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

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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 y = y+2e™, 6 If y,=e™ and y,=xe™, then yi=-2e™, yra4e™, yi =e™-2xe*, and yị=—4c”?+4xe”?", Hence M+Ayt4y, = (4e7)+4(-2e™)+4(e*) = 0 and ya+4y;+4y; = (T—4221+4xe?)+A(e?r~2xe?')+4(xe?) =0 8 If y,=cosx—cos2x and y; =sinx-cos2x, then y/=—sinx+2sin2x,

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yt y, = (-cosx+4cos2x)+(cosx—cos2x) = 3cos 2x and

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) 0 = (_5x2+5x?)+(6x2—10x?+4x”)Inx = 13 Substtuionof y=e” into 3y=2y gives theequation 3re” = 2£” that simplifies to 3r=2 Thus r=23 14 Substitution of y=e" into 4y"=y gives the equation 4r’e" = e™ that simplifies to 4r=1 Thus r=+1/2

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

18 C=3

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

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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

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39 yx) = x? 40, yx) = 1 or yx) = -1 41 y(z) = £ 1/2

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)

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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

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11 12 13 14 15 16 17 If a(t) = 50 then v(t) = [304i = 50+vạ = 50/+10 Hence x(t) = J0r+10)dt= 2517 +10t+x) = 2517 +102+10 If a(t) = —20 then v(t) = [(-20)de = —20/+wy = —20/—15 Hence x(t) = 201-15) at= —10/?—15f+x¿ = ~100”~15f+5 If a(t) = 3¢ then v(t) = [ra = i+y = 4°+5 Hence x(t} = [&?+5)4:= ‡È+51+x, = ‡+5t, If a(t) = 2¢+1 then v(t) = j@i+ba = P4tty = 41-7 Hence x(t) = [d?+t-= 1 +‡fT7i+xy = 3 +‡11— 7+4

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

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19 20 21 22 23 24 C=0 sothat v(0)=—-10) Hence xứ) = [(—l0cos5)4t= ~2sin5r+€C = ~2sin5r+10 (taking C=—10 so that x(0)=8)

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 a = -9.8 m/s2 so v = ~9.81-10 and y = —4.92— 10+ yạ The ball hits the ground when y = 0 and v= -9.81-10 = -60, so t=5.10s Hence yo = 4.9(5.10)? + 10(5.10) = 178.57 m

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

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27 28 29 26 is traveling at 70 ft/sec Taking xo = 0 and vo = 60 mph = 88 ft/sec, we get y= —at+ 88, 176 in and v = O yields ¢ = 88/a Substituting this value of t and x 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 a = 20+, x = -10P + vợ It stops when v = 0 (so vo = 201), and hence when x= 75 = 10° +(200 = 107 Thus ¢ = V7.5 sec so vo = 20V7.5 = 54.77 m/sec = 197 km/hr

Starting with xo = 0 and vo = 50 km/h = 5x10* m/h, we find by the method of Problem 24 that the car's deceleration is a = (25/3)x10" nvh* Then, starting with x9 =

0 and vo = 100km/h = 10° m/h, we substitute t = vo/a into

x = -at’ + vot

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

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30 31 32 33 34

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 y = -l6f + vọt, we solve for vọ = 96 ft/sec as the initial speed with which the person can throw a ball straight upward On Planet Gzyx: From Problem 27, the surface gravitational acceleration on planet Gzyx is a = 10 fUsec”, so y= -10t+96 and y = -St +960

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

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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 y = 3x- (48/5) + C, and the initial condition y(—1/2) = 0 gives C = 6/5 Hence the swimmer’s trajectory 1s y(x) = (1/5)(15x — 48x? + 6), so his downstream drift is y(1/2) = 2.4 miles SECTION 1.3

SLOPE FIELDS AND SOLUTION CURVES

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

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slope fields in Problems 1-10 are shown in the answers section of the textbook and will not be duplicated here 11 12 13 14 15 16 17 18 19 20 21 22 23

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 y-axis if C<0 Each isocline xy = C is a rectangular hyperbola that opens along the line y = x if C>0, along y = -x if C<0 Each isocline x— y’ = C, or y* = x~C, isa translated parabola that opens along the x-axis Each isocline y~x? = C, or x” = y—C, is a translated parabola that opens along the yraxis

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

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24 25 26 27 28 29 30 31 32 35 36 12

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

x= 0

Neither f(x,y) = @—i)/y nor of /dy = -(Œ~ 1)/y? is continuous near (1, 0), so the existence-uniqueness theorem guarantees nothing

Both f(x,y) = In(1 + y*) and dof /dy = 2y/(1+ y*) are continuous near (0, 0), so the theorem guarantees the existence of a unique solution near x = 0

Both ƒ(x,y) = x?- y and af /dy = ~2y are continuous near (0, 1), so the theorem guarantees both existence and uniqueness of a solution in some neighborhood of x = 0 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

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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

