CHAPTER ---------------------------------------------------- 5. ----
Chapter Five
Section 5.1 1. Apply the ratio test : lim aB $b8" k a B $b 8 k
Hence the series converges absolutely for kB $k " . The radius of convergence is 3 oe " . The series diverges for B oe # and B oe % , since the n-th term does not approach zero. 3. Applying the ratio test, k8x B#8# k B# oe lim oe ! 8 _ ka8 "bx B#8 k 8_ 8 " lim
8_
oe lim kB $k oe kB $k
8_
The series converges absolutely for all values of B . Thus the radius of convergence is 3 oe _. 4. Apply the ratio test : k#8" B8" k oe lim #kBk oe #kBk 8 _ k#8 B8 k 8_ lim
Hence the series converges absolutely for #kBk, or kBk "# . The radius of convergence is 3 oe "# . The series diverges for B oe ,,"# , since the n-th term does not approach zero. 6. Applying the ratio test,
8 _ k a8
lim
8aB B! b8"
Hence the series converges absolutely for kaB B! bk " . The radius of convergence is 3 oe " . At B oe B! " , we obtain the harmonic series, which is divergent. At the other endpoint, B oe B! " , we obtain "
_
"baB B! b k
8
oe lim
8 kaB B! bk oe kaB B! bk 8_ 8 "
a "b 8 , 8 8oe"
which is conditionally convergent. 7. Apply the ratio test :
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---------------------------------------------------- CHAPTER 5. ---- $ 8 a8 "b# aB #b8" k$ 8" 8# aB #b8 k a 8 "b # " oe lim k a B # b k oe k aB # b k # 8_ $8 $
Hence the series converges absolutely for " kB #k ", or kB #k $ . The radius of $ convergence is 3 oe $ . At B oe & and B oe " , the series diverges, since the n-th term does not approach zero. 8. Applying the ratio test, lim k88 a8 "bx B8" k 88 " oe lim 8 kBk oe kBk, 8" 8 _ a 8 "b 8 _ a 8 "b / 8x B8 88 8 oe 8 _ a 8 "b lim lim OE" 8_ " 8 " oe/ . 8
8_
lim
since
Hence the series converges absolutely for kBk /. The radius of convergence is 3 oe / . At B oe ,, /, the series diverges, since the n-th term does not approach zero. This follows from the fact that
8_
lim
8x /8 oe ". 88 # 1 8
10. We have 0 aBb oe /B , with 0 a8b aBb oe /B , for 8 oe " # . Therefore 0 a8b a!b oe ". Hence the Taylor expansion about B! oe ! is /B oe "
_
B8 . 8x 8oe!
Applying the ratio test,
k8xB8" k " lim oe lim kBk oe !. 8 _ ka8 "bx B8 k 8_ 8 "
The radius of convergence is 3 oe _ . 11. We have 0 aBb oe B , with 0 w aBb oe " and 0 a8b aBb oe ! , for 8 oe # . Clearly, 0 a"b oe " and 0 w a"b oe " , with all other derivatives equal to zero. Hence the Taylor expansion about B! oe " is B oe " aB "b
Since the series has only a finite number of terms, the converges absolutely for all B . 14. We have 0 aBb oe "a" Bb, 0 w aBb oe "a" Bb# , 0 ww aBb oe #a" Bb$ , with 0 a8b aBb oe a "b8 8xa" Bb8" , for 8 " . It follows that 0 a8b a!b oe a "b8 8x ________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ---- for 8 ! . Hence the Taylor expansion about B! oe ! is
_ " oe " a "b8 B8 . " B 8oe!
Applying the ratio test,
The series converges absolutely for kBk " , but diverges at B oe ,, " .
kB8" k oe lim kBk oe kBk. lim 8 _ k B8 k 8_
15. We have 0 aBb oe "a" Bb, 0 w aBb oe "a" Bb# , 0 ww aBb oe #a" Bb$ , with 0 a8b aBb oe 8xa" Bb8" , for 8 " . It follows that 0 a8b a!b oe 8x, for 8 ! . Hence the Taylor expansion about B! oe ! is
_ " oe " B8 . " B 8oe!
Applying the ratio test,
The series converges absolutely for kBk " , but diverges at B oe ,, " .
kB8" k oe lim kBk oe kBk. 8 _ k B8 k 8_ lim
16. We have 0 aBb oe "a" Bb, 0 w aBb oe "a" Bb# , 0 ww aBb oe #a" Bb$ , with 0 a8b aBb oe 8xa" Bb8" , for 8 " . It follows that 0 a8b a#b oe a "b8" 8x for 8 ! . Hence the Taylor expansion about B! oe # is
_ " oe " a " b 8 aB # b 8 . "B 8oe!
Applying the ratio test, lim
The series converges absolutely for kB #k " , but diverges at B oe " and B oe $ . 17. Applying the ratio test, lim ka8 "bB8" k 8" oe lim kBk oe kBk. 8k 8_ 8_ k8B 8
8_
aB #b8" k a B #b 8 k
oe lim kB #k oe kB #k.
8_
The series converges absolutely for kBk " . Term-by-term differentiation results in
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---------------------------------------------------- CHAPTER 5. ---- C oe " 8# B8" oe " %B *B# "' B$
_ w 8oe"
C oe " 8# a8 "b B8# oe % ")B %)B# "!!B$
_ ww 8oe#
Shifting the indices, we can also write C w oe " a8 "b# B8 and C ww oe " a8 #b# a8 "b B8
_ _ 8oe! 8oe!
20. Shifting the index in the second series, that is, setting 8 oe 5 " , " +5 B5" oe " +8" B8 .
_ _ 8oe"
5oe!
Hence " +5" B " +5 B
_ _ 5 5oe! 5"
5oe!
oe " +5" B " +5" B5
_ _ 5
oe +" " a+5" +5" bB5"
_ 5oe"
5oe!
5oe"
21. Shifting the index by # , that is, setting 7 oe 8 # , " 8a8 "b+8 B8# oe " a7 #ba7 "b+7# B7
_ _
8oe#
oe " a8 #ba8 "b+8# B8
8oe!
7oe! _
22. Shift the index down by # , that is, set 7 oe 8 # . It follows that " +8 B8# oe " +7# B7
_ _
8oe!
oe " +8# B8
8oe#
7oe# _
24. Clearly,
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---------------------------------------------------- CHAPTER 5. ---- ^" B# " 8a8 "b+8 B8# oe " 8a8 "b+8 B8# " 8a8 "b+8 B8
_ _ _ 8oe# 8oe# 8oe#
Shifting the index in the first series, that is, setting 5 oe 8 # , " 8a8 "b+8 B8# oe " a5 #ba5 "b+5# B5
_ _ 5oe! _
8oe#
oe " a8 #ba8 "b+8# B8
8oe!
Hence
^" B
#
8oe#
" 8a8 "b+8 B
_
8#
oe " a8 #ba8 "b+8# B " 8a8 "b+8 B8 .
_ _ 8 8oe! 8oe#
Note that when 8 oe ! and 8 oe " , the coefficients in the second series are zero. So that
^" B# " 8a8 "b+8 B8#
_ 8oe#
oe " ca8 #ba8 "b+8# 8a8 "b+8 dB8
_ 8oe!
26. Clearly, " 8+8 B8" B " +8 B8 oe " 8+8 B8" " +8 B8"
_ _ _ _ 8oe! 8oe" 8oe!
8oe"
Shifting the index in the first series, that is, setting 5 oe 8 " , " 8+8 B8" oe " a5 "b+5" B5
_ _ 5oe!
8oe"
Shifting the index in the second series, that is, setting 5 oe 8 " , " +8 B
_ 8"
8oe!
oe " +5" B5
_ 5oe"
Combining the series, and starting the summation at 8 oe " , " 8+8 B8" B " +8 B8 oe +" " ca8 "b+8" +8" dB8
_ _ _ 8oe! 8oe"
8oe"
27. We note that
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---------------------------------------------------- CHAPTER 5. ----
B " 8a8 "b+8 B8# " +8 B8 oe " 8a8 "b+8 B8" " +8 B8
_ _ _ _ 8oe# 8oe! 8oe# 8oe!
Shifting the index in the first series, that is, setting 5 oe 8 " , " 8a8 "b+8 B8" oe " 5 a5 "b+5" B5
_ _ 5oe" _ 5oe!
8oe#
oe " 5 a5 "b+5" B5 ,
since the coefficient of the term associated with 5 oe ! is zero. Combining the series, B " 8a8 "b+8 B
_ 8oe# 8#
" +8 B oe " c8a8 "b+8" +8 dB8
_ _ 8 8oe! 8oe!
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---------------------------------------------------- CHAPTER 5. ---- Section 5.2 1. Let C oe +! +" B +# B# +8 B8 . Then C ww oe " 8a8 "b+8 B8# oe " a8 #ba8 "b+8# B8
_ _ 8oe# 8oe!
Substitution into the ODE results in " a8 #ba8 "b+8# B8 " +8 B8 oe !
_ _ 8oe!
8oe!
or " ca8 #ba8 "b+8# +8 dB8 oe !
_
8oe!
Equating all the coefficients to zero,
a8 #ba8 "b+8# +8 oe ! , +8# oe +8 , a8 "ba8 #b
8 oe ! " #
We obtain the recurrence relation 8 oe ! " #
The subscripts differ by two, so for 5 oe " # +#5# +#5% +! +#5 oe oe oeoe a#5 "b#5 a#5 $ba#5 #ba#5 "b#5 a#5 bx and +#5" oe Hence C oe +! "
_ _ B#5 B#5" +" " a#5 bx a#5 "bx 5oe! 5oe!
+#5" +#5$ +" oe oeoe #5 a#5 "b a#5 #ba#5 "b#5 a#5 "b a#5 "bx
The linearly independent solutions are C" oe +! OE" C# oe +" OEB B# B% B' oe +! -9=2 B #x %x 'x B$ B& B( oe +" =382 B . $x &x (x
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---------------------------------------------------- CHAPTER 5. ---- 4. Let C oe +! +" B +# B# +8 B8 . Then C ww oe " 8a8 "b+8 B8# oe " a8 #ba8 "b+8# B8
_ _ 8oe# 8oe!
Substitution into the ODE results in " a8 #ba8 "b+8# B8 5 # B# " +8 B8 oe ! .
_ _ 8oe!
8oe!
Rewriting the second summation, " a8 #ba8 "b+8# B8 " 5 # +8# B8 oe ! ,
_ _ 8oe#
8oe!
that is, #+# $ # +$ B " a8 #ba8 "b+8# 5 # +8# `B8 oe !
_ 8oe#
Setting the coefficients equal to zero, we have +# oe ! , +$ oe ! , and a8 #ba8 "b+8# 5 # +8# oe ! , for 8 oe # $ %
The recurrence relation can be written as +8# oe 5 # +8# , 8 oe # $ % . a8 #ba8 "b
The indices differ by four, so +% , +) , +"# , are defined by 5 # +! 5 # +% 5 # +) +% oe , +) oe , +"# oe , . %$ )( "# "" Similarly, +& , +* , +"$ , are defined by 5 # +" 5 # +& 5 # +* +& oe , +* oe , +"$ oe , . &% *) "$ "# The remaining coefficients are zero. Therefore the general solution is 5# % 5% 5' ) C oe + ! "" B B B"# %$ )(%$ "# "" ) ( % $ # % 5 5 5' + " "B B& B* B"$ . &% *)&% "$ "# * ) % % Note that for the even coefficients,
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---------------------------------------------------- CHAPTER 5. ----
+%7 oe and for the odd coefficients, +%7" oe
5 # +%7% , 7 oe " # $ a%7 "b%7 5 # +%7$ , 7 oe " # $ . %7a%7 "b
Hence the linearly independent solutions are C" aBb oe " "
_ _
7" a "b7" ^5 # B% $ % ( ) a%7 $ba%7 %b 7oe!
7" a "b7" ^5 # B% C# aBb oe B" " % & ) * a%7 %ba%7 &b -- 7oe!
6. Let C oe +! +" B +# B# +8 B8 . Then C w oe " 8+8 B8" oe " a8 "b+8" B8
_ _ 8oe" 8oe!
and C ww oe " 8a8 "b+8 B8# oe " a8 #ba8 "b+8# B8
_ _ 8oe# 8oe!
Substitution into the ODE results in ^# B# " a8 #ba8 "b+8# B8 B" a8 "b+8" B8 %" +8 B8 oe ! .
_ _ _ 8oe! 8oe! 8oe!
Before proceeding, write B# " a8 #ba8 "b+8# B8 oe " 8a8 "b+8 B8
_ _ 8oe! 8oe#
and B" a8 "b+8" B8 oe " 8 +8 B8
_ _ 8oe! 8oe"
It follows that
%+! %+# a$+" "#+$ bB " c#a8 #ba8 "b+8# 8a8 "b+8 8 +8 %+8 dB
_ 8oe# 8
oe !.
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---------------------------------------------------- CHAPTER 5. ---- Equating the coefficients to zero, we find that +# oe +! , +$ oe +" % , and +8# oe 8# #8 % +8 , 8 oe ! " # . #a8 #ba8 "b a#5 b# %5 % +#5 #a#5 #ba#5 "b
The indices differ by two, so for 5 oe ! " # +#5# oe and +#5$
a#5 "b# %5 # oe +#5" #a#5 $ba#5 #b
Hence the linearly independent solutions are B% B' C" aBb oe " B ' $!
