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

ISBN 0-471-31999-6

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k. k

k=1

(-1)k t k!

and use mathematical induction to verify this form for fn(t). Using Eq.(7) again we have:

n

fn+1(t) = -?[1 + fn(s) ] ds = -t - X

k^k+1

0

k+1 k+1

(-1) t (k+1)!

(-1)k+"t (k+1)!

n+1

(-1) V

i!

k=1

, where i = k+1. Since this

k=0 i=1

is the same form for f n+i(t) as derived from fn(t) above, we have verified by mathematical induction that fn(t) is as given.

4c. From part a, let f(t) = lim fn(t) = X

2

k=1

3

(-1)ktk k!

t t

2 3! ... .

Since this is a power series, recall from calculus that:

at

e

k k 2 2 3 3

a t a t a t

= 1 + at + + +

k!

3!

k=0

If we let

= 1 +f (t).

-t t2 t3

a = -1, then we have e = 1 - t + — - — +

2 3!

Hence f(t) = e-t -1.

7. As in Prob.4,

f 1(t) = !fc(sf 0(s) +1)ds = s|0 = t

t 2 s3 t t3

f2(t) = I (s +1)ds = (— + s) |0 = t + —

2 0 3 0 3

J 4 3 5 ^3 ^5

t 2 s s s t t t

(s + — +1)ds = (— + ---------------------- +s) |0 = t + — + -------------.

0 3 3 3-5 0 3 3-5

t<-2 s_

3

Based upon these we hypothesize that:

3

4

2

Section 2.9

31

"V t

f n(t) = I ------------------- and use mathematical induction

1-3-5-(2k-1)

k = 1

to verify this form for fn(t). Using Eq.(7) again we have:

2k

fn+1(t) = J0'( I

n

= I

k=1 n

- I

s

+ 1)ds

1-3-5- (2k-1)

k = 1

t2k+1

------------------ + t

1-3-5- (2k + 1) k=1

n t2k+1

1-3-5- (2k + 1) k = 0

t2i-1

n+1

= I -------------, where i = k+1. Since this is

1-3-5- (2i-1) i=1

the same form for f n+i(t) as derived from f n(t) above, we have verified by mathematical induction that fn(t) is as given.

35

x 7

— + O(x ). Thus, for 5!

7

(t ) we have

(t - —)3 5

2! t5 7

+ --- + O(t ).

x

. sinx = x 3! +

t2 + 2! t4 t6 + O

4! 6!

_ + t4 t6

' 2! 4! 6!

3! 5!

Section 2.9, Page 124

2. Using the given difference equation we have for n=0, yi = y0/2; for n=1, y2 = 2y1/3 = y0/3; and for n=2,

y3 = 3y2/4 = y0/4. Thus we guess that yn = y0/(n+1), and

n+1

the given equation then gives yn+1 = yn = y0/(n+2),

n+1 n+2 n 0

which, by mathematical induction, verifies yn = y0/(n+1) as the solution for all n.

5. From the given equation we have y1 = .5y0+6.

32

Section 2.9

10.

13.

14.

17.

2 1

y2 = .5y1 + 6 = (.5) y0 + 6(1 + —) and

3 11

y3 = .5y2 + 6 = (.5) y0 + 6(1 + — + —). In general, then

3 2 0 2 4

n 11

yn = (.5) y0 + 6(1 + - + - + —-)

2 2

n 1 - (1/2)n

= (.5) ny0 + 6(---------^)

0 1 - 1/2

= (.5)ny0 + 12 - (.5)n12

= (.5) n(y0-12) + 12. Mathematical induction can now be used to prove that this is the correct solution.

The governing equation is yn+2 = pyn-b, which has the n 1-pn

solution yn = p y0 - b (Eq.(14) with a negative b).

1-P

Setting y360 = 0 and solving for b we obtain

~\~360

(1-P)P y0

b = , where p = 1.0075 for part a.

1-p360

You must solve Eq.(14) numerically for p when n = 240, y240 = 0, b = -$900 and y0 = $95,000.

p-1

Substituting un = ------- + vn into Eq.(21) we get

n p n p-1 p-1 p-1

+ vn+1 = p(-------- + vn)(1 - ----- - vn) or

p p p

p-1 1

vn+1 = - ----- + (p-1 + pvn)( - - vn)

p p

1-p p-1 2 2

= ----- +---------- - (p-1)vn + vn- pvn = (2-p)vn - pvn-

p p

15a. For u0 = .2 we have u2 = 3.2u0(1-u0) = .512 and u2 = 3.2u1(1-u1) = .7995392. Likewise u3 = .51288406, u4 = .7994688, u5 = .51301899, u6 = .7994576 and u7 = .5130404. Continuing in this fashion, u14 = u16 = .79945549 and u15 = u17 = .51304451.

For both parts of this problem a computer spreadsheet was used and an initial value of u0 = .2 was chosen. Different initial values or different computer programs may need a slightly different number of iterations to reach the limiting value.

Miscellaneous Problems

33

17a.The limiting value of .65517 (to 5 decimal places) is

reached after approximately 100 iterations for p = 2.9. The limiting value of .66102 (to 5 decimal places) is reached after approximately 200 iterations for p = 2.95. The limiting value of .66555 (to 5 decimal places) is reached after approximately 910 iterations for p = 2.99.

17b. The solution oscillates between .63285 and .69938 after approximately 400 iterations for p = 3.01. The solution oscillates between.59016 and .73770 after approximately 130 iterations for p = 3.05. The solution oscillates between .55801 and .76457 after approximately 30 iterations for p = 3.1. For each of these cases additional iterations verified the oscillations were correct to five decimal places.

18. For an initial value of .2 and p = 3.448 we have the solution oscillating between .4403086 and .8497146.

After approximately 3570 iterations the eighth decimal place is still not fixed, though. For the same initial value and p = 3.45 the solution oscillates between the four values: .43399155, .84746795, .44596778 and

.85242779 after 3700 iterations.. For p = 3.449, the solution is still varying in the fourth decimal place after 3570 iterations, but there appear to be four values.

Miscellaneous Problems, Page 126

Before trying to find the solution of a D.E. it is necessary to know its type. The student should first classify the D.E. before reading this section, which indentifies the type of each equation in Problems 1 through 32.

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