hence,

Finally, we substitute for y and solve for y to get

as an explicit solution to equation (6). Also note that is a solution. ◆

Equations of the Form dy/dx � G(ax � by)

When the right-hand side of the equation can be expressed as a function of thecombination ax � by, where a and b are constants, that is,

then the substitution

transforms the equation into a separable one. The method is illustrated in the next example.

Solve

(8)

The right-hand side can be expressed as a function of that is,

y � x � 1 � Ax � y � 2B�1 � � Ax � yB � 1 � 3 Ax � yB � 2 4�1 ,

x � y,

dydx

� y � x � 1 � Ax � y � 2B�1 .

z � ax � by

dydx

� G Aax � byB ,dy/dx � f Ax, yB

x � 0

y � x tan Aln 0 x 0 � CBy/xy � tan Aln 0 x 0 � CB .

arctan y � ln 0 x 0 � C .

� 1y2 � 1

dy � � 1x

dx ,

y � x dydx

� y � y2 � 1 .

dy/dx � y � x Ady/dxB.y � y/x

y/x

dy

dx�

xy � y2 � x2

x2 �y

x� ay

xb 2

� 1 ,

Axy � y2 � x2B dx � x2 dy � 0 .


Example 1

Solution

Example 2

Solution

so let To solve for , we differentiate with respect to x to obtainand so Substituting into (8) yields

or

Solving this separable equation, we obtain

from which it follows that

Finally, replacing z by yields

as an implicit solution to equation (8). ◆

Bernoulli Equations

Ax � y � 2B2 � Ce2x � 1

x � y

Az � 2B2 � Ce2x � 1 .

12

ln 0 Az � 2B2 � 1 0 � x � C1 ,

� z � 2Az � 2B2 � 1 dz � �dx ,

dzdx

� Az � 2B � Az � 2B�1 .

1 �dzdx

� �z � 1 � Az � 2B�1 ,

dy/dx � 1 � dz/dx.dz/dx � 1 � dy/dx,z � x � ydy/dxz � x � y.


Bernoulli Equation

Definition 5. A first-order equation that can be written in the form

(9)

where and are continuous on an interval and n is a real number, is called aBernoulli equation.†

Aa, bBQ AxBP AxBdydx

� P AxBy � Q AxByn ,

†Historical Footnote: This equation was proposed for solution by James Bernoulli in 1695. It was solved by his brotherJohn Bernoulli. (James and John were two of eight mathematicians in the Bernoulli family.) In 1696, Gottfried Leibnizshowed that the Bernoulli equation can be reduced to a linear equation by making the substitution y � y1�n.

Notice that when n � 0 or 1, equation (9) is also a linear equation and can be solved by the method discussed in Section 2.3. For other values of n, the substitution

transforms the Bernoulli equation into a linear equation, as we now show.

Y � y1�n

Dividing equation (9) by yields

(10)

Taking we find via the chain rule that

and so equation (10) becomes

Because is just a constant, the last equation is indeed linear.

Solve

(11)

This is a Bernoulli equation with To transform (11)into a linear equation, we first divide by y3 to obtain

Next we make the substitution Since the transformed equation is

(12)

Equation (12) is linear, so we can solve it for y using the method discussed in Section 2.3.When we do this, it turns out that

Substituting gives the solution

Not included in the last equation is the solution that was lost in the process of dividing(11) by . ◆

Equations with Linear Coefficients

We have used various substitutions for y to transform the original equation into a new equationthat we could solve. In some cases we must transform both x and y into new variables, say, u andy. This is the situation for equations with linear coefficients—that is, equations of the form

(13) Aa1x � b1 y � c1B dx � Aa2x � b2 y � c2B dy � 0 ,

y3y � 0

y�2 �x2

�120

� Ce�10x .

y � y�2

y �x2

�120

� Ce�10x .

dydx

� 10y � 5x .

� 12

dydx

� 5y � � 52

x ,

dy/dx � �2y�3 dy/dx,y � y�2.

y�3

dydx

� 5y�2 � �

52

x .

n � 3, P AxB � �5, and Q AxB � �5x/2.

dydx

� 5y � �

52

xy3 .

1/ A1 � nB1

1 � n dydx

� P AxBy � Q AxB .dydx

� A1 � nBy�n

dydx

,

y � y1�n,

y�n

dydx

� P AxBy1�n � Q AxB .yn


Example 3

Solution

where the ai’s, bi’s, and ci’s are constants. We leave it as an exercise to show that whenequation (13) can be put in the form which we solved via

the substitution Before considering the general case when let’s first look at the special situa-

tion when Equation (13) then becomes

which can be rewritten in the form

This equation is homogeneous, so we can solve it using the method discussed earlier in thissection.