#
C# aBb oe B
B$ (B& "*B( % "'! "*#!
7. Let C oe +! +" B +# B# +8 B8 . Then C oe " 8+8 B8" oe " a8 "b+8" B8
_ _ w 8oe" 8oe!
and C ww oe " 8a8 "b+8 B8# oe " a8 #ba8 "b+8# B8
_ _ 8oe# 8oe!
Substitution into the ODE results in " a8 #ba8 "b+8# B B" a8 "b+8" B #" +8 B8 oe ! .
_ _ _ 8 8 8oe! 8oe!
8oe!
First write B" a8 "b+8" B8 oe " 8 +8 B8
_ _ 8oe! 8oe"
We then obtain #+# #+! " ca8 #ba8 "b+8# 8 +8 #+8 dB8 oe !
_ 8oe"
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---------------------------------------------------- CHAPTER 5. ---- It follows that +# oe +! and +8# oe +8 a8 "b , 8 oe ! " # . Note that the indices differ by two, so for 5 oe " # +#5 and +#5" +#5# +#5% a "b 5 + ! oe oe oeoe #5 " a#5 $ba#5 "b " $ & a#5 "b +#5" +#5$ a "b 5 + " oe oe oeoe #5 a#5 #b#5 # % ' a#5 b
Hence the linearly independent solutions are
_ B# B% B' a "b8 B#8 " C" aBb oe " oe" " "$ "$& " $ & a#8 "b 8oe" _ B$ B& B( a "b8 B#8" " C# aBb oe B oeB # #% #%' # % ' a#8b 8oe"
9. Let C oe +! +" B +# B# +8 B8 . Then C oe " 8+8 B
_ w 8oe" 8"
oe " a8 "b+8" B8
_ 8oe!
and C ww oe " 8a8 "b+8 B8# oe " a8 #ba8 "b+8# B8
_ _ 8oe# 8oe!
Substitution into the ODE results in ^" B# " a8 #ba8 "b+8# B8 %B" a8 "b+8" B8 '" +8 B8 oe ! .
_ _ _ 8oe! 8oe! 8oe!
Before proceeding, write B# " a8 #ba8 "b+8# B8 oe " 8a8 "b+8 B8
_ _ 8oe! 8oe#
and B" a8 "b+8" B8 oe " 8 +8 B8
_ _ 8oe! 8oe"
It follows that
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---------------------------------------------------- CHAPTER 5. ----
'+! #+# a#+" '+$ bB " ca8 #ba8 "b+8# 8a8 "b+8 %8 +8 '+8 dB
_ 8oe#
8
oe !.
Setting the coefficients equal to zero, we obtain +# oe $+! , +$ oe +" $ , and +8# oe a8 #ba8 $b +8 , 8 oe ! " # . a8 "ba8 #b
Observe that for 8 oe # and 8 oe $ , we obtain +% oe +& oe ! . Since the indices differ by two, we also have +8 oe ! for 8 % . Therefore the general solution is a polynomial C oe +! +" B $+! B# +" B$ $ . Hence the linearly independent solutions are C" aBb oe " $B# and C# aBb oe B B$ $ .
10. Let C oe +! +" B +# B# +8 B8 . Then C ww oe " 8a8 "b+8 B8# oe " a8 #ba8 "b+8# B8
_ _ 8oe# 8oe!
Substitution into the ODE results in ^% B# " a8 #ba8 "b+8# B8 #" +8 B8 oe !
_ _ 8oe! 8oe!
First write B# " a8 #ba8 "b+8# B8 oe " 8a8 "b+8 B8 .
_ _ 8oe! 8oe#
It follows that #+! )+# a#+" #%+$ bB " c%a8 #ba8 "b+8# 8a8 "b+8 #+8 dB8 oe ! .
_ 8oe#
We obtain +# oe +! % , +$ oe +" "# and
%a8 #b+8# oe a8 #b+8 , 8 oe ! " # .
Note that for 8 oe # , +% oe ! . Since the indices differ by two, we also have +#5 oe ! for 5 oe # $ . On the other hand, for 5 oe " # , +#5" oe +" a#5 $b+#5" a#5 &ba#5 $b+#5$ oe oeoe 5 # a#5 "ba#5 "b %a#5 "b % % a#5 "ba#5 "b
Therefore the general solution is ________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ----
_ B# B#8" " 8 C oe +! + " B + ! + " . % % a#8 "ba#8 "b 8oe"
Hence the linearly independent solutions are C" aBb oe " B# % and C# aBb oe B
_ B#8" B$ B& B( . oeB" 8 % a#8 "ba#8 "b "# #%! ##%! 8oe"
11. Let C oe +! +" B +# B# +8 B8 . Then C w oe " 8+8 B8" oe " a8 "b+8" B8
_ _ 8oe" 8oe!
and C oe " 8a8 "b+8 B
_ ww 8oe# 8#
oe " a8 #ba8 "b+8# B8
_ 8oe!
Substitution into the ODE results in ^$ B# " a8 #ba8 "b+8# B8 $B" a8 "b+8" B8 " +8 B8 oe ! .
_ _ _ 8oe! 8oe! 8oe!
Before proceeding, write B and B" a8 "b+8" B8 oe " 8 +8 B8
_ _ 8oe! 8oe" _ #"
8oe!
a8 #ba8 "b+8# B oe " 8a8 "b+8 B8
_ 8 8oe#
It follows that
8 '+# +! a %+" ")+$ bB " c$a8 #ba8 "b+8# 8a8 "b+8 $8 +8 +8 dB oe !. _ 8oe#
We obtain +# oe +! ' , #+$ oe +" * , and
$a8 #b+8# oe a8 "b+8 , 8 oe ! " # .
The indices differ by two, so for 5 oe " # +#5 oe and
$ & a#5 "b +! a#5 "b+#5# a#5 $ba#5 "b+#5% oe oeoe 5 # a#5 #ba#5 b $a#5 b $ $ # % a#5 b
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---------------------------------------------------- CHAPTER 5. ---- a#5 b+#5" a#5 #ba#5 b+#5$ # % ' a#5 b +" oe # oeoe 5 $a#5 "b $ a#5 "ba#5 "b $ $ & a#5 "b
+#5" oe
Hence the linearly independent solutions are
_ B# B% &B' $ & a#8 "b B#8 C" aBb oe " oe"" 8 ' #% %$# $ # % a#8b 8oe"
C# aBb oe B
_ #B$ )B& "'B( # % ' a#8b B#8" oeB" 8 * "$& *%& $ $ & a#8 "b 8oe"
12. Let C oe +! +" B +# B# +8 B8 . Then C w oe " 8+8 B8" oe " a8 "b+8" B8
_ _ 8oe" 8oe!
and C oe " 8a8 "b+8 B8# oe " a8 #ba8 "b+8# B8
_ _ ww 8oe# 8oe!
Substitution into the ODE results in a" Bb" a8 #ba8 "b+8# B8 B" a8 "b+8" B8 " +8 B8 oe ! .
_ _ _ 8oe! 8oe! 8oe!
Before proceeding, write B " a8 #ba8 "b+8# B8 oe " a8 "b8 +8" B8
_ _ 8oe! 8oe"
and B" a8 "b+8" B8 oe " 8 +8 B8
_ _ 8oe! 8oe"
It follows that #+# +! " ca8 #ba8 "b+8# a8 "b8 +8" 8 +8 +8 dB8 oe !.
_ 8oe"
We obtain +# oe +! # and
a8 #ba8 "b+8# a8 "b8 +8" a8 "b+8 oe !
for 8 oe ! " # . Writing out the individual equations, ________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ----
The coefficients can be calculated successively as +$ oe +! a# $b, +% oe +$ # +# "# oe +! #% , +& oe $+% & +$ "! oe +! "#! , . We can now see that for 8 # , +8 is proportional to +! . In fact, for 8 # , +8 oe +! a8xb . Therefore the general solution is C oe +! + " B Hence the linearly independent solutions are C# aBb oe B and C" aBb oe " "
_
$ # +$ # " + # % $ +% $ # + $ + # & % +& % $ + % # + $ ' & +' & % + & $ + %
oe! oe! oe! oe!
+! B# +! B$ + ! B% . #x $x %x
B8 8x 8oe#
13. Let C oe +! +" B +# B# +8 B8 . Then C oe " 8+8 B
_ w 8oe" 8"
oe " a8 "b+8" B8
_ 8oe!
and C ww oe " 8a8 "b+8 B8# oe " a8 #ba8 "b+8# B8
_ _ 8oe# 8oe!
Substitution into the ODE results in # " a8 #ba8 "b+8# B8 B" a8 "b+8" B8 $" +8 B8 oe ! .
_ _ _ 8oe! 8oe! 8oe!
First write B" a8 "b+8" B8 oe " 8 +8 B8
_ _ 8oe! 8oe"
We then obtain %+# $+! " c#a8 #ba8 "b+8# 8 +8 $+8 dB8 oe !
_ 8oe"
It follows that +# oe $+! % and
#a8 #ba8 "b+8# a8 $b+8 oe !
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---------------------------------------------------- CHAPTER 5. ---- for 8 oe ! " # . The indices differ by two, so for 5 oe " # +#5 oe a#5 "b+#5# a#5 "ba#5 "b+#5% oe # oe #a#5 "ba#5 b # a#5 $ba#5 #ba#5 "ba#5 b a "b5 $ & a#5 "b oe +! #5 a#5 bx a#5 #b+#5" a#5 ba#5 #b+#5$ oe # oe #a#5 ba#5 "b # a#5 #ba#5 "ba#5 ba#5 "b a "b5 % ' a#5 ba#5 #b oe +" #5 a#5 "bx
and
+#5" oe
Hence the linearly independent solutions are
_ " " " ( a "b8 % ' a#8 #b #8" C# aBb oe B B$ B& B oeB" B $ #! #"! #8 a#8 "bx 8oe"
_ $ & ( ' a "b8 $ & a#8 "b #8 C" aBb oe " B# B% B oe " B % $# $)% #8 a#8bx 8oe!
15a+b. From Prob. # , we have C" aBb oe "
_
Since +! oe C a!b and +" oe C w a!b , we have C aBb oe # C" aBb C# aBb . That is, " " " " CaBb oe # B B# B$ B% B& B' . $ % "& #% :% oe # B B# B$ $ :& oe # B B# B$ $ B% %
B#8 #8 8x 8oe!
and
C# aBb oe "
_
#8 8x B#8" . a#8 "bx 8oe!
The four- and five-term polynomial approximations are
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---------------------------------------------------- CHAPTER 5. ---- a, b .
a- b. The four-term approximation :% appears to be reasonably accurate awithin "!%b on the interval kBk !( . 17a+b. From Prob. (, the linearly independent solutions are C" aBb oe " "
_
a "b8 B#8 " $ & a#8 "b 8oe" a "b8 B#8" # % ' a#8b 8oe"
_
Since +! oe C a!b and +" oe C w a!b , we have C aBb oe % C" aBb C# aBb . That is, " % " % CaBb oe % B %B# B$ B% B& B' . # $ ) "& " :% oe % B %B# B$ # " % :& oe % B %B# B$ B% # $
C# aBb oe B "
The four- and five-term polynomial approximations are
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---------------------------------------------------- CHAPTER 5. ---- a, b .
a- b. The four-term approximation :% appears to be reasonably accurate awithin "!%b on the interval kBk !& . 18a+b. From Prob. "# , we have C" aBb oe " "
_
Since +! oe C a!b and +" oe C w a!b , we have C aBb oe $ C" aBb # C# aBb . That is, $ " " " " ' CaBb oe $ #B B# B$ B% B& B . # # ) %! #%! $ :% oe $ #B B# # $ :& oe $ #B B# # " $ B # " $ " % B B # )
B8 8x 8oe#
and
C# aBb oe B .
The four- and five-term polynomial approximations are
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---------------------------------------------------- CHAPTER 5. ---- a, b .
a- b. The four-term approximation :% appears to be reasonably accurate awithin "!%b on the interval kBk !* . 20. Two linearly independent solutions of Airy's equation aabout B! oe !b are C" aBb oe " "
_ _
B$8 # $ a$8 "ba$8b 8oe"
Applying the ratio test to the terms of C" aBb , lim
C# aBb oe B "
B$8" $ % a$8ba$8 "b 8oe"
Similarly, applying the ratio test to the terms of C# aBb ,
8 _ k$
k# $ a$8 "ba$8b B$8$ k " oe lim kBk$ oe ! $8 k 8 _ k# $ a$8 #ba$8 $b B 8 _ a$8 "ba$8 #ba$8 $b $ % a$8ba$8 "b B$8% % a$8 $ba$8 %b B$8" k " kBk$ oe ! 8 _ a$8 #ba$8 $ba$8 %b
lim
oe lim
Hence both series converge absolutely for all B . 21. Let C oe +! +" B +# B# +8 B8 . Then C oe " 8+8 B
_ w 8oe" 8"
oe " a8 "b+8" B8
_ 8oe!
and
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---------------------------------------------------- CHAPTER 5. ---- C oe " 8a8 "b+8 B8# oe " a8 #ba8 "b+8# B8
_ _ ww 8oe# 8oe!
Substitution into the ODE results in " a8 #ba8 "b+8# B8 #B" a8 "b+8" B8 - " +8 B8 oe ! .
_ _ _ 8oe! 8oe!
8oe!
