# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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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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