Books
in black and white
Main menu
Share a book About us Home
Books
Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics
Ads

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
Previous << 1 .. 422 423 424 425 426 427 < 428 > 429 430 431 432 433 434 .. 609 >> Next

6 1 6 2 2
t/6 2 ï
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
y(t) = e (2co^5 t + _ sin y5 t).
-1 / a+2
25b. y(1) = e (2cosy5 + ,_ sin 5 ) = 0 and hence
V 5
2 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 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 ]_/2 on the
interval a < t < b. Note y2(a) 0 and y2(b) 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
ӳ 12 - 2ӳ "(ӳ,2)
()' = ------------- = -----------, which cannot be zero
y 2 2
2 2 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
Previous << 1 .. 422 423 424 425 426 427 < 428 > 429 430 431 432 433 434 .. 609 >> Next