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Unformatted text preview: 25 x"“ . 4 A 5 2.1 Differential Equations and Solutions
equilibrium solutions are the solutions that can be "’seen in the direction ﬁeld 1n
Figure 8. They are shown plotted 1n bluez' 1n Figure 10. Next we notice that f (y) — 1 — y2 is positive if —1 < y < 1 and negative
otherwise. Thus, if y(t) IS a solution to equation (1.24), and —1 < y < 1, then ’=l——y2>0. Having a positive derivative, y is an increasing function. How large can a solution y(t) get? if it gets larger than 1, then y’__  1 —— y2 < 0,
so y(t) will be decreasing. We cannot complete this line of reasoning at this point,
but in Section 2. 9 we will develop the argument, and we will be able to conclude
that 1f y(O) yo > 1, then y(t) 1s decreasing and y(t) —> 1 ast —> 00. On the other hand, if y(O)_ — yo satisﬁes —1 < yo < 1, then y’=1—y2> 0, so y(t) will be increasing. We will again conclude that y(t) 1ncreases and approaches
1 as t —> 00. Thus any solution to the equation y’ # __1 — y2 with an initial value
yo > —1 approaches 1 as t ——> 00. Finally, if we consider a solution y(t) with y(0) = yo < —1, then a similar
analysis shows that y ’(t)— _ 21 — y2 < 0, so y(t) IS decreasing. As y(t) decreases,
its derivative y (t)— _ 1 — y2 gets more and more negative. Hence, y(t) decreases
faster and faster and must approach —00 as 1‘ increases. Typical solutions to equa
tion (1.24) are shown in Figure 11. These solutions were found with a computer,
but their qualitative nature can be found simply by looking at the equation. ,,/ \\\»~\\\\ Ia.)—
_—,/ \\~\\\\\\ )1,..
———«z \\\\\\\\\ Iza——‘
—,,/ \\\\\\\ /,
/// \\\\~\\\\ ,I,..
,// \\\~\\\ //———
_,,., \\\\\\\ 11,—
,,/ \\\\\\\ xz
—»z/ \\\\\\\\\ ,i,
..a;/ \\\\\\\\\ /,,—
.,/ \\~\\\\\\ /,..
/// \\\\~\~\\ /,—
_,// \\\\\\\\\ /,,_
,—,,/ \\\\\\\ //,
,/ \\\~\\\ //,,
_——,/ \\\\\~\\\ /,,.
_,,/ \\\~\\\\\ /,,_
.,a/ \\\x\\\\\ /,,, IO. Equilibrium2 solutions to
ation y’ — 1 — y. ////’
////z
/ / / I.’
////,
////z
////z
/ /',~ ,.— \ \ \ \.\ \ \ \ \
\ \ \ \ \ \.\ \ \
\ \ \ \.\ \.\ \ \
\ \ \ \ \.\ \ \
\ \ \ \ \.\.\ \ \
\ \ \ \ \ \.\ \ \
\ \ \ \ \ \.\ \ \
\\\\\x.\\\
\ \ \.\ \ x \ \ \
\ \ \ \.\ \ \ \ \
\ \ \.\ \ \ \ \ \‘
\ \ \ \.\ x \ \ \
\ \ \ \.\ \ \ \ \ f/ / /
», / / /
,.,., / /
, / / / / i ~—//// 
LA Figure I 1. Typical solutions to the equation y’ = 1 — yz. EXERQESES In Exercises 1 and 2 , given the function ¢, place the ordinary
d1fferential equation ¢(t, y, y/) = 0 in normal form. 1 (MIC, y,z) = x22 + (1 +x)y
2 $06,350 =xz —2y —x2 111 Exercises 3~6, show that the given solution is a general solu—
tion of the differential equation. Use a computer or calculator
to sketch the solutions for the given values of the arbitrary con~
stant. Experiment with different intervals for 1 until you have a plot that shows what you consider to be the most important
behavior of the family. 3 y’ = —ry. ya) = Cer<1/Z>’2,C = —3, —2, . .. ,3
4 y’+y=2t,y(t)=2t—2+Cer’,C=—3,—2,... ,3 5. y’ + (1/2)y = 2cos t, y(t) = (4/5) cost + (8/5) sint i—
Ce‘WZ)’, C = ~5, —4, . .. ,5 6. y' = y<4 — y), ya) = 4/(1+ Cr“). C =1.2,... , i
,.
i.
z.
