Bear H.S. Understanding Calculus

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160

Understanding

Calculus

Therefore,

for 0 < x

sinx [ x

]

I x

---

1--+-

<-.

x 3! 5! - 7!

Use thisto

approximate

Si~x

dx to withinfo

.00003.

Hint:

Since

Si~X

0+, youcanregardthis as a proper

integral.

Separable Differential

Equations

Many physical facts attain their mathematical expression as differential equations. In the

simplest case, a differential

equation

is an equation expressing a relationship between two

variables and the rate of change (derivative) of one variable with respect to the other. For

example,

= 3x

is a simple differential equation, and the solutions are all the functions

y = x

+ c. All the indefinite integration problems we have treated can be considered

differential equations in this way. Thus, "find

f(x)

dx"

means the same as "solve the

differential equation

= f (x)." The solutions are all the functions y = f f (x) dx + c.

Notice that now we want

all solutions, so the arbitrary constant is necessary. In general, the

solutions of a differential equation represent «family of curves, and not just a single curve. To

specify a specific solution, we must specify an initial condition like

y(xo) =

Yo.

EXAMPLE 27.1

Find all solutions of the differential equation, and find the specific solution that satisfies the given initial

di . dy - 1 (0) - 1

con ition: -;h -

~,y

Solution

4+x

1 1 1 X

Y = 4 + x

= 2

tan-

+ c.

This is the family of all solutions. To find c so that the initial condition is satisfied, substitute 0 for

x and

1 for y:

1 0

1 = -

tan-

- + C' c = 1.

2 2 '

The solution that satisifies the initial condition is

1 1 X

Y = 2

tan-

+ 1.

A first-order differential equation is one that involves only the first derivative. A first-

order differential equation generally has a one-parameter family of solutions; that is, the

solutions depend on a single arbitrary constant. In this chapter, we will study the following

161

162

Understanding Calculus

useful type of first-order equation:

!(x)

-=--,

g(y)

or equivalently,

g(y)

f(x).

(27.1)

An equation of this type is called separable, and we say the variables separate because we

can write all the x's on one side and all the y's on the other. Webend the notationa little and

agree that (27.1) can also be written

g(y)dy

!(x)dx.

(27.2)

As this last form suggests,the solutionsof (27.1) or (27.2)are obtainedsimply by integrating:

fg(y) dy =f

f(x)

dx +c. (27.3)

That is, the solutions

y of (27.2) will be the functionsthat are definedimplicitlyby one of the

equations

G(y)

F(x)

+ c

where G and F are antiderivatives of g and

j',

respectively.

EXAMPLE 27.2

dy y2

Solve-

--.

x + 1

Solution

We separatethe variablesand integrate:

1 1

-dy=

--dx,

y2 x + 1

- - = log

+ e.

Since e is arbitrary, we can writeloge insteadof c to simplifythe expression:

= log [x +

+ loge = log c]x +

11,

-1

y-----

-loge\x+ll"

The process of separating the variables and integrating appears to be a purely formal

one, but it really does give all the solutions. It is easy to see by differentiating both sides that

y satisfies an equation

G(y)

F(x)

+c,

(27.4)

where

G'(y)

g(y)

and

F'(x)

!(x),

then y also satisfiesthe differentialequation

g(y)

f(x).

(27.5)

It is also easy to show(Problem27.19) that there are no solutionsof the differential equation

(27.5)other than functions that satisfy (27.4).

One verycommonseparabledifferential equation is the exponentialchangeequation

=ay, (27.6)

Chapter 27 • SeparableDifferential Equations

163

(27.7)

where

a is a given constant. The equation expresses the fact that y increases (a > 0) or

decreases

(a < 0) at a rateproportional to y. Thisequationcharacterizes a numberof physical

phenomena, rangingfromthegrowthofbacteriacoloniestothedecayofradioactive substances.

Wecan separatevariables and integrate to solve (27.6),but it is simplerjust to noticethat the

functions

y =

ce"

all satisfy the differential equation.

Moreover,

if y is any solution, (i.e.,

any functionsuch that

= ay), then

.!!..-(ye-aX)

= dy .

_ aye-aX

dx dx

= ay .

aye?"

Since 1x

(ye-

)

= 0,

ye-

is constant, and

y = ce

That is, everyfunction (27.7) satisfies the differential equation, and any function that satisfies

the differential equationhas the form (27.7).

EXAMPLE

27.3

Let y bethe numberof bacteriain a colonyat timet, andassume

= ay, so y = ce" for someconstants

c and

a. If y =500 at t =0 and y = 2000at t = 2 hours, what is y at t =4 hours?

Solution

Since y = 500 at t = 0, we have 500 =

ceo

= c and c = 500. Noticethat c will alwaysbe the valueof

y at t =0 in exponential change problems. Now we use the fact that y =2000 when t = 2 to finda, or

more usefully, to find

e":

2000 =500e

=500(e

)2,

)2 =4,

e" =2, e" = 2

•

The final solutionis y =

:SOO

. 2'. When t =4, y =500 . 2

= 8000.

EXAMPLE

27.4

Solve - = 2xy + 2x.