The above discussion suggests the following procedure for solving (13). If then we seek a translation of axes of the form

where h and k are constants, that will change and changeSome elementary algebra shows that such a transformation

exists if the system of equations

(14)

has a solution. This is ensured by the assumption , which is geometrically equiva-lent to assuming that the two lines described by the system (14) intersect. Now if satisfies(14), then the substitutions transform equation (13) into the homo-geneous equation

(15)

which we know how to solve.

Solve

(16)

Since we will use the translation of axes ,, where h and k satisfy the system

Solving the above system for h and k gives Hence, we let and. Because dy � dy and dx � du, substituting in equation (16) for x and y yields

dy

du�

3 � (y/u)

1 � (y/u) .

A�3u � yB du � Au � yB dy � 0

y � y � 3x � u � 1h � 1, k � �3.

h � k � 2 � 0 .

�3h � k � 6 � 0 ,

y � y � kx � u � ha1b2 � A�3B A1B � A1B A1B � a2b1,

A�3x � y � 6B dx � Ax � y � 2B dy � 0 .

dy

du� �

a1u � b1y

a2u � b2y� �

a1 � b1 Ay/uBa2 � b2 Ay/uB ,

x � u � h and y � y � kAh, kBa1b2 � a2b1

a1h � b1k � c1 � 0 ,a2h � b2k � c2 � 0

a2x � b2y � c2 into a2u � b2y.a1x � b1y � c1 into a1u � b1y

x � u � h and y � Y � k ,

a1b2 � a2b1,

dy

dx� �

a1x � b1y

a2x � b2y� �

a1 � b1 A y/xBa2 � b2 A y/xB .

Aa1x � b1yB dx � Aa2x � b2yB dy � 0 ,

c1 � c2 � 0.a1b2 � a2b1,

z � ax � by.dy/dx � G Aax � byB,a1b2 � a2b1,


Example 4

Solution

The last equation is homogeneous, so we let . Then and, sub-stituting for , we obtain

Separating variables gives

from which it follows that

When we substitute back in for z, u, and y, we find

This last equation gives an implicit solution to (16). ◆

A y � 3B2 � 2 Ax � 1B A y � 3B � 3 Ax � 1B2 � C .

y2 � 2uy � 3u2 � C ,

Ay/uB2 � 2 Ay/uB � 3 � Cu�2 ,

z2 � 2z � 3 � Cu�2 .

12

ln 0 z2 � 2z � 3 0 � �ln 0u 0 � C1 ,

� z � 1z2 � 2z � 3

dz � �� 1u

du ,

z � u dzdu

�3 � z1 � z

.

y/udy/du � z � u Adz/duB,z � y/u


In Problems 1–8, identify (do not solve) the equation ashomogeneous, Bernoulli, linear coefficients, or of theform

1.2.3.4.

5.

6.

7.

8.

Use the method discussed under “Homogeneous Equa-tions” to solve Problems 9–16.

9.10.11.12.

13.

14.dy

du�u sec A y/uB � y

u

dxdt

�x2 � t2t2 � x2

tx

Ax2 � y2B dx � 2xy dy � 0

A y2 � xyB dx � x2 dy � 0

A3x2 � y2B dx � Axy � x3y�1B dy � 0

Axy � y2B dx � x2 dy � 0

A y3 � uy2B du � 2u2y dy � 0

cos Ax � yB dy � sin Ax � yB dx

A ye�2x � y3B dx � e�2x dy � 0

u dy � y du � 2uy du

At � x � 2B dx � A3t � x � 6B dt � 0

dy/dx � y/x � x3y2

A y � 4x � 1B2 dx � dy � 0

2tx dx � At2 � x2B dt � 0

y¿ � G Aax � byB. 15. 16.

Use the method discussed under “Equations of the Form” to solve Problems 17–20.

17. 18.19. 20.

Use the method discussed under “Bernoulli Equations”to solve Problems 21–28.

21.

22.

23.

24.

25. 26.

27. 28.dydx

� y3x � y � 0dr

du�

r2 � 2ruu2

dydx

� y � exy�2dxdt

� tx3 �xt

� 0

dy

dx�

yx � 2

� 5 Ax � 2By1/2

dydx

�2yx

� x2y2

dydx

� y � e2xy3

dydx

�yx

� x2y2

dy/dx � sinAx � yBdy/dx � Ax � y � 5B2 dy/dx � Ax �y�2B2dy/dx � 2x � y � 1

dy/dx � G Aax � byBdy

dx�

y Aln y � ln x � 1Bx

dydx

�x2 � y2

3xy

2.6 EXERCISES

Use the method discussed under “Equations with LinearCoefficients” to solve Problems 29–32.