First write B" a8 "b+8" B8 oe " 8 +8 B8
_ _ 8oe! 8oe"
We then obtain #+# - +! " ca8 #ba8 "b+8# #8 +8 - +8 dB8 oe !
_ 8oe"
Setting the coefficients equal to zero, it follows that +8# oe
a#8 -b +8 a8 "ba8 #b
for 8 oe ! " # . Note that the indices differ by two, so for 5 oe " # +#5 oe
a%5 % -b+#5# a%5 ) -ba%5 % -b+#5% oe oe a#5 "b#5 a#5 $ba#5 #ba#5 "b#5 - a- %5 )ba- %5 %b oe a "b 5 +! a#5 bx a%5 # -b+#5" a%5 ' -ba%5 # -b+#5$ oe oe #5 a#5 "b a#5 #ba#5 "b#5 a#5 "b a- #b a- %5 'ba- %5 #b oe a "b 5 +" a#5 "bx - # -a- %b % -a- %ba- )b ' B B B #x %x 'x
and
+#5" oe
Hence the linearly independent solutions of the Hermite equation aabout B! oe !b are C" aBb oe "
C# aBb oe B
- # $ a- #ba- 'b & a- #ba- 'ba- "!b ( B B B $x &x (x
a,b. Based on the recurrence relation ________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ---- a#8 -b +8 , a8 "ba8 #b
+8# oe
the series solution will terminate as long as - is a nonnegative even integer. If - oe #7, then one or the other of the solutions in Part a,b will contain at most 7# " terms. In particular, we obtain the polynomial solutions corresponding to - oe ! # % ' ) "! -oe! -oe# -oe% -oe' -oe) - oe "! C" aBb oe " C# aBb oe B C" aBb oe " #B# C# aBb oe B #B$ $ C" aBb oe " %B# %B% $ C# aBb oe B %B$ $ %B& "&
a- b. Observe that if - oe #8 , and +! oe +" oe " , then +#5 oe a "b5
#8 a#8 %5 )ba#8 %5 %b a#5 bx
and +#5" oe a "b5
for 5 oe " # c8#d. It follows that the coefficient of B8 , in C" and C# , is +8 oe 5 %5 5x a "b a#5"bx for 8 oe #5 "
5
a#8 #b a#8 %5 'ba#8 %5 #b a#5 "bx
% a "b5 a#55xx for 8 oe #5 b
Then by definition, L8 aBb oe
a#5 bx a#5 bx a "b5 #8 %5 5x C" aBb oe a "b5 5x C" aBb for 8 oe #5 a#5"bx # a#5"bx a "b5 #8 %5 5x C# aBb oe a "b5 C# aBb for 8 oe 5x
#5 "
Therefore the first six Hermite polynomials are
L! aBb oe " L" aBb oe #B L# aBb oe %B# # L$ aBb oe )B) "#B L% aBb oe "'B% %)B# "# L& aBb oe $#B& "'!B$ "#!B
23. The series solution is given by ________________________________________________________________________
page 189
---------------------------------------------------- CHAPTER 5. ---- " " " " CaBb oe " B# # B% $ B' % B) # # #x # $x # %x
24. The series solution is given by C aBb oe " B# B% B' B) . ' $! "#!
25. The series solution is given by C aBb oe B B$ B& B( B* . # #% #%' #%')
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page 190
---------------------------------------------------- CHAPTER 5. ----
26. The series solution is given by C aBb oe B B$ B& B( B* "# #%! ##%! "'"#)
27. The series solution is given by CaBb oe " B% B) B"# . "# '(# ))(!%
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---------------------------------------------------- CHAPTER 5. ----
28. Let C oe +! +" B +# B# +8 B8 . Then C oe " 8+8 B8" oe " a8 "b+8" B8
_ _ w 8oe" 8oe!
and C ww oe " 8a8 "b+8 B8# oe " a8 #ba8 "b+8# B8
_ _ 8oe# 8oe!
Substitution into the ODE results in a" Bb" a8 #ba8 "b+8# B8 B" a8 "b+8" B8 # " +8 B8 oe ! .
_ _ _ 8oe! 8oe! 8oe!
After appropriately shifting the indices, it follows that #+# #+! " ca8 #ba8 "b+8# a8 "b8 +8" 8 +8 # +8 dB8 oe !.
_ 8oe"
We find that +# oe +! and
a8 #ba8 "b+8# a8 "b8 +8" a8 #b+8 oe ! $ # +$ # " + # + " % $ +% $ # + $ & % +& % $ + % + $ ' & +' & % + & # + % oe! oe! oe! oe!
for 8 oe " # . Writing out the individual equations,
Since +! oe ! and +" oe " , the remaining coefficients satisfy the equations
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page 192
---------------------------------------------------- CHAPTER 5. ----
$ # +$ " oe ! % $ +% $ # + $ oe ! & % +& % $ + % + $ oe ! ' & +' & % + & # + % oe ! That is, +$ oe "' , +% oe ""# , +& oe "#% , +' oe "%& , . Hence the series solution of the initial value problem is " " " " "$ ( CaBb oe B B$ B% B& B' B ' "# #% %& "!!)
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page 193
---------------------------------------------------- CHAPTER 5. ---- Section 5.3 2. Let C oe 9aBb be a solution of the initial value problem. First note that C ww oe a=38 BbC w a-9= BbC Differentiating twice,
Given that 9a!b oe ! and 9 w a!b oe " , the first equation gives 9 ww a!b oe ! and the last two equations give 9 www a!b oe # and 9 3@ a!b oe ! . 3. Let C oe 9aBb be a solution of the initial value problem. First write C ww oe Differentiating twice, C www oe " B w $ 68 B C C B# B#
C www oe a=38 BbC ww #a-9= BbC w a=38 BbC C 3@ oe a=38 BbC www $a-9= BbC ww $a=38 BbC w a-9= BbC
" ^B B# C ww a$B 68 B B #bC w a$ ' 68 BbC ` B$ " '^B# B$ C www ^$B# 68 B #B# %BC ww B% a' )B "#B 68 BbC w a") 68 B "&bC "
C 3@ oe
Given that 9a"b oe # and 9 w a"b oe ! , the first equation gives 9 ww a"b oe ! and the last two equations give 9 www a!b oe ' and 9 3@ a!b oe %# . 4. Let C oe 9aBb be a solution of the initial value problem. First note that C ww oe B# C w a=38 BbC Differentiating twice,
Given that 9a!b oe +! and 9 w a!b oe +" , the first equation gives 9 ww a!b oe ! and the last two equations give 9 www a!b oe +! and 9 3@ a!b oe %+" .
C www oe B# C ww a#B =38 BbC w a-9= BbC C 3@ oe B# C www a%B =38 BbC ww a# #-9= BbC w a=38 BbC
5. Clearly, :aBb oe % and ; aBb oe 'B are analytic for all B . Hence the series solutions converge everywhere. 7. The zeroes of T aBb oe " B$ are the three cube roots of " . They all lie on the unit circle in the complex plane. So for B! oe ! , 3738 oe " . For B! oe # , the nearest ________________________________________________________________________
page 194
---------------------------------------------------- CHAPTER 5. ---- root is /31$ oe S" 3$ <# , hence 3738 oe $ .
9a,b. :aBb oe B and ; aBb oe " are analytic for all B . a- b. :aBb oe B and ; aBb oe " are analytic for all B . a. b. :aBb oe ! and ; aBb oe 5B# are analytic for all B . a/b. The only root of T aBb oe " B is " . Hence 3738 oe " . a1b. :aBb oe B and ; aBb oe # are analytic for all B . a3b The zeroes of T aBb oe " B# are ,, 3 . Hence 3738 oe " . a4b The zeroes of T aBb oe % B# are ,,# . Hence 3738 oe # . a5 b The zeroes of T aBb oe $ B# are ,,$ . Hence 3738 oe $ . a6b The only root of T aBb oe " B is " . Hence 3738 oe " . a7b. :aBb oe B# and ; aBb oe $# are analytic for all B . a8b. :aBb oe a" Bb# and ; aBb oe $# are analytic for all B . 12. The Taylor series expansion of /B , about B! oe ! , is / oe"
_ B
8. The only root of T aBb oe B is zero. Hence 3738 oe " .
B8 8x 8oe!
Let C oe +! +" B +# B# +8 B8 . Substituting into the ODE, "
_ _ _ B8 " a8 #ba8 "b+8# B8 -- B " +8 B8 oe ! 8 x --8 oe ! 8oe! 8oe!
First note that B " +8 B8 oe " +8" B8 oe +! B +" B# +# B$ +8" B8 .
_ _ 8oe! 8oe"
The coefficient of B8 in the product of the two series is -8 oe #+#
" " " '+$ "#+% a8 "b8 +8" a8 #ba8 "b+8# a 8 "b x a8 #b x 8x
Expanding the individual series, it follows that
#+# a#+# '+$ bB a+# '+$ "#+% bB# a+# '+$ "#+% #!+& bB$ + ! B + " B# + # B $ oe !
Setting the coefficients equal to zero, we obtain the system #+# oe !, #+# '+$ +! oe !, +# '+$ "#+% +" oe ! , +# '+$ "#+% #!+& +# oe ! , . Hence the general solution is
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---------------------------------------------------- CHAPTER 5. ---- CaBb oe +! +" B +! B$ B% B& B' % a+! +" b a#+" +! b OE +! #+" . $ ' "# %! "#! C" aBb oe " C# aBb oe B
We find that two linearly independent solutions are B$ B% B& ' "# %! B% B& B' "# #! '!
Since :aBb oe ! and ; aBb oe B/B converge everywhere, 3 oe _ . 13. The Taylor series expansion of -9= B , about B! oe ! , is -9= B oe "
_
a "b8 B#8
8oe!
a#8b x
Let C oe +! +" B +# B# +8 B8 . Substituting into the ODE,
8 #8 " a "b B -- " a8 #ba8 "b+8# B8 -- " 8+8 B8 #" +8 B8 oe ! a#8b x
_
_
_
_
8oe!
8oe!
8oe"
8oe!
The coefficient of B8 in the product of the two series is
-8 oe #+# ,8 '+$ ,8" "#+% ,8# a8 "b8 +8" ," a8 #ba8 "b+8# ,! , #+# #+! " -8 B8 " a8 #b+8 B8 oe !
_ _ 8oe" 8oe"
in which -9= B oe ,! ," B ,# B# ,8 B8 . It follows that
Expanding the product of the series, it follows that
#+# #+! '+$ B a +# "#+% bB# a $+$ #!+& bB$ +1 B +$ B$ #+% B% oe !
Setting the coefficients equal to zero, +# +! oe ! , '+$ +" oe ! , +# "#+% oe ! , $+$ #!+& +$ oe ! , Hence the general solution is CaBb oe +! +" B +! B# +" B$ B% B& B' B( +! +" +! +" . ' "# '! "#! &'! B% B' "# "#!
We find that two linearly independent solutions are C" aBb oe " B#
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page 196
---------------------------------------------------- CHAPTER 5. ---- C# aBb oe B
The nearest zero of T aBb oe -9= B is at B oe ,,1# Hence 3738 oe 1# 14. The Taylor series expansion of 68a" Bb , about B! oe ! , is 68a" Bb oe "
_
B$ B& B( ' '! &'!
a "b8" B8
8oe"
8
Let C oe +! +" B +# B# +8 B8 . Substituting into the ODE, "
_
a "b8 B8
_
"
8oe!
a "b
8x
8 -- " a8 #ba8 "b+8# B
_
8oe!
8"
B8
8oe"
8
8 8 -- " a8 "b+8" B B " +8 B oe ! _ _ 8oe! 8oe!
The first product is the series
#+# a #+# '+$ bB a+# '+$ "#+% bB# a +# '+$ "#+% #!+& bB$ .
The second product is the series
+" B a#+# +" #bB# a$+$ +# +" $bB$ a%+% $+$ # #+# $ +" %bB$
Combining the series and equating the coefficients to zero, we obtain #+# oe ! #+# '+$ +" +! oe ! "#+% '+$ $+# $+" # oe ! #!+& "#+% *+$ $+# +" $ oe ! Hence the general solution is B$ B% (B& B' & CaBb oe +! +" B a+! +" b a#+! +" b +" OE +" + ! . $ ' #% "#! "#! We find that two linearly independent solutions are C" aBb oe " C# aBb oe B B$ B% B' ' "# "#! B$ B% (B& ' #% "#!
The coefficient :aBb oe /B 68a" Bb is analytic at B! oe ! , but its power series has a radius of convergence 3 oe " . ________________________________________________________________________
page 197
---------------------------------------------------- CHAPTER 5. ----
15. If C" oe B and C# oe B# are solutions, then substituting C# into the ODE results in Setting B oe ! , we find that T a!b oe ! . Similarly, substituting C" into the ODE results in Ua!b oe ! . Therefore T aBbUaBb and V aBbT aBb may not be analytic. If they were, Theorem $#" would guarantee that C" and C# were the only two solutions. But note that an arbitrary value of Ca!b cannot be a linear combination of C" a!b and C# a!b Hence B! oe ! must be a singular point. 16. Let C oe +! +" B +# B# +8 B8 . Substituting into the ODE, " a8 "b+8" B8 " +8 B8 oe !
_ _ 8oe!
# T aBb #B UaBb B# V aBb oe ! .
8oe!
That is, " ca8 "b+8" +8 dB8 oe !
_
8oe!