Solution

2x(y

+ 1),

=2xdx

y + 1 '

log

= x

+ loge.

Wewrite the arbitraryconstant in the form logc and use e

Jog c

= c to simplifythe expression:

Iy +

= e

x 2

•

10g

c = ce

x 2

.y+ 1 = ±ce

x 2

•

Wenow allow c to be positiveor negative and we drop the ± sign, so

finally

y =

-1

ce'

EXAMPLE

27.5

Newton'slaw of coolingsaysthat a hot bodywillcooloff at a rate proportional to thedifferencebetween

its temperature and the temperature of the surroundings. Suppose a cup of coffee has temperature

T = 130°F at t = 0 in a room at temperature

70°F.

If T = 100°F at t = 3 minutes, what is the

temperature at

t = 5 minutes? at t =

°6

minutes?

164

Understanding Calculus

Solution

The coolinglawgivesthe differential equation

-k(T

- 70),

where

T is the temperature in degreesFahrenheit and k is a positive constantof proportionality.

Weseparatethe variables and solve.

--=-kdt

-70

10g(T - 70) =

-kt

+logc,

T - 70 =

ce:".

Since T = 130at t = 0, c = 60 and

=70 + 60e-

•

Nowwe use the condition T = 100when t = 3 to find«",

100 = 70 + 60e-

(e-

)3,

2-!,

= 2-

/3.

Finally, we have

= 70 +60 . 2-

•

Whent = 5 minutes, this gives

T = 70 + -

70 + 19 = 89.

When t = 6 minutes,

T = 70 +

=70 + 15 = 85.

A body falling in a vacuum accelerates

a constant rate

32 ft/sec", Since tall

vacuums are rare, air resistance usually plays a significant role, with the acceleration usually

being retarded at a rate proportional to the speed. Thus, a more realistic differential equation

for the velocity

v of a falling object is

- =32 - kv,

(27.8)

where the constant

k depends on the shape of the object. In Problem 27.18 you are asked to

solve (27.8) and show that

v =

-(1

) ,

from which we conclude that v approaches a terminal velocity of ¥ft/sec.

PROBLEMS

Solvethe differential equation. Find the constantif an initialconditionis

given.

dy 2

27.1 - =

xcosx;

y(O) = 2

27.2

= sec

x; y

(~)

= 0

Chapter 27 • SeparableDifferential Equations

165

dy 3 rr

27.3

dx = x2 +9; y(3) =

dy x

27.4

dx = x

+1;

yeO)

27.5

dy x

+ 1

yeO)

= 2

dx y

27.6

dx = e

;

y(O)

=log 2

27.7

1+ x

27.8

= Y

JI=X2

27.9 dx - 5y = 0

27.10

dx - 5y = 10

27.11 - -

2xy

= 0

27.12 - -

2xy

= 2x

dy Y

27.13 - = - + y

dx x

27.14

= Y +2

x-I

27.15 Suppose a beaker of boiling water at 100°C cools to 40°C in five minutes in a room at

20°C.What is its temperature

T at time t? at time t = 10 minutes?

27.16 Suppose there are 100bacteriain a colonyat t = 0 and 500 three hours later. When will

there be 1000bacteriain the colony? Whenwill there be 2500?

27.17 Show that the solutionsof *=ay can be written y =

yoA'

where

is the initial value

and A is a constant.

27.18 Solve equation (27.8) for a falling body with air resistance proportional to speed, and

v =0 whent = 0, and showthat v

¥as t

00.

(A skydiverwithunopened chute

reaches a terminalvelocityof about

120 mph, or about 176ft/sec, which givesk = .18.)

27.19 Showthat if y satisfiesg

(y)

*= f

(x),

theny must satisfyan equationG

(y)

= F (x) +c,

with G' = g and

Hint:

Define

H(x)

G(y(x))

and show that

H'(x)

F'(x),

H(x)

F(x)

+ c.

(28.2)

(28.4)

First-Order

Linear

Equations

A first-order

linear

differential equation is one that can be written in the following form:

p(x)y

q(x),

(28.1)

where

p(x)

and

q(x)

are given functions. The equation is called

linear

because if we let

L(y)

denote the left side of (28.1), then for any constant c and any two functions Yl and

we have:

L(CYl) = CL(Yl),

L(YI

Y2)

L(Yl)

+ L(Y2).

If q (x)

0 the equation (28.1) is called homogeneous, and the equation looks like this:

L(y)

= dx +

p(x)y

(28.3)

The homogeneous equation (28.3) is called the reduced form of equation (28.1). Notice that

is a solution

(28.1), so that L(yo) =

q(x),

and Yl is a solution

the reduced equation

(28.3), so that

L(Yl)

= 0, then

CYI

is again a solution

(28.1), since

Lty« +CYl) = L(yo) +CL(Yl) =

q(x)

+C •

The process of solving the linear equation (28.1) consists

finding one solution

and adding

to it all solutions

CYI

the reduced equation.

The following observation is the key to finding the solutions

(28.1) and (28.3). Let

P(x)

= f

p(x)

dx,

so that

P'(x)

p(x),

and check that

d dy

- [eP(X)y] =

eP(x)

- +

eP(x)

p(x)y

dx dx

eP(x)

[~~

p(x)y

This calculation shows that the left side

(28.1) becomes exactly the derivative

eP(x)

Y if

we multiply by

eP(x).

