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# Sect1.1 - SECTVON 1.1 Phone/n ‘1> with(DEtools>...

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Unformatted text preview: SECTVON 1.1 Phone/n ‘1 > with (DEtools) : > eqn:=diff(y(t)}t)=—1-2*y(t); 8 ‘(t)=-1-2)I(t) e :=- qn Qty > DEplot (eqn, [t, y] ,t=-2 . .2, arrows=THIN,y=-2 . .2) ; i‘ivs\v\.v\v\\v\\ _ «.Ekuuufviu.\\‘\v\ \1 _ tri‘.1§.‘1\\n1\\\ \vivk‘iv..\\1\.\\u IE5\\\¥\V\4\V\\‘\ ‘illi‘mv\V\\-\\v\‘\‘ 511\$\.v\.v\v\\r\\\ svkiviviumqiu\\\\ tL.rLT\\1\~\.V\V\.\\ ...\......v\.v...\\v.\1\‘\:\\ \1 1{\\V\V\V\V\\\ .......\1\l.\\\1.\‘\w\v\ \1 iv..L.\.v\.v\v1.v\.v\.\\. \.v..\\r\v\v\\\\.\\1 {\\r\v\v\v\\\. ivivivi»\v\\v\\\. inti\$N\\\v\\..\u\\ a“. x x 4W» i \ x i x \ l \ aaanwaﬁw-‘a—‘a Reﬁaraea’a,’ ////////// ////////// “MMNHHHWH HH‘N‘“H“‘ //«/4/¢lol 13/1/17]. //f/II ///..l./rr //f/Il ./a//.,IG/4.I //ff..1.r.. /c/4/4I1..ul v/u/c/a/frl! l/ff/I. zl/llh ////Il //./.t/«lfr //¢/../I.I. o/a/y/IL/cl //¢/4/414l. ///f.f.f. ////if ///f.1.¢l //f.4/.II. As t —> infinity the solution y-> -1/2 for all initial conditions . , [JP-1 PROBLE" 0 > with (DEtools) :. > eqnz=diff (y (t) ,t)=y(t) +2; 3 eqn .= atym —y(t) + 2 > DEplot (eqn, [t\,y] ,t=-2 . .2, arrows=THIN,y=-4 . .0) ; > ' ' ///’/////// \\\\\\\\\\' \\\\\\\\\.\ \\\\\\\\\\ \\\\\\\\\\ \\\\\\\\\\ \\\\\\\\\\. As t —> infinity the solution y—> infinity, negative infinity or -2 depending on the initial condition. If y(0)>-2 then y¥> infinity as t-> infinity. If y(0)=-2 then y=-2 for all t.‘ If y(0) <—2 then‘y-> negative infinity as t-> infinity. T.’ pa \ Section 1.1, Problem 9 Problem. Write down a differential equation of the form (711% = ay + b Whose solutions have the required behavior as t -> 00: All other solutions diverge from y = 2; Solution. y = 2 must be an equilibrium solution; when y = 2, j—lﬁ = 0. Therefore 01(2) + b i 0 and b = —2a (see Example 3). The solutions must diverge; ‘ so we want a to be positive; for example, let a,= 2. Then an (not the only) equation is In Maple, one can check the solution by entering: > with(DEtools): > eqn := diff(y(t), t) = 2 * y(t) - 4; > DEplot(eqn, y(t)«, t = -2.4V.2, arrows = THIN, y = -2. .4); (Tim Section 1.1, Problem 12 Problem. Based on the direction ﬁeld, determine the behavior of y as t ’—-+ 00. If this behavior depends on the initial value of y at t = 0, describe this dependency: M=-w®—y) Solution. To draw the direCtion ﬁeld, in Maple enter: > with(DEtools)': L > eqn := diff(y(t), t) = -y(t) * (5 - y_(t)); > DEplot(eqn, y(t), t = -2.-.2? arrows = THIN, y = -2‘.-.4); exxxx““xx\x\\\\x 5\\\ii. ii\\x\x;i killil ;5llllllél ‘ \vhv‘lr »‘ I‘lt‘v )1 \xallx eiixlxliir \\»\\\ xxxrlyxxx “WNW \HMHH \\xxxx xxxrix\\\ f‘illll 'ii‘l‘lilili \‘l\:\‘:\4\i\v‘-\.\i\:\v\\\:\v\ ‘\\\\\\\ \\\\\\\\\ ‘ e ‘ .\\\\\:T\ \\\\\\\\\\\‘\ ‘2" ' a",/ ‘//’/‘ w .. .- . ///’4“ /d ////////////V/{// / .ZZZI/f/7/{JIIZ4ZI g ;/.;l!7-fiJ/,izii/;Z / flillfliléiézllli- l ,liill—emmlml There appear to be two equilibria, one at y = 0 and another at y = 5 (as solving for y’ = 0 indicates). If the initial condition is 31(0) < 0, then y(t) will increase and approach y = 0. If the initial condition is 31(0) L> 0 but also 31(0) < 5, then y(t) decreases and approaches 3;. = 0. If [the initial condition is y(0) >‘5, then y(t) grows without bound. 