Setting the coefficients equal to zero, we obtain +8" oe for 8 oe ! " # . It is easy to see that +8 oe +! a8 xb . Therefore the general solution is CaBb oe +! "" B oe +! / B B# B$ #x $x +8 8"
The coefficient +! oe C a!b, which can be arbitrary. 17. Let C oe +! +" B +# B# +8 B8 . Substituting into the ODE, " a8 "b+8" B8 B " +8 B8 oe !
_ _ 8oe!
8oe!
That is, " a8 "b+8" B8 " +8" B8 oe !
_ _ 8oe"
8oe!
Combining the series, we have
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page 198
---------------------------------------------------- CHAPTER 5. ---- +" " ca8 "b+8" +8" d B8 oe !
_
Setting the coefficient equal to zero, +" oe ! and +8" oe +8" a8 "b for 8 oe " # . Note that the indices differ by two, so for 5 oe " # +#5# +#5% +! +#5 oe oe oeoe a#5 b a#5 #ba#5 b # % a#5 b and +#5" oe ! Hence the general solution is CaBb oe +! "" B# B% B' B#8 # $ 8 # # #x # $x # 8x oe +! /B:^B# #.
8oe"
The coefficient +! oe C a!b, which can be arbitrary.
_
19. Let C oe +! +" B +# B# +8 B8 . Substituting into the ODE, a" Bb" a8 "b+8" B8 " +8 B8 oe !
_ 8oe! 8oe!
That is, " a8 "b+8" B8 " 8 +8 B8 " +8 B8 oe !
_ _ _ 8oe" 8oe!
8oe!
Combining the series, we have +" +! " ca8 "b+8" 8 +8 +8 d B8 oe !
_ 8oe"
Setting the coefficients equal to zero, +" oe +! and +8" oe +8 for 8 oe ! " # . Hence the general solution is CaBb oe +! " B B# B$ B8 ` " oe +! "B
The coefficient +! oe C a!b, which can be arbitrary.
21. Let C oe +! +" B +# B# +8 B8 . Substituting into the ODE,
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page 199
---------------------------------------------------- CHAPTER 5. ---- " a8 "b+8" B8 B " +8 B8 oe " B
_ _ 8oe!
8oe!
That is, " a8 "b+8" B8 " +8" B8 oe " B
_ _ 8oe"
8oe!
Combining the series, and the nonhomogeneous terms, we have a+" "b a#+# +! "bB " ca8 "b+8" +8" d B8 oe !
_ 8oe#
Setting the coefficients equal to zero, we obtain +" oe " , #+# +! " oe ! , and +8# +8 oe , 8 oe $ % . 8 The indices differ by two, so for 5 oe # $ +#5 oe +#5# +#5% a "b5" +# a " b 5 a+ ! " b oe oeoe oe , % ' a#5 b # % ' a#5 b a#5 b a#5 #ba#5 b +#5" +#5$ a "b 5 oe oeoe $ & a#5 "b a#5 "b a#5 "ba#5 "b " +! # B$ B% B& B' B +! # +! $ # $ # #x $ & # $x
and for 5 oe " # +#5" oe oe
Hence the general solution is CaBb oe +! B CaBb oe +! ""
Collecting the terms containing +! , B# B% B' # $ # # #x # $x # $ B B B% B& B' B( "B # $ # $ # #x $ & # $x $ & ( B# B$ B% B& B' B( # $ # $ # #x $ & # $x $ & (
#
Upon inspection, we find that CaBb oe +! /B:^ B# # "B
Note that the given ODE is first order linear, with integrating factor .a>b oe /B # . The general solution is given by
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page 200
---------------------------------------------------- CHAPTER 5. ---- CaBb oe /B # ( /? # .? aC a!b "b/B # "
#
B
#
#
!
23. If ! oe ! , then C" aBb oe " . If ! oe #8 , then +#7 oe ! for 7 8 " . As a result, C" aBb oe " " a "b7
8 7oe"
#7 8a8 "ba8 7 "ba#8 "ba#8 $ba#8 #7 "b #7 B a#7bx
!oe! !oe# !oe% C# aBb oe B " a "b7
8 7oe"
" " $B# " "!B#
$& % $ B
If ! oe #8 " , then +#7" oe ! for 7 8 " . As a result,
#7 8a8 "ba8 7 "ba#8 $ba#8 &ba#8 #7 "b #7" B a#7 "bx
!oe" !oe$ !oe& 24a+b. Based on Prob. #$, !oe! !oe# !oe%
B B & B$ $ B "% B$ $
#" & & B
" " $B# " "!B# T! aBb oe "
$& % $ B
C" a"b oe " C" a"b oe # C" a"b oe ) $
Normalizing the polynomials, we obtain
T# aBb oe
" $ # B # # $ "& $& T% aBb oe B# B% ) % ) !oe" !oe$ !oe& Similarly, B B & B$ $ B "% B$ $ C# a " b oe " C# a"b oe ) C# a"b oe "&
# $
#" & & B
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page 201
---------------------------------------------------- CHAPTER 5. ---- T" aBb oe B
a, b .
$ & T$ aBb oe B B$ # # "& $& $ '$ & T& aBb oe B B B ) % )
a- b. T! aBb has no roots. T" aBb has one root at B oe ! . The zeros of T# aBb are at B oe ,, "$ The zeros of T$ aBb are B oe !, ,,$& . The roots of T% aBb are given by B# oe S"& #$!<$& , S"& #$!<$& . The roots of T& aBb are given by B oe ! and B# oe S$& #(!<'$ , S$& #(!<'$ . 25. Observe that T8 a "b oe
But T8 a"b oe " for all nonnegative integers 8. 27. We have ^B# "8 oe "
8
oe a " b 8 T 8 a "b .
a "b 8 a "b5 a#8 #5 bx " #8 5 xa8 5 bxa8 #5 bx 5oe!
8#
a "b85 8 x #5 B , 5 x a 8 5 bx 5oe!
which is a polynomial of degree #8. Differentiating n times,
in which the lower index is . oe c8#d " . Note that if 8 oe #7 ", then . oe 7 " . ________________________________________________________________________
page 202
8 .8 # a "b85 8 x ^B "8 oe " a#5 ba#5 "ba#5 8 "bB#58 , .B8 5 xa8 5 bx 5oe.
---------------------------------------------------- CHAPTER 5. ---- Now shift the index, by setting 5 oe 8 4. Hence a "b 4 8 x .8 # ^B "8 oe " a#8 #4ba#8 #4 "ba8 #4 "bB8#4 a8 4bx4 x .B8 4oe!
c8#d
oe 8x "
c8#d
a "b4 a#8 #4bx 8#4 B a8 4bx4 xa8 #4bx 4oe!
Based on Prob. #&,
.8 # ^B "8 oe 8x #8 T8 aBb 8 .B
29. Since the 8 " polynomials T! , T" , , T8 are linearly independent, and the degree of T5 is 5 , any polynomial, 0 , of degree 8 can be expressed as a linear combination 0 aBb oe " +5 T5 aBb .
8 5oe!
Multiplying both sides by T7 and integrating, ( 0 aBbT7 aBb.B oe " +5 ( T5 aBbT7 aBb.B .
" 8 " " 5oe! "
Based on Prob. #), ( T5 aBbT7 aBb.B oe
" "
# $57 . #7 "
Hence ( 0 aBbT7 aBb.B oe
" "
# +7 #7 "
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---------------------------------------------------- CHAPTER 5. ---- Section 5.4 2. We see that T aBb oe ! when B oe ! and " . Since the three coefficients have no factors in common, both of these points are singular points. Near B oe !,
B!
lim B :aBb oe lim B
B!
B# a"
#B
Bb# % Bb#
oe #
B!
lim B# ; aBb oe lim B#
B!
B# a"
oe %
The singular point B oe ! is regular. Considering B oe ",
B"
lim aB "b:aBb oe lim aB "b
B"
B# a" Bb#
#B
The latter limit does not exist. Hence B oe " is an irregular singular point. 3. T aBb oe ! when B oe ! and " . Since the three coefficients have no common factors, both of these points are singular points. Near B oe !,
B!
lim B :aBb oe lim B
B!
B# B# a" Bb B# a" B# oe " Bb
The limit does not exist, and so B oe ! is an irregular singular point. Considering B oe ",
B"
lim aB "b:aBb oe lim aB "b
B"
B"
lim aB "b# ; aBb oe lim aB "b#
B"
$B oe ! B# a" Bb
Hence B oe " is a regular singular point. 4. T aBb oe ! when B oe ! and ,, " . Since the three coefficients have no common factors, both of these points are singular points. Near B oe !,
B!
lim B :aBb oe lim B
B!
B$ a"
# B# b
The limit does not exist, and so B oe ! is an irregular singular point. Near B oe ",
B"
lim aB "b:aBb oe lim aB "b
B"
# oe " B$ a" B# b
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---------------------------------------------------- CHAPTER 5. ---- lim aB "b# ; aBb oe lim aB "b#
B"
B"
B$ a"
# oe ! B# b
Hence B oe " is a regular singular point. At B oe " ,
B"
lim aB "b:aBb oe lim aB "b
B"
B$ a"
# oe " B# b # oe ! B# b
B"
lim aB "b# ; aBb oe lim aB "b#
B"
B$ a"
Hence B oe " is a regular singular point. 6. The only singular point is at B oe ! . We find that B lim B :aBb oe lim B # oe " B! B! B lim B# ; aBb oe lim B#
B!
B!
B# / # oe / # B#
Hence B oe ! is a regular singular point. 7. The only singular point is at B oe $ . We find that
B$
lim aB $b:aBb oe lim aB $b
B$
#B oe ' B$ " B# oe ! B$
B$
lim aB $b# ; aBb oe lim aB $b#
B$
Hence B oe $ is a regular singular point. 8. Dividing the ODE by Ba" B# b , we find that
$
:aBb oe
" # and ; aBb oe . Ba" B# b Ba" Bb# a" Bb$ lim B :aBb oe lim B
B!
The singular points are at B oe ! and ,, " . For B oe !,
B!
" oe " Ba" B# b # oe !
B!
lim B# ; aBb oe lim B#
B!
Ba" Bb# a" Bb$
Hence B oe ! is a regular singular point. For B oe ", ________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ---- lim aB "b:aBb oe lim aB "b
B"
B"
" " oe #b Ba" B # #
# $
B"
lim aB "b# ; aBb oe lim aB "b#
B"
Ba" Bb a" Bb
oe
" %
Hence B oe " is a regular singular point. For B oe ",
B"
lim aB "b:aBb oe lim aB "b
B"
" " oe #b Ba" B # Ba" Bb# a" Bb$ #
B"
lim aB "b# ; aBb oe lim aB "b#
B"
The latter limit does not exist. Hence B oe " is an irregular singular point. 9. Dividing the ODE by aB #b# aB "b, we find that :aBb oe a B #b $
#
and ; aBb oe
# . aB #baB "b a B #b # $ $
The singular points are at B oe # and " . For B oe #,
B#
lim aB #b:aBb oe lim aB #b
B#
The limit does not exist. Hence B oe # is an irregular singular point. For B oe ",
B"
lim aB "b:aBb oe lim aB "b
B"
a B #b #
oe !
B"
lim aB "b# ; aBb oe lim aB "b#
B"
# oe ! aB #baB "b
Hence B oe " is a regular singular point. 10. T aBb oe ! when B oe ! and $ . Since the three coefficients have no common factors, both of these points are singular points. Near B oe !,
B!
lim B :aBb oe lim B
B!
B" " oe Ba$ Bb $ # oe ! Ba$ Bb
B!
lim B# ; aBb oe lim B#
B!
Hence B oe ! is a regular singular point. For B oe $, ________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ---- lim aB $b:aBb oe lim aB $b
B$
B$
B" % oe Ba$ Bb $ # oe ! Ba$ Bb
B$
lim aB $b# ; aBb oe lim aB $b#
B$
Hence B oe $ is a regular singular point. 11. Dividing the ODE by aB# B #b, we find that :aBb oe B" # and ; aBb oe . aB #baB "b aB #baB "b lim aB #b:aBb oe lim B" " oe B# B " $
The singular points are at B oe # and " . For B oe #,
B#
B#
lim aB #b# ; aBb oe lim
#aB #b oe ! B# B "
Hence B oe # is a regular singular point. For B oe ",
B"
lim aB "b:aBb oe lim
B" # oe B" B # $ #aB "b oe ! a B #b
B"
lim aB "b# ; aBb oe lim
B"
Hence B oe " is a regular singular point. 13. Note that :aBb oe 68kBk and ; aBb oe $B . Evidently, :aBb is not analytic at B! oe ! . Furthermore, the function B :aBb oe B 68kBk does not have a Taylor series about B! oe !. Hence B oe ! is an irregular singular point. 14. T aBb oe ! when B oe ! . Since the three coefficients have no common factors, B oe ! is a singular point. The Taylor series of /B ", about B oe !, is Hence the function B :aBb oe #a/B "bB is analytic at B oe ! . Similarly, the Taylor series of /B -9= B , about B oe !, is The function B# ; aBb oe /B -9= B is also analytic at B oe ! . Hence B oe ! is a regular singular point. /B -9= B oe " B B$ $ B% ' . /B " oe B B# # B$ ' .
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---------------------------------------------------- CHAPTER 5. ---- 15. T aBb oe ! when B oe ! . Since the three coefficients have no common factors, B oe ! is a singular point. The Taylor series of =38 B , about B oe !, is Hence the function B :aBb oe $=38 BB is analytic at B oe ! . On the other hand, ; aBb is a rational function, with
B!
