henry edwards, david e penney elementary differential equations with boundary value problems 2003

FIRST-ORDER DIFFERENTIAL EQUATIONS SECTION 1.1 DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELING The main purpose of Section 1.1 is simply to introduce the basic notation and terminolog

FIRST-ORDER DIFFERENTIAL EQUATIONS

SECTION 1.1

DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELING

The main purpose of Section 1.1 is simply to introduce the basic notation and terminology of

differential equations, and to show the student what is meant by a solution of a differential

equation Also, the use of differential equations in the mathematical modeling of real-world

phenomena is outlined

Problems 1-12 are routine verifications by direct substitution of the suggested solutions into the

given differential equations We include here just some typical examples of such verifications

3 If y,=cos2x and y,=sin 2x, then y/=-—2sin2x and y, =2cos 2x so

yy = —4c0s2x = -4y, and yy = —4sin2x = —4y,

Thus yƒ+4y, = 0 and yz+4y, = 0

4 If v y,=e™ and y,=e™, then y,=3e and y,=-3e™ so 2 1 2

+

vị = 9e” = 9y, and y2 = 9e?” = 0y,

5 If y=e*-e™, then y'=e*+e™* so y'-y = (et +e")-(e*-e*) = 2e Thus

8 If y,=cosx—cos2x and y; =sinx-cos2x, then y/=—sinx+2sin2x,

y=—cosx+4cos2x, and y; =cosx+2sin2x, y; =—sinx+4cos2x Hence

Section 1.1

ylty, = (-sinx+4cos2x)+(sinx—cos2x) = 3cos2x

1L of y=y,=x7 then y'=-2x7 and y"=6x", so

xy +5xy+4y = x2(6x°)+5x(-2x”)+4(x?) = 0

If y=y;=x”lnx then y=x2-~2x *lnx and y=-5x”+6x “lnx, so

x?y +5xy +4y = x'(-5x*+6x*lnx)+5x(x?~2x'Inx)+4(xInx)

15 Substitution of y=e™ into y"+y’—2y = 0 gives the equation r’e™ +re™—2e" =0

that simplifies to r?+r—2 = (r+2)(r—1) = 0 Thus r=-2 or r= 1

16 Substitution of y=e™ into 3y’+3y'~4y = 0 gives the equation

arte +3re” —4e* =0 that simplifies to 3r7+3r—4 = 0 The quadratic formula then

gives the solutions r = (3+ 457)/6

The verifications of the suggested solutions in Problems 17-36 are similar to those in Problems 1-12 We illustrate the determination of the value of C only in some typical cases

1, C=2

19 If y(x) = Ce*-1 then y(0)=5 gives C-1 = 5, so C=6.,

20 If y(x) = Ce*+x-1 then y(0) = 10 gives C-1 = 10, so C= Il

21 C#7

Chapter 1

C=-56

24 C= 17

25 If y(x) = tan(x?+C) then y(0)=1 gives the equation tan C = 1 Hence one value

of C is C=2/4 (as is this value plus any integral multiple of 7)

26 Substitution of x= and y=0 into y = (x+C)cosx yields the equation

0 = (n+C)(-1), so C = -2

27 y =xty

28 The slope of the line through (x,y) and (x/2,0) is y’ = (y-O)Ax—x/2) = 2y/x,

so the differential equation is xy’ = 2y

29 If m=y’ is the slope of the tangent line and m’ is the slope of the normal line at (x, y),

then the relation mm’=—1 yields m’ = 1/y’ = (y-1)x—0) Solution for y’ then

gives the differential equation (I-y)y’ = x

30 Here m=y’ and m'=D,(x7+k) = 2x, so the orthogonality relation mm’ =—1 gives

the differential equation 2x y’ = ~1

31 The slope of the line through (x,y) and (~y,x) is y’ = (x-y)A~y—x), so the

differential equation is (x+y)y’ = y—x

In Problems 32-36 we get the desired differential equation when we replace the "time rate of

change” of the dependent variable with its derivative, the word "is" with the = sign, the phrase

“proportional to" with k, and finally translate the remainder of the given sentence into symbols

42 y(x) = cosx or y(x) = sinx

43 {a) y(10)=10 yields 10=14C—10), so C=101/10

(b) There is no such value of C, but the constant function y(x)=0 satisfies the

conditions y’=y? and y(0)=0

(c) It is obvious visually that one and only one solution curve passes through each

point (a,b) of the xy-plane, so it follows that there exists a unique solution to the initial

value problem y’=y’, y(a)=b

44, (b) Obviously the functions u(x) =— x* and v(x)=+.x* both satisfy the differential

equation xy’ = 4y But their derivatives u’(x)=—4x' and v'(x)=+4x° match at

x = 0, where both are zero Hence the given piecewise-defined function y(x) is differentiable, and therefore satisfies the differential equation because u(x) and v(x) do

so (for x<0 and x20, respectively)

