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Unformatted text preview: 18.03 PSI Solutions—Spring 09 PartB: {1. 6pts=2+2+2];[2. 9pts=2+1+1+2+3]; [3. 5pts=2+2+1] 1a. dT dT .
ﬁ—k(30—'T) => m——kdt => ln(T—30)=—kt+c => T—30=Ae“'“ (A :6“) T(0)=70 => ; T(24)=50 => k=—2111n2~.o3 1b. T’ = k(30 + q/k — T). Same type of equation as (a) with 30 + q/k in place of 30: T=30+%—Ae‘kt As t —> oo, 6"“ ——> 0, so steady state T: 30 + q/k
1c. 30+%=60,30+q—;=70 => q1=30k,q2=40k => 3—1:
2 r5105 2a. See Figure 1.
2b. Answer: .73; 31(0) 2 .74 = curve goes up; 31(0) S .72 => curve goes down. Thus .72<A<.74
y’=0 => or b=:l:\/E,a20 2d. f(w) = —\/5 works (other answers are ok). Direction ﬁeld is horizontal but f’ (x) =
—%a:‘3/2 < 0 so direction ﬁeld points into y > —\/E and solutions can’t cross from above. See Figure 2. 2c. 29. Same idea as (d). We will check that z is a subsolution (2’ < z2 —a:) and hence solutions
that begin above the graph of 2 stay above it. dz—lda: => 2—1—1:r+c : z— 1 '
22—2 ‘2 _c+a:/2 ’
1 2
2(0)—2 = C—"'2' => 21—1—3: Notice that z —> 00 as x —> 1, so we can focus on a: < 1. Comparing slopes: 2 2 2 2
I 2_ __ _ 1__ 2 2
z<z z ¢=> (l—m)2<<1—x> a: 4:: $<(1——x)2 4:» :c( as) < which is true for 0 < a: < 1, hence z’ < 22 — a3, i.e., z is a subsolution. The graph of this
inequality (Figure 3) shows direction ﬁelds that point into the region above 2(w). Solution
curves are trapped in y > 2(15). They can enter, but can’t leave. In symbols, y(0)>2 => y(m)>z(m)forallm ==> y—>ooasm—>x1forsomea:1$1. 3a. Given y = ace2 — 3x, a = (y + Saw/$2. ' 2 ‘3
y’=2a,z—3=2(”+23”)x—3=2§+3 = 3/: 9+ ”3
{I} {I} ~ (I! 3b. Y=ky=k(am2—3m)=%k2w2—3k¢ =>v Y=§XL3X ...
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