Multiplying both sides

(28.1) by

eP(x),

we get the equivalent equation

eP(x)

[~~

P(X)y]

=eP(x)q(x),

167

(28.5)

(28.6)

168 Understanding Calculus

which because of (28.4) can be written

- [eP(X)y] = eP(x)q(x).

The solutionsof (28.5),andhencethe solutionsof (28.1), since(28.1)and (28.5)areequivalent,

can now be obtainedby integratingboth sides of (28.5):

eP(x)y =feP(x)q(x)

+C,

y =e-P(x) feP(X)q(x)dx +c.>-:

q(x)

0, then (28.6) gives the general solution y = Ce-P(x) of the reduced equation. The

expression

= e-P(x) f

eP(x)q(x)dx

is the one particular solution of (28.1) that we need to go with the general solution ce-P(x) of

the reduced equation.

Rather than memorizingthe formula (28.6), it is much easier to multiply both sides of

the equation by

eP(x),and integrate.

EXAMPLE

28.1

Solve dy + 3x

y =

Solution

Here

p(x)

= 3x

, P

(x)

and we multiply both sides by e

x 3

3 dy 3 2

- +

. 3x y = 0,

y =

ce-

x 3

•

We could also have solved the equation by separating the variables, but the above method is simpler and

cleaner.

EXAMPLE

28.2

Solve dy +

,JX.

Solution

Here

p(x)

P(x)

logx,

and we multiply both sides by e

10gx

= x:

dy 3

+ y = x

d 3

-[xy]

2 5

Y=Sx

+ C,

2 3

Y =

+Cx-

Chapter28 • First-OrderLinear

Equations

EXAMPLE

28.3

Solve dy + y =x

•

Solution

multiply

bothsidesby

eX:

eXy

= x

_[eXy]

= x

eXy=

xdx+C.

Integrating by parts

twice,

or more

efficiently,

consulting an

integral

table,we get

eXy =x

- 2xe

+2e

+C,

y = x

2x +2 +Ce-

•

The general formulafor the particularsolution in

Example

28.3is

dx =x

2x +2.

For anyequation

169

(28.7)

+ay =

q(x),

with

p(x)

a constant, a, the particularsolution will be

yo=e-

feaxq(x)dx. (28.8)

Ifq(x) isapolynomial, then(28.8)willagainbeapolynomial ofthesame

degree.

Forinstance,

in Example 28.3, we had

q(x)

= x

and

= x

2x + 2. Rather than

fight

through the

integration in (28.8), wecanjust substitute a generalpolynomial in the equation and see what

the coefficients mustbe.

Second

Solution

Example

28.3.

Sinceq(x) = x

we try

y=A+Bx+Cx

-=B+2Cx.

Substituting in the equation, we get

(B +

2Cx)

+ (A +

+ Cx

)

(A + B) + (B +

2C)x

+ Cx

•

Hence,

we musthave

A+B=O

+2C

C = 1.

Thus, C = 1, B = - 2, A =2, and

=2 - 2x

is the same as the particular solution given by the integral (28.7).

170

Understanding Calculus

It is importantto emphasizehere that the substitution trick worksonly if the coefficient

p(x)

is a constant, so the equationlooks like this:

+ay

q(x).

However,

p(x)

is constant,the substitution techniqueworksnotjust for polynomials q

(x),

but also if

q(x)

= K

cosbx

q(x)

= K sinbx or

q(x)

= K

cosbx

+ K2sinbx. In this

case, try

y = A cosbx + B sin

bx.

You

must includeboth sine and cosine terms in your trial

function, evenif

q(x) hasonlyoneterm. Ifq (x) = Ke

, thentry y = Ae

providedb

-a.

If b =

-a,

is a solutionof the reducedequation,and the particularsolutionhas the form

y =Axe-ax (Problem28.14).

EXAMPLE

28.4

Solve - - 4y = 5 cos2x.

Solution

Weknowthe solutionsof the reducedequationare y = Ce

To finda particularsolution,we try

y = A cos2x + B sin 2x

dy =

-2A

sin2x + 2B cos2x.

Substitutingin the equationgives

(-2A

sin2x + 2B cos 2x) -

4(A

cos2x + B sin 2x) = 5cos 2x,

(2B - 4A)

cos2x +

(-2A

- 4B) sin2x = 5cos 2x.

Therefore, y will be a solutionif

-4A+2B

-2A

-4B

=0.

Multiplythe firstequationby two and add:

(-8A

- 2A) + (4B -

4B)

= 10,

-lOA

= 10,

=-1.

From the secondequationB =

-1

A, so B =

and the particularsolutionis

1 . 2

= - cos 2x +

sin x.

The general solutionis

-cos2x

sin2x +cr-,

EXAMPLE

28.5

dy _ y =4e

•

Solution

Wetry

y = Ae

, -

= 3Ae

•

Substitutionyields the condition

3Ae

= 4e

2A = 4, A = 2.

The general solutionis