[a
i.
v « 141g t‘wﬁm N.‘ WNW .m .. .. n V...“ W . ‘tmmi‘mfﬂmﬂr .4 31.:LLC__LKHJHLU~L'£ 1.. . « . wt ”.22“ .11... 26 CHAPTER 2 First—Order Equations 2:“. \ 5;”; 7. A‘general solution may fail to produce all solutions of a ‘ é, fferential equation. In Exercise 6, show that y = 0 is a w . . . . .
solution of the d1fferent1al equation, but no value of C 1n
the given general solution will produce this solution. 8. (a) Use implicit differentiation to show that t2 + y2 = C 2
implicitly deﬁnes solutions of the differential equa
tion I + yy’ : 0.
(b) Solve t2 + y2 = C2 for y in terms of t to provide
explicit solutions. Show that these functions are also
solutions of t + yy’ = 0. (0) Discuss the interval of existence for each of the solu—
tions in part (b). (d) Sketch the solutions in part (b) for C = l, 2, 3, 4. 9. (a) Use implicit differentiation to show that 12—4322 : C 2
implicitly deﬁnes solutions of the differential equa—
tion I — 4yy’ = O.
(b) Solve z‘2 — 4y2 2 C 2 for y in terms of t to provide
explicit solutions. Show that these functions are also
solutions oft  4yy’ = O. (c) Discuss the interval of existence for each of the solu
tions in part (b).
(d) Sketch the solutions in part (b) for C : l, 2, 3, 4. 10. Show that y(t) = 3/(61‘ 2 11) is a solution of y’ = —2y2,
y(2) : 3. Sketch this solution and discuss its interval of
existence. Include the initial condition on your sketch. 11'. Show that y(t) : 4/ (l — 524”) is a solution of the initial
value problem y’ = y(4 — y), y(0) = —1. Sketch this solution and discuss its interval of existence. Include the initial condition on your sketch. In Exercises 12—15, use the given general solution to ﬁnd a so—
lution of the differential equation having the given initial con
dition. Sketch the solution, the initial condition, and discuss
the solution’s interval of existence. 12. y'+4y : cost, y(t) : (4/17) cost—I—(l/l7) sint+Ce“4’,
y(0) = —1 13. ty’ + y = t2, y(t) = (1/3)t2 + C/t, 32(1) : 2 14. ty’ + (r + 1)y : 2te“, y(t) = e"(t + C/r), 32(1) : l/e 15. y' = y(2 + y), W) = 2/(—I + Ce”). y(0) = —3 16. Maple, when asked for the solution of the initial value
problem y’ = f, y(0) = 1, returns two solutions:
y(t) : (l/4)(r + 2)2 and y(t) = (l/4)(t — 2)2. Present
a thorough discussion of this response, including a check
and a graph of each solution, interval of existence, and so on. Him: Remember that «/a2 = lal. In Exercises 17—20, plot the direction ﬁeld for the differential
equation by hand. Do this by drawing short lines of the appro—
priate slope centered at each of the integer valued coordinates @Where—25t52and—lfyfl.
