HW 1 Problems

HW 1 Problems - | l l i i i 1' HS '- [p- 0) ins u) 12) v0,...

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Unformatted text preview: | l l i i i 1' HS '- [p- 0) ins u) 12) v0, l3) :1), in la— ed Ell- lu- ed 2 I Linea,- Equations; Method of Integrating Factors 39 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 +ye-Idy : 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 arm-112, 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 Zx-l-y d 3 @4335. 14% 3’ @‘52/36.(x2+3xy+y2)dx-x2dy=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 time-consuming 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 E l l r l (r 23 Modeling with First Order Equations 59 and Eq. (26) is replaced by 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 well-stirred 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 well-stirred 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 E E 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 i E e i l l E E tof the 10h his se ,al 13 is, is w E g E i 1: '7' i i i ; 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 3f the :l HOWE lepth f; .cient able. ffish ition why :iny d in 1ght 3W6 ant, s of (i) vho sly. .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. : l l l l l l l l l ‘r i j 92 24. (a) Determine y at any time t by solving Eq. (i) subject to the initial condition y(0) 2 yo, (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~year-olds 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 II E ; r i l t E l l 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. dx-i—(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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This note was uploaded on 01/16/2011 for the course MECH&AE 182A taught by Professor Chatterjee during the Summer '10 term at UCLA.

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HW 1 Problems - | l l i i i 1' HS '- [p- 0) ins u) 12) v0,...

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