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21, 22 23 24 25 26 27, 2 r" _ xdx 2_ 2 2ydy = Tere: y? =Ax?~l6+€Œ y(5)=2 implies C=1 so y? = 1+vx°-16 ˆ ® li Iny =xÍ-x+lnC; y() = Cexp@“—z) ỳ

y)=—3 implies C=-3 so y(x) = —3exp(x*-x)

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 Jy sinx y(4)=% implies C=% so y(x) = Ssinx - wy JE+>} Iny = Inx+x?+lnC; y() = Cxexp(x’) y x yŒ)=1 implies C=e7 so y(x) = xexp(x?-1) dy 1 c3 = = | (2x+3x7); y [c ) -— = x? 4°4+C; y xZ)=—— 142 ++ yQ=—-1 implies C=~1 so y(x) = 1-2? -x?

| = [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

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29 30 31 32 33 34 35 36 16

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 dollars When the book has been overdue for ¢ years, the fine owed is given in dollars by A(t) = 0.30 e°° Hence the amount owed after 100 years is given by AQ00) = 0.300%" = 44.52 dollars

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

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37 38 39 40 41

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 time ft is AW) = $1Q-O0)] and also that A(T) = Q(7) Thus Q,~ Q(T) = 9 Q(T) After substituting Q(T) = Q,e" we readily solve for T = (Inl0/In2)(1.28x10") = 4.25x10°

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

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42 43 44 45 46 18

(a) The pressure at an altitude of x miles is given by p(x)= 29,.92e°** Hence the pressure at altitude 10000 ft is p(10000/5280) = 20.49 inches, and the pressure at altitude 30000 ft is p(30000/5280) = 9.60 inches

{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”,

gow = Ne 2 4377, N,

5

Finally we solve this last equation for ? = (In137.7)/(c~k) ~ 5.99 Thus we get an estimate of about 6 billion years for the age of the universe

The cake's temperature will be 100° after 66 min 40 sec; this problem is just like Example 6 in the text

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

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

With P(t) = 0 this yields

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47 48 49 50 c= rPoe'l(e- 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

Trang 20

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

Trang 21

54 35 56, With g = 32 ft/sec? and a = Z(1/12)”, Equation (24) simplifies to dy nt Ay) = -= fy 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

Trang 22

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 16 oa _ 2 ssn = —kt+— 3 5 15° When we substitute y = 2 (ft) and ¢ = 1800 (sec, that is, 30 min), we find that k = 0.009469 Finally, y = 0 when _ 448 = = 3154 sec = 53 min 34 sec 15k t

Thus the tank is empty at 1:53:34 pm (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

Trang 23

60 If ¢ = a@ at 12 noon, then we know that T(t) = 70+286e™ = 80, T(a+1) = 70+286e"% = 75, Hence 28.6e™ =10 and 2866 “2t = 5, It follows that e* = 1/2, so k = In2 Finally the first of the previous two equations yields a = (In 2.86)/n 2) = 1.516hr = 1 br 31 min,

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

t= ne™

We are given that x = 2 when ¢ = fg9+1 and x = 4 when 7 = to +3, so it follows

that

ott =me* and ñg+3 = tết Elimination of fo yields the equation

ef 30% 42 = (e*—1y(e%-2) = 0,

so it follows (since & > 0) that e* = 2 Hence fo+ 1 = 2%, so t = 1 Thus it began

to snow at 6 a.m

Trang 24

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

Trang 25

10 11 12 13 14 15 16 17 1⁄3 y(Œœ) = 3x+Œx”

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

p=exp([(3/x)dr)=6™ = 2°; D,(y-x')=227; yrattic

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 +€

3 2

Trang 26

18 19 20 21 22 23 24, 25 26 p=exp([(-2/a)dr)=e =x"; D,(y-x*) =cos.x, y:x” =sinx+C y(x) = x? (sinx+C)

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

p=exp([(-2x)4x)=e”: D,(y-e* )=30; yen =x4C

Trang 27

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

1, 1€

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

Trang 28

30 31 3 33 34 After division of the given equation by 2x, multiplication by the integrating factor p= x” yields “V2 2 = 1 xy dy v2 y = x? cosx, DĐ 2y) = x cosx, xy = c+] 1”? cost dt

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

) y= cel] {lock ar }el* acl = -Py,+@

(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

Trang 29

36

37

38

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

Trang 30

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, aT a= ka, TC, and since q, and k are both constants, it follows that aT Ay S = Constant,

‘That is, the product of the cross-sectional area normal to the heat rate and temperature gradient remains a constant and independent of distance x It follows that since Ay increases with x, then dT'/dx must decrease with x Hence, the temperature distribution appears as shown above Note the gradient decreases with increasing x

Trang 31

FIND: Sketch temperature distribution and give brief explanation to justify shape SCHEMATIC: Hot water pi a 7 a Tea 5 Thsulation %

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial) conduction, (3) No internal heat generation, (4) Insulation has uniform properties independent of temperature and position

