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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
Download (direct link): elementarydifferentialequat2001.pdf
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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
4-r -3 '
S1
VS2 J
'0'
The determinant of the
coefficients set equal to zero yields r = 0, -2. For r = 0 we find 4^1 = 3^2. Choosing S2 = 4 we find Si. = 3
and thus S
(i)
3
Similarly for r = -2 we have
S(2) = ' 1 ' ' 3 ' ' 1N
and thus x = c1 + c2
V 2 y V 4, V 2 y
-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(r-2) = 0. For r= 0 we have
1r i
-i 1r
1 i Y Si
. i 1 JvS2
= (1r) 2 + i2 =
= 0 or
iS1 + S2 = 0 and thus
1
is one eigenvector. Similarly
1
is the eigenvector for r = 2.
^8 -6-r)
K.0 )
\ 4 /
i
14. The eigenvalues and eigenvectors of the coefficient
Section 7.5
139
16.
matrix satisfy
1-r -1 4
3 2-r -1 2 1 -1-r
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
\ '%1 ' ' 0 N
%2 = 0
) v%3, V 0 ,
. The determinant
' 0 -1 4 " '%1 ' ' 0 N
corresponding to r1 must satisfy 3 1 -1 %2 = 0
V 2 1 -2, v%3 , V 0 ,
Using row reduction we obtain the equivalent system + ^3 = 0, ^2 - 4^3 = 0. Letting t,1 = 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
( 1 ' ' 1N
(2) = -1 and %(3) = 2
t1 1 V 1,
( 1 1 ( 1 1 ( 11
x = c1 -4 et + c2 -1 e-2t + c3 2
T 1 1 T 1 1 T 1
,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 = ci
e-t + c2
1
5
,3t
' 1 ' ( 11 ( 11
system of equations c1 + c21 =
V 1 , V 5 J CO
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 -1
3 -2
written as
V
2-r -1
3 -2-r
eigenvectors are r1 = 1, E
For t * 0
\ ^1 ' ' 0 "
/ v^2 y v 0 ,
= 1, E (1) =
. The eigenvalues and and r2 = -1,
1
:(2)
1
Substituting these in the assumed form we
' 1 ^ ' 1 ^
obtain the general solution x = c1 t + c2 t
V 1 , v 3 ,
25.
27.
r
a ;
v3 ^
Section 7.6
141
31c. The eigevalues are given by
-1-r -1
-a -1-r
r2 + 2r + 1 - a = 0. Thus r1/2 = -1Va
Note that in Part (a) the eigenvalues are both negative while in Part (b) they differ in sign. Thus, in this part, if we choose a = 1, then one eigenvalue is zero, which is the transition of the one root from negative to positive. This is the desired bifurcation point.
Section 7.6, Page 390
1. We assume a solution of the form x = ? ert thus r and ?
. The determinant of
'3-r -2" ^1 ' ' 0 '
are solutions of =
,4 -1-ry ^2 y v 0 ,
coefficients is (r2-2r-3) + 8 = r2
eigenvalues are r = 1 2i. The eigenvector
corresponding to 1 + 2i satisfies
2-2i -2
4 -2-2i
\ ^1 ' ' 0 '
/ v^2 y v 0 ,
or (2-2i)^1 - 2?2 = 0. If ^1 = 1, then ?2 = 1-i and
:(1)
1 1
1-i
and thus one
complex-valued solution of the D.E. is
x(1)(t) =
1
>(1+2i)t
To find real-valued solutions (see Eqs.8 and 9) we
take the real and imaginary parts, respectively of
x(1) (t). Thus x(1) (t)
1-i
t(cos2t + isin2t)
= e
cos2t + isin2t
cos2t + sin2t + i(sin2t - cos2t)
cos2t cos2t + sin2t j
' sin2t vsin2t - cos2t^
Hence the general solution of the D.E. is
x = c1e
cos2t cos2t + sin2t j
+ c2e
sin2t sin2t - cos2t j
The
solutions spiral to ^ as t ^ ^ due to the et terms.
1
t
t
= e
+ ie
142
Section 7.6
7. The eigenvalues and eigenvectors of the coefficient
matrix satisfy
1-r 0 0
2 1-r -2
3 2 1-r
\ ' 0N
^2 = 0
) v^3 , V 0 ,
The
determinant of coefficients reduces to (1-r)(r2 - 2r + 5) so the eigenvalues are ^ = 1, r2 = 1 + 2i, and r3 = 1 - 2i. The eigenvector corresponding to ^ satisfies
' 0 0 0 '
2 0 -2 ^2
ro 2 0 , ?3)
3^1 + 2^2 = 0.
0
; hence - ^3 = 0 and
2
so one solution of the D.E. is
-3
et. The eigenvector
corresponding to r2 satisfies
^-2i 0 2 -2i -2 v3 2 -2i
\ ' 0N
^2 = 0
) v^3 , V 0 ,
Hence = 0 and i^2 + ^3 = 0. If we let = 1, then = -i. Thus a complex-valued solution is
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