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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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c^3+ c2t! = 0. These equations have a nontrivial solution for c1 and c2 only if the determinant of coefficients is
zero. But the determinant of coefficients is -2t^t1 n 0 for t0 and t1 as specified. Hence f(t) and g(t) are linearly independent on -1 < t < 1.
Section 3.4, Page 158 1. exp(1+2i) = e1+21= ee21 = e(cos2 + isin2).
5. Recall that 21_i = eln(21-1) = e(1-i)ln2
rt
7. As in Section 3.1, we seek solutions of the form y = e . Substituting this into the D.E. yields the characteristic 2
equation r - 2r + 2 = 0, which has the roots r1 = 1 + i
44
Section 3 . 4
11.
14.
18.
22.
23a.
and r2 = 1 - i, using the quadratic formula. Thus l = 1 and m = 1 and from Eq.(17) the general solution is y = c1etcost + c2efcsint.
2
The characteristic equation is r + 6r + 13 = 0, which
-6+v-16
has the roots r =
-3±2i. Thus l = -3 and
m = 2, so Eq.(17) becomes y = c1e 3tcos2t + c2e 3tsin2t.
The characteristic equation is 9r + 9r - 4, which has the real roots -4/3 and 1/3. Thus the solution has the
same form as in Section 3.1, y(t) = c1et/3 + c2e-4t/3.
The characteristic equation is
r + 4r + 5 = 0, which has the roots r1,r2
-2t
-2 ± i. Thus
2t
'cost + c2e
sint and
-2t
cost +
y = c1e
y' = (-2c1+c2)e
-2t
(-c1-2c2)e sint, so that y(0)= c1= 1 and y'(0) = -2c1+c2= 0,
2t
or c2 = 2.
Hence y = e
cost + 2sint)
o.a 04 o§ 0.8 (*ia i.^ij 22I14 sb as 3
The characteristic equation is
2
r + 2r + 2 = 0, so r1,r2 = -1 ± i.
Since the I.C. are given at p/4 we want to alter Eq.(17) by
^p/4
d1 and c2
letting c1
Thus, for l = -1 and m = 1 we
e-(t-p/4)
-(t-p/4)
P/4,
ed
have y = e (t p/4) (d1cost + d2sint)
? (t-P/4)
so y = -e (d1cost + d2sint) +e (-d1sint +d2cost)
Thus ^/2d1/2 + \J~2d2/2 = 2 and -^/"2d1 = -2 and hence y = \f2 e (t p/4) (cost + sint).
The characteristic equation is 3r
1 . a/23 .
the roots r1,r2 =  ±
i.
6
t/6 ( V23
e (c.cos----
16
t + c2sin
6
2 r + 2 Thus u(t) =
0, which has
6
t) and we obtain u(0) = c1 = 2
2
Section 3.4
45
, 1 /23"
and u (0) =  c + c2 = 0. Solving for c2 we find
6 1 6 2 2
t/s 323 2 323
u(t) = e / (2cos 1 - sin 1).
6 y23 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 ± ^J~5i and y(t) = e-t(c1cosV~51 + c2sin33t).
Thus y(0) = c1 = 2 and y'(0) = -c1 + \/~5 c2 = a and hence
-t / a+2 1
y(t) = e (2cosy 5 t + sin y5t).
-1 / a+2 I
25b. y(1) = e (2cosy5 + ,__ sin y 5 ) = 0 and hence
V 5
2y 5
a = -2 - ---------- = 1.50878.
tanV 5
t a+2 t
25c. For y(t) = 0 we must have 2cosy5t +__________________ siny5t = 0 or
V 5
I -2 V 5
tany 5 t = --------. For the given a (actually, for a > -2)
a+2
I 2 y 5
this yields V 5 t = p - arctan since arctan x < 0 when
a+2
x < 0.
2\f~5
2 5d. From part (c) arctan ----------  0 as a ? , so t ? p
a+2
31. -d[e1t(cosmt + isinmt)] = 1e1t(cosmt + isinmt) dt
+ e1t(-msinmt + imcosmt) = ie1t(cosmt + isinmt)
+ ime1t(isinmt + cosmt) = e1t(i+im) (cosmt + isinmt).
_ 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 y1/y2 on the
interval a < t < b. Note y2(a) n 0 and y2(b) n 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
yi y1y2 - y2yi -W(yi,y2)
(  )' = ----------- = , which cannot be zero
y 2 2 y2 y2 y2
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 d y/dx , the D.E. d y/dx - y = 0. Hence the
general solution of the original D.E. is y = c1cosx + c2sinx,
f -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.
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