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W in Chapter 8', that proeeed directly from the differential equation and need no expression
for the solution. Software packages such as Maple and Mathematica readily execute such
procedures and produce graphs of solutions of differential equations. ‘ FIGURE 2.1.4' mesa: curves of 2y + ty =" 2. I Figure 2.1.4 displays graphs of the solution (47) for several values of c. From the ﬁgure it
may be plausible to conjecture that all solutions approach a limit as r w) 00. The limit can be
found analytically (see Problem 32). In each of Problems 1 through 12:
(a) Draw a direction ﬁeld for the given differential equation.
(b) Based on an inspection of the direction ﬁeld, describe how solutions behave for large r. (c) Find the general solution of the given differential equation, and use it to determine how
solutions behave as I —> oo. 1'. y’ + By :2 t + e'2r 94?, 2. y’ — 2y : 3262’ 3. y’+y=te"—1 .. 93.62, 4. y’+(1/r)y=30052t, z>0
5.y’—2y=3e‘ 9'0, 6. ty’+2y=sint, r>0 7. y’ +213» 2 Zia—’2 9% s. (r + ﬂy“ +45» = (1 + er? 9. 2y’+y=32‘ $52,10ty’—y=tze—', r>0 11. y’ +y =5siu2r 9Q, 12. 231* +y = 3:2 In each of Problems 13 through 20 ﬁnd the solution of the given initial value problem. 13. y’ —— y : 2:822 ya» = 1
14' y' + 2y = Ie‘z', 31(1) = 0 777;; ‘
15. t_):'.{e2y:r2_t+11 y(1}:%’ {>0 .¥,._f, 15 3”+ (My = (cosr)/t2, y(7t) = 0, t> 0  17. y' H 2): 2 e2” ):(0) = 2 Chapter 2. First Order Differential Eqnatgoi 48 dy x i 3"“ dy x2
7. — = 8. —— 2
dx y + 6” dx 1 + y2  In each of Problems 9 through 20:
i (3) Find the solution of the given initial vaiue problem in explicit form.
(to) Plot the graph of the solution.
(0) Determine (at least approximately) the interval in which the solution is deﬁned.
$2 9. y’ z (1 w my, 31(0): —1/6 91% 10. y’ .—_ (1 w 2x)/y, y(1) = —2
935?, 11. xdx +yeIdy : 0, y(0) = 1 $0, 12. dr/de = rZ/e, r(1) = 2
$0, 13. y’ = Zx/(y +x2y), y(0) = —2 $3 14. y’ = xy3(1 arm112, y(0) = 1
s52, 15. y' a 2m: + 2y), yo) = 0 ea 16. y' = x(x2 + 1W3. yo) = 4142
no 17. y’ = (3x2 — was/(2y — 5). yo) = 1
e‘Q 18 y“ = (N — ewe +4y). no) : 1
375?, 19. sin2xdx+cos3ydy=0. ﬂit/2} =Ir/3
935?, 20. y2(1 — xlimdy = arcsinxdx, y(0) = 1
Some of the results requested in Problems 21 through 28 can be obtained either by 501' the given equations analyticaily or by plotting numerically generated approximations to
solutions. Try to form an opinion as to the advantages and disadvantages of each approm  I I $3 21. Solve the initial value problem
y! = (1 + 3x2)/(3y2 _ 63’), ND I: 1 .1 and determine the intervai in which the solution is valid.
Hint: To ﬁnd the interval of deﬁnition, look for points where the integral curve 11 vertical tangent.
g3 22. Solve the initial value problem if = act/(3y? — 4). ya) = 0 i H i _ and determine the intervai in which the solution is valid.
Hint: To ﬁnd the interval of deﬁnition, took for points where the integral curve 1 vertical tangent.
$2, 23. Solve the initial value problem i
y’ i 2112 + xyz, rm) :1
and determine where the solution attains its minimum value.
@‘Q 24. Solve the initial value problem
y” = (2 — ewe + 2y), ya» = 0 and determine where the solution attains its maximum value.
