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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.
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6 1 6 2 2
t/6 Ëð¿¿ 2 ëÃ¯¯
u(t) = e ' (2cos--------1 - sin-----1).
6 V23 6
23b. To estimate the first time that | u(t) | = 10 plot the graph of u(t) as found in part (a). Use this estimate in an appropriate computer software program to find t = 10.7598.
2
25a.The characteristic equation is r + 2r + 6 = 0, so
r1,r2 = -1 ± yfb i and y(t) = e-t(c1cos1 + c2sin^/"5t).
Thus y(0) = c1 = 2 and y'(0) = -c1 + \/~5 c2 = a and hence
-t /— a+2 ³—
y(t) = e (2co^5 t + _ sin y5 t).
-1 /— a+2 ³—
25b. y(1) = e (2cosy5 + ,_ sin ó 5 ) = 0 and hence
V 5
2Ó 5
a = -2 - -------- = 1.50878.
tanv 5
t— a+2 t—
25c. For y(t) = 0 we must have 2cosy 5 t + sl^5 t = 0 or
V 5
I— -2 V 5
tany 5 t = ---------. For the given a (actually, for a > -2)
a+2
I— 2 ó 5
this yields V 5 t = p - arctan - since arctan x < 0 when
a+2
x < 0.
2\J~5
2 5d. From part (c) arctan --------- ^ 0 as a ^ •, so t ^ p
a+2
31. -d[e1t(cosmt + isinmt)] = 1e1t(cosmt + isinmt)
dt
+ eN-msinmt + imcosmt) = ie1t(cosmt + isinmt)
+ ime1t(isinmt + cosmt) = e1t(i+im) (cosmt + isinmt).
d rt rt
Setting r = l+im we then have---------e = re .
dt
33. Suppose that t = a and t = b (b>a) are consecutive zeros of y1. We must show that y2 vanishes once and only once
in the interval a < t < b. Assume that it does not
46
Section 3 . 4
35.
38.
vanish. Then we can form the quotient Ó]_/Ó2 on the
interval a < t < b. Note y2(a) Ï 0 and y2(b) Ï 0,
otherwise y1 and y2 would not be linearly independent
solutions. Next, y1/y2 vanishes at t = a and t = b and
has a derivative in a < t < b. By Rolles theorem, the derivative must vanish at an interior point. But
Ó³ ó1ó2 - Ó2Ó³ "ì(Ó³,Ó2)
(—)' = ------------- = -----------, which cannot be zero
y 2 2
Ó2 Ó 2 Ó 2
since y1 and y2 are linearly independent solutions.
Hence we have a contradiction, and we conclude that y2 must vanish at a point between a and b. Finally, we show that it can vanish at only one point between a and b. Suppose that it vanishes at two points c and d between a and b. By the argument we have just given we can show that y1 must vanish between c and d. But this contradicts
the hypothesis that a and b are consecutive zeros of y1.
-t2
We use the result of Problem 34. Note that q(t) = e > 0 for -• < t < •. Next, we find that (q' + 2pq)/q3/2 = 0. Hence the D.E. can be transformed into an equation with
r -t2/2
constant coefficients by letting x = u(t) = J e dt. Substituting x = u(t) in the differential equation found in part (b) of Problem 34 we obtain, after dividing by the
2 2 2 2 coefficient of dy/dx , the D.E. dy/dx - y = 0. Hence the
general solution of the original D.E. is y = c1cosx + c2sinx,
-t2/2 = J e dt.
Rewrite the D.E. as y" + (a/t)y' + (b/t2)y = 0 so that p = a/t and q = b/t2, which satisfy the conditions of parts (c) and (d) of Problem 34. Thus 2 1/2
x = J(1/t ) ' dt = lnt will transform the D.E. into
dy2/dx2 + (a-1)dy/dx + by = 0. Note that since b is constant, it can be neglected in defining x.
39. By direct substitution, or from Problem 38, x = lnt will
22
transform the D.E. into dy/dx + y = 0, since a = 1 and b = 1. Thus y = c1cosx + c2sinx, with x = lnt, t > 0.
Section 3.5
Section 3.5, Page 166
14.
rt
Substituting y = e
2
r - 2r + 1 = 0, which gives r1
into the D.E., we find that
1 and r
2
1. Since the
roots are equal, the second linearly independent solution is tefc and thus the general solution is y = c1et + c2tefc.
The characteristic equation is 25r - 20r + 4 = 0, which
2
may be written as (5r-2 ) r1,r2 = 2/5. Thus y = c1e
0 and hence the roots are
2t/5
+ c2 te
2t/5
12. The characteristic equation is r2 - 6r + 9 = (r-3)2, which has
the repeated root r = 3. Thus 3t 3t , . ,
y = c1e + c2te , which gives
y(0)
0, y'(t)
3t 3t
c2 (e +3te )
and y'(0) = c2
3t
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