=38 B oe B B$ $x B& &x .
lim B# ; aBb oe lim B#
B!
" B# oe " B#
Hence B oe ! is a regular singular point. 16. T aBb oe ! when B oe ! . Since the three coefficients have no common factors, B oe ! is a singular point. We find that
B!
Although the function V aBb oe -9> B does not have a Taylor series about B oe ! , note that B# ; aBb oe B -9> B oe " B# $ B% %& #B' *%& . Hence B oe ! is a regular singular point. Furthermore, ;aBb oe -9> BB# is undefined at B oe ,, 81 . Therefore the points B oe ,, 81 are also singular points. First note that
B,,81
lim B :aBb oe lim B
B!
" oe " B
lim aB...81b:aBb oe lim aB...81b
B,,81
" oe ! B
Furthermore, since -9> B has period 1 ,
; aBb oe -9> BB oe -9>aB ... 81bB " oe -9>aB ... 81b a B ... 81 b ,, 8 1 a B ... 81 b a B ... 81 b ,, 8 1
Therefore aB ... 81b# ; aBb oe aB ... 81b-9>aB ... 81b" From above,
aB ... 81b-9>aB ... 81b oe " aB ... 81b# $ aB ... 81b% %& .
Note that the function in brackets is analytic near B oe ,, 81. It follows that the function aB ... 81b# ; aBb is also analytic near B oe ,, 81 . Hence all the singular points are regular. 18. The singular points are located at B oe ,, 81 , 8 oe ! " . Dividing the ODE by B =38 B , we find that B :aBb oe $ -=- B and B# ; aBb oe B# -=- B . Evidently, B :aBb is not even defined at B oe !. Hence B oe ! is an irregular singular point. On the other hand, the Taylor series of B -=- B, about B oe !, is
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---------------------------------------------------- CHAPTER 5. ----
Noting that -=- aB ... 81b oe a "b8 -=- B ,
B -=- B oe " B# ' (B% $'! .
aB ... 81b:aBb oe $a "b8 aB ... 81b-=- aB ... 81bB oe $a "b8 aB ... 81b-=- aB ... 81b"
It is apparent that aB ... 81b:aBb is analytic at B oe ,, 81 . Similarly,
" a B ... 81 b ,, 8 1
aB ... 81b# ; aBb oe aB ... 81b# -=- B oe a "b8 aB ... 81b# -=- aB ... 81b,
which is also analytic at B oe ,, 81 . Hence all other singular points are regular. 20. B oe ! is the only singular point. Dividing the ODE by #B# , we have :aBb oe $a#Bb and ; aBb oe B# a" Bb# It follows that
B!
lim B :aBb oe lim B
B!
$ $ oe , #B #
B!
lim B# ; aBb oe lim B#
B!
a" Bb " oe # #B #
Hence B oe ! is a regular singular point. Let C oe +! +" B +# B# +8 B8 . Substitution into the ODE results in #B# " a8 #ba8 "b+8# B8 $B" a8 "b+8" B8 a" Bb " +8 B8 oe ! .
_ _ _ 8oe! 8oe! 8oe!
That is, #" 8a8 "b+8 B $" 8 +8 B " +8 B " +8" B8 oe ! .
_ _ _ _ 8 8 8 8oe# 8oe" 8oe! 8oe"
It follows that +! a#+" +! bB " c#8a8 "b+8 $8 +8 +8 +8" dB8 oe !
_ 8oe#
Equating the coefficients to zero, we find that +! oe ! , #+" +! oe ! , and We conclude that all the +8 are equal to zero. Hence CaBb oe ! is the only solution that can be obtained. 22. Based on Prob. #", the change of variable, B oe "0 , transforms the ODE into the ________________________________________________________________________
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a#8 "ba8 "b+8 oe +8" , 8 oe # $ .
---------------------------------------------------- CHAPTER 5. ---- form 0% .# C .C #0$ C oe !. # .0 .0
Evidently, 0 oe ! is a singular point. Now :a0b oe #0 and ; a0b oe "0% . Since the value of lim 0# ; a0b does not exist, 0 oe ! , that is, B oe _ , is an irregular singular point.
0 !
24. Under the transformation B oe "0 , the ODE becomes 0 % OE" that is, ^0% 0# : a0 b oe It follows that
0!
" .# C " " .C "#0$ OE" # #0# !a! "bC oe ! , # . 0# 0 0 0 .0 .# C .C #0$ !a! "bC oe ! . # .0 .0 #0 ! a ! "b and ; a0b oe # # . 0 a0 "b " #0 oe !, "
Therefore 0 oe ! is a singular point. Note that 0#
lim0 :a0b oe lim 0
0!
0#
Hence 0 oe ! aB oe _b is a regular singular point.
0!
lim0# ; a0b oe lim 0#
0!
! a ! "b oe ! a ! "b 0 # a 0 # "b
26. Under the transformation B oe "0 , the ODE becomes 0% that is, 0% .# C .C #^0$ 0 -C oe !. . 0# .0 #a0# "b and ; a0b oe % . $ 0 0
0!
.# C " .C "#0$ #0# -C oe !, # .0 0 .0
Therefore 0 oe ! is a singular point. Note that : a0 b oe
It immediately follows that the limit lim0 :a0b does not exist. Hence 0 oe ! aB oe _b ________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ---- is an irregular singular point. 27. Under the transformation B oe "0 , the ODE becomes 0% .# C .C " #0$ C oe !. # .0 .0 0 # " and ; a0b oe & . 0 0
Therefore 0 oe ! is a singular point. Note that : a0 b oe We find that
0!
lim0 :a0b oe lim 0
0!
# oe #, 0 a "b . 0&
but
0!
The latter limit does not exist. Hence 0 oe ! aB oe _b is an irregular singular point.
lim0# ; a0b oe lim 0#
0!
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---------------------------------------------------- CHAPTER 5. ---- Section 5.5 1. Substitution of C oe B< results in the quadratic equation J a<b oe ! , where J a<b oe <a< "b %< # oe <# $< #
The roots are < oe # , " . Hence the general solution, for B ! , is C oe -" B# -# B" 3. Substitution of C oe B< results in the quadratic equation J a<b oe ! , where J a<b oe <a< "b $< % oe <# %< % C oe a-" -# 68kBkb B#
The root is < oe # , with multiplicity two . Hence the general solution, for B ! , is
5. Substitution of C oe B< results in the quadratic equation J a<b oe ! , where J a < b oe < a < "b < " oe <# #< " C oe a-" -# 68kBkb B
The root is < oe " , with multiplicity two . Hence the general solution, for B ! , is
6. Substitution of C oe aB "b< results in the quadratic equation J a<b oe ! , where J a<b oe <# (< "# The roots are < oe $ , % . Hence the general solution, for B " , is C oe -" aB "b$ -# aB "b% J a<b oe <# &< " -# kBk
7. Substitution of C oe B< results in the quadratic equation J a<b oe ! , where The roots are < oe S& ,, #*<# . Hence the general solution, for B ! , is C oe -" kBk
S& #*<# S& #*<#
8. Substitution of C oe B< results in the quadratic equation J a<b oe ! , where ________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ---- J a<b oe <# $< $
The roots are complex, with < oe S$ ,, 3$ <# . Hence the general solution, for B !, is C oe -" kBk$# -9=OE
$ #
68kBk -# kBk$# =38OE
$ #
68kBk
10. Substitution of C oe aB #b< results in the quadratic equation J a<b oe ! , where J a<b oe <# %< ) The roots are complex, with < oe # ,, #3 . Hence the general solution, for B #, is C oe -" aB #b# -9=a# 68kB #kb -# aB #b# =38a# 68kB #kb J a< b oe < # < %
11. Substitution of C oe B< results in the quadratic equation J a<b oe ! , where The roots are complex, with < oe S" ,, 3"& <# . Hence the general solution, for B !, is C oe -" kBk"# -9=OE
"& #
68kBk -# kBk"# =38OE
"& #
68kBk
12. Substitution of C oe B< results in the quadratic equation J a<b oe ! , where J a<b oe <# &< % C oe - " B - # B% 14. Substitution of C oe B< results in the quadratic equation J a<b oe ! , where J a<b oe %<# %< "( The roots are complex, with < oe "# ,, #3 . Hence the general solution, for B !, is C oe -" B"# -9=a# 68 Bb -# B"# =38a# 68 Bb The roots are < oe " , % . Hence the general solution, for B ! , is
Invoking the initial conditions, we obtain system the of equations
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---------------------------------------------------- CHAPTER 5. ----
-" oe # " -" #-# oe $ # Hence the solution of the initial value problem is CaBb oe # B"# -9=a# 68 Bb B"# =38a# 68 Bb
As B p ! , the solution decreases without bound. 15. Substitution of C oe B< results in the quadratic equation J a<b oe ! , where J a<b oe <# %< % The root is < oe # , with multiplicity two . Hence the general solution, for B ! , is C oe a-" -# 68 kBkb B# -" oe # #-" -# oe $ Hence the solution of the initial value problem is CaBb oe a# ( 68 kBkb B#
Invoking the initial conditions, we obtain the system of equations
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---------------------------------------------------- CHAPTER 5. ----
We find that CaBb p ! as B p ! . 18. Substitution of C oe B< results in the quadratic equation <# < " oe ! . The roots are <oe " ,, " %" . #
If " "% , the roots are complex, with <"# oe ^" ,, 3%" " # . Hence the general solution, for B ! , is " " C oe -" kBk"# -9=OE %" " 68kBk -# kBk"# =38OE %" " 68kBk # #
Since the trigonometric factors are bounded, CaBb p ! as B p ! . If " oe "% , the roots are equal, and Since lim kBk 68 kBk oe ! , C aBb p ! as B p ! . If " "% , the roots are real, with <"# oe ^" ,, " %" # . Hence the general solution, for B ! , is
B!
C oe -" kBk"# -# kBk"# 68 kBk
Evidently, solutions approach zero as long as "# " %" # ! That is, ! " "% . Hence all solutions approach zero, for " ! . 19. Substitution of C oe B< results in the quadratic equation <# < # oe ! . The roots are < oe " , # . Hence the general solution, for B ! , is C oe -" B" -# B# Invoking the initial conditions, we obtain the system of equations ________________________________________________________________________
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C oe -" kBk"#
"%" #
-# kBk"#
"%" #
---------------------------------------------------- CHAPTER 5. ----
-" - # oe " -" #-# oe # Hence the solution of the initial value problem is CaBb oe # # " " # # B B $ $
The solution is bounded, as B p ! , if # oe # . 20. Substitution of C oe B< results in the quadratic equation <# a! "b< &# oe ! . Formally, the roots are given by < oe " ! ,, ! # #! * #
a3b The roots <"# will be complex, if k" !k "! . For solutions to approach zero, as B p _ , we need "! " ! ! a33b The roots will be equal, if k" !k oe "! In this case, all solutions approach zero as long as " ! oe "! . a333b The roots will be real and distinct, if k" !k "! . It follows that For solutions to approach zero, we need " ! !# #! * ! That is, " ! "! . Hence all solutions approach zero, as B p _ , as long as ! " 23a+b. Given that B oe /D , C aBb oe C a/D b oe AaD b. By the chain rule, .C . .A .D " .A oe A aD b oe oe . .B .B .D .B B .D <7+B oe " ! ! # #! * . #
oe
" ! ,, S! " "!<S! " "!< #
Similarly, .# C . " .A " .A " . # A .D oe " oe # .B# .B B .D B .D B .D # .B " .A " .#A oe # # B .D B .D # a,b. Direct substitution results in ________________________________________________________________________
page 216
---------------------------------------------------- CHAPTER 5. ---- B# " that is, .# A .A a ! "b "A oe !. # .D .D " .# A " .A " .A # ! B" "A oe !, B# .D # B .D B .D
The associated characteristic equation is <# a! "b< " oe !. Since D oe 68 B , it follows that CaBb oe Aa68 Bb a- b. If the roots <"# are real and distinct, then C oe - " / <" D - # / < # D oe -" B<" -# B<#
a. b. If the roots <"# are real and equal, then
C oe - " / <" D - # D / < " D oe -" B<" -# B<" 68 B
a/b. If the roots are complex conjugates, then < oe - ,, 3. , and
C oe /-D a-" -9= .D -# =38 .D b oe B- c-" -9=a. 68 Bb -# =38a. 68 Bbd
24. Based on Prob. #$, the change of variable B oe /D transforms the ODE into . # A .A #A oe ! . .D # .D The associated characteristic equation is <# < # oe ! , with roots < oe " , # . Hence AaD b oe -" /D -# /#D , and C aBb oe -" B" -# B# 26. The change of variable B oe /D transforms the ODE into .# A .A ' &A oe /D . # .D .D The associated characteristic equation is <# ' < & oe ! , with roots < oe & , " . Hence A- aD b oe -" /D -# /&D Since the right hand side is not a solution of the homogeneous equation, we can use the method of undetermined coefficients to show that a particular solution is [ oe /D "# . Therefore the general solution is given by AaD b oe -" /D -# /&D /D "# , that is, C aBb oe -" B" -# B& B"# 27. The change of variable B oe /D transforms the given ODE into ________________________________________________________________________
page 217
---------------------------------------------------- CHAPTER 5. ----
.# A .A $ #A oe $/#D #D . .D # .D The associated characteristic equation is <# $ < # oe ! , with roots < oe " , # . Hence A- aD b oe -" /D -# /#D Using the method of undetermined coefficients, let [ oe E/#D FD/#D GD H . It follows that the general solution is given by AaD b oe -" /D -# /#D $D/#D D $# , that is, CaBb oe -" B -# B# $ B# 68 B 68 B $#
28. The change of variable B oe /D transforms the given ODE into .# A %A oe =38 D . .D #
The solution of the homogeneous equation is A- aD b oe -" -9= #D -# =38 #D The right hand side is not a solution of the homogeneous equation. We can use the method of undetermined coefficients to show that a particular solution is [ oe " =38 D . Hence $ the general solution is given by AaD b oe -" -9= #D -# =38 #D " =38 D , that is, $ CaBb oe -" -9=a# 68 Bb -# =38a# 68 Bb " =38a68 Bb $ 29. After dividing the equation by $ , the change of variable B oe /D transforms the ODE into .# A .A $ $A oe ! . # .D .D The associated characteristic equation is <# $ < $ oe ! , with complex roots < oe S$ ,, 3$ <# . Hence the general solution is AaD b oe /$D# '-" -9=S$ D#< -# =38S$ D#<",