$2 25. Solve the initial vaiue problem y’ : Zoos 2x/(3 +2y), 31(0) :2 #1
and determine where the solution attains its maximum value.
$42, 26. Solve the initial value problem
y’=2(1+x)(1+y2). KO) =0 and determine where the solution attains its minimum value. 50 Chapter 2. First Order Bifferential Equations (d) Solve Eq. (iii), obtaining u implicitly in terms of x.
(e) Find the solution of Eq. (i) by replacing u by y/x in the solution in part (d). (f) Draw a direction ﬁeld and some integral curves for Eq. (i). Recall that the right side
of Eq. (i) actually depends only on the ratio y/x. This means that integral curves have the
same slope at all points on any given straight line through the origin, although the slope
changes from one line to another. Therefore the direction ﬁeld and the integral curves are
symmetric with respect to the origin. Is this symmetry property evident from your plot? The method outlined in Problem 30 can be used for any homogeneous equation. That is,
the substitution )1 = xv (x) transforms a homogeneous equation into a separable equation.
The latter equation can be solved by direct integration, and then replacing v by y [it gives
the solution to the original equation. In each of Problems 31 through 38: (a) Show that the given equation is homogeneous.
(b) Solve the differential equation. (c) Draw a direction ﬁeld and some integral curves. Are they symmetric with respect to
the origin? .  dy Jc2+xy+y2 4, dy xzmt~3y2
$2, 31. —:———— 32, ——=
dx x2 QQ dx 2x)!
dy 4y—3x dy 4x+3y
3. — 3 34, M : ~—
gz 3 dx way $2 mix Zxly
d 3
@4335. 14% 3’ @‘52/36.(x2+3xy+y2)dxx2dy=0
dx x—y
dy X2 —3y2 dy 3y2exZ
3 . —m 2 33 —— m
$2 7 (ix 2,5}, é‘?’ dx 2xy Differential equations are of interest to nonmathematicians primarily because of the
possibility of using them to investigate a wide variety of problems in the physical,
biological, and social sciences. One reason for this is that mathematical models and
their solutions lead to equations relating the variables and parameters in the prob
lem. These equations often enable you to make predictions about how the natural
process will behave in various circumstances. It is often easy to vary parameters in
the mathematical model over wide ranges, whereas this may be very timeconsuming
or expensive, if not impossible, in an experimental setting. Nevertheless, mathemat
ical modeling and experiment or observation are both critically important and have
somewhat complementary roles in scientiﬁc investigations. Mathematical models are
validated by comparison of their predictions with experimental results. On the other
hand, mathematical analyses may suggest the most promising directions to explore
experimentally, and they may indicate fairly precisely what experimental data will
be most helpful. In Sections 1.1 and 1.2 we formulated and investigated a few simple mathemati
cal models. We begin by recapitulating and expanding on some of the conclusions
reached in those sections. Regardless of the speciﬁc ﬁeld of application, there are
three identiﬁable steps that are always present in the process of mathematical mod
cling. i
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do _ gR2 (28) Equation (28) is separabie but not linear, so by separating the variables and integrating, we
obtain + c. . (29) 2 R+x Sincex = 0 when I : 0, the initial condition (27) at t = O can be replaced by the condition that
v 2 on when x :,0. Hence c = (113/2) — gR and ‘ / 2R2
v=i ﬁwEﬁbtgix. am Note that Eq. (30) gives the velocity as a function of altitude rather than as a function of time.
The plus Sign must be chosen if the body is rising, and the minus sign if it is falling back to earth.
To determine the maximum altitude a: that the body reaches, we set a z 0 and x = S in Eq. (30) and then solve for E, obtaining 2
UOR :m. 31 s MW, ( ) Solving Eq. (31) for no, we ﬁnd the initial velocity required to lift the body to the altitude E,
namely, _ 9?
use naﬁxg. . on The escape velocity 11., is then found by letting; —> oo. Consequentiy,
v6 = ZgR. (33) The numerical value of v,E is approximately 6.9 mils, or 11.1 kmls.
