Elementary Differential Equations and Boundary Value Problems  Boyce W.E.
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5. Proceeding as in Problem 1 we assume a solution of the form x = ? ert, where r, ?x, ?2 must now satisfy
Evaluating the determinant of the
coefficients set equal to zero yields r = 1, 3 as the
eigenvalues. For r = 1 we find ^ = ?2 and thus
rH '?1' ' 0 '
r

2

1 v?2 j v 0 ,
1
ISJ
1
È
?(1)
?(2)
1
1
and for r = 3 we find ?2 = ?i and hence The general solution is then
' 1N ' 1 N
x = c1 e
1
Ã
+
c
Þ
V 1 Ó V1 ó
3t
Since there are two negative
eigenvalues, we would expect the trajectories to be similar to those of Example 2.
Setting c2 = 0 and eliminating t (as in Problem 1) we find
that
1
1
e t approaches the
1
origin along the line x2 = x^ Similarly
the origin along the line
e3t approaches
K1)
e
138
Section 7.5
x2 = x!. As long as ci Ô 0 (since e is the dominant term as t^0), all trajectories approach the origin asymptotic to x2 = x1. For c1 = 0, the trajectory approaches the origin along x2 = x1, as shown in the graph.
The characteristic equation is (5/4  r)2  9/16 = 0, so r = 2,1/2. Since the roots are of the same size, the behavior of the solutions is similar to Problem 5, except the trajectories are reversed since the roots are positive.
Again assuming x = %ert we find that r, %1, %2 must satisfy
¥ \ ¥Ã \
4r 3
%1 V%2 Ó
0
The determinant of the
coefficients set equal to zero yields r = 0, 2. For r = 0 we find 4^1 = 3%2. Choosing %2 = 4 we find ^1 = 3
and thus %
(1)
3
Similarly for r = 2 we have
= ' 1ë ' 3 ë ' 1 ^
and thus x = c1 + c2
V 2 Ó 4 20
2t
To sketch the
trajectories, note that the general solution is equivalent to the simultaneous equations x1 = 3c1 + c2e
2t
and x2 = 4c1 + 2c2e
2t
'2t. Solving the first equation for c2e and substituting into the second yields x2 = 2x1  2c1 and thus the trajectories are parallel straight lines.
The eigvalues are given by r(r2) = 0. For r= 0 we have
1—r i
i 1—r
1 i Y%1
. —i 1 J\%2
= (1—r) 2 + i2 =
= 0 or
—i%1 + %2 = 0 and thus
/14
is one eigenvector. Similarly
is the eigenvector for r = 2.
^8 6r)
K.0 )
—i
14. The eigenvalues and eigenvectors of the coefficient
Section 7.5
139
16.
matrix satisfy
1r 1 4
3 2r 1
2 1 1r
of coefficients set equal to zero reduces to
3 2
r  2r  5r + 6 = 0, so the eigenvalues are r1 = 1, r2 = 2, and r3 = 3. The eigenvector
\ ×1ë Î
%2 = 0
I v%3 , V 0 ,
. The determinant
' 0 1 4" ×1 ' ' 0 '
corresponding to r1 must satisfy 3 1 1 %2 = 0
V 2 1 v%3 , V 0 ,
1
I
Using row reduction we obtain the equivalent system + %3 = 0, %2  4%3 = 0. Letting = 1, it follows that
1
%3 = 1 and %2 = 4, so %
(1)
. In a similar way the
eigenvectors corresponding to r2 and r3 are found to be
, respectively. Thus the
general solution of the given D.E. is
' 11 ' 1 N
(2) = 1 and %(3) = 2
1 V 1,
I
Ã 11 ' 1 ' Ã 11
x = c1 4 et + c2 1 e2t + c3 2
1 v1, V 1,
I
,3t
Notice that the
"trajectories" of this solution would lie in the x1 x2 x3 three dimensional space.
The eigenvalues and eigenvectors of the coefficient
and r2 = 3,
1
(2)
1
matrix are found to be r1 = 1, %(1) =
Thus the general solution of the given D.E.
The I.C. yields the
. The augmented
1
is x = c1
et + c2
1
5I
,3t
Ã 11 Ã11 ã 11
system of equations c1 + c21 =
1I V 5 J 3I
140
Section 7.5
matrix of this system is
1 1 . 1
1 5 . 3
and by row reduction
we obtain
1 1 . 1
. . Thus c2 = 1/2 and c1 = 1/2.
. 0 1 .1/2y
Substituting these values in the general solution gives the solution of the I.V.P. As t ^ the solution
becomes asymptotic to x = —
2
e3t, or x2 = 5x1.
20. Substituting x = Etr into the D.E. we obtain
r?t
r
'2 14
3 2,