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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.
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(x  2)2 cos(2 ln |x  2|) + c2(x  2)2 sin(2ln |x  2|)
1e 2
|x| 1/2cos(tv/t5ln |x|) + c2|x| 1/2 sin(tv/t5ln |x|)
x + c2 x
13.
1/2
y
= 2x3/2 - x1
(2lnx)  x 1/2sin(2lnx) 15. y  2x2  7x2ln |x|
x 1 cos(2 ln x)
0
1
เ < 1 and โ > 0
เ < 1 and โ > 0, or เ  1 and โ > 0
เ > 1 and โ > 0
เ > 1 and โ > 0, or เ  1 and โ > 0
เ  1 and โ > 0
17.
19.
เ < 1 ส  2
c,x 1 + c2 x2
27. y 
28. y 
29. y  31. x >
+ c2 x 5 + x
25. y  c1 x2 + c2x2 lnx + 4 lnx + 4
x + c2x2 + 3x2 ln x + ln x + 2 cos(2lnx) + c2 sin(2lnx) + 1 sin(ln x) x 3/2 cos(3 lnx) + c2x 3/2 sin(2 lnx)
c1  k1, c2  k2; x < 0 : c1(  1)r1  k1, c2  k2
700
See SSM for detailed solutions to 2
Section 5.6, page 271
1. r(2r  1) = 0; an = 
(n + r )[2(n + r)  1]
r, = 2, r2 = 0
x1/2
x2
1   +
2-5 2  4  5  9 2-4-6-5-9-13
+
+
(1)nx2 n
2nn!5  9  13    (4n + 1)
+
x2
K2(x) = 1  7-ร +
2-3 2-4-3-7 2-4-6-3-7- 11
+
+
( 1)nx2 n
2nn!3  7  11    (4n  1)
+
2. r2  9 = 0;
n2
(n + r)2 
rl = 3 , r2 = 3
ำ,(x) = x
x1/3
1 I x\2
1------------+D (ฟ) +
1
าฟ( 2)'
+
1!(1 + f) V2/ ' 2!(1 + f)(2 + 3) V2
(1) / x\2m
+
m!(1 + 3)(2 + 3)    (m + 1)
(2)
+
ำ2(ี) = X
_ x1/3
1 / x \ 2
1--------IT) ( 2) +
1
d( D4
+
1! (1  f) V2! ' 2! (1  -)(2  ณ) V2,
(1) / x \2m
+
! (1  3)(2  3)    (  f)
(2)
+
Hint: Let n = 2m in the recurrence relation, m = 1, 2, 3,....
a
3. r(r  1) = 0; an = 
n1
ำ,(x) = x
(n + r)(n + r  1)  1
(1)n
x x2
1  -------- + --------- +    +
1!2! 2!3! n! (n + 1)!
r1 = 1 , r2 = 0
xn +
a
n1
4. r2 = 0; an = 9,
(n + r)2
r1 = r2 = 0
y, (x) = 1 + 2 +
x2
(l!)2 (2!)2
+
xn
 -----๎
(n!)2
an2
+
5. r(3r  1) = 0; a =- ,
n (n + r)[3(n + r)  1]
r = 1 r = 0
i1 3, 2 U
1/3
y,(x) = x
-  _L(?\+ _^ x:\ +.
1!7 \ 2 2!7  13 \ 2 /
(1)m
m!7  13    (6m + 1) \ 2
+
y2(x) 1 1!5 \ 2 ) + 2!5  11 \ 2
xr +
1 x2
x2
-ณ +   +
(1)m
m!5  11    (6m  1)
Hint: Let n = 2m in the recurrence relation, m = 1, 2, 3,....
a
n2
4
6
4
6
9
701
See SSM for detailed solutions to 11
14
15
16ab
6. r2 - 2 = 0; a =------------a"-า---; r = 42, r = 2
n (n + r)2 - 2 1 2
71 (x) = x^
72(x) = x-^
x x2
1(1 + 2V2) 2! (1 + 2v/2')(2 + 2^2)
I (-1)n M I '
n!(1 + 2\/2)(2 + 2^2) (n + 2V2) .
x x2
1(1 - 2V2) 2! (1 - 2\/2)(2 - 2^2)
I (-1)n n I "
n! (1 - 2\f2)(2 - 2^2) (n - 2^2) .
7. r2 = 0; (n + r)an = an-1; r1 = r2 = 0
x2 x3 xn
71(x) = 1 + x + - + - +  + - +  = ex
8. 2r2 + r  1 = 0; (2n + 2r  1)(n + r + 1)an + 2an-2 = 0; r1 = 1, r2 = 1
_ 1/2 I , x2 : x4 : (-1)mx2m
y1 (x) = x1/2 1-------1------------- I-------------------------1 
^  7 2!7  11 m!7  11  (4m + 3)
(x4 ( 1)mx2 m
1 - x2 + ---------------+--------------(-^x----+ ฆ
2!5 m!5  9  (4m - 3)
9. r2  4r + 3 = 0; (n + r  3)(n + r  1)an  (n - r  2)an-1 = 0; r1 = 3, r2 = 1
3 2 x2 2 xn
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