$ #
and therefore CaBb oe B$# "-" -9=OE
68 B -# =38OE
$ #
68 B.
30. Let B !. Setting C oe a Bb< , successive differentiation gives C w oe <a Bb<" and C ww oe <a< "ba Bb<# . It follows that Since B# oe a Bb# , we find that Pca Bb< d oe <a< "bB# a Bb<# ! < Ba Bb<" " a Bb<
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page 218
---------------------------------------------------- CHAPTER 5. ---- Pca Bb< d oe <a< "ba Bb< ! <a Bb< " a Bb< oe a Bb< c<a< "b ! < " d
Given that <" and <# are roots of J a<b oe <a< "b ! < " , we have Pca Bb<3 d oe !. Therefore C" oe a Bb<" and C# oe a Bb<# are linearly independent solutions of the differential equation, PcCd oe ! , for B ! , as long as <" <# .
________________________________________________________________________
page 219
---------------------------------------------------- CHAPTER 5. ---- Section 5.6 1. T aBb oe ! when B oe ! . Since the three coefficients have no common factors, B oe ! is a singular point. Near B oe !,
B!
lim B :aBb oe lim B
B!
" " oe #B # " oe ! #
_
B!
lim B# ; aBb oe lim B#
B!
Hence B oe ! is a regular singular point. Let C oe B< ^+! +" B +# B# +8 B8 oe " +8 B<8 .
8oe!
Then C w oe " a< 8b+8 B<8"
_ 8oe!
and C oe " a< 8ba< 8 "b+8 B<8#
_ ww 8oe!
Substitution into the ODE results in # " a< 8ba< 8 "b+8 B<8" " a< 8b+8 B<8"
_ _ 8oe!
" +8 B<8" oe ! .
8oe!
8oe! _
That is, # " a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8 " +8# B<8 oe ! .
_ _ _ 8oe! 8oe! 8oe#
It follows that
+! c#<a< "b <dB< +" c#a< "b< < "dB<"
_ 8oe#
" c#a< 8ba< 8 "b+8 a< 8b+8 +8# dB<8 oe !
Assuming that +! ! , we obtain the indicial equation #<# < oe !, with roots <" oe "#
________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ---- and <# oe ! . It immediately follows that +" oe ! . Setting the remaining coefficients equal to zero, we have +8# , 8 oe # $ . +8 oe a< 8bc#a< 8b "d For < oe "# , the recurrence relation becomes +8# +8 oe , 8 oe # $ . 8a" #8b Since +" oe ! , the odd coefficients are zero. Furthermore, for 5 oe " # , +#5 oe +#5# +#5% a "b5 + ! oe oe 5 #5 a" %5 b a#5 #ba#5 ba%5 $ba%5 "b # 5x & * "$ a%5 "b
For < oe ! , the recurrence relation becomes +8# +8 oe , 8 oe # $ . 8a#8 "b Since +" oe ! , the odd coefficients are zero, and for 5 oe " # , +#5 oe +#5# +#5% a "b5 + ! oe oe 5 #5 a%5 "b # 5x $ ( ""a%5 "b a#5 #ba#5 ba%5 &ba%5 "b
_ a "b5 B#5 C" aBb oe B " " 5 # 5x & * "$ a%5 "b -- 5oe"
The two linearly independent solutions are
C# aBb oe " "
_
a "b5 B#5 #5 5x $ ( ""a%5 "b 5oe"
3. Note that B :aBb oe ! and B# ; aBb oe B , which are both analytic at B oe ! . Set C oe B< a+! +" B +# B# +8 B8 b. Substitution into the ODE results in
8oe!
" a< 8ba< 8 "b+8 B<8" " +8 B<8 oe ! ,
_ _ 8oe!
and after multiplying both sides of the equation by B , " a< 8ba< 8 "b+8 B<8 " +8" B<8 oe ! .
_ _ 8oe"
8oe!
It follows that
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---------------------------------------------------- CHAPTER 5. ---- +! c<a< "bdB " ca< 8ba< 8 "b+8 +8" dB<8 oe !
_ < 8oe"
Setting the coefficients equal to zero, the indicial equation is <a< "b oe ! . The roots are <" oe " and <# oe ! . Here <" <# oe " . The recurrence relation is +8" , 8 oe " # . +8 oe a< 8ba< 8 "b For < oe " , +8 oe Hence for 8 " , +8 oe +8" , 8 oe " # . 8a8 "b
+8" +8# a "b8 + ! oe oeoe . a8 "b8# a8 "b 8a8 "b 8xa8 "bx C" aBb oe B "
_
Therefore one solution is
a "b8 B8 . 8xa8 "bx 8oe!
5. Here B :aBb oe #$ and B# ; aBb oe B# $ , which are both analytic at B oe ! . Set C oe B< a+! +" B +# B# +8 B8 b. Substitution into the ODE results in $" a< 8ba< 8 "b+8 B
_ 8oe! <8
#" a< 8b+8 B
_ 8oe!
<8
" +8 B<8# oe !
_ 8oe!
It follows that
+! c$<a< "b #<dB< +" c$a< "b< #a< "bdB<"
_ 8oe#
" c$a< 8ba< 8 "b+8 #a< 8b+8 +8# dB<8 oe ! .
Assuming +! ! , the indicial equation is $<# < oe ! , with roots <" oe "$ , <# oe ! Setting the remaining coefficients equal to zero, we have +" oe ! , and +8# +8 oe , 8 oe # $ . a< 8bc$a< 8b "d It immediately follows that the odd coefficients are equal to zero. For < oe "$ , +8# , 8 oe # $ . +8 oe 8a" $8b So for 5 oe " # , ________________________________________________________________________
page 222
---------------------------------------------------- CHAPTER 5. ---- +#5# +#5% a "b5 + ! oe oe oe 5 . #5 a'5 "b # 5x ( "$ a'5 "b a#5 #ba#5 ba'5 &ba'5"b +8 oe So for 5 oe " # , +#5 oe +8# , 8 oe # $ . 8a$8 "b
+#5
For < oe ! ,
a "b5 + ! +#5# +#5% . oe oe 5 a#5 #ba#5 ba'5 (ba'5 "b #5 a'5 "b # 5x & "" a'5 "b C" aBb oe B " "
_ _ 5 a "b 5 B# OE -- 5x ( "$ a'5 "b # 5oe"
The two linearly independent solutions are
"$
C# aBb oe " "
5 a "b 5 B# OE 5x & ""a'5 "b # 5oe"
6. Note that B :aBb oe " and B# ; aBb oe B #, which are both analytic at B oe ! . Set C oe B< a+! +" B +# B# +8 B8 b. Substitution into the ODE results in
8oe!
" a< 8ba< 8 "b+8 B
_
<8
" a< 8b+8 B<8
_ 8oe! _
" +8 B<8" #" +8 B<8 oe !
_ 8oe! 8oe!
After adjusting the indices in the second-to-last series, we obtain +! c<a< "b < #dB " ca< 8ba< 8 "b+8 a< 8b+8 # +8 +8" dB<8 oe !
_ < 8oe"
Assuming +! ! , the indicial equation is <# # oe ! , with roots < oe ,, # . Setting the remaining coefficients equal to zero, the recurrence relation is +8" , 8 oe " # . +8 oe a < 8b # # First note that a< 8b# # oe S< 8 # <S< 8 # <. So for < oe # , +8 oe +8" , 8 oe " # . 8S8 ## <
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---------------------------------------------------- CHAPTER 5. ---- It follows that
For < oe # ,
a "b 8 + ! , 8 oe " # . +8 oe 8xS" ## <S# ## <S8 ## < +8" , 8 oe " # , 8S8 ## <
+8 oe and therefore
a "b 8 + ! , 8 oe " # . +8 oe 8xS" ## <S# ## <S8 ## < _ a "b8 B8 " " # <S# ## <S8 ## < 8 oe " 8xS" #
The two linearly independent solutions are C" aBb oe B
#
C# aBb oe B
#
_ a "b8 B8 "" 8xS" ## <S# ## <S8 ## < 8oe"
7. Here B :aBb oe " B and B# ; aBb oe B , which are both analytic at B oe ! . Set C oe B< a+! +" B +# B# +8 B8 b. Substitution into the ODE results in
8oe!
" a< 8ba< 8 "b+8 B<8" " a< 8b+8 B<8"
_ _ 8oe! _
" a< 8b+8 B<8 " +8 B<8 oe !
_ 8oe! 8oe!
After multiplying both sides by B , " a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8
_ _ 8oe!
8oe!
" a < 8 b+ 8 B
_ 8oe!
<8+1
" +8 B<8+1 oe !
_ 8oe!
After adjusting the indices in the last two series, we obtain
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---------------------------------------------------- CHAPTER 5. ---- +! c<a< "b <dB " ca< 8ba< 8 "b+8 a< 8b+8 a< 8b+8" dB<8 oe !
_ < 8oe"
Assuming +! ! , the indicial equation is <# oe ! , with roots <" oe <# oe ! . Setting the remaining coefficients equal to zero, the recurrence relation is +8" +8 oe , 8 oe " # . <8 With < oe ! , +8 oe Hence one solution is C" aBb oe " B B# B8 oe /B . "x #x 8x +8" , 8 oe " # . 8
8. Note that B :aBb oe $# and B# ; aBb oe B# "#, which are both analytic at B oe ! . Set C oe B< a+! +" B +# B# +8 B8 b. Substitution into the ODE results in #" a< 8ba< 8 "b+8 B<8 $" a< 8b+8 B<8
_ _ 8oe! 8oe! _
#" +8 B<8# " +8 B<8 oe !
_ 8oe! 8oe!
After adjusting the indices in the second-to-last series, we obtain
_
+! c#<a< "b $< "dB< +" c#a< "b< $a< "b "d
8oe#
" c#a< 8ba< 8 "b+8 $a< 8b+8 +8 # +8# dB<8 oe !
Assuming +! ! , the indicial equation is #<# < " oe ! , with roots <" oe "# and <# oe " . Setting the remaining coefficients equal to zero, the recurrence relation is +8 oe #+8# , 8 oe # $ . a< 8 "bc#a< 8b "d
Setting the remaining coefficients equal to zero, we have +" oe ! , which implies that all of the odd coefficients are zero. With < oe "# , +8 oe So for 5 oe " # , #+8# , 8 oe # $ . 8a#8 $b
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---------------------------------------------------- CHAPTER 5. ---- +#5# +#5% a "b5 + ! oe oe oe . 5 a%5 $b 5x ( "" a%5 $b a5 "b5 a%5 &ba%5 $b #+8# , 8 oe # $ . 8a#8 $b
+#5
With < oe " , +8 oe So for 5 oe " # , +#5 oe
+#5# +#5% a "b5 + ! oe oe . 5 a%5 $b a5 "b5 a%5 ""ba%5 $b 5x & * a%5 $b C" aBb oe B"# " "
_
The two linearly independent solutions are
a "b8 B#8 8x ( "" a%8 $b -- 8oe"
_
C# aBb oe B" " "
a "b8 B#8 8x & * a%8 $b -- 8oe"
9. Note that B :aBb oe B $ and B# ; aBb oe B $, which are both analytic at B oe ! . Set C oe B< a+! +" B +# B# +8 B8 b. Substitution into the ODE results in
8oe!
" a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8" $" a< 8b+8 B<8
_ _ _ 8oe!
" +8 B<8" $" +8 B<8 oe !
_ _ 8oe! 8oe!
8oe!
After adjusting the indices in the second-to-last series, we obtain +! c<a< "b $< $dB<
_ 8oe"
" ca< 8ba< 8 "b+8 a< 8 #b+8" $a< 8 "b+8 dB<8 oe !
Assuming +! ! , the indicial equation is <# %< $ oe ! , with roots <" oe $ and <# oe " . Setting the remaining coefficients equal to zero, the recurrence relation is +8 oe With < oe $ , a< 8 #b+8" , 8 oe " # . a< 8 "ba< 8 $b
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---------------------------------------------------- CHAPTER 5. ---- a8 "b+8" , 8 oe " # . 8a8 #b
+8 oe It follows that for 8 " , +8 oe
+8# # +! a8 "b+8" oe oeoe . 8a8 #b 8x a8 #b a8 "ba8 #b # B8 C" aBb oe B " " . 8x a8 #b --
_ $ 8oe"
Therefore one solution is
10. Here B :aBb oe ! and B# ; aBb oe B# "% , which are both analytic at B oe ! . Set C oe B< a+! +" B +# B# +8 B8 b. Substitution into the ODE results in
8oe!