(e) If a20 (for instance), chose C, so that C, a’ = b Then the function

(x) = Ca“ if x<0,

satisfies the given differential equation for every value of C

SECTION 1.2 INTEGRALS AS GENERAL AND PARTICULAR SOLUTIONS

Chapter 1

1; Integration of y’=2x+1 yields y(x) = JQx+Dae = x°+x+C Then substitution

of x=0, y=3 gives 3 = 0+0+C = C, so y(x) = x? 4x43

2 Integration of y=(x-2)” yields y(x) = [a-2? = ‡(x-2)`+CŒ Then

substitution of x=2, y=l gives l = Ủ+C = C, so y(x) = 1-2)

3 Integration of y’=Vx yields y(x) = jW& = 23°C Then substitution of

x=4, y=0 gives O=¥%+C, so y(x) = ‡@”?—8),

4 Integration of yv=x? yields y(x) = fra = —l/x+C Then substitution of

x=1, y=5 gives 5=-1+C, so y(x) = -1/x+6

5 Integration of y’=(x+2)"? yields y(x) = f+ 2)? dx = 2Vx+2+C Then

substitution of x=2, y=—1 gives -1=2-2+C, so y(x) = 2Vx+2-5

6 Integration of y’=x(x?+9)'? yields y(x) = xG2 +9)!" de = 10749)? +,

Then substitution of x=—4, y=O gives 0=1(5) +C, so y(x) = 4[ (a? +9)? - 125],

Is Integration of y’=10Ax"+1) yields y(x) = 104? +1) de = 10tan'x+C Then

substitution of x=0, y=0 gives 0=10-0+C, so y(x) = 10tan’ x

8, Integration of y’=cos2x yields y(x) = foos2xdx = tsin2x+C Then substitution

of x=0, y=1 gives 1=O+C, so y(x) = ÿsin2x+l

9 Integration of y’=1/Vl—x? yields y(x) = fu l-x? dx = sin’ x+C Then

substitution of x=0, y=0 gives O=0+C, so y(x) = sin™ x

1Q ‘Integrationof y=xe” yields

y(x) = fuera = fuet du = (u~-De" = —(x+)De*+C

If a(t) = 2t+1 then v(t) = [@i+Ða =Pttty =P+t-7 Hence

xt) = f@+i-Dar= TẾ +‡!†T TH+ờa = {TP +31—TI+A

W a0) = L/A¿+4 then vữ) = [uw+4 dt = Wr+4+C = 2144-5 (taking

C=-5 so that v(0)=-1) Hence

xũ) = [(F+4-54r= 30+4)")~5r+C = 444)? -51-2 (taking C=—29/3 sothat x(0)=1)

If a(@) = @4+1)" then v(t) = Jư+p? đi = -‡+)”+C = -‡ữứ+J ”+‡ (taking

C=‡ sothat v(0) =0) Hence

a(t) = f[-t@+D7 +4 ]dr= Fey tye = Ha+n t+]

Chapter 1

v = —9.8/+ 49, so the ball reaches s maximum height (v= 0) after /= 5 seconds Is

maximum height then is y(5) = 4.9657 + 49(5) = 122.5 meters

= -32? and y = -167 +400, so the ball hits the ground (y = 0) when

t = Ssec, andthen vy = —32(5) =—160 ft/sec

a = -10 m/s and vo = 100 km/h = 27.78 m/s, so v = —10f+ 27.78, and hence x(t) = —5¢ + 27.781 The car stops when v = 0, t ~ 2.78, and thus the distance

traveled before stopping is x(2.78) = 38.59 meters

v = -9.81+ 100 and y = 4.97? + 100¢4 20

(a) v = 0 when ¢= 100/9.8 so the projectile’s maximum height is

y(100/9.8) = —4.9(100/9.8)” + 100(100/9.8) + 20 = 530 meters

(b) It passes the top of the building when y(t) = -4.927 + 100r+ 20 = 20,

and hence after t= 100/4.9 = 20.41 seconds

{c) The roots of the quadratic equation y(t) = -4.97° + 100r+ 20 = 0 are

t = -0.20, 20.61 Hence the projectile is in the air 20.61 seconds

v = -32t-40 and y = -162- 40/ + 555 The ball hits the ground (y = 0)

when ¢ = 4.77 sec, with velocity v = v(4.77) ~ -192.64 fUsec, an impact speed of about 131 mph

