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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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-e -e
1
ã 3(t-s) 2(t-s) \ n
[e - e ]g(s)ds.
The complete solution is then obtained by adding
3t 2t
c1e + c2e to Y(t).
14. That t and tet are solutions of the homogeneous D.E. can be verified by direction substitution. Thus we assume
Y = tu1(t) + tetu2(t). Following the pattern of earlier
problems we find tu^t) + tetu"2(t) = 0 and
56
Section 3 . 7
u—(t) + (t+1)etu2 = 2t. [Note that g(t) = 2t, since the D.E. must be put into the form of Eq.(16)]. The solution of these equations gives u—(t) = -2 and u^(t) = 2e-t.
Hence, u—(t) = -2t and u2(t) = -2e-t, and
t -t 2
Y(t) = t(-2t) + te (-2e ) = -2t - 2t. However, since t
is a solution of the homogeneous D.E. we can choose as
2
our particular solution Y(t) = -2t .
18. For this problem, and for many others, it is probably easier to rederive Eqs.(26) without using the explicit form for y—(x) and y2(x) and then to substitute for y—(x) and y2(x) in Eqs.(26). In this case if we take y— = x-1/2 sinx and y2 = x-1/2cosx, then W(y—,y2) = -1/x.
If the D.E. is put in the form of Eq.(16), then g(x) = 3x-1/2sinx and thus u— (x) = 3sinxcosx and u2(x) = -3sin x = 3(-1 + cos2x)/2. Hence
2
u—(x) = (3sin x)/2 and u2(x) = -3x/2 + 3(sin2x)/4, and
2
3 sin x sinx 3x 3 sin2x cosx
Y(x) =---------------— + ( - — + --------)
2
3 sin x sinx 3x 3 sinx cosx cosx
+ ( - — + ----------------------)
2 Vx 2
3 sinx 3\/^xcosx
2 V x 2
The first term is a multiple of y—(x) and thus can be neglected for Y(x).
22. Putting limits on the integrals of Eq.(28) and changing the integration variable to s yields
•t y2(s)g(s)ds ft y—(s)g(s)ds
fty2(s)g(s)ds , , ft
Y(t) = -y—(t) ³ ------------- + y2(t) ³
1 Jt„ W(y ,yj(s) 2 Jt,
Jt0 W(y—,y2)(s) JtQ W(y—,y2)(s)
³-t -y—(t)y2(s)g(s)ds j-
Jt0 W(y ,yj(s) Jt
Jt
t -y—(t)y2(s)g(s)ds ft y2(t)y—(s)g(s)ds
W(y—,y2)(s) W(y—,y2)(s)
t [y—(s)y2(t) - y—(t)y2(s)]g(s)ds
-----------------------------------. To show that
t0
y—(s)y2(s) - y —(s)y2(s)
Y(t) satisfies L[y] = g(t) we must take the derivative of
Y using Leibnitz's rule, which says that if
Y(t) = JtG(t,s)ds, then Y'(t) = G(t,t) + Jt —(t,s)ds.
Jt0 dt
Letting G(t,s) be the above integrand, then G(t,t) = 0
2
4
2
Section 3.8
57
dG y1(s)y2(t) - Ó³(t)y2(s)
and — = ----------------------------- g(s). Likewise
dt W(y1,y2)(s)
3G(t,t) ft d2G
1t
+ ---(t,s)ds
dt Jt0 at2
J‘ty1(s)y2(t) - yi (t)y2 (s)
-------------------------ds.
t0 W(y1,y2)(s)
Since y1 amd y2 are solutions of L[y] = 0, we have L[Y] = g(t) since all the terms involving the integral will add to zero. Clearly Y(t0) = 0 and Y'(t0) = 0.
25. Note that y1 = eXtcosmt and y2 = eXtsin^t and thus
2 It
W(yi,y2) = me . From Problem 22 we then have:
m t Is It ¦ ^ It ^ Is ¦
fc e cosmse sinmt - e cosmte sinms
Y(t) = ------------------------------------ g(s)ds
jt0 me2ls
= m"1 I e1(t-s)[cosms sinmt - cosmt sinms]g(s)ds t0
= m-1 1 e1(t-s) [sinm(t-s)]g(s)ds. t0
29. First, we put the D.E. in standard form by dividing by
t2: y" - 2y'/t + 2y/t2 = 4. Assuming that y = tv(t) and
substituting in the D.E. we obtain tv" = 4. Hence v (t) = 4lnt + c2 and v(t) = 4 I lnt dt + c2t =
4(tlnt - t) + c2t. The general solution is
ooo î
c1y1(t) + tv(t) = c1t + 4(t lnt - t ) + c2t . Since -4t
2
is a multiple of y2 = c2t we can write
22 y = c1t + c2t + 4t lnt.
Section 3.8, Page 197
2. From Eq.(15) we have Rcosd = -1, and Rsind = ó/13 . Thus
R = V i + 3 = 2 and 5 = tan-1 (-ä/3) + p = 2p/3 @ 2.09440. Note
that we have to "add" p to the inverse tangent value since 5 must be a second quadrant angle. Thus u = 2cos(t-2p/3).
6. The motion is an undamped free vibration. The units are in the CGS system. The spring constant k = (100 gm)(98 0cm/sec2)/5cm. Hence the D.E. for the motion is 100u" + [(100 • 980)/5]u = 0 where u is measured in cm and time in sec. We obtain u" + 196u = 0
58
Section 3 . 8
so
u = Acos14t + Bsin14t. The I.C. are u(0) = 0 ^ A = 0 and u'(0) = 10 cm/sec ^ B = 10/14 = 5/7. Hence u(t) = (5/7)sin14t, which first reaches equilibrium when 14t = p, or t = p/14.
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