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# Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
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27a. By sketching the direction field and using the D.E.we note that y' < 0 for y > 4 and y' approaches zero as y approaches 4. For 0 < y < 4, y' > 0 and again approaches zero as y approaches 4. Thus limy = 4 if y0 > 0. For
t
y0 < 0, y' < 0 for all y and hence y becomes negatively unbounded (-) as t increases. If y0 = 0, then y' = 0
for all t, so y = 0 for all t.
27b. Separating variables and using a partial fraction
1 1 4
expansion we have ( - )dy =  tdt. Hence
y y-4 3
ln | y | =  t2 + c1 and thus | | = ec1e2t2/3 = ce2t2/3,
y-4 3 1 y-4
where c is positive. For y0 = .5 this becomes
y 2t2/3 .5 1
= ce and thus c = =  . Using this value
4 -y 3.5 7
12
Section 2.2
for c and solving for y yields y(t) = ---------------------.
1 + 7e2t2/3
Setting this equal to 3.98 and solving for t yields t = 3.29527.
cy+d
29. Separating variables yields dy = dx. If a n 0 and
ay+b
ay+b n 0 then dx = ( +-------------)dy. Integration then
a a(ay+b)
dy dv
3 0c. If v = y/x then y = vx and  = v + x and thus the
dx dx
dv v-4
D.E. becomes v + x = -----------. Subtracting v from both
dx 1-v
dv v2-4
sides yields x =
dx 1-v
1-v 1
30d. The last equation in (c) separates into dv = dx. To
v2-4 x
integrate the left side use partial fractions to write
1-v A B ,
= ------ + -----, which yields A = -1/4 and B = -3/4.
v-4 v-2 v+2
13
Integration then gives - ln|v-2| - ln|v+2| = ln|x| - k,
44
ln|x4||v-2||v+2|3 = 4k after manipulations using properties of the ln function.
31a. Simplifying the right side of the D.E. gives
dy/dx = 1 + (y/x) + (y/x) so the equation is homogeneous.
31b. The substitution y = vx leads to
dv 2 dv dx
v + x = 1 + v + v or ----------- = . Solving, we get
dx 1 + v2 x
arctanv = ln|x| + c. Substituting for v we obtain arctan(y/x) - ln|x| = c.
33b. Dividing the numerator and denominator of the right side
dv 4v - 3
by x and substituting y = vx we get v + x = ----------
dx 2-v
2
dv v2 + 2v - 3
which can be rewritten as x = --------------. Note that
dx 2 - v
v = 3 and v = 1 are solutions of this equation. For
v n 1, 3 separating variables gives
or
Section 2.3
13
2 - v 1
------------- dv = dx. Applying a partial fraction
(v+3)(v-1) x
decomposition to the left side we obtain
11 5 1 dx
[------------ ]dv = , and upon integrating both sides
4 v-1 4 v+3 x
15 we find that ln|v-1| - ln|v+3| = ln|x| + c.
44
Substituting for v and performing some algebraic
manipulations we get the solution in the implicit form
|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.
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