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3. Compare your final solution with the one provided to see whether yours is more or less efficient than the guide, since there is frequently more than one correct way to solve a problem.
4. Ask yourself why that particular problem was assigned.
The use of a symbolic computational software package, in many cases, would greatly simplify finding the solution to a given problem, but the details given in this solutions manual are important for the student's understanding of the underlying mathematical principles and applications. In other cases, these software packages are essential for completing the given problem, as the calculations would be overwhelming using analytical techniques. In these cases, some steps or hints are given and then reference made to the use of an appropriate software package.
In order to simplify the text, the following abbreviations have been used:
D.E. differential equation(s)
0.D.E. ordinary differential equation(s)
P.D.E. partial differential equation(s)
1.C. initial condition(s)
I.V.P. initial value problem(s)
B.C. boundary condition(s)
B.V.P. boundary value problem(s)
I wish to express my appreciation to Mrs. Susan A. Hickey for her excellent typing and proofreading of all stages of the manuscript. Dr. Josef Torok has also provided invaluable assistance with the preparation of the figures as well as
assistance with many of the solutions involving the use of
symbolic computational software.
Charles W. Haines Professor of Mathematics and Mechanical Engineering Rochester Institute of Technology Rochester, New York June 2000
CHAPTER 1 1
CHAPTER 2 6
CHAPTER 3 35
CHAPTER 4 64
CHAPTER 5 ............................................... 73
CHAPTER 6 .............................................. 103
CHAPTER 7 121
CHAPTER 8 ............................................. 16 0
CHAPTER 9 .............................................. 174
CHAPTER 10 .............................................. 2 03
Section 1.1, Page 8
2. For y > 3/2 we see that y' > 0
and thus y(t) is increasing there. For y < 3/2 we have y' < 0 and thus y(t) is decreasing there. Hence y(t) diverges from 3/2 as t^ro.
4. Observing the direction field, we see that for y>-1/2 we have y'<0, so the solution is decreasing here. Likewise, for y<-1/2 we have y'>0 and thus y(t) is increasing here.
Since the slopes get closer to zero as y gets closer to -1/2, we conclude that y^-1/2 as t^ro.
If all solutions approach 3, then 3 is the equilibrium
solution and we want ---------- < 0 for y > 3 and > 0 for
y < 3. Thus -------- = 3-y.
11. For y = 0 and y = 4 we have y' = 0 and thus y = 0 and
y = 4 are equilibrium solutions. For y > 4, y' < 0 so if y(0) > 4 the solution approaches y = 4 from above. If 0 < y(0) < 4, then y' > 0 and the solutions "grow" to y = 4 as t. For y(0) < 0 we see that y' < 0 and the solutions diverge from 0.
13. Since y' = y2, y = 0 is the equilibrium solution and y' > 0 for all y. Thus if y(0) > 0, solutions will diverge from 0 and if y(0) < 0, solutions will aproach y = 0 as tro.
15a. Let q(t) be the number of grams of the substance in the
water at any time. Then = 300(.01)
300(10 2- 10 6q).
15b. The equilibrium solution occurs when q = 0, or c = 10 gm,
independent of the amount present at t = 0 (all solutions
approach the equilibrium solution).
16. The D.E. expressing the evaporation is ---------------- = - aS, a > 0.
4 3 2
Now V = pr and S = 4pr , so S = 4P 3
Thus --------- = - kV , for k > 0.
Section 1.2 Page 14
1b. dy/dt = -2y+5 can be rewritten as = -2dt. Thus
ln I y-5/2 I = -2t + c1, or y-5/2 = ce-2t. y(0) = y0 yields
c = y0 - 5/2, so y = 5/2 + (y0-5/2)e-2t.
If y0 > 5/2, the solution starts above the equilibrium solution and decreases exponentially and approaches 5/2 as t. Conversely, if y < 5/2, the solution starts below 5/2 and grows exponentially and approaches 5/2 from
below as t<^>.
dy/dt | |
4a. Rewrite Eq.(ii) as ---------- = a and thus ln | y | = at+c1(- or
y = ceat.
4b. If y = y1(t) + k, then ------ = ----. Substituting both
these into Eq.(i) we get = a(y1+k) - b. Since
= ay1, this leaves ak - b = 0 and thus k = b/a.
Hence y = y1(t) + b/a is the solution to Eq(i).
6b. From Eq.(11) we have p = 900 + cet/2. If p(0) = p0, then
c = p0 - 900 and thus p = 900 + (p0 - 900)et/2. If p0 < 900, this decreases, so if we set p = 0 and solve for T (the time of extinction) we obtain eT/2 = 900/(900-p0), or T = 2ln[900/(900-p0)].
8a. Use Eq.(26).
8b. Use Eq.(29).
dQ dQ/dt I I
10a. = -rQ yields ---------- = -r, or ln | Q | = -rt + c1. Thus
dt Q 1
Q = ce-rt and Q(0) = 100 yields c = 100. Hence Q = 100e-rt.
Setting t = 1, we have 82.04 = 100e-r, which yields
r = .1980/wk or r = .02828/day.
13a. Rewrite the D.E. as -------- = , thus upon integrating and