# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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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

1—r i

-i 1—r

1 i Y Si

. —i 1 JvS2

= (1—r) 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

t—1 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 = -1±Va

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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