Section 1.2

x = ~at’/2 + 88,

we solve for a = 22 fi/sec” Hence the car skids for 1 = 88/22 = 4 sec

If a = —20m/sec? and xp = O then the car's velocity and position at time ¢ are given

and find that x = 60m whenv = 0 Thus doubling the initial velocity quadruples the

distance the car skids

If vo = O and yo = 20 then

y= -at and y = ~‡aŸ +20

Substitution of t = 2, y = 0 yields a = 10 ft/sec’ If vo = 0 and

yo = 200 then

-10r and y = -57 +200

= 1

Chapter 1

Hence y = 0 when £ = x40 = 210 sec and v = -20V10 = -63.25 ft/sec

OnEarth: vy = —32f+ vo, so ¢ = vo/32 at maximum height (when v = 0)

Substituting this value of ? and y = 144 in

Therefore v = 0 yields ¢ = 9.6 sec, andthence Ymax = y(9.6) = 460.8 ft is the

height a ball will reach if its initial velocity is 96 ft/sec

If vo = O and yo = A then the stone’s velocity and height are given by

veg, ysO5 gf +h

Hence y = 0 when t = J2h/g so

-gf2h/g = —J2gh

The method of solution is precisely the same as that in Problem 30 We find first that, on

Earth, the woman must jump straight upward with initial velocity vo = 12 ft/sec to

reach a maximum height of 2.25 ft Then we find that, on the Moon, this initial velocity

yields a maximum height of about 13.58 ft

v

We use units of miles and hours If xo = vo = 0 then the car‘s velocity and position

after t hours are given by Tạ 2

35 = (0.5)(60/002) = 301,

whence ¢ = 7/6hr, that is, 1:10 p.m

35, Integration of y’ = (9/vs)(1 —4x°) yields

y = Givs(3x-4x°)+ C,

and the initial condition y(-1/2) = 0 gives C = 3/vs Hence the swimmer’s trajectory

1s

yœ) = (3/)(3x— 4x + 1)

Substitution of y(1⁄2) = I nowgives v; = 6 mph

36 — Integration of y’ = 3(1 — 16x‘) yields

As pointed out in the textbook, the instructor may choose to delay covering Section 1.3 until later

in Chapter 1 However, before proceeding to Chapter 2, it is important that students come to

grips at some point with the question of the existence of a unique solution of a differential equation — and realize that it makes no sense to look for the solution without knowing in advance that it exists The instructor may prefer to combine existence and uniqueness by

simplifying the statement of the existence-uniqueness theorem as follows:

Suppose that the function f(x,y) and the partial derivative of / dy are both

continuous in some neighborhood of the point (a, b) Then the initial value

problem

đ

i = f(xy), ya) = b

slope fields in Problems 1-10 are shown in the answers section of the textbook and will not be

Each isocline x- 1 = C isa vertical straight line

Each isocline x+ y = C isa straight line with slope m = ~1

Each isocline y = C20, thatis, y = Vc or y= -wC, is a horizontal straight

line

Each isocline aly = C, thatis, y = CỔ, isa horizontal straight line

Each isocline yx = C, or y = Cx, isa straight line through the origin

Each isocline x*- y’ = C isa hyperbola that opens along the x-axis if C > 0, along the

Each isocline is an exponential graph of the form y = Ce’*,

Because both f(x,y) = 2x’y’ and of /dy = 4x’y are continuous everywhere, the

existence-uniqueness theorem of Section 1.3 in the textbook guarantees the existence of a

unique solution in some neighborhood of x = 1

Both f(x,y) = xIny and of /dy = x/y are continuous in a neighborhood of

(1, 1), so the theorem guarantees the existence of a unique solution in some

neighborhood of x = 1

Both f(x,y) = y'® and af/dy = (1/3)y~? are continuous near (0, 1), so the

theorem guarantees the existence of a unique solution in some neighborhood of x = 0

ƒŒ,y) = y!? is continuous in a neighborhood of (0, 0), bụt 3ƒ/2y = (1/3)y ”” is

not, sO the theorem guarantees existence but not uniqueness in some neighborhood of x= 0

ƒŒ,y) = Œ— yy? is not continuous at (2,2) because it is not even defined if y>x

Hence the theorem guarantees neither existence nor uniqueness in any neighborhood of the point x = 2

ƒ#Œœ,y) = (x y)'? and af /dy = -(1/2)@- yy are continuous in a neighborhood

of (2, 1), so the theorem guarantees both existence and uniqueness of a solution in some neighborhood of x = 2