29.30.31.32.

In Problems 33–40, solve the equation given in:

33. Problem 1. 34. Problem 2.




41. Use the substitution to solve equa-tion (8).

42. Use the substitution to solve

43. (a) Show that the equation ishomogeneous if and only if [Hint: Let .]

(b) A function is called homogeneous oforder n if Show that theequation

is homogeneous if are bothhomogeneous of the same order.

44. Show that equation (13) reduces to an equation ofthe form

when [Hint: If , thenso that .]

45. Coupled Equations. In analyzing coupled equa-tions of the form

dxdt

� ax � by ,

dydt

� ax � by ,

a2 � ka1 and b2 � kb1a2/a1 � b2/b1 � k,a1b2 � a2b1a1b2 � a2b1.

dydx

� G Aax � byB ,

M Ax, yB and N Ax, yBM Ax, yB dx � N Ax, yB dy � 0

H Atx, tyB � tnH Ax, yB.H Ax, yBt � 1/xf Atx, tyB � f Ax, yB.dy/dx � f Ax, yB

dydx

�2yx

� cos A y/x2B .y � yx2

y � x � y � 2

A2x � y � 4B dx � Ax � 2y � 2B dy � 0

A2x � yB dx � A4x � y � 3B dy � 0

Ax � y � 1B dx � A y � x � 5B dy � 0

A�3x � y � 1B dx � Ax � y � 3B dy � 0


†Historical Footnote: Count Jacopo Riccati studied a particular case of this equation in 1724 during his investigation ofcurves whose radii of curvature depend only on the variable y and not the variable x.

where are constants, we may wish todetermine the relationship between x and y ratherthan the individual solutions For thispurpose, divide the first equation by the second toobtain

(17)

This new equation is homogeneous, so we can solveit via the substitution We refer to the solu-tions of (17) as integral curves. Determine the inte-gral curves for the system

46. Magnetic Field Lines. As described in Problem20 of Exercises 1.3, the magnetic field lines of adipole satisfy

Solve this equation and sketch several of these lines.

47. Riccati Equation. An equation of the form

(18)

is called a generalized Riccati equation.†

(a) If one solution—say, —of (18) is known,show that the substitution reduces(18) to a linear equation in y.

(b) Given that is a solution to

use the result of part (a) to find all the other solu-tions to this equation. (The particular solution

can be found by inspection or by usinga Taylor series method; see Section 8.1.)u AxB � x

dydx

� x3 A y � xB2 �yx ,

u AxB � x

y � u � 1/yu AxB

dydx

� P AxB y2 � Q AxB y � R AxB

dydx

�3xy

2x2 � y2 .

dxdt

� 2x � y .

dydt

� �4x � y ,

y � y/x.

dydx

�ax � byax � by

.

x AtB, y AtB.a, b, a, and b

In this chapter we have discussed various types of first-order differential equations. The mostimportant were the separable, linear, and exact equations. Their principal features and methodof solution are outlined below.

Separable Equations: Separate the variables and integrate.

Linear Equations: . The integrating factor reducesthe equation to

Exact Equations: Solutions are given implicitly by . If , then is exact and F is given by

or

When an equation is not separable, linear, or exact, it may be possible to find an integrat-ing factor or perform a substitution that will enable us to solve the equation.

Special Integrating Factors: is exact. If dependsonly on x, then

is an integrating factor. If depends only on y, then

is an integrating factor.

Homogeneous Equations: . Let . Then and the transformed equation in the variables y and x is separable.

Equations of the Form: . Let z � ax � by. Then and the transformed equation in the variables z and x is separable.

Bernoulli Equations: . For or 1, let . Thenand the transformed equation in the variables y and x is linear.

Linear Coefficients: For , letand , where h and k satisfy

Then the transformed equation in the variables u and y is homogeneous.

a2h � b2k � c2 � 0 .

a1h � b1k � c1 � 0 ,

y � y � kx � u � ha1b2 � a2b1Aa1x � b1 y � c1B dx � Aa2x � b2 y � c2B dy � 0.

dy/dx � A1 � nBy�n Ady/dxB, y � y1�nn � 0dy/dx � P AxBy � Q AxB yn

dz/dx � a � b Ady/dxB,dy/dx � G Aax � byBdy/dx � y � x Ady/dxB,y � y/xdy/dx � G A y/xB

m A yB � exp c�a0N/ 0x � 0M/ 0yM

bdy dA0N/ 0x � 0M/ 0yB /Mm AxB � exp c�a0M/ 0y � 0N/ 0x

Nbdx d

A0M/ 0y � 0N/ 0xB /NMM dx � MN dy � 0

F � �N dy � h AxB , where h¿ AxB � M �00x �N dy .

