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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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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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