The preceding caiculation of the escape velocity neglects the effect of air resistance, so the actuai escape velocity (including the effect of air resistance) is somewhat higher. On the other
hand, the effective escape velocity can be signiﬁcantly reduced if the body is transported a
considerabie distance above sea level before being launched. Both gravitational and frictional
forces are thereby reduced; air resistance, in particular, diminishes quite rapidly with increasing
aititude. You should keep in mind also that it may well be impractical to impart too large an
initial velocity instantaneously; space vehicles, for instance, receive their initial acceleration during a period of a few minutes. PROBiéMS‘W.WMMHHsMW.W mum“ 1. Consider a tank used in certain hydrodynamic experiments. After one experiment the
tank contains 200 L of a dye solution with a concentration of 1 gfL. To prepare for the
next experiment, the tank is to be rinsed with fresh water ﬂowing in at a rate of 2 L/min,
the wellstirred solution ﬂowing out at the same rate. Find the time that will elapse before
the concentration of dye in the tank reaches 1% of _i_t&riginaj value. " ' ' s 2. A tank initially contains 120 L of pure water. A mixture containing a concentration of
7/ g/L of salt enters the tank at a rate of 2 Limin, and the wellstirred mixture leaves the
tank at the same rate. Find an expression in terms of y for the amount of salt in the tank
at any time I. Also ﬁnd the limiting amount of salt in the tank as t —> 00. i
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i Chapter 2. First Order Differential Equations (c) Suppose that certain remains are discovered in which the current residual amount of
carbon—14 is 20% of the original amount. Determine the age of these remains. ' 13; The population of mosquitoes in a certain area increases at a rate proportional to the
7 current population, and in the absence of other factors, the population doubles each
week. There are 200,000 mosquitoes in the area initially, and predators (birds, bats, and so forth) eat 20,000 mosquitoes/day. Determine the population of mosquitoes in the area
at any time. $2 14. Suppose that a certain population has a growth rate that varies with time and that this population satisﬁes the differential equation dy/dt = (0.5 + sin r)y/5. (a) Ify(0) : 1, ﬁnd (or estimate) the time I at which the population has doubled. Choose other initial conditions and determine whether the doubling time 1: depends on the initial population. (b) Suppose that the growth rate is replaced by its average value 1/10. Determine the
* doubling time t in this case. '_ (0) Suppose that the term sint in the differential equation is replaced by sin 2m; that is,
the variation in the growth rate has a substantially higher frequency. What eﬁect does this
have on the doubling time I? (d) Plot the solutions obtained in parts (a), (b), and (c) on a single set of axes. $53 15. Suppose that a certain population satisﬁes the initial value problem dy/dt : r(t)y m k, y(0) = yo, where the growth rate r0) is given by r(r) = (1 + sint)/5, and k represents the rate of
predation. (a) Suppose that k 2 1/5. Plot y versus I for several values of yn between 1/2 and 1. (b) Estimate the critical initial population yc below which the population will become
extinct. (c) Choose other values of k and ﬁnd the corresponding yc for each one.
(d) USe the data you have found in parts (b) and (c) to plot yc versus k. 16. Newton‘s law of cooling states that the temperature of an object changes at a rate pro—
portional to the difference between its temperature and that of its surroundings. Suppose
that the temperature of a cup of coffee obeys Newton’s law of cooling. If the coffee has a
temperature of 200T when freshly poured, and 1 min later has cooled to 190°F in a room
at 70°F, determine when the coffee reaches a temperature of 150°F. Q, 17. Heat transfer from a body to its surroundings by radiation, based on the Stefanﬂ Boltzmann5 law, is described by the differential equation a!
d—L: = —oc(u4 — T4), (i)
where MU) is the absolute temperature of the body at time t, T is the absolute temperature of the surroundings, and at is a constant depending on the physical parameters of the body. . 5Jozef Stefan (1835—1893), professor of physics atVienna, stated the radiation law on empirical grounds
in 1879. His student Ludwig Boltzmann (1844—1906) derived it theoretically from the principles of ther
modynamics in 1884. Boltzmann is best known for his pioneering work in statistical mechanics. l
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; 2.3 Modeling with First Order Equations 63 However, if u is much larger than T, then solutions of Eq. (1) are well approximated by
solutions of the simpler equation
a! ..
