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Unformatted text preview: SECTVON 1.1 Phone/n ‘1 > with (DEtools) : > eqn:=diff(y(t)}t)=—12*y(t); 8 ‘(t)=12)I(t) e :=
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//f.4/.II. As t —> infinity the solution y> 1/2 for all initial conditions . , [JP1 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) ;
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\\\\\\\\\\ \\\\\\\\\\. 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
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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 differentialequation .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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 Spring '04
 Yoon
 Differential Equations, Equations

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