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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 5-y'=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 First-Order 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 half-life. We now know that the half-life is closer to 5730 years, called the Cambridge half-life, but radio— carbon dating labs still use the Libby half-life 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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