I; = waif. (n)
Suppose that a body with initial température 2000°K is surrounded by a medium with
temperature 300°K and that o: = 2.0 X 10‘12 0K’s'l's.
(a) Determine the temperature of the body at any time by solving Eq. (ii).
(b) Plot the graph of u versus t. (c) Find the time t at which u(t) = 600, that is, twice the ambient temperature. Up to this
time the error in using Eq. (ii) to approximate the solutions of Eq. (i) is no more than 1%. . Consider an insulated box (a building, perhaps) with internal temperature MU). According to Newton’s law of cooling, it satisﬁes the differential equation % = ek[u — T(t)L (i)
where T0) is the ambient (externai) temperature. Suppose that TU) varies sinusoidally;
for example, assume that T(t) = To + T1 cos cut. (a) Solve Eq. (i) and express u(t) in terms of t, k, To, T}, and m. Observe that part of
your solution approaches zero as it becomes large; this is called the transient part. The
remainder of the solution is called the steady state; denote it by 50‘). (b) Suppose that t is measured in hours and that to : it / 12, corresponding a period of 24 h
for T0). Further, let Ta = 60“F, T1 = 15°F, and k = 0.2/h. Draw graphs of SO) and T0)
Versus t on the same axes. From your graph estimate the amplitude R of the oscillatory
part of S (t). Also estimate the time lag 1: between corresponding maxima of TU) and S (I).
(e) Let It, T9, T1, and a) now be unspeciﬁed. Write the oscillatory part of S (t) in the form
R cos[cu(t — r}]. Use trigonometric identities to ﬁnd expressions for R and 1:. Let T1 and
to have the values given in part (b), and plot graphs of R and r versus k. 19. Consider a lake of constant volume V containing at time 2‘ an amount Q{t) of pollutant, evenly distributed throughout the lake with a concentration 60‘), where C(t) : Q{I)/ V.
Assume that water containing a concentration 16 of pollutant enters the take at a rate r,
and that water ieaves the lake at the same rate. Suppose that poliutants are also added
directly to the lake at a constant rate P. Note that the given assumptions neglect a num
ber of factors that may, in some cases, be iniportantﬁfor example, the water added or
lost by precipitation, absorption, and evaporation; the stratifying effect of temperature
differences in a deep lake; the tendency of irregularities in the coastline to produce shei
tered bays; and the fact that pollutants are not deposited evenly throughout the lake but
(usually) at isoiated points around its periphery. The results below must be interpreted in
the light of the neglect of such factors as these. (a) If at time t = 0 the concentration of pollutant is Go, ﬁnd an expression for the concen
tration c(r) at any time. What is the limiting concentration as t —> 00? (b) If the addition of pollutants to the lake is terminated (k = t) and P : O for t > 0),
determine the time interval T that must elapse before the concentration of pollutants is
reduced to 50% of its original value; to 10% of its original value. (c) Table 2.3.2 contains data6 for severai of the Great Lakes. Using these data, determine 7, _ i 6This problem is based on R. H. Rainey, “Natural Displacement of Pollution from the Great Lakes,”
Science 155 {1967), pp. 124%1243; the information in the table was taken from that source. tions icaﬂy 20 the
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.s a 2.5 Autonomous Equations and Population Dynamics 91' The assumption of a constant catch rate It may be reasonable when y is large but becomes
less so when y is small. (a) If h < rK/4, show that Eq. (i) has two equilibrium points y1 and y; with yi < y;;
determine these points. (b) Show that yl is unstable and y2 is asymptotically stable. (c) From a plot of f (y) versus y, show that if the initial population ya > y}, then y —+ y;
as t —> 00, but that if yo < y}, then )1 decreases as 1‘ increases. Note that y z 0 is not an
equilibrium point, so if yo < y}, then extinction will be reached in a ﬁnite time. (d) If h > rK/4, show that y decreases to zero as it increases regardless of the value of ya.