F � �M dx � g A yB , where g¿ A yB � N �00y �M dx

M dx � N dy � 00N/ 0x0M/ 0y �F Ax, yB � CdF Ax, yB � 0.

d AmyB /dx � mQ, so that my � �mQ dx � C.m � exp 3�P AxB dx 4dy/dx � P AxBy � Q AxB

dy/dx � g AxBp A yB.


Chapter Summary

In Problems 1–30, solve the equation.

1. 2.

3.

4.

5.6.7.

8.

9.10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.22.23. Ay � xB dx � Ax � yB dy � 0

A2x � 2y � 8B dx � Ax � 3y � 6B dy � 0

A y � 2x � 1B dx � Ax � y � 4B dy � 0

dydu

�yu

� �4uy�2

Ax2 � 3y2B dx � 2xy dy � 0

dydx

� A2x � y � 1B2dydu

� 2y � y2

dydx

� y tan x � sin x � 0

dydx

� 2 � 22x � y � 3

dxdt

�x

t � 1� t2 � 2

dydx

�yx

� x2sin 2x

A y3 � 4exyB dx � A2ex � 3y2B dy � 0

dxdt

� 1 � cos2 At � xB� 3 y�1/2 � A1 � x2 � 2xy � y2B�1 4 dy � 031 � A1 � x2 � 2xy � y2B�1 4 dx

Ax2 � y2B dx � 3xy dy � 0

dydx

�2yx

� 2x2y2

t3y2 dt � t4y�6 dy � 0

2xy3 dx � A1 � x2B dy � 0

3sin AxyB � xy cos AxyB 4 dx � 31 � x2cos AxyB 4 dy � 0

dydx

�3yx

� x2 � 4x � 3

Ax2 � 2y�3B dy � A2xy � 3x2B dx � 0

dydx

� 4y � 32x2dydx

�ex�y

y � 1

Review Problems 77

REVIEW PROBLEMS

24.25.

26.

27.

28.

29.

30.


31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41. Express the solution to the following initial valueproblem using a definite integral:

Then use your expression and numerical integrationto estimate to four decimal places.y(3)

dydt

�1

1 � t2 � y , y A2B � 3 .

dydx

� 4y � 2xy2 , y A0B � �4

dydx

�2yx

� x�1y�1 , y A1B � 3

2y dx � Ax2 � 4B dy � 0 , y A0B � 4

A2x � yB dx � Ax � y � 3B dy � 0 , y A0B � 2

y A1B � 0� 3 cos A2x � yB � ey 4 dy � 0 , 32 cos A2x � yB � x2 4 dx

A2y2 � 4x2B dx � xy dy � 0 , y A1B � �2

dydx

�2yx

� x2cos x , y ApB � 2

At � x � 3B dt � dx � 0 , x A0B � 1

dydx

� axy

�yxb , y A1B � �4

(x 3 � y) dx � x dy � 0 , y(1) � 3

dydx

� Ax � y � 1B2 � Ax � y � 1B2� A3x2y2 � 6xy � 2x2yB dy � 0A4xy3 � 9y2 � 4xy2B dx

dydx

�x � y � 1x � y � 5

A3x � y � 5B dx � Ax � y � 1B dy � 0

dydx

� xy � 0

y Ax � y � 2B dx � x Ay � x � 4B dy � 0A2y/x � cos xB dx � A2x/y � sin yB dy � 0

1. An instructor at Ivey U. asserted: “All you need toknow about first-order differential equations is howto solve those that are exact.” Give arguments thatsupport and arguments that refute the instructor’sclaim.

2. What properties do solutions to linear equationshave that are not shared by solutions to either sepa-rable or exact equations? Give some specific exam-ples to support your conclusions.


TECHNICAL WRITING EXERCISES

3. Consider the differential equation

where a, b, and c are constants. Describe what hap-pens to the asymptotic behavior as of thesolution when the constants a, b, and c are varied.Illustrate with figures and/or graphs.

xS�q

dydx

� ay � be�x , y A0B � c ,

Fundamentals of Differential Equations 8th Nagel

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