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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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/
1 1 1
2 -1 -1 V -3 2 2
\ ' 1 N
Z2 = 1
) v^3 , v 0 ,
T 1 1 T—1 1 T 1 '-1 N
equivalent system 0 1 1 Z2 = 3
o o o v^3 , V 0 ,
By row reduction we find the
If we let
1
^0 j
+
Section 7.8
155
17e.
17f.
19a.
19b.
Z2 = 0, then Z3 = 3 and Z1 = 2, so Z =
x(3) (t) =
' 0 ' ' 1 ' ' 2 N
1 (t2/2)e2t + 1 te2t + 0
t—1 1 V 0, co
and
,2t
? is the matrix with x(1) as the first column, x the second column and x(3) as the third column.
(2)
as
1 1 0 v-1 0 3,
and using row operations on T and I, or a
computer algebra system, T
-3 3 2
3 -2 -2
and thus
' 2 1 0 '
T-1AT = 0 2 1 = J
v 0 0 2 y
J2 = JJ = 'l 1 ' 'l 1 ' 2l ^
=
V 0 l, V 0 V V 0 l2 ,
J3 = JJ2 = 'l 1 ' V 2l" 3l2 N
=
V 0 V V 0 l2 , V 0 l3 ,
Based upon the results of part a, assume
/ ZS
ln nln-1
J
V 0 l
n+1 = jJ =
, then
rl 1?fln nln-1^
v
0 l
0
ln
V+1 (n+1)ln
, which is the same as Jn with n
v 0 Xn
replaced by n+1. Thus, by mathematical induction, Jn has the desired form.
0
13 ^
T
1-1 1 17
n
19c. From Eq.(23), Section 7.7, we have
156
Section 7.9
exp(Jt) = I +
= I +
n=1
n=1
n!
Xntn nXn-1tn
1 +
n!
0
Xntn
n!
X ntn
n!
X — X
n!
n=1
n=1
Xn-1tn
(n-1)!
1 +
X
n=1
X ntn
n!
(eXt teXt" V 0 eXt y
Since
X
n=1
Xn-1tn
(n-1)!
= t( 1 + X
n=1
nn
— ) = teXt.
n!
19d. From Eq. (28), Section 7.7, we have
x = exp(Jt)x0 =
V
O 0 'i ( 0 A
= x1 eXt + x2
v x2 , v 0 ,
Section 7.9, Page 417
1. From Section 7.5
(eXt teXtYx1x 0 eXt x2.
(x0eXt+x2teXt^ x02eXt
te
Xt
x(c) = ci
g (t) =
1
et +
1
c2
-t
Note that
' 1 ' et + ' 0N

V 0, t—1
t and that r = 1 is an eigenvalue of
the coefficient matrix. Thus if the method of undetermined coefficients is used, the assumed form is given by Eq.(18).
2. Using methods of previous sections, we find that the
eigenvalues are rj_ = 2 and r2 = -2, with corresponding
V3
.-73.
eigenvectors
Thus
jntn
0
Section 7.9
157
x(c) = c1
2t
+ C2
1
-2t
Writing the
' 1 ' ' 0 N
term as et + V a/"3 j
V 0 ,
can assume x(p) = aet + be D.E., we obtain
e t we see that we Substituting this in the
ael - be p = Aae1 + Abe p + ' 1 ' et + ' 0 N
V a/"3 j
V 0,
' 1 ' ' 0N
V 0 , V 0 ,
e , where A
is the given coefficient matrix. All the terms involving
et must add to zero and thus we have Aa - a +
This is equivalent to the system
V3" a2 = -1 and V3a1 - 2a2 = 0, or a2 = -2/3 and
a2 = -1/V"3. Likewise the terms involving e-t must add
The solution
of this system is b1 = -1 and b2 = 2/^/3. Substituting these values for a and b into x(p) and adding x (p) to
' 0 ' ' 0 N
zero, which yields Ab + b + V V3 J =
V 0 ,
x
(c)
yields the desired solution.
3. The method of undetermined coefficients is not straight forward since the assumed form of x(p) = acost + bsint leads to singular equations for a and b. From Problem 3 of Section 7.6 we find that a fundamental matrix is
?(t) =
5cost 5sint
v2cost + sint -cost + 2sint matrix is
The inverse
?-1(t) =
cost - 2sint
2cost + sint 5
sint
cost
, which may be found as
in Section 7.2 or by using a computer algebra system. Thus we may use the method of variation of parameters where x = ?(t) u(t) and u(t) is given by u'(t) = ?-1(t)g(t) from Eq.(27). For this problem
g (t) =
f 4-1
-cost
and thus
e
5
158
Section 7.9
u'(t) =
cost - 2sint
2cost + sint 5
sint
-cost
1 4-1
-cost
V sint y
1 2 - 3cost2t + sin2t
5 ^ -1 - cos2t - 3sin2t J after multiplying and using appropriate trigonometric identities. Integration and multiplication by V yields the desired solution.
5
4. In this problem we use the method illustrated in Example 1. From Problem 4 of Section 7.5 we have the
transformation matrix T =
1 1
Inverting T we find
that T
-1
-1
1
If we let x = Ty and substitute
into the D.E., we obtain
y' =
1 ' 1 -1 ' ' 1 1 ' / -, - ^ 1 1 1 7 -, -, 1 1 -1 -2t e 2t
— y + — v -2e y
5 V 4 1y V 4 -2y t—1 1 5 v 4 1 ,
'-3 0 ' 1 / e -2t + 2et 1
y + — 2t - 2e t y This corresponds
V 0 2 y 5 V 4e -
the two scalar equations
yi + 3yi = (1/5)e-2t + (2/5)et, y2 - 2y2 = (4/5)e-2t - (2/5)et, which may be solved by the methods of Section 2.1. For
the first equation the integrating factor is e3t and we obtain (e3tyi)' = (1/5)et + (2/5)e
4t
so
e yi = (1/5)e + (1/10)e + ci. For the second equation
the integrating factor is e (e-2ty2)' = (4/5)e-4t - (2/5)e
•2t
so
Hence
-2t.
Y2 = -(1/5)e-4t + (2/5)e
-t
y =
1/5
-1/5
e-2t +
^1/10^
2/5
et +
+ c2. Thus
Finally,
/ -3t ^
cxe 3t
c2e2t /
multiplying by T, we obtain
x = Ty = ' 0 ' e-2t + ' 1/2 1 et + c1 ' 11 e-3t + c2 ' 11
V -1 y V 0 , V-4 y v 1 y
2t
The last two terms are the general solution of the corresponding homogeneous system, while the first two terms constitute a particular solution of the nonhomogeneous system.
4-4 1 y
e
Section 7.9
159
12. Since the coefficient matrix is the same as that of Problem 3, use the same procedure as done in that problem, including the ?-1 found there. In the interval n/2 < t < n sint > 0 and cost < 0; hence |sint| = sint, but |cost| = -cost.
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