(e) If h :: rK/4, show that there is a single equilibrium point 32 : K/2 and that this point
is semistable (see Problem 7). Thus the maximum sustainable yield is hm = rK/4, corre sponding to the equilibrium value y = K /2. Observe that hm has the same value as Ym
in Problem 20(d). The ﬁshery is considered to be overexploited if y is reduced to a level below K/2. Epidemics. The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli in 1760 on smallpox. In more recent years
many mathematical models have been proposed and studied for many different diseases. 16 Problems 22 through 24 deal with a few of the simpler models and the conclusions that can be
drawn from them. Similar models have also been used to describe the spread of rumors and of consumer products. 22. 23. Suppose that a given population can be divided into two parts: those who have a given
disease and can infect others, and those who do not have it but are susceptible. Let x be the
proportion of susceptible individuals and y the proportion of infectious individuals; then
x l y = 1. Assume that the disease spreads by contact between sick and well members
of the population and that the rate of spread dy/dt is proportional to the number of such contacts. Further, assume that members of both groups move about freely among each
other, so the number of contacts is proportional to the product of x and y. Sincex : I — y,
we obtain the initial value problem dy/dt = 0M1 — y), W) = yo. (i)
where (x is a positive proportionality factor, and yo is the'initial proportion of infectious
individuals. 4 ‘ (a) Find the equilibrium points for the differential equation (i) and determine whether
each is asymptotically stable, semistable, or unstable. (b) Solve the initial value problem (i) and verify that the conclusions you reached in
part (a) are correct. Show that y(t) —+ 1 as t —> 00, which means that ultimately the dis ease spreads through the entire population.
Some diseases (such as typhoid fever) are spread largely by carriers,individuals who can transmit the disease but who exhibit no overt symptoms. Let x and y, respectively, denote
the proportion of susceptibles and carriers in the population. Suppose that carriers are identiﬁed and removed from the population at a rate )3, so dy/dr = —ﬁy  (i) 15A standard source is the book by Bailey listed in the references. The models in Problems 22 through 24
are discussed by Bailey in Chapters 5, 10, and 20, respectively. :
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(b) Use the result of part (a) to ﬁnd I at any time t by solving Eq. (ii) subject to the initial condition x(0) = x0.
(0) Find the proportion of the population that escapes the epidemic by ﬁnding the limiting value ofx as t —> oo. Daniel Bernoulli’s work in 1760 h
versial inoculation program against smallpox, which at that. time wa public health. His model applies equally well to any other disease that, once contracted
and survived, confers a lifetime immunity. number of t
this cohort who have not had smallpox by year t and who are therefore still susceptible. Let ,6 bathe rate at which susceptibles contract 5
people 'who contract smallpox die from the disease.
all causes other than smallpox. Then dx/dt, the rat declines, is given by Chapter 2. First Order preferential Equatiorts Suppose also that the disease spreads at a rate proportional to the product of x and y; thus ' (ix/d1 = ——u:xy. (a) ad the goal of appraising the effectiveness of a contro '
s a major threat to Consider the cohort of individuals born in a given year (t = O), and let MI) be the
hese individuals surviving t years later. Let x0) be the number of members of mallpox, and let u be the rate at which
Finally, let Mt) be the death rate from
c at which the number of susceptibles dwmz—u+umu. m i The ﬁrst term on the right side of Eq. (i) is the rate at which susceptibles contract smallpox,
and the second term is the rate at which they die from all other causes. Also dn/dt = —v,8x — “(0:1, (ii) where tin/d: is the death rate of the entire cohort, and the two terms on the right side are
the death rates due to smallpox and to all other causes, respectively. (a) Let 2 2 x/n and show that 2 satisﬁes the initial value problem dz/dt = w,82(1 — vz), z(0) 2: 1. (iii) Observe that the initial value problem (iii) does not depend on lu(t). (b) Find z(t) by solving Eq. (iii).