" a< 8ba< 8 "b+8 B<8 " +8 B<8#
_ _ 8oe!
" _ " +8 B<8 oe ! % 8oe!
After adjusting the indices in the second series, we obtain " " +! "<a< "b B< +" "a< "b< B<" % %
_ " " "a< 8ba< 8 "b+8 +8 +8# B<8 oe ! % 8oe#
Assuming +! ! , the indicial equation is <# < " oe !, with roots <" oe <# oe "# % Setting the remaining coefficients equal to zero, we find that +" oe ! . The recurrence relation is +8 oe With < oe "# , +8 oe +8# , 8 oe # $ . 8# a#< #8 "b# %+8# , 8 oe # $ .
Since +" oe ! , the odd coefficients are zero. So for 5 " , +#5 oe Therefore one solution is
+#5# +#5% a "b5 + ! oe oeoe . %5 # %# a5 "b# 5 # %5 a5xb#
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---------------------------------------------------- CHAPTER 5. ---- C" aBb oe B " "
_
a "b8 B#8 ##8 a8xb#
8oe"
--.
12a+b. Dividing through by the leading coefficient, the ODE can be written as C ww For B oe " , :! oe lim aB "b:aBb oe lim
B"
B !# Cw C oe !. " B# " B# B " oe B" B " #
;! oe lim aB "b# ; aBb oe lim
B"
!# a" Bb oe ! B" B"
For B oe " , :! oe lim aB "b:aBb oe lim
B" #
B " oe B" B " #
! # a B "b ;! oe lim aB "b ; aBb oe lim oe ! B" B" a" Bb Hence both B oe " and B oe " are regular singular points. As shown in Example " , the indicial equation is given by < a < "b : ! < ; ! oe ! .
a,b. Let > oe B " , and ?a>b oe C a> "b Under this change of variable, the differential equation becomes Based on Part a+b, > oe ! is a regular singular point. Set ? oe ! +8 ><8 Substitution
_
In this case, both sets of roots are <" oe "# and <# oe ! .
^># #>? ww a> "b? w !# ? oe !
into the ODE results in
_ _
8oe!
8oe!
" a< 8ba< 8 "b+8 ><8 #" a< 8ba< 8 "b+8 ><8" " a< 8b+8 ><8 " a< 8b+8 ><8" !# " +8 ><8 oe !
_ _ _ 8oe! 8oe! 8oe! 8oe!
Upon inspection, we can also write ________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ---- " a< 8b# +8 ><8 #" a< 8bOE< 8 " +8 ><8" !# " +8 ><8 oe !
_ _ _ 8oe!
8oe!
#
8oe!
After adjusting the indices in the second series, it follows that +! "#<OE< ><" " "a< 8b# +8 #a< 8 "bOE< 8 +8" !# +8 ><8 oe ! # #
"
_
"
8oe!
Assuming that +! ! , the indicial equation is #<# < oe ! , with roots < oe ! , "# . The recurrence relation is " a< 8b# +8 #a< 8 "bOE< 8 +8" !# +8 oe ! , 8 oe ! " # . # With <" oe "# , we find that for 8 " , %!# a#8 "b# +8 oe +8" %8a#8 "b # # # #` 8 c" %! dc* %! d a#8 "b %! oe a "b +! #8 a#8 "bx
With <# oe ! , we find that for 8 " ,
! # a8 "b # +8 oe +8" 8a#8 "b !a !bc" !# dc% !# da8 "b# !# ` oe a "b 8 +! 8x $ & a#8 "b
The two linearly independent solutions of the Chebyshev equation are C" aBb oe kB "k"# " " a "b8
_ 8oe"
c" %!# dc* %!# da#8 "b# %!# ` aB "b 8 -- #8 a#8 "bx
C# aBb oe " " a "b8
_ 8oe"
!a !bc" !# dc% !# da8 "b# !# ` aB " b 8 8x $ & a#8 "b
13. Here B :aBb oe " B and B# ; aBb oe - B , which are both analytic at B oe ! . In fact, Hence the indicial equation is <a< "b < oe ! , with roots <"# oe ! . Set ________________________________________________________________________
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:! oe lim B :aBb oe " and ;! oe lim B# ; aBb oe ! .
B! B!
---------------------------------------------------- CHAPTER 5. ----
C oe +! + " B + # B# + 8 B 8 . Substitution into the ODE results in " 8a8 "b+8 B8" " 8+8 B8"
_ _ 8oe" _
8oe#
" 8+8 B8 -" +8 B8 oe !
_ 8oe! 8oe!
That is, " 8a8 "b+8" B " a8 "b+8" B8
_ _ 8 8oe! _ _
8oe"
" 8+8 B8 -" +8 B8 oe !
8oe" 8oe!
It follows that +" - +! " a8 "b# +8" a8 -b+8 `B8 oe ! .
_ 8oe"
Setting the coefficients equal to zero, we find that +" oe -+! , and +8 oe That is, for 8 # , +8 oe a8 " - b +8" , 8 oe # $ . 8#
a8 " - b a -ba" -ba8 " -b +8" oe oe +! . 8# a8xb# C" aBb oe " "
_
Therefore one solution of the Laguerre equation is
a -ba" -ba8 " -b a8xb
#
B8
8oe"
Note that if - oe 7 , a positive integer, then +8 oe ! for 8 7 " . In that case, the solution is a polynomial C" aBb oe " "
7
a -ba" -ba8 " -b a8xb
#
B8
8oe"
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---------------------------------------------------- CHAPTER 5. ---- Section 5.7 2. T aBb oe ! only for B oe ! . Furthermore, B :aBb oe # B and B# ; aBb oe # B# It follows that :! oe lim a # Bb oe # ;! oe lim ^# B# oe #
B! B!
and therefore B oe ! is a regular singular point. The indicial equation is given by <a< "b #< # oe ! ,
4. The coefficients T aBb , UaBb , and V aBb are analytic for all B - ` . Hence there are no singular points. 5. T aBb oe ! only for B oe ! . Furthermore, B :aBb oe $ =38 B and B# ; aBb oe # It B follows that =38 B oe$ B! B ;! oe lim # oe # :! oe lim $
B!
that is, <# $< # oe ! , with roots <" oe # and <# oe " .
and therefore B oe ! is a regular singular point. The indicial equation is given by that is, <# #< # oe ! , with roots <" oe " $ and <# oe " $ . <a< "b $< # oe ! ,
6. T aBb oe ! for B oe ! and B oe # . We note that :aBb oe B" aB #b" # , and ; aBb oe aB #b" # For the singularity at B oe ! , " " oe B! #aB #b % # B ;! oe lim oe! B! #aB #b :! oe lim and therefore B oe ! is a regular singular point. The indicial equation is given by " < a < "b < oe ! , % that is, <# $ < oe ! , with roots <" oe %
$ %
and <# oe ! . For the singularity at B oe # ,
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page 231
---------------------------------------------------- CHAPTER 5. ---- :! oe lim aB #b:aBb oe lim
" " oe B# B# #B % aB #b ;! oe lim aB #b# ; aBb oe lim oe! B# B# # and therefore B oe # is a regular singular point. The indicial equation is given by " < a < "b < oe ! , % that is, <# & < oe ! , with roots <" oe % 7. T aBb oe ! only for B oe ! . Furthermore, B :aBb oe follows that
B! & %
and <# oe ! .
" #
=38 B #B
and B# ; aBb oe " It
:! oe lim B:aBb oe "
B!
;! oe lim B# ; aBb oe " and therefore B oe ! is a regular singular point. The indicial equation is given by < a < "b < " oe ! ,
that is, <# " oe ! , with complex conjugate roots < oe ,, 3 . 8. Note that T aBb oe ! only for B oe " . We find that :aBb oe $aB "baB "b , and ; aBb oe $aB "b# . It follows that :! oe lim aB "b:aBb oe lim $aB "b oe ' ;! oe lim aB "b ; aBb oe lim $ oe $
B" B" # B" B"
and therefore B oe " is a regular singular point. The indicial equation is given by that is, <# (< $ oe ! , with roots <" oe S( $( <# and <# oe S( $( <# . <a< "b '< $ oe ! ,
10. T aBb oe ! for B oe # and B oe # . We note that :aBb oe #BaB #b# aB #b" , and ; aBb oe $aB #b" aB #b" For the singularity at B oe # ,
B#
lim aB #b:aBb oe lim
B# B#
#B , %
which is undefined. Therefore B oe ! is an irregular singular point. For the singularity at B oe # ,
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---------------------------------------------------- CHAPTER 5. ---- :! oe lim aB #b:aBb oe lim
B#
a B #b $aB #b ;! oe lim aB #b# ; aBb oe lim oe! B# B# B#
B# #
#B
oe
" %
and therefore B oe # is a regular singular point. The indicial equation is given by " < a < "b < oe ! , % that is, <# & < oe ! , with roots <" oe % 11. T aBb oe ! for B oe # and B oe # . We note that :aBb oe #Ba% B# b , and ; aBb oe $a% B# b For the singularity at B oe # , :! oe lim aB #b:aBb oe lim #B oe " B# B# B # $a# Bb ;! oe lim aB #b# ; aBb oe lim oe! B# B# B# and therefore B oe # is a regular singular point. The indicial equation is given by < a < "b < oe ! ,
& %
and <# oe ! .
that is, <# #< oe ! , with roots <" oe # and <# oe ! . For the singularity at B oe # , :! oe lim aB #b:aBb oe lim #B oe " B# B# # B $aB #b ;! oe lim aB #b# ; aBb oe lim oe! B# B# #B and therefore B oe # is a regular singular point. The indicial equation is given by < a < "b < oe ! ,
that is, <# #< oe ! , with roots <" oe # and <# oe ! . 12. T aBb oe ! for B oe ! and B oe $ . We note that :aBb oe #B" aB $b" , and ; aBb oe "aB $b# For the singularity at B oe ! , :! oe lim B :aBb oe lim # # oe B! B! B $ $ B# ;! oe lim B# ; aBb oe lim oe! B! B! aB $b# and therefore B oe ! is a regular singular point. The indicial equation is given by
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page 233
---------------------------------------------------- CHAPTER 5. ---- # < a < "b < oe ! , $ that is, <# & < oe ! , with roots <" oe $ :! oe lim aB $b:aBb oe lim
B$ & $
and <# oe ! . For the singularity at B oe $ ,
# # oe B$ B$ B $ # ;! oe lim aB $b ; aBb oe lim a "b oe "
B$
and therefore B oe $ is a regular singular point. The indicial equation is given by # < a < "b < " oe ! , $
13a+b. Note the :aBb oe "B and ; aBb oe "B . Furthermore, B :aBb oe " and B# ; aBb oe B It follows that :! oe lim a"b oe "
B! B!
that is, <# " < " oe ! , with roots <" oe S" $( <' and <# oe S" $( <' . $
;! oe lim a Bb oe ! and therefore B oe ! is a regular singular point. a,b The indicial equation is given by < a < "b < oe ! ,
that is, <# oe ! , with roots <" oe <# oe ! .
a- b. Let C oe +! +" B +# B# +8 B8 . Substitution into the ODE results in
8oe!
" a8 #ba8 "b+8# B8" " a8 "b+8" B8 " +8 B8 oe ! .
_ _ _ 8oe! 8oe!
After adjusting the indices in the first series, we obtain +" +! " c8a8 "b+8" a8 "b+8" +8 dB8 oe ! .
_ 8oe"
Setting the coefficients equal to zero, it follows that for 8 ! , +8 +8" oe . a 8 "b # So for 8 " ,
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---------------------------------------------------- CHAPTER 5. ----
+8 oe
+8" +8# " oe +! # oe oe # 8 8# a 8 " b a8xb#
With +! oe " , one solution is
For a second solution, set C# aBb oe C" aBb 68 B ," B ,# B# ,8 B8 . Substituting into the ODE, we obtain PcC" aBbd 68 B # C" aBb P " ,8 B8 -- oe ! .
_ w 8oe"
" " " 8 C" aBb oe " B B# B$ # B . % $' a8xb
Since PcC" aBbd oe ! , it follows that
P " ,8 B8 -- oe # C"w aBb .
_ 8oe"
More specifically, ," " c8a8 "b,8" a8 "b,8" ,8 dB8 oe
_ 8oe"
" " " % oe # B B# B $ B . ' (# "%%! Equating the coefficients, we obtain the system of equations ," %,# ," *,$ ,# "',% ,$ Solving these equations for the coefficients, ," oe #, ,# oe $%, ,$ oe """!), ,% oe #&$%&' , . Therefore a second solution is $ "" $ #& % C# aBb oe C" aBb 68 B " #B B# B B % "!) $%&' 14a+b. Here B :aBb oe #B and B# ; aBb oe ' B/B Both of these functions are analytic at B oe ! , therefore B oe ! is a regular singular point. Note that :! oe ;! oe ! a,b The indicial equation is given by oe oe oe oe # " "' "(#
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page 235
---------------------------------------------------- CHAPTER 5. ---- < a < "b oe ! , that is, <# < oe ! , with roots <" oe " and <# oe ! . a- b. In order to find the solution corresponding to <" oe " , set C oe B ! +8 B8 Upon
_
substitution into the ODE, we have
_ _
8oe!