Both ƒ(x,y) = z~ l⁄ and df /dy = 4x- 1)/y? are continuous near (0, 1), so the

theorem guarantees both existence and uniqueness of a solution in some neighborhood of

If fa.y) =-(i- x2 then of /dy = yA — y’y is not continuous when y = 1,

so the theorem does not guarantee uniqueness

The two solutions are yi(x) = O (constant) and y2(x) = x

The isoclines of y' = y/x are the straight lines y = Cx through the origin, and y’ = C atpointsof y = Cx, so it appears that these same straight lines are the solution curves of xy'= y Then we observe that there is

(i) a unique one of these lines through any point not on the y-axis;

(it) no such line through any point on the y-axis other than the origin; and (iii) infinitely many such lines through the origin

ƒ(x.y) = Axy? and af /dy = 2x7!” are continuous if y>0, so forall a and all

b>O there exists a unique solution near x = a such that y(a) = ở If 6 = 0 then the theorem guarantees neither existence nor uniqueness For any a, both yi(x) = 0 and yo(x) = œ —đ?? are solutions with y(a) = 0 Thus we have existence but not

uniqueness near points on the x-axis

Chapter 1

Of course it should be emphasized to students that the possibility of separating the variables is

the first one you look for The general concept of natural growth and decay is important for all

differential equations students, but the particular applications in this section are optional

Torricelli's law in the form of Equation (24) in the text leads to some nice concrete examples and

Wi _ fa, 4im@y-) = x+4inc; 2y-1 = Ce® 2y-1

yQ)=1 implies C=e? so y(x) = ‡(+e”*?)

CO IS lny = InGinx)+lnŒ, y(+) = Csinx

| = [60 dx, e = 3e*+C; y(x) =In(3e*+C)

y(0)=0 implies C=-2 so y(x) =In(3e*-2)

Jeese = In tany = V¥x+C; y(x) =tan? (vz+c)

x

The population growth rate is k = InG0000/25000)/10

of the city ¢ years after 1960 is given by P(f) = 25000992, The cxpected year 2000

population is then P(40) = 25000e°992% = 51840

The population prowth rate is k = In(6)/10 = 0.17918, so the population after t

hours is given by P(t) = Pe" To find how Jong it takes for the population to double, we therefore need only solve the equation 2P = % ects for

ft = (In2)/0.17918 = 3.87 hours

As in the textbook discussion of radioactive decay, the number of “C atoms after 7

years is given by N(t) = No 2 90001216! Hence we need only solve the equation

IN, = No eP000I216 For ¢ = (In6)/0.0001216 = 14735 years to find the age of the skull

As in Problem 31, the number of “C atoms after ¢ years is given by

N(t) = 5.0x10" 00012167 Hence we need only solve the equation

4.6x10° = 5.0x1019999?9! for the age f= (in (5.0/4.6))/0.0001216 = 686 years

of the relic Thus it appears not to be a genuine relic of the time of Christ 2000 years ago

The amount in the account after f years is given by A(t) = 5000e°™ Hence the

amount in the account after 18 years is given by A(20) = 5000/92 ~ 21,103.48

To find the decay rate of this drug in the dog's blood stream, we solve the equation

1 = e™ (half-life 5 hours) for k = (In2)/5 = 0.13863 Thus the amount in the dog's bloodstream after ¢ hours is given by AQ) = Aye 99%, WWe therefore solve the

equation AQ) = Aye = 50x45 = 2250 for A, = 2585 mg, the amount to

anesthetize the dog properly

To find the decay rate of radioactive cobalt, we solve the equation + = e°* (half-life

§.27 years) for k =(In2)/5.27 = 0.13153 Thus the amount of radioactive cobalt left after ¢ years is given by A(t) = Aero We therefore solve the equation

Chapter 1

Taking ¢ = 0 when the body was formed and ¢ = T now, the amount Q(t) of ”®U in

the body at time ¢ (in years) is given by Q(t) = Que™, where k = (In 2)/(4.51x10°)

The given information telis us that

_6Œ) _ 99

Q,~ Q(T)

After substituting Q(7) = One”, we solve readily for ef = 19/9, so

T = (1/)ln(19/9) = 4.86x10” Thus the body was formed approximately 4.86 billion

years ago

Taking ¢ = 0 when the rock contained only potassium and t = T now, the amount

Q(1) of potassium in the rock at time / (in years) is given by Q(t) = Quế”, where

k = (In2)/(1.28x10”?) The given information tells us that the amount A(f) of argon at