(c) Bernoulli estimated that v = ,8 = $15. Using these values, determine the proportion of 20~yearolds who have not had smallpox. Note: On the basis of the model just described and the best mortality data available at
the time, Bernoulli calculated that if deaths due to smallpox could be eliminated (u z 0),
then approximately 3 years could be added to the average life expectancy (in 1760) of 26 years, 7 months. He therefore supported the inoculation program. Bifurcation Points. For an equation of the form where a is a real parameter, the critical points (
value of a. As a steadily increases or decreases,
called a bifurcation point, critical points come together, or separate,
may either be lost or gained. Bifurcation points are of great interest in many app ctr/d3 : f (a, y), (i) equilibrium solutions) usually depend on the
it often happens that at a certain value of a,
and equilibrium solutions
lications, because near them the nature of the solution of the underlying differential equation is under~ going an abrupt change. For example,
up and become turbulent. Or an axia
large lateral displacement. Or, as th in ﬂuid mechanics a smooth (laminar) ﬂow may break
11y loaded column may suddenly buckle and exhibit a
e amount of one of the chemicals in a certain mixture is in creased, spiral wave patterns of varying color may suddenly emerge in an originally quiescent ‘t\3 Ur. t
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i 26 Exact Equations and Integrating Factors  '99 EXAMPLE 4 Thus, if (nil/1)), is to equal (,u.N)x, it is necessary that do My —Nx
ax  “T” (27) If (M y — Nx) /N is a function of x only, then there is an integrating factor a that also
depends only on x; further, pix) can be found by solving Eq. (27), which is both linear and separable.
A similar procedure can be used to determine a condition under which Eq. (23) has an integrating factor depending only on y; see Problem 23. Find an integrating factor for the equation
(3xy + yz) + (x2 + xyly’ = 0 (19) and then SOEVe the equation. In Example 3 we showed that this equation is not exact. Let us determine whether it has an
integrating factor that depends on x only. On computing the quantity (M y — N1) [N , we ﬁnd
that My(x:y)_Nx{xsy) = = N035?) x2+xy x Thus there is an integrating factor it that is a function of it only, and it satisﬁes the differential
equation (in M __ = _. 29 dx x ( )
Hence M0?) = x. (30) Multiplying Eq. (19) by this integrating factor, we obtain
(3ny + XJ‘Z) + (x3 + x230)“ = 0. (31)
The latter equation is exact, and its solutions are given implicitly by
x3y + %x2y2 : c. I‘ (32) Solutions may also be found in explicit form since Eq. (32) is quadratic in y.
You may also verify that a second integrating factor for Eq. (19) is 1
WW +y) ’ and that the same solution is obtained, though with much greater difﬁculty, if this integrating
factor is used (see Problem 32). Way} = Determine whether each of the equations in Problems 1 through 12 is exact. If it is exact,ﬁnd
the solution. ._. g ,z: 1. {23: +3) + (2y w2)y' = 0
3. (3x2—2xy+2}dx+{6y? —x2+3)dy = 0
4. (2x);2 + 2y) + (sz3: + 236))” = 0 2. (2x + 4y) + (2x — my : o 3699 990:4 .Ln {Ch npter First @rdezr Differential Equations ﬂ__ax+by 6 31—y__ax¥by
nix bx+cy 'dx_ bx—cy
{ex siny w 2ysinx) six + (a‘ cosy +2cosx) dy = 0
(exsiny+3y)dx— (3x— exsiny)dy :0
(yet? cos 2t — 23")” sin 2x + 2x) dx + (raw (2052): — 3) dy = 0
10. (y/x+6x)dx+(h1x—2)dy:0, x>0
ll. (x1ny+xy)dx+(ylnx+xy)dy=0; x>0, y>0 x dx d
12' + r 0
In each of Problems 13 and 14 solve the given initial value problem and determine at least
approximately where the solution is valid. '13. (2x—y)cix+(2y—x)dy=0, y(1)=3 14. (9x2+y—1)c£x—(4y—x)dy=0, y(1):0 In each of Problems 15 and 16 ﬁnd the value of b for which the given equation is exact, and
then solve it using that value of b. 15. (ch2 + bxzy) dx + (x + y)x2 dy = O 16. (yam + x) dx ~t— bxezx’ dy = 0 17. Assume that Eq. (6) meets the requirements of Theorem 2.6.1 in a rectangle R and is
therefore exact. Show that a possible function 1110:, y) is , x Y
mm): f M(S.yo)d3+ Newt. yo
where (x0, yo) is a point in R.
