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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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0
^3 )
T
1,-1 1 1 j
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!
I — X
n!
n=1
n=1
Xn-1tn
(n-1)!
1 +
I
n=1
X ntn
n!
feXt teXt4 V 0 eXt ó
Since
I
n=1
Xn-1tn
(n-1)!
= t( 1 + I
n=1
nn
— ) = teXt.
n!
19d. From Eq. (28), Section 7.7, we have
x = exp(Jt)x0 =
v
f v 0 1 f 01
= x1 + x2
t
X
e
v x2 , v 0 ,
Section 7.9, Page 417
1. From Section 7.5
feXt teXtYx11 0 eXt x2|
fx0eXt+x2teXt 1 x02eXt
te
Xt
x(c) = ci
g (t) =
1
et +
1
c2
-t
Note that
' 1 ' e ' 0 ë

V 0, 1I
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 r1 = 2 and r2 = -2, with corresponding
V3
ßÒ1Ï
.-^ I
eigenvectors
Thus
Jntn
0
Section 7.9
157
x(c) = c1
V3
2t
+ C2
1
-àÄ
-2t
Writing the
' 1ë ' 0 N
term as et + V a/"3 c
V 0 ,
can assume x(p) = aet + be D.E., we obtain
e t we see that we Substituting this in the
aet - be t = Aaet + Abe t + ' 1 ^ et + ' 0 N
V 0,
' 1ë ' 0 ë
V 0 , V 0 ,
e t, 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
\f3 a2 = -1 and V"3a1 - 2a2 = 0, or a1 = -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/ä/3. Substituting these values for a and b into x(p) and adding x (p) to
' 0 ë 0
zero, which yields Ab + b + V a/3 c =
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) =
-cost
and thus
e
5
158
Section 7.9
u'(t) =
cost - 2sint
2cost + sint 5
sint
-cost
V sint
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 ^ ¥ - - Ë 1 1 -1 e
1 1 2
rt
--- y + --- V 2e ó
5 V 4 1, V 4 -2, 1 5 4
1 1
'-3 0 ë 1 / -2t + 2eb 1
e
y + --- ee t This corresponds
V 0 2 , 5 V 4e -

the two scalar equations
Ó1 + 3y1 = (1/5)e-2t + (2/5)et,
Ó2 - 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 (e3ty^)' = (1/5)et + (2/5)e
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