1 ’o'ﬁq Section 1.1, Problem 16 Problem. A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time. Solution. The most likely best choice for the independent variable in this problem is time, as one is told to write a differential equation for the volume of the raindrop as a function of time. The most likely best choice for dependent variable is the volume of the raindrop, as that is the quantity whose rate of change one is ’ trying to model. .At a given time t its volume would be written as V(t). Given that one is modeling a raindrop, the time t is probably best measured in seconds or minutes; let us take seconds. The volume is probably best measured in cubic centimeters. (The units turn out not to affect this particular question, which is only setting up the model, but they can affect the Solution of the differential equation, and selecting the units is worth practicing.)- By assumption, one knows that the rate of change of the volume V(t) of the raindrop is proportional to its surface area A(t), and its volume is decreasing; so one may immediately write 3% = —kA(t) (with it understood that k, the rate of evaporation, is positive). ' To complete the model, one needs to know the area A(t) of the raindrop in terms of the dependent variable V(t). As the raindrop is by assumption spherical, one knows that the volume V = 117W3, and the surface area A = 47(7’2. 3 One can then solve the ﬁrst expression for r, and ﬁnd that r = 3 34%. Sub- stltutmg that expressmn for r 1n the area expressron gives one A = 47r(%)§, or A = 3%(47r)%v%. Then, ﬁnally, that turns the expression for % into: 2—? = —k3%(47r)%v, (Of course, since 3§(47r)% is just a number, one could replace the constant k with the constant K 2. 3% (470% k, and simplify the equation further to 43:— : —-KV§ .) calm Ill”; Section 1.1, Problem 17 Problem. A certain drug is being administeredintravenously to a hospital patient. Fluid containing 52% of the drug enters the patient’s bloodstream at arate of 100%. The drug is absorbed by bodytis’sues or otherwise leaves the bloodstream at a rate proportional to the amount present, with ‘a rate constant of 0.4%. a. Assuming that the drug is always uniformly present throughout the blood- stream, write a differentialeq-uation .for the amount of the drug that is present in the bloodstream at any time. , Solution. The independent variable is'tinie 't, measured in hours (based on the problem’s statement). The dependent variable is the amount of drug in the system; call it M (t) and measure it in milligrams. At any given time t, the amount of drugfin the system is increasing at the rate of Syn—3; - 100%";3 = 500%}; and it is decreaSing at the rate of 0.4M (wig—f A differential equation describing the rate of change is therefore ’ dM 73? = —0.4M(t) + 500 b. How much of the drug is present in the bloodstream after a long time? Solution. The onlyequilibrium of the system is when —0.4M + 500 = 0, or M = 1250. The question is then whether solutions converge to, or diverge from, this equilibrium. As the equation has the form 4%: TM + k, with r = —0.4.a negative number, it seems likely solutions will converge to M = 12507719 This can be checked: In Maple, enter the commands: > with(DEtools) :_ l _ > eqn := diff (M(t) , t) = -- 0.4 * M(t) ‘+' 500; ‘ > DEplot(eqn, M(t) , t = 0. .10, arrows = THIN, M' =’ 0. .2000); So after a long time there will be. about 1250mg in the bloodstream. ' . 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Sect1.1 - SECTVON 1.1 Phone/n ‘1> with(DEtools>...

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