Thus the age of the rock is about 1.25 billion years

Because A = 0 the differential equation reduces to T' = kT, so T(t) = 25e“ The

fact that 7(20) = 15 yields k = (1/20)In(5/3), and finally we solve

5=25£/” for t= (n5)/k = 63 min

The amount of sugar remaining undissolved after ¢ minutes is given by A(t) = Ae;

we find the value of & by solving the equation A(1) = Aye = 0.75A, for

& =—1n0.75 = 0.28768 To find how long it takes for half the sugar to dissolve, we solve

the equation A(t) = Ae“ =‡Áy for f=(In2)/0.28768 ~2.41 minutes

(a) The light intensity at a depth of x meters is given by I(x)=J,e'" We solve

the equation I(x)=f,e'** =41, for x=(In2)/1.4 = 0.495 meters

{b) To find the altitude where p= 15 in., we solve the equation 29.92e°” =15 for

x = (In 29.92/15)/0.2 = 3.452 miles = 18,200 ft

(a) A’ =rA+Q {b) The solution of the differential equation with A(O) = 0 is given by

rA+Q = Qe"

When we substitute A = 40 (thousand), r = 0.11, and ¢ = 18, we find that

Q = 0.70482, that is, $704.82 per year

Let W,Œ) and N,(t) be the numbers of 23817 and *U atoms, respectively, al time ¢ (in 8 P y

—kt billions of years after the creation of the universe) Then N,(f)=N,e™' and

N,(t)=N,e, where N, is the initial number of atoms of each isotope Also,

k =(1n2)/4.51 and c=(In2)/0.71 from the given half-lives We divide the equations

for N, and N, and find that when rf has the value corresponding to “now”,

(b) By separating the variables we solve the differential equation for

c-rP() = (e-r Po) e"

With P(t) = 0 this yields

Chapter 1

With Po = 10,800, t = 60, and r = 0.010 we get $239.37 for the monthly payment

at 12% annual interest With r = 0.015 we get $272.99 for the monthly payment at 18% annual interest

f Nữ) denotes the number of people (in thousands) who have heard the rumor after f

days, then the initial value problem is

N’ = k(100-N), N(O) = 0 and we are given that N(7) = 10 When we separate variables (dN /(00-N) =k dt) and integrate, we get In(l00O-N)= —kt+C, and the initial condition N(0)=0 gives C=In 100 Then 100-N =100e™, so N(t)=100(1-e™) We substitute r= 7,

N= 10 and solve for the value k =In(100/90)/7 = 0.01505 Finally, 50 thousand

people have heard the rumor after ¢ = (In 2)/k = 46.05 days

With A(y) constant, Equation (19) in the text takes the form

dy Sek 7 AY

We readily solve this equation for 2 = kt+C Thecondition y(0) = 9 yields

C = 6, andthen y(1) = 4 yields k = 2 Thus the depth at time 7 (in hours) is

yt) = 3 ` and hence it takes 3 hours for the tank to empty

With A = (3)? and a = #(1/12)”, and taking g = 32 ft/sec”, Equation (20) reduces

to 162y = - The solution such that y = 9 when ¢ = 0 is given by 324jy = -t+972 Hence y = 0 when ¿ = 972sec = 16min 12 sec

The radius of the cross-section of the cone at height y is proportional to y, so A(y) is proportional to y Therefore Equation (20) takes the form

52

53

20

2ÿ = -kt+C

The initial condition (0) = (the height of the cylinder) yields C= 2Vh Then

substitution of t= 7, y=0 gives k= @xh 3⁄7 It follows that

y=hq—17)

If r denotes the radius of the cylinder, then

Vy) = ary = arhQ-t/T) = Yd-t/TY

Since x = ys, the cross-sectional areais A(y) = 7x? = y*?, Hence the general equation A(y)y’ = —ay2gy reduces to the differential equation yy’ =—k with general solution

(2y? = t+ C

The initial condition y(0) = 12 gives C = 72, andthen y(1) = 6 yields k = 54

Upon separating variables and integrating, we find that the the depth at time 7 is

y(t) = xJI44-108: y0)

Hence the tank is empty after ¢ = 144/108 hr, that is, at 1:20 p.m

(a) Since x° = by, the cross-sectional areais A(y) = ax’ = xby Hence the equation A(y)y’ = -aJ2gy reduces to the differential equation

y?y = —k = -(3/#b)jJ2g with the general solution

(2/4)y?2 = -kt + C

The initial condition y(0) = 4 gives C = 16/3, andthen y(1) = 1 yields k = 14/3

It follows that the depth at time 7 is

yf) = (8-79

(b) The tank is empty after ¢ = 8/7 hr, that is, at 1:08:34 p.m

Chapter 1

o& at 18 °

If z denotes the distance from the center of the cylinder down to the fluid surface, then

y = 3-z and A(Qy) = 109-27)", Hence the equation above becomes

dz 7 109-2)? = 2(3_ Hs @-z) a 18t 2)