W I ,2
1. «:y+t 1§;fy\‘§——y'—t 19. y’ = ttan(y/2) 120. Ry 2: (22 y) / (1 + yl) In Exercises 21—24, use a computer to draw a direction ﬁeld
for the given ﬁrst—order differential equation. Use the indi—
cated bounds for your display window. Obtain a printout and
use a pencil to draw a number of possible solution trajectories
on the direction ﬁeld. If possible, check your solutions with a computer. 21. y’:—ty,R={(t,y):—3ft§3,—~5fySS} 22. y’=yZ—t,R={(t,y):«25t510,—4gy54}
23. y’:t—y+1,R:{(t,y):—6§t§6,—6:y§6}
24 y’=(y+t)/(y—t).R={(t,y)2—55t55,—5: For each of the initial value problems in Exercises 25—28 use
a numerical solver to plot the solution curve over the indicated
inte1val. Try different display windows by experimenting with
the bounds on y. Note: Your solver might require that you ﬁrst
place the differential equation in normal form. 25. y + y’ = 2, y(0) = 0,1: 6 {—2, 10] 26. y’ + ty = r2, y(0) = 3,: e [—4, 4] 27. y’ — 3y = sint, y(0) = —3,t E [—671, JT/4]
28. y’ + (cos 0)) = sint, y(0) = 0, r e [—10, 10] Some solvers allow the user to choose dependent and indepen—
dent variables. For example, your solver may allow the equa— tion r’ = —25r + 6”, but other solvers will insist that you
change variables so that the equation reads y’ = —2ty + e“ ,
or y’ = —2x y + e“, should your solver require t or x as the independent variable. For each of the initial value problems
in Exercises 29 and 30, use your solver to plot solution curves
over the indicated interval. 29. r’ +xr = cos(2x), r(0) : —3, x e [—4, 4]
30. T’ + T = s, T(—3) = 0, s e [—5, 5] In Exercises 31234, plot solution curves for each of the initial
conditions on one set of axes. Experiment with the different
display windows until you ﬁnd one that exhibits what you feel
is all of the important behavior of your solutions. Nate: Se—
lecting a good display window is an art, a skill developed with
experience. Don’t become overly frustrated in these ﬁrst at tempts.
31. y’ = y(3 — y), y(0) = —2, —1,0, 1,2, 3,4,5 32. x’ — x2 = r, x(0) : —2, 0, 2, x(2) = 0, x(4) = —3, 0, 3,
x(6) = 0 33 17’: 5111063)), y(0) : 0.5, 1.0, 1.5, 2.0, 2.5
34. x’ = —tx, x(0) : #3,~2,’1,0, 1,2,3 35. Bacteria in a petri dish is growing according to the equa—
tion
at P —— : 0.44P,
dz where P is the mass of the accumulated bacteria (mea—
sured in milligrams) after t days. Suppose that the initial
mass of the bacterial sample is 1.5 mg. Use a numerical
solver to estimate the amount of bacteria after 10 days. ' Exeaessss In Exercises 1—12, ﬁnd the general solution of the indicated
differential equation. If possible, ﬁnd an explicit solution. @' 2 xy My 2 2y
3;}; _ 6x1), 4. yI : (1 + y2)ex 5y'=xy+y 6. y’zyeX—ZeX—l—y—2
7. y’=x/(y+2) 8.
9. xzy’zylny—y’ 10. 12. y’ = (2xy+2x)/(x2—1) y’ = xy/(x — 1)
xy’ — y = 2x2y
11. y3y' : x + 2);/ In Exercises 13—18, ﬁnd the exact solution of the initial value
problem. Indicate the interval of existence. “ “RM 7 {13.32 = y/x, y(1) = —2 “r4 y’ = 270 + y2)/y, y(0) =1
we = (shun/y, yer/2) =1
16. y’ 2 ex”, y(0) = 0 17. y': (1+ yz), y(0) =1 18 y’ = x/(1+ 2y).y<—1>= 0 5) 6) n0
; is In Exercises 19—22, ﬁnd exact solutions for each given initial
COndrtion. State the interval of existence in each case. Plot
each exact solution on the interval of existence. Use a numeri— cal Solver to duplicate the solution curve for each initial value
problem. 19. y' = x/y, y(0) :1, y(0) : _1
20 y' = —X/y. y(0) = 2. W» = ‘2
21. y’ : 2 — y, y(0) = 3, y(0) :1 The integral on the left 22.