18. Show that any separable equation MOE) +Nty)y' = 0
is also exact. In each of Problems 19 through 22 show that the given equation is not exact but becomes exact
when multiplied by the given integrating factor. Then solve the equation. 19. nyB + x(1 + y2)y’ 2 0, Mm) 2 1/ch3 u 2 "K
20. —2€"‘sinx) dx + dy = 0, Mm) =ye"
y 21 ydx + (2x  yet) dy = 0, Way) = y
22. (x+2) siny dx+xcosy dyzﬂ, p,(x,y) =xe‘
23. Show that if (N: — Mﬂ/M = Q, where Q is a function of y only, then the differential equation a
M + Ny’ = 0 has an integrating factor of the form
#0) = exp f Qty) dy. . Show that iftNx — My)/(xM — yN) : R, where R depends on the quantity xy only, then
the differential equation
M + Ny’ 2 O has an integrating factor of the form Mxy). Find a general formula for this integrating
factor. 1.1} 2' 7 Numerical Approximations: Euler’s Method 101
W In each of Problems 25 through 31 ﬁnd an integrating factor and solve the given equation.
25. (3ny +ny +y3)dx + (x2 +y2)dy = 0 26. y’ : emf); n i
27. dxi—(x/y—Siuy)dy=0 28. ydx+(2xy—e'23')dy=0
29. exdx+tercoty+2ycscy)dy:0 ,
30 [4(x3/y2) + (3/301 dx + [30502) + 4y] 013’ ﬂ 0
31. (3x+ 9) + (5: +33) 53 = 0
y y x dx Hint: See Problem 24. 32. Solve the differéntial equation (3):)» + yz) + (x2 + xy)y' = 0 using the integrating factor ,u.(x, y) : [xy(2x + y)]“1. Verify that the solution is the same
as that obtained in Example 4 with a different integrating factor. Recall two important facts about the ﬁrst order initial value problem _‘ _f(r1y)a yUO) = yo. (L) dt First, if f and Elf/3y are continuous, then the initial value problem (1) has a unique
solution y = $0) in some interval surrounding the initial point t = to. Second, it is
usually not possible to ﬁnd the solution (f) by symbolic manipulations of the differ—
ential equation. Up to now we have considered the main exceptions to the latter
statement: differential equations that are linear, separable, or exact, or that can be
transformed into one of these types. Nevertheless, it remains true that solutions of
the vast majority of ﬁrst order initial value problems cannot be found by analytical
means, such as those considered in the ﬁrst part of this chapter. Therefore it is important to be able to approach the problem in other ways. As we
have already seen, one of these ways is to draw a direction ﬁeld for the differential
equation (which does not involve solving the equation) and then to visualize the
behavior of solutions from the direction ﬁeld. This has the advantage of being a
relatively simple process, even for complicated differential equations. However, it
does not lend itself to quantitative computations or comparisons, and this is often a
critical shortcoming. For example, Figure 2.7.1 shows a direction ﬁeld for the differential equation w=3—ZI—%y. (2) From the direction ﬁeld you can visualize the behavior oiusohgions on the rectangle
shown in the ﬁgure. A solution starting at a point on the y—axis initially increases with
3‘, but it soon reaches a maximum value and then begins to decrease as t increases
further. 7' ...
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 Summer '10
 Chatterjee
 Boundary value problem

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