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

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dw 2 2

= (s+1) + (w+2) , upon substitution into the given ds

D.E.

4a. Following Ex. 1 of the text, from Eq.(7) we have

fn+1(t) = I tf(s,f(s))ds, where f(t,f) = -1 - f for this

0

problem. Thus if f0(t) = 0, then f1(t) = -I tds = -t;

0

ft t2

f2(t) = -I (1-s)ds = -t + ---------;

2 0 2

30

Sect ion

2 . 8

2 t2 .3

J't s t t

(1-s + ----------)ds = -t + -------------- - ---------;

0 2 2 2-3

-f

³

2

s

3

s

t

2

f4(t) = -| (1 - s + - )ds = -t + - + .

3!

3!

4!

Based upon these we hypothesize that: fn(t) = X,

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+1t (k+1)!

ο+1

(-1)Σ i!

k=1

, where i = k+1. Since this

k=0 i=1 is the same form for f n+1(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

0<-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) = It( I

n

= I

k=1 n

-1

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+1(t) as derived from f n(t) above, we have verified by mathematical induction that fn(t) is as given.

3 5

x 7

+ O(x ). Thus, for 5!

7

(t ) we have

(t - )3 5

2! t5 7

+ ---- + O(t ).

x

. sinx = x 3! +

t2 t4 t6 + O

4! 6!

_ t* + t4 t6

' 2! 4! 6!

3! 5!

Section 2.9, Page 124

2. Using the given difference equation we have for n=0, y1 = 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+2

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

Σο = (.5) y0 + 6(1 + - + - +

2 2

ο 1 - (1/2)ο

= (.5) ΟΣ0 + 6(------------/^)

0 1 - 1/2

= (.5)Οσ0 + 12 - (.5)Ο12

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