23. 24. f 1’26. . Lang. 27. 2S:
M 2.2 Solutions to Separable Equations 35 contains the expression y’ (I) dt. This is inviting us to change
the variable of integration to y, since when we do that, we use the equation dy =
y’(t) dt. Making the change of variables leads to / mm = / gmdr. Notice the similarity between (2.36) and (2.37). Equation (2.36), which has no
meaning by itself, acquires a precise meaning when both sides are integrated. Since
this is precisely the next step that we take when solving separable equations, we can
be sure that our method is valid. We mention in closing that the objects in (2.36), h(y) dy and g(t) dt, can be
given meaning as formal objects that can be integrated. They are called dﬁeren
lial forms, and the special cases like dy and dt are called differentials. The basic
formula connecting differentials d y and dt when y is a function of I is (2.37) dy = y’(t)dt, the change—of—variables formula in integration. These techniques will assume greater
importance in Section 2.6, where we will deal with exact equations. The use of dif
ferential forms is very important in the study of the calculus of functions of several
variables and especially in applications to geometry and to parts of physics. y’ = (y2 + 1)/y. y(1) = 2 Suppose that a radioactive substance decays according to
the model N’ = Noe‘M . Show that the half—life of the
radioactive substance is given by the equation T1/2 = —.—. (2.38) The half—life of 238U is 4.47 X 107 yr. (a) Use equation (2.38) to compute the decay constant A
for 238U. (b) Suppose that 1000 mg of 238U are present initially.
Use the equation N = Noe’“ and the decay constant
determined in part (a) to determine the time for this
sample to decay to 100 mg. ﬁritium, 3H, is an isotope of hydrogen that is sometimes
. sed as a biochemical tracer. Suppose that 100 mg of 3H
decays to 80 mg in 4 hours. Determine the half—life of 3H. 'hje isotope Technetium 99m is used in medical imag—
It has a half—life of about 6 hours, a useful feature
for radioisotopes that are injected into humans. The Tech—
netium, having such a short half—life, is created artiﬁcially
on scene by harvesting from a more stable isotope, 99Mb.
If 10 g of 99’“Tc are “harvested” from the Molybdenum,
how much of this sample remains after 9 hours? The isotope Iodine 131 is used to destroy tissue in an over—
active thyroid gland. It has a half—life of 8.04 days. If a
hospital receives a shipment of 500 mg of 1311, how much
of the isotope will be left after 20 days? 36 CHAPTERZ FirstOrder Equations 28. A substance contains two Radon isotopes, 210Rn [11/2 : 2.42 h] and 211Rn [rm = 15 h]. At ﬁrst, 20% of the decays
come from 211Rn. How long must one wait until 80% do
so? Suppose that a radioactive substance decays according to the model N : Noe‘”. (a) Show that after a period of TA = 1 /)t, the material has
decreased to 6‘1 of its original value. TA is called the
time constant and it is deﬁned by this property. (b) A certain radioactive substance has a half—life of 12
hours. Compute the time constant for this substance. (c) If there are originally 1000 mg of this radioactive sub
stance present, plot the amount of substance remain—
ing over four time periods TA. 29.. In the laboratory, a more useful measurement is the decay rate
R, usually measured in disintegrations per second, counts per minute, etc. Thus, the decay rate is deﬁned as R = ~dN /d t.
