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

ISBN 0-471-31999-6

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.(1)

.(2)

CiX'"' + c2x solutions, then x = cix

Assume that if x (1)

(1)

.(2)

x

(k)

are

+ + ckx

(k)

is a solution.

Then use Theorem 7.4.1 to conclude that x + c^+1x(k+1) is

also a solution and thus c1x(1) +

+ ck+1x(k+1) is a

solution if x(1), ..., x(k+1) are solutions. 2a. From Eq.(10) we have W =

x11)

x(2)

x21)

x2

(2)

= x11) x(2)

x(1) x12)

Taking the

derivative of these two products yields four terms which may be written as

Section 7.4

135

dW — = [ dt

dx11(

x22)

xi1»

dx12)

] + [ x1

(1)

dx(2)

dx(1)

x(2)].

dt dt dt dt

The terms in the square brackets can now be recognized as the respective determinants appearing in the desired solution. A similar result was mentioned in Problem 20 of Section 4.1.

2b.

2c.

If x

(1)

is substituted into Eq.(3) we have = P11 x11) + P12 x21)

dx11) dt dx21) dt

Substituting the first equation above and its counterpart

= P21 x11) + P22 x21)

for x

(2)

into the first determinant appearing in dW/dt

and evaluating the result yields p11

x11)

x21)

x(2)

x(2)

= puW.

Similarly, the second determinant in dW/dt is evaluated as p22W, yielding the desired result.

dW

From prt b we have — = [p11(t) + p22(t)]dt which gives

W

W(t) = c expl[pn(t) + p22 (t)]dt.

6a. W =

= 2t

2

t2 = t2.

t t2

1 2t

6b. Pick t = t0, then c1x(1) (t0) + c2x(2) (t0) = 0 implies

, which has a non-zero solution

rT O ( 12 ^ t0 ' 0 N

c1 + c2 = 0 V J

V 1 , v 2t0 ,

t0 t20

= 2t0

t02 = t20 = 0.

for c1 and c2 if and only if

1 2t0

Thus x(1) (t) and x(2) (t) are linearly independent at each point except t = 0. Thus they are linearly independent on every interval.

6c. From part a we see that the Wronskian vanishes at t = 0, but

not at any other point. By Theorem 7.4.3, if p(t), from

Eq.(3), is continuous, then the Wronskian is either identically zero or else never vanishes. Hence, we conclude that the D.E. satisfied by x(1) (t) and x(2) (t) must have at

least one discontinuous coefficient at t = 0.

6d. To obtain the system satisfied by x(1) and x

(2) we

136

Section 7.5

consider

x = c1x(1) + c2x(2), or

/ \ x1

x2

= c1

+ c2

t2

2t

( ' 3 / - ~\ / ^ \

x1 111 2t

= ci 1 + c2

v x2, V 0 V 2 y

Taking the derivative we obtain

Solving this last system for ci and c2 we find ci = xi - tx2 and c2 = x2/2. Thus

, which yields

xi = tx1 - — x2 and x2 = x2. Writing this system in 2

/ \ x1 ^ t ^ x2 2 t2

. CN X V i . II + —

V x2 y V 1 y 2 V 2t y

matrix form we have x =

t - t2/2

x'. Finding the

inverse of the matrix multiplying x' yields the desired solution.

Section 7.5, Page 381

1. Assuming that there are solutions of the form x = ?ert, we substitute into the D.E. to find

r%e

rt

'3 -2"

%ert. Since % = I% =

1 0

0 1

write this equation as

'3 -2"

thus we must solve

3-r -2

% - r V

%1 0

%2 y V 0

%, we can % = 0 and

for r, %i, %2-

1 0 0 1/

The determinant of the coefficients is (3-r)(-2-r) + 4 = r2 - r - 2, so the eigenvalues are r = -1, 2. The eigenvector corresponding to r = -1

satisfies

Thus x(1)(t) = %(1)e

' 4 -2 1 ' 0 N

V 2 t—1 1 V%2 y II V 0 y

(1) ^-t

1

, which yields 2%1 - %2 = 0.

e , where we have set %1 = 1.

(Any other non zero choice would also work). In a similar fashion, for r = 2, we have

or ^1 - 2^2 = 0. Hence x(2)(t) = ?(2)e2t = e2t by

/ . . \ it 3 / ~ \

1 -2 %1 0

V 2 -4 y V%2 y V 0 y

2

11J

2

i2 -2 j

i2 -2 )

i 2 -2-r)

K 2 )

Section 7.5

137

setting = 1* The general solution is then

x = c^(1) (t) + c2x(2 ^ (t)* To sketch the trajectories we follow the steps illustrated in Examples 1 and 2*

/ \ x1 ' 1 '

Setting c2 = 0 we have x = = c1

V X2 y V 2 y

e or x1

c1e

-t

and x2 = 2c1e-t and thus one asymptote is given by x2 = 2x!. In a similar

fashion c! = 0 gives x2 = (l/2)x1 as a second

asymptote. Since the roots differ in sign, the trajectories for this problem are similar in nature to those in Example 1. For c2 ^ 0, all solutions will be

asymptotic to x2 = (l/2)x1 as t ^ ^. For c2 = 0, the

solution approaches the origin along the line x2 = 2x1.

5. Proceeding as in Problem 1 we assume a solution of the form x = ? ert, where r, ^1, ?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 't)1 = ^2 and thus

' -2-r 1^ '?1 ' ' 0 '

, 1 -2-r, v?2 y V 0 y

?(1)

?(2)

1

1

and for r = -3 we find ^2 = -^1 and hence The general solution is then

' 1N ' 1 N

x = c1 e-t + c2

V 1 y v-1 y

-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 = x1* Similarly

the origin along the line

e 3t approaches

e

138

Section 7.5

x2 = -x2. As long as ci ^ 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.

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