180(3+z)'?dz = dt,

and integration yields

120342)? = mt+C

Now z= 0 when rt = 0, so C = 1200132 The tank is empty when z = 3 (that is,

when y = 0) and thus after

+ = (120/8(62-3?2) = 362.90 sec

It therefore takes about 6 min 3 sec for the fluid to drain completely

A(y) = x(§y—y”) asin Example 7 in the text, butnow @ = 2/144 in Equation (24),

so the initial value problem is

188y- yy ==Jy, yO) = 8

We seck the value of £ when y = 0 The answer is t= 869 sec = 14 min 29 sec

The cross-sectional area function for the tank is A = #(1—y”) and the area of the

bottom-hole is @ = 10“z, so Eq (24) in the text gives the initial value problem

zu-y) = -10°z/2x9ãy, y(0) = I

The initial condition y(0)=1 implies that C = 2 - 2/5 = 8/5, so y=0 after

p= (8/5)/(1.4x10% 10) = 3614 seconds Thus the tank is empty at about 14 seconds after 2 pm

57 (a) As in Example 8, the initial value problem is

2%

z@y~y)T = =#lxy, 3(0)=4

where & = 0.6r2./2g = 4.8r” Integrating and applying the initial condition just in the

Example 8 solution in the text, we find that

(b) The radius of the bottom-hole is

r= Jk/48 ~ 0.04442 ft = 0.53 in, thus about a half inch

58 The given rate of fall of the water level is dy/dt = -4 in/hr = -(1/10800) ft/sec With

A = me? and a = m7’, Equation (24) is

(7e?)(1/10800) = —(ar’)/2ey = —8ar?.fy

Hence the curve is of the form y = ket, and in order that it pass through (1,4) we

must have k = 4 Comparing Jy = 2x’ with the equation above, we see that

(820800) = 1⁄2,

so the radius of the bottom hole is z = 1/(24043) ft = 1/35in

59 Letr = 0 at the time of death Then the solution of the initial value problem

so the death occurred at 10:29 a.m

Let ¢ = 0 when it began to snow, and t = t at 7:00 a.m Let x denote distance along

the road, with x = 0 where the snowplow begins at 7:00 a.m If y = ct is the snow

depth at time #, w is the width of the road, and v = dx/dt is the plow’s velocity, then

“plowing at a constant rate" means that the product wyv is constant Hence our

differential equation is of the form

The solution with x = 0 when ¢ = fo is

at 8 a.m and 9 a.m., respectively Elimination of f gives the equation

2e#—e*-1 = 0,

which we solve numerically for k = 0.08276 Using this value, we finally solve one of

the preceding pair of equations for f = 2.5483 hr ~ 2 hr 33 min Thus it began to snow at 4:27 a.m

SECTION 1.5 LINEAR FIRST-ORDER EQUATIONS

1 ø=exp([I+)=e": D, (y-e*)=2e%; yret=2e™+C, y(x) = 24+Ce™

y(0)=0 implies C=-2 so y(x) = 2-2£”

+ p=exp([C24:)=e”: Dye )=3, y:e”=3x+Ci yx) = (3x+C)e™

y(0)=0 implies C=0 so y(x) = 3xe™

3 p=exp([34x)=e" D, (y:)=2> yết =x?+C: sey & G2+@)ˆ*

+ p=exp([(-2x)dx)=e"s D,(y-e* )=1; yer =xtC; yx) = (xt Ce"

5, paexp([Q/adx)=e™ =x"; D(yx')=3x, yates ec

yœ) = x+C/z”; y()=5 implies C=4 so yx) = vedi?

