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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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|y-x| = c|y+3x|5. v = 1 and v = -3 yield y = x and y = -3x, respectively, as solutions also.
35b. As in Prob.33, substituting y = vx into the D.E. we get
dv 1+3v dv (v+1)2
v + x = --------, or x = ----------. Note that v = -1 (or
dx 1-v dx 1-v
y = -x) satisfies this D.E. Separating variables yields
1-v dx
------dv = . Integrating the left side by parts we
(v+1) 2 x
v-1 I I I I y
obtain ----- - ln | v+1 | = ln | x | + c. Letting v = then
v+1 x
y-x y+x y-x
yields------ - ln | --- | = ln | x | + c, or -- - ln | y+x | = c.
y+x x y+x
This answer differs from that in the text. The answer in the text can be obtained by integrating the left side, above, using partial fractions. By differentiating both answers, it can be verified that indeed both forms satisfy the D.E.
Section 2.3, Page 57
1. Note that q(0) = 200 gm, where q(t) represents the amount of dye present at any time.
2. Let S(t) be the amount of salt that is present at any time t, then S(0) = 0 is the original amount of salt in the tank,
2 is the amount of salt entering per minute, and 2(S/120) is the amount of salt leaving per minute (all amounts measured in grams). Thus
dS/dt = 2 - 2S/120, S(0) = 0.
3. We must first find the amount of salt that is present after
10 minutes. For the first 10 minutes (if we let Q(t) be the amount of salt in the tank):
= (2) - 2, Q(0) = 0. This I.V.P. has the solution: Q(10)
= 50(1-) 9.063 lbs. of salt in the tank after the first 10
14
Section 2.3
minutes. At this point no more salt is allowed to enter, so the new I.V.P. (letting P(t) be the amount of salt in the tank) is:
= (0)(2) - 2, P(0) = Q(10) = 9.063. The solution of this problem is P(t) = 9.063, which yields P(10) = 7.42 lbs.
4. Salt flows into the tank at the rate of (1)(3) lb/min. and it flows out of the tank at the rate of (2) lb/min. since the volume of water in the tank at any time t is 200 +
(1)(t) gallons (due to the fact that water flows into the tank faster than it flows out). Thus the I.V.P. is dQ/dt =
3 - Q(t), Q(0) = 100.
7a. Set = 0 in Eq.(16) (or solve Eq.(15) with S(0) = 0).
7b. Set r = .075, t = 40 and S(t) = $1,000,000 in the answer to
(a) and then solve for k.
7c. Set k = $2,000, t = 40 and S(t) = $1,000,000 in the answer
to (a) and then solve numerically for r.
9. The rate of accumulation due to interest is .1S and the rate
of decrease is k dollars per year and thus
dS/dt = .1S - k, S(0) = $8,000. Solving this for S(t) yields S(t) = 8000 - 10k(-1). Setting S = 0 and substitution of t = 3 gives k = $3,086.64 per year. For 3 years this totals $9,259.92, so $1,259.92 has been paid in interest.
10. Since we are assuming continuity, either convert the monthly
payment into an annual payment or convert the yearly
interest rate into a monthly interest rate for 240 months. Then proceed as in Prob. 9.
12a. Using Eq. (15) we have - S = 800(1) or
S (800t), S(0) 100,000. Using an integrating factor and
integration by parts (or using a D.E. solver) we get S(t) t c. Using the I.C. yields c . Substituting this value into S, setting S(t) 0, and solving numerically for t yields t 135.363 months.
16a. This problem can be done numerically using appropriate D.E. solver software. Analytically we have
(.1.2sint)dt by separating variables and thus y(t) cexp(.1t.2cost). y(0) 1 gives c , so
y(t) exp(.2.1t.2cost). Setting y 2 yields ln2 .2 .1 .2cos, which can be solved numerically to give
2.9632. If y(0) , then as above,
y(t) exp(.2.1t.2cost). Thus if we set y 2 we get the
Section 2.3
15
same numerical equation for and hence the doubling time has not changed.
18. From Eq.(26) we have 19 = 7 + (20-7) or
k = -ln = ln(13/12). Hence if (T) = 15 we get: 15 = 7 +
13. Solving for T yields
T = ln(8/13)/-ln(13/12) = ln(13/8)/ln(13/12) min.
19. Hint: let Q(t) be the quantity of carbon monoxide in the
room at any time t. Then the concentration is given by x(t) = Q(t)/1200.
20a. The required I.V.P. is dQ/dt = kr + P - r,
Q(0) = V. Since c = Q(t)/V, the I.V.P. may be rewritten Vc(t) = kr + P - rc, c(0) = , which has the solution c(t) = k + + ( - k - ).
20b. Set k = 0, P = 0, t = T and c(T) = .5 in the solution found
in (a).
21a.If we measure x positively upward from the ground, then
Eq.(4) of Section 1.1 becomes m = -mg, since there is no air resistance. Thus the I.V.P. for v(t) is dv/dt = -g, v(0) = 20. Hence = v(t) = 20 - gt and x(t) = 20t - (g/2) + c. Since x(0) =
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