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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
Download (direct link): elementarydifferentialequat2001.pdf
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dt
2
therefore y(t) = —gt /2 + (usinA)t + d2. Since y(0) = h we have d2 = h and y(t) = —gt /2 + (usinA)t + h.
31d. Let t,, be the time the ball reaches the wall. Then
w
L
x(tw) = L = (ucosA)tw and thus tw = --------------. For the ball
w w w ucosA
to clear the wall y(tw) > H and thus (setting L
tw = --------, g = 32 and h = 3 in y) we get
w ucosA
—16L2
--------- + LtanA + 3 > H.
22 u cos A
—161.98 sinA
31e. Setting L = 350 and H = 10 we get --------------- + 350------ > 7
cos2A cosA
2
or 7cos A — 350cosAsinA + 161.98 < 0. This can be solved numerically or by plotting the left side as a function of A and finding where the zero crossings are.
31f. Setting L = 350, and H = 10 in the answer to part d
18
Section 2.4
2
-16(350) 2
yields ------------ + 350tanA = 7, where we have chosen the
22 u cos A
equality sign since we want to just clear the wall.
2 2 1,960,000
Solving for u we get u = ----------------------. Now u will
175sin2A-7cos2A
have a minimum when the denominator has a maximum. Thus
350cos2A + 7sin2A = 0, or tan2A = -50, which yields
A = .7954 rad. and u = 106.89 ft./sec.
Section 2.4, Page 72
I. If the equation is written in the form of Eq.(1), then p(t) = (lnt)/(t-3) and g(t) = 2t/(t-3). These are defined and continuous on the intervals (0,3) and (3,<^>), but since the initial point is t = 1, the solution will be continuous on 0 < t < 3.
2
4. p(t) = 2t/(2-t)(2+t) and g(t) = 3t /(2-t)(2+t).
8. Theorem 2.4.2 guarantees a unique solution to the D.E.
22
through any point (t0,y0) such that t0 + y0 < 1 since df 221/2
— = -y(1-t -y ) is defined and continuous only for dy
1-t2-y2 > 0. Note also that f = (1-t2-y2)1/2 is defined
and continuous in this region as well as on the boundary 22
t +y = 1. The boundary can't be included in the final
df
region due to the discontinuity of — there.
dy
1+t2 df 1+t2 1+t2
II. In this case f = ------------ and — = --- - ---------,
y(3-y) dy y(3-y)2 y2 (3-y)
which are both continuous everywhere except for y = 0 and y = 3.
13. The D.E. may be written as ydy = -4tdt so that 2
y 2 2 2
— = -2t +c, or y = c-4t . The I.C. then yields 2
2 2 2 2 2 2 y0 = c, so that y = y0 - 4t or y = ±v y0-4t , which is
defined for 4t2 < y0 or |t| < |y0|/2. Note that y0 n 0 since Theorem 2.4.2 does not hold there.
17. From the direction field and the given D.E. it is noted
that for t > 0 and y < 0 that y' < 0, so y ^ -• for
y0 < 0. Likewise, for 0 < y0 < 3, y' > 0 and y' ^ 0 as
Section 2.4
19
y ^ 3, so y ^ 3 for 0 < y0 < 3 and for y0 > 3, y' < 0 and again y' ^ 0 as y^ 3, so y^3 for y0 > 3. For y0 = 3, y' = 0 and y = 3 for all t and for y0 = 0, y' = 0 and y = 0 for all t.
22a. For y1 = 1-t, y1 = -1 =
-t+[t2+4(1-t) ] 1/2 2
-t+[(t-2)2] 1/2 2
-t+|t-2|
= -1 if 2
(t-2) > 0, by the definition of absolute value. Setting t = 2 in y1 we get y1(2) = -1, as required.
22b. By Theorem 2.4.2 we are guaranteed a unique solution only
-t+(t2+4y)1/2 2 _!/2
where f(t,y) = ----------------- and fy(t,y) = (t +4y) ' are
2 y
continuous. In this case the initial point (2,-1) lies
2 df
in the region t + 4y < 0, in which case — is not
3y
continuous and hence the theorem is not applicable and there is no contradiction.
22
22c. If y = y2(t) then we must have ct + c = -t /4, which is
not possible since c must be a constant.
23a. To show that f(t) = e2t is a solution of the D.E., take
its derivative and substitute into the D.E.
24. [c f(t)]' + p(t)[cf(t)] = c[f'(t) + p(t)f(t)] = 0 since f(t) satisfies the given D.E.
25. [y1(t) + y2(t)]' + p(t)[y1(t) + y2(t)] =
y'i(t) + p(t)yi(t) + y2 (t) + p(t)y2(t) = 0 + g(t).
27a. For n = 0,1, the D.E. is linear and Eqs.(3) and (4) apply.
_ T ^ 1-n ^ dv , -ndy dy 1 ndv
27b. Let v = y then --------- = (1-n)y so = y ,
dt dt dt 1-n dt
which makes sense when n n 0,1. Substituting into the n
y dv n
D.E. yields --------- + p(t)y = q(t)y or
1-n dt
20
Section 2.4
v' + (1-n)p(t)y1 n = (1-n)q(t). Setting v = y1n then yields a linear D.E. for v.
-2 n dv „ -3dy dy 1 3dv
28. n = 3 so v = y and --------------- = -2y — or = -— y ----.
dt dt dt 2 dt
Substituting this into the D.E. gives
1 3 dv 2 1 3
- — y — + —y = —y . Simplifying and setting
2 dt t t2
-2
y = v then gives the linear D.E.
4 2 1
v' - —v = -—, where m (t) = — and
t t2 t4
4 2 2+5ct 5 1/2
v(t) = ct + ---- = -----------. Thus y = ±[5t/(2 + 5ct )] / .
5t 5t
-1 dv 2 dv
29. n = 2 so v = y and ----- = -y —. Thus the D.E.
dt dt
2 dv 2 dv
becomes -y ------ - ry = -ky or — + rv = k. Hence
dt dt
rt -rt
m(t) = e and v = k/r + ce . Setting v = 1/y then
yields the solution.
32. Since g(t) is continuous on the interval 0 < t ? 1 we may solve the I.V.P.
y1 + 2y1 = 1, y1(0) = 0 on that interval to obtain
y1 = 1/2 - (1/2)e-2t, 0 < t < 1. g(t) is also continuous
/
for 1 < t; and hence we may solve y2 + 2y2 = 0 to obtain
-2t
y2 = ce , 1 < t. The solution y of the original I.V.P.
must be continuous (since its derivative must exist) and hence we need c in y2 so that y2 at 1 has the same value as y1 at 1. Thus
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