8oe!
" a8 #ba8 "b+8" B8" #" a8 "b+8 B8" ' /B " +8 B8" oe ! .
_ 8oe! 8oe!
After adjusting the indices in the first two series, and expanding the exponential function, " 8a8 "b+8 B #" 8 +8" B8 ' +! B a'+! '+" bB#
_ _ 8 8oe" $
8oe"
a'+# '+" $+! bB a'+$ '+# $+" +! bB% oe ! . #+" #+! '+! '+# %+" '+! '+" "#+$ '+# '+# '+" $+! #!+% )+$ '+$ '+# $+" +! oe! oe! oe! oe!
Equating the coefficients, we obtain the system of equations
Setting +! oe " , solution of the system results in +" oe %, +# oe "($, +$ oe %("#, +% oe "*""#! , . Therefore one solution is C" aBb oe B %B# "( $ %( % B B . $ "#
The exponents differ by an integer. So for a second solution, set
C# aBb oe + C" aBb 68 B " -" B -# B# -8 B8 .
_ C" aBb P" " -8 B8 -- oe ! . B 8oe"
Substituting into the ODE, we obtain + PcC" aBbd 68 B #+ C"w aBb #+ C" aBb +
Since PcC" aBbd oe ! , it follows that
_ 8oe"
P" " -8 B8 -- oe #+ C"w aBb #+ C" aBb +
C" aBb . B
More specifically,
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---------------------------------------------------- CHAPTER 5. ---- " 8a8 "b-8" B8 #" 8 -8 B8 ' a' '-" bB
_ _
8oe"
a'-# '-" $bB# oe + "!+B
8oe"
'" # "*$ $ +B +B . $ "#
Equating the coefficients, we obtain the system of equations 'oe + #-# )-" ' oe "!+ '" '-$ "!-# '-" $ oe + $ "*$ "#-% "#-$ '-# $-" " oe + "# Solving these equations for the coefficients, + oe ' . In order to solve the remaining equations, set -" oe ! . Then -# oe $$, -$ oe %%*' , -% oe "&*&#% , Therefore a second solution is C# aBb oe ' C" aBb 68 B "" $$B# %%* $ "&*& % B B ' #%
15a+b. Note the :aBb oe 'BaB "b and ; aBb oe $B" aB "b" . Furthermore, B :aBb oe 'B# aB "b and B# ; aBb oe $BaB "b It follows that 'B# oe! B! B " $B ;! oe lim oe! B! B " :! oe lim and therefore B oe ! is a regular singular point. a,b The indicial equation is given by < a < "b oe ! ,
_
that is, <# < oe ! , with roots <" oe " and <# oe ! . a- b. In order to find the solution corresponding to <" oe " , set C oe B ! +8 B8 Upon substitution into the ODE, we have
8oe!
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page 237
---------------------------------------------------- CHAPTER 5. ---- " 8a8 "b+8 B8" " 8a8 "b+8 B8
_ _ 8oe" _
8oe"
'" a8 "b+8 B8# $" +8 B8" oe ! .
_ 8oe! 8oe!
After adjusting the indices, it follows that " 8a8 "b+8" B8 " 8a8 "b+8 B8
_ _ 8oe" _
8oe#
'" a8 "b+8# B8 $" +8" B8 oe ! .
_ 8oe# 8oe"
That is, #+" $+! " c 8a8 "b+8 ^8# 8 $+8" 'a8 "b+8# dB8 oe !
_ 8oe#
Setting the coefficients equal to zero, we have +" oe $+! # , and for 8 # , 8a8 "b+8 oe ^8# 8 $+8" 'a8 "b+8# .
If we assign +! oe " , then we obtain +" oe $# , +# oe *% , +$ oe &""' , . Hence one solution is $ * &" """ & C" aBb oe B B# B$ B% B . # % "' %! The exponents differ by an integer. So for a second solution, set C# aBb oe + C" aBb 68 B " -" B -# B# -8 B8 .
_ C" aBb P " " - 8 B8 -- oe ! , B 8oe"
Substituting into the ODE, we obtain #+B C"w aBb #+ C"w aBb '+B C" aBb + C" aBb +
since PcC" aBbd oe ! . It follows that
_ 8oe"
P" " -8 B8 -- oe #+ C"w aBb #+B C"w aBb + C" aBb '+B C" aBb +
C" aBb . B
Now P" " -8 B8 -- oe $ a #-# $-" bB a '-$ &-# '-" bB# 8oe" a "#-% *-$ "#-# bB$ a #!-& "&-% ")-$ bB%
_
Substituting for C" aBb , the right hand side of the ODE is
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page 238
---------------------------------------------------- CHAPTER 5. ----
( $ $$ )'( % %%" & + +B +B# +B$ +B +B . # % "' )! "! Equating the coefficients, we obtain the system of equations $oe+ ( #-# $-" oe + # $ '-$ &-# '-" oe + % $$ "#-% *-$ "#-# oe + "' We find that + oe $. In order to solve the second equation, set -" oe !. Solution of the remaining equations results in -# oe #"% , -$ oe "*% , -% oe &*('% , . Hence a second solution is C# aBb oe $ C" aBb 68 B "" #" # "* $ &*( % B B B % % '%
16a+b. After multiplying both sides of the ODE by B , we find that B :aBb oe ! and B# ; aBb oe B . Both of these functions are analytic at B oe ! , hence B oe ! is a regular singular point. a,b. Furthermore, :! oe ;! oe ! . So the indicial equation is <a< "b oe ! , with roots <" oe " and <# oe ! . a- b. In order to find the solution corresponding to <" oe " , set C oe B ! +8 B8 Upon
_
substitution into the ODE, we have " 8a8 "b+8 B8 " +8 B8" oe ! .
_ _ 8oe!
8oe!
8oe"
That is, " c8a8 "b+8 +8" d B8 oe ! .
_
8oe"
Setting the coefficients equal to zero, we find that for 8 " , +8" +8 oe . 8a8 "b It follows that
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page 239
---------------------------------------------------- CHAPTER 5. ---- a "b8 + ! +8" +8# . oe oeoe a8 "b8# a8 "b 8a8 "b a8xb# a8 "b
+8 oe
Hence one solution is " " " % " & C" aBb oe B B# B$ B B . # "# "%% #))! The exponents differ by an integer. So for a second solution, set C# aBb oe + C" aBb 68 B " -" B -# B# -8 B8 .
_ C" aBb P" " -8 B8 -- oe ! . B 8oe"
Substituting into the ODE, we obtain + PcC" aBbd 68 B #+ C"w aBb +
Since PcC" aBbd oe ! , it follows that
_ 8oe"
P" " -8 B8 -- oe #+ C"w aBb +
C" aBb . B
Now P" " -8 B8 -- oe " a#-# -" bB a'-$ -# bB# a"#-% -$ bB$ 8oe" a#!-& -% bB% a$!-' -& bB&
_
Substituting for C" aBb , the right hand side of the ODE is
$ & ( " + +B +B# +B$ +B% . # "# "%% $#! Equating the coefficients, we obtain the system of equations "oe + $ #-# -" oe + # & '-$ -# oe + "# ( "#-% -$ oe + "%% Evidently, + oe " . In order to solve the second equation, set -" oe ! . We then find that -# oe $% , -$ oe ($' , -% oe $&"(#) , . Therefore a second solution is $ ( $& % C# aBb oe C" aBb 68 B "" B# B$ B % $' "(#) ________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ---- 19a+b. After dividing by the leading coefficient, we find that :! oe lim B :aBb oe lim
B!
# a " ! " bB oe # B! "B !" B oe ! "B
Hence B oe ! is a regular singular point. The indicial equation is <a< "b # < oe ! , with roots <" oe " # and <# oe ! . a,b For B oe ", :! oe lim aB "b:aBb oe lim
B"
;! oe lim B# ; aBb oe lim
B!
B!
B"
# a" ! " bB oe "#!" B !"aB "b oe ! B" B
;! oe lim aB "b# ; aBb oe lim
B"
Hence B oe " is a regular singular point. The indicial equation is < # a# ! " b < oe ! ,
with roots <" oe # ! " and <# oe ! . a- b. Given that <" <# is not a positive integer, we can set C oe ! +8 B8 Substitution
_
into the ODE results in
_ _
8oe!
Ba" Bb" 8a8 "b+8 B8# c# a" ! " bBd" 8 +8 B8" !" " +8 B8 oe !
_ 8oe# 8oe" 8oe!
That is, " 8a8 "b+8" B8 " 8a8 "b+8 B8 # " a8 "b+8" B8
_ _ _ 8oe# 8oe! _ _
8oe"
a" ! " b" 8 +8 B8 !" " +8 B8 oe !
8oe" 8oe!
Combining the series, we obtain # +" !" +! ca# ## b+# a" ! " !" b+" dB " E8 B8 oe ! ,
_ 8oe#
in which
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---------------------------------------------------- CHAPTER 5. ---- E8 oe a8 "ba8 # b+8" c8a8 "b a" ! " b8 !" d+8 . a8 !ba8 " b +8 a8 "ba8 # b
Note that 8a8 "b a" ! " b8 !" oe a8 !ba8 " b . Setting the coefficients equal to zero, we have # +" !" +! oe ! , and +8" oe for 8 " . Hence one solution is C" aBb oe "
Since the nearest other singularity is at B oe " , the radius of convergence of C" aBb will be at least 3 oe " . a. b. Given that <" <# is not a positive integer, we can set C oe B"# ! ,8 B8 Then
_
!" !a! "b" a" "b # B B # "x # a# "b #x !a! "ba! #b" a" "ba" #b $ B # a# "ba# #b $x
Substitution into the ODE results in Ba" Bb" a8 " # ba8 # b+8 B8# "
_ 8oe! _
8oe!
c# a" ! " bBd" a8 " # b+8 B8# !" " +8 B8"# oe !
_ 8oe! 8oe!
That is,
8oe!
" a8 " # ba8 # b+8 B8# " a8 " # ba8 # b+8 B8"#
_ _ 8oe! _ _
# " a8 " # b+8 B8# a" ! " b" a8 " # b+8 B8"# !" " +8 B8"# oe !
_ 8oe! 8oe! 8oe!
After adjusting the indices,
8oe!
" a8 " # ba8 # b+8 B8# " a8 # ba8 " # b+8" B8#
_ _ 8oe" _ _
# " a8 " # b+8 B8# a" ! " b" a8 # b+8" B8# !" " +8" B8# oe !
_ 8oe! 8oe" 8oe"
Combining the series, we obtain " F8 B8# oe ! ,
_
8oe"
in which ________________________________________________________________________
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---------------------------------------------------- CHAPTER 5. ---- F8 oe 8a8 " # b,8 ca8 # ba8 # ! " b !" d,8" . a8 ! # ba8 " # b ,8" 8a8 " # b
Note that a8 # ba8 # ! " b !" oe a8 ! # ba8 " # b . Setting F8 oe !, it follows that for 8 " , ,8 oe Therefore a second solution is C# aBb oe B"# ""
a" ! # ba" " # b B a# # b"x a" ! # ba# ! # ba" " # ba# " # b # B a# # ba$ # b# x
a/b. Under the transformation B oe "0 , the ODE becomes
" " .# C " " " .C 0% OE" # oe#0$ OE" 0# "# a" ! " b !" C oe ! . .0 0 0 .0 0 0 0 ^0 $ 0 # .# C .C #0# # 0# a " ! " b0` !" C oe ! . . 0# .0 a# # b 0 a " ! " b !" and ; a0b oe $ . #0 0 0 0# a# # b 0 a " ! " b oe "!", 0! 0" !" oe !" 0"
That is,
Therefore 0 oe ! is a singular point. Note that : a0 b oe It follows that
0!
:! oe lim0 :a0b oe lim
Hence 0 oe ! aB oe _b is a regular singular point. The indicial equation is or <# a! " b< !" oe ! Evidently, the roots are < oe ! and < oe " . 21a+b. Note that :aBb oe ! " and ; a0b oe > . = B B <a< "b a" ! " b< !" oe ! ,
;! oe lim0# ; a0b oe lim
0!
0!
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page 243
---------------------------------------------------- CHAPTER 5. ---- It follows that
B!
lim B :aBb oe lim ! B"= ,
B! 0!
lim0# ; a0b oe lim " B#=
0!
Hence if = " or > # , one or both of the limits does not exist. Therefore B oe ! is an irregular singular point. a- b. Let C oe +! B< +" B<" +8 B<8 Write the ODE as B$ C ww ! B# C w " C oe ! . Substitution of the assumed solution results in " a8 <ba8 < "b+8 B8<" !" a8 <b+8 B8<" " " +8 B8< oe !
_ _ _ 8oe! 8oe!
8oe!
Adjusting the indices, we obtain " a8 " <ba8 < #b+8" B8< !" a8 " <b+8" B8< " " +8 B8< oe !
_ _ _ 8oe" 8oe!
8oe"
Combining the series, " +! " E8 B8< oe ! ,
_
in which E8 oe " +8 a8 " <ba8 < ! #b+8" . Setting the coefficients equal to zero, we have +! oe ! . But for 8 " , +8 oe a8 " <ba8 < ! #b +8" "
8oe"
Therefore, regardless of the value of <, it follows that +8 oe ! , for 8 oe " # .
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page 244
---------------------------------------------------- CHAPTER 5. ---- Section 5.8 3. Here