6 p=exp([G/dr)=e™ =x"; Đ.(y+)=72°: yx sx’ +

yx) = 22 +C/ x; y(2)=5 implies C=32 so y(x) = x7 +32/x°

7 Pp =exp(J(1/2x) de) =e" =vx; D,(y-vx)=5; y:x=5x+C

p=exp([(Ux)ar)=e Six, Đ,(y-/x)=lx, y-1/x=lnz+C

yx) = xinx+ŒCx; y()=7 Implies C=7 so y(x) = xInx+7x

p =exp([(-3/2x)dr) =? =x?) D, (y-x x)= 9x72 y.x3?2=3xv2+C

32

y(œ) = 3x +Œry

p =exp{f(/x~-3)dx)=e"* =xe™: D, (y.xe”)=0; y.xe =€

yœ) = Cx 2z”; y()=0 implies C=O so y(x) = 0 (constant)

3

yx) = 42° +Cx"; y(2)=1 implies C=56 so y(x) = 1x°4+56x7

p=exp([1dx)=e"; D,{y-e") =e"; ye ate +E

x mx

yx) = ‡e°+C£”; y(0)=[ implies C=4L so y(x) = te t+te

p=exp([(-3/x)de)=e™ = x7, D(y x? )ax"; yx =Inx+C

ya) = xinx+Cr; yA) =10 implies C=10 so y(x) = x*Inx+10x°

p=exp([2xdx) =e"; D,(y-e" J=xe"; viet ate +€

yx) =44Ce™; —y(0)=-2 implies C=~$ so y(x) = 1-Se°

p=exp (feos xdx)=e at DD, (y : en) =e cosx, ye =e 4C

p =exp(Í cot xáx)= chen9 =sinx; D,(y-sinx)=sin xcosx

y-sinxz=‡sin°x+C; y(z) = †sinx+Ccscx

p=exp([(~!~x)&)=e””? D, (yer? aden?

yer sa”? LỚ yx) = -14¢08"""

y(0)=0 implies C=1 so y(x) = -i¢e*"?

p=exp([(-3/x) de) =e = x7; D,{y-x") =cos x; yex? ssinx+C

y(x) = xÌsinx+C+x”; y(2#)=0 implies C=O so y(x) = x? sinx

yx) = (x+€)e”: y(0=5 implies C=5 so y(x) = (x°+5)e”

p =exp(Í(2—3/+) dr) sere ye) D, (y-xe"* ) =4e"

ye? $1)” exp3x?/2) = - 200? +4) 4¢, ya) = —2exp(3x?/2)+C (x? +1) exp(-3x? /2)

Finally, y(0)= 1 implies that C =3 so the desired particular solution is

y(+) = —2exp(3x7/2)+3(x? +1)? exp(—3x? /2)

26 With x’=dx/dy, the differential equation is y'x’+4y’x=1 Then with y asthe

independent variable we calculate

pty) = exp(f(4/y)dy) = ef” = yD, (x-y') = y

xY 3 +CŒ; x(y) = ats ayy

H Ne II

27 With x’=dx/dy, the differential equation is _x’—x= ye’ Then with y as the

independent variable we calculate

ply) = exp(fiDdy) =D, (xe) = y

we? = ty tC, xy) = ($y? +C)e

28 With x’=dx/dy, the differential equation is (1+ y?)x’-2yx=1 Then with y as the

independent variable we calculate

p(y) = exp(f(-2y A+ yay) = eM = de yy

D,(x-(+y*y") = + y?)?

An integral table (or trigonometric substitution) now yields

x(y) = 1[y+(I+3°)(en” y+€)]

The initial condition y(1) = 0 implies that C = 0, so the desired particular solution is

yx) = x? Ỉ £?2 cos? ái

@) yf = cel) =-Py,, so y+Py, = 0

(a) If y=Acosx+Bsinx then

y'+y = (A+B)cosx+(B-A)sinx = 2sinx

provided that A=~1 and B=1 These coefficient values give the particular solution

yx) = sin x ~ cos x

(b) The general solution of the equation y'+y=0 is y(x) = Ce™ so addition to the particular solution found in part (a) gives y(x) = Ce™* + sin x — cos x

(c) The initial condition y(0)= 1 implies that C= 2, so the desired particular

solution is y@) = 2e* + sin x—cos x

The amount xứ) of salt (in kg) after 7 seconds satisfies the differential equation

x=—x/200, so xứ) = 10072, Hence we need only solve the equation

10 = 100e”™ for t =461 sec =7 min 41 sec (approximately)

Let x(t) denote the amount of pollutants in the Jake after ¢ days, measured in millions of cubic feet Then x(t) satisfies the linear differential equation dx/d¡=1/4—x/16 with

solution xŒ)=4+l6e””® satisfying x(0)=20 The value of ? such that x=8 is

Chapter 1

36

37

38

The only difference from the Example 4 solution in the textbook is that V = 1640 km?

and r=410 kn/yr for Lake Ontario, so the time required is

such that x(0) = 50 The tank is full after t = 150 min, at which time x(150) = 393.75 Ib

KNOWN: Steady-state, one-dimensional heat conduction through an axisymmetric shape