Using the equation d N / dt = —)tN, it is easily seen that R =
AN. Furthermore, differentiating the solution N = Noe‘“
with respect to t reveals that R = Roe'“, (2.39) in which R0 is the decay rate at t = 0. That is, because R
and N are proportional, they both decrease with time accord—
ing to the same exponential law. Use this idea to help solve Exercises 30—3 1. 30. Jim, working with a sample of 131l in the lab, measures the
decay rate at the end of each day. TrME COUNTS TEME COUNTS
(nave) (COUNTS/DAY) (DAYS) (moms/nu)
1 , 938 6 5 87
2 822 7 536
3 753 8 494
4 738 9 455
5 647 10 429 Like any modern scientist, Jim wants to use all of the data
instead of only two points to estimate the constants R0 and
A in equation (2.39). He will use the technique of regres
sion to do so. Use the ﬁrst method in the following list that
your technology makes available to you to estimate A (and
R0 at the same time). Use this estimate to approximate the ,
half—life of 1311. (a) Some modern calculators and the spreadsheet Excel
can do an exponential regression to directly estimate
R0 and 2». Taking the natural logarithm of both sides of equa
tion (2.39) produces the result lnR : ~M +lnR0. 0)) Now lnR is a linear function of 1‘. Most calcula—
tors, numerical Software such as MATLAB®, and
computer algebra systems such as Mathematica and
Maple will do a linear regression, enabling you to esti—
mate ln R0 and A (e.g., use the MATLAB® command polyfit). 31. 32. 3S. (c) If all else fails, plotting the natural logarithm of the
decay rates versus the time will produce a curve that
is almost linear. Draw the straight line that in your
estimation provides the best ﬁt. The slope of this line provides an estimate of ~—)t. A 1.0 g sample of Radium 226 is measured to have a decay
rate of 3.7 x 1010 disintegrations/s. What is the half—life of
226Ra in years? Note: A chemical constant, called Avo
gadro’s number, says that there are 6.02 x 1023 atoms per
mole, a common unit of measurement in chemistry. Fur—
thermore, the atomic mass of 226Ra is 226 g/mol. Radiocarbon dating. Carbon 14 is produced naturally
in the earth’s atmosphere through the interaction of cos
mic rays and Nitrogen 14. A neutron comes along and
strikes a 14N nucleus, knocking off a proton and creating
a 14 C atom. This atom now has an afﬁnity for oxygen and
quickly oxidizes as a 14C02 molecule, which has many
of the same chemical properties as regular C02. Through
photosynthesis, the 14C02 molecules work their way into
the plant system, and from there into the food chain. The
ratio of 14C to regular carbon in living things is the same as
the ratio of these carbon atoms in the earth’s atmosphere,
which is fairly constant, being in a state of equilibrium.
When a living being dies, it no longer ingests 14C and the
existing 14C in the now defunct life form begins to de—
cay. In 1949, Willard F. Libby and his associates at the
University of Chicago measured the half—life of this de
cay at 5568 :1: 30 years, which to this day is known as the
Libby halflife. We now know that the halflife is closer
to 5730 years, called the Cambridge halflife, but radio—
carbon dating labs still use the Libby halflife for technical
and historical reasons. Libby was awarded the Nobel prize
in chemistry for his discovery. (a) Carbon 14 dating is a useful dating tool for organisms
that lived during a speciﬁc time period. Why is that? 3
Estimate this period. 2%
ii (u) Suppose that the ratio of 14C to carbon in the charcoal on a cave wall is 0.617 times a similar ratio in living
wood in the area. Use the Libby half—life to estimate the age of the charcoal. i it” 33. A urder victim is discovered at midnight and the tem—
“Nwmwsmperature of the body is recorded at 31°C. One hour later, the temperature of the body is 29°C. Assume that the sur
rounding air temperature remains constant at 21°C. Use
Newton’s law of cooling to calculate the victim’s time of
death. Note: The “normal” temperature of a living human
. being is approximately 37°C.
. Su" pose a cold beer at 40°F is placed into a warm room
, °F. Suppose 10 minutes later, the temperature of the
beer is 48°F. Use Newton’s law of cooling to ﬁnd the tem—
perature 25 minutes after the beer was placed into the
room. Referring to the previous problem, suppose a 50° bottle of
beer is discovered on a kitchen counter in a 70° room. Ten
minutes later, the bottle is 60°. If the refrigerator is kept ...
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 Spring '09
 Staufeneger
 Differential Equations, Equations

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