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 4; = Constant ‘That is, q, is independent of radius (r), Since the thermal conductivity is also constant, it follows that aT ar | 7 Constant r

This relation requires that the product of the radial temperature gradient, dT/dr, and the radius, r, remains constant throughout the insulation For our situation, the temperature distribution must appear as shown in the above, right sketch

Trang 32

KNOWN: A spherical shell with prescribed geometry and surface temperatures FIND: Sketch temperature distribution and explain shape of the curve SCHEMATIC: ri 1 Spherical ‘shell te To 5 1 5 gan ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in radial (spherical coordinates) direction, (3) No internal generation, (4) Constant properties, ANALYSIS: Fourier's coordinate) system has the form at 2) aT fe = Ay = ak (Are?)

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

in = Gout = # ae(t) -

That is, q, is a constant, independent of the radial coordinate Since the thermal conductivity is constant, it follows that

+ |dT

2 (2) cama

‘This relation requires that the product of the radial temperature gradient, dT/dr, and the radius squared, r, remains constant throughout the shell Hence, the temperature distribution appears as shown in the above, right sketch

Trang 33

distribution and heat rate

FIND: Expression for the thermal conductivity, k SCHEMATIC: Units Aw) =(I-x) T-K T@)=.500(1-2x-x3) ÿ X~”” A-ma

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduetion in x- direction, (3) No internal heat generation

ANALYSIS: Application of the energy balance relation, Eq 1.11a, to the system, it follows that since Ej, = Eout, Ge = Constant # f(x) Using Fourier's law, Eq 2.1, with appropriate expressions for Ay and T, yields aT SEI ae =k: (1—xjmt> —#x - x) % 600W = —k «(1 —x)m* » F- (s00(1 ~ 2x ~ xy) K Solving for k and recognizing its units are W/m K, 8000 20 A (1 —x)(300(—2 — 3x")| uê S (1—x)(@+ 3x?) 4 COMMENTS: (1) Note that at x=0, k=10W/mK and that indeed the units are correctly obtained

Trang 34

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 SCHEMATIC: đớn,

ASSUMPTIONS: (1) One-dimensional conduction in x (negligible temperature gradients along y), (2) Steady-state conditions, (3) Adiabatic sides, (4) No internal heat generation

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

Trang 35

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

oak Ba- (ky ¢ar) 2 6)

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

Trang 36

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 SCHEMATIC: „ tt} i L:05m = ffs Riumdibuedier k=25 WmkK merle

ASSUMPTIONS: (1) One-dimensional heat flow, (2) No internal heat generation, (3) Steady-state conditions, (4) Constant properties ANALYSIS: The rate equation and temperature gradient for this system are * T T.-T;

ant and get t2)

Trang 37

thickness, FIND: Unknowns for various temperature conditions and sketch distribution, SCHEMATIC: at L025 v gf, Temperature gradient ke80 0c = :

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

Trang 38

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 SCHEMATIC: T t 4 ; † T T,=400K 5 + %-600K T, Te : k=100W/mK h U Lri00nm 0L *xX xen xen @ () ()

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

Trang 39

cylinder surface FIND: Expressions for heat rate at cylinder surface and fluid temperature SCHEMATIC: ———— Tir)=e+br2 fe — —> h,e—>

ASSUMPTIONS: (1) One-dimensional, radial condueton, (2) Steady-state conditions, (3) Constant properties

ANALYSIS: The heat rate from Fourier's law for the radial (cylindrical) system has the form

av —-—kA, FS % dr

Using the expression for the temperature distribution, T(r) =a + br’, evaluate the temperature gradient, dT /dr, to find the heat rate, ty =—k(2zrL) #br = —4zkbLrẺ At the outer surface (r — rạ), the conduction heat rate is Ger, = —AtkbLrd a

From a surface energy balance at r=

Arar, = đáy = h(2ar¿L) [T(rs) — Tx| ,

Trang 40

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

SCHEMATIC:

Zit _/B,Ty=100°C

ASSUMPTIONS: (1) Two-dimensional conduction, (2) Steady-state conditions, (3) No heat generation, (4) Constant properties ANALYSIS: At the surface A, the temperature gradient in the x-direction must be zero That is, (@T/dx), = 0 This follows from the requirement that the heat flux

‘vector must be normal to an isothermal surface, The heat rate at the surface A is given by Fourier's law written as

+ =-ewy, BF) wT = 0 x 2m x 0X = K hn

Shae ema By] mos Bx BOE = AO

On the surface B, it follows that

(01 /2y)n =0 4 in order to satisfy the requirement that the heat flux vector be normal to the isothermal

surface B Using the conservation of energy requirement, Eq 1.11a, on the body, find SA —9xn =0 or an = 45,4 + ote that, dup =e › and hence | (ØT/Bxìn = IA a {000 W/m) _

(E/Eh Tung “TGẤY/miKxiim T60 Km: a

COMMENT: Note that in using the conservation requirement,

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