FIND: Sketch temperature distribution and explain shape of curve

SCHEMATIC:

_ HT, Sipe of curve

—

ASSUMPTIONS: (1) Steady-state, one-dimensional conduiction, (2) Constant properties, (3) No internal heat generation

ANALYSIS: Performing an energy balance on the object according to Eq 1.lla,

Ejx — Bou = 0, it follows that

Big = Bou = a

and that q, # qu(x) That is, the heat rate within the object is everywhere constant,

From Fourier's law,

appears as shown above Note the gradient decreases with increasing x

COMMENTS: (1) Be sure to recognize that dT/dx is the slope of the temperature distribution (2) What would the distribution be when T, >? (3) Show on the above plot how the heat flux, qy, varies with distance

FIND: Sketch temperature distribution and give brief explanation to justify shape

ANALYSIS: Fourier's law, Eq 2.1, for this one-dimensional (cylindrical) radial system has the form

where A, = 2zr£ and £ is the axial length of the pipe-insulation system Recognize that

for steady-state conditions with no internal heat generation, an energy balance on the

system requires Ej, = Egy: since Ey = Ey =0 and hence

COMMENTS: (1) Note that while q is a constant and independent of r, q; ïs not a constant How does q;(r) vary with r? (2) Recognize that the radial temperature gradient, dT/dr, decreases with increasing radius

KNOWN: A spherical shell with prescribed geometry and surface temperatures

FIND: Sketch temperature distribution and explain shape of the curve

where Ay is the surface area of a sphere given as A, = 471? For steady-state conditions,

an energy balance on the system requires that since Éy = Ey, = 0, Big =2Egy, and thus

Eq 2.1, for this one-dimensional, radial (spherical

COMMENTS: Note that for the above conditions, q #a,(r); that is, q is everywhere constant But how does q; vary asa function of radius?

distribution and heat rate

FIND: Expression for the thermal conductivity, k

(2) Recognize that the 1-D assumption is an approximation which is more appropriate

as the area change with distance x is less

KNOWN: End-face temperatures and temperature dependence of k for a truncated cone

FIND: Variation with axial distance along the cone of q„, qx, k, and dT /dx

ANALYSIS: For the prescribed conditions, it follows from conservation of energy, Eq 1.11a, that for a differential control volume, Ê¡y = Esu or qy = qx+ay Henee

x Is independent of x

Since A(x) increases with increasing x, it follows that qi =qx/A(x) decreases with

increasing x Since T decreases with increasing x, k increases with increasing x Hence, from Fourier's law, Eq 2.2,

aT

wake

it follows that | dT/dx | decreases with increasing x.

transfer through a plane wall

FIND: Effect of k(T) on temperature distribution, T(x)

ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3)

No internal heat generation

ANALYSIS: From Fourier’s law and the form of k(T),

ˆ aT aT

The shape of the temperature distribution may be inferred from knowledge of

a? /dx* = d(dT /dx)/dx Since q; is independent of x for the prescribed conditions,

KNOWN: Thermal conductivity and thickness of a one-dimensional system with no internal heat generation and steady-state conditions,

FIND: Unknown surface temperatures, temperature gradient or heat flux

Using Eqs (1) and (2), the unknown quantities can be determined

thickness,

FIND: Unknowns for various temperature conditions and sketch distribution,

v gf, Temperature gradient ke80 0c = :

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) No internal heat generation, (4) Constant properties,

ANALY#0R: ‘Tha ets oqstion and temperate rodent fort jee are

KNOWN: Plane wall with prescribed thermal conductivity, thickness, and surface temperatures

FIND: Heat flux, qj, and temperature gradient, dT/dx, for the three different

coordinate systems shown

ASSUMPTIONS: (1) One-dimensional heat flow, (2) Steady-state conditions, (3) No

internal generation, (4) Constant properties

ANALYSIS: The rate equation for conduetion heat transfer is

From a surface energy balance at r=

Arar, = đáy = h(2ar¿L) [T(rs) — Tx| ,

and solving for T, using the heat rate at r =r, find

KNOWN: Two-dimensional body with specified thermal conductivity and two

‘isothermal surfaces of prescribed temperatures; one surface, A, has a prescribed temperature gradient,

FIND: Temperature gradients, 9T/x and AT/éy, at the surface B

‘vector must be normal to an isothermal surface, The heat rate at the surface A is given

by Fourier's law written as

On the surface B, it follows that

(E/Eh Tung “TGẤY/miKxiim T60 Km: a

COMMENT: Note that in using the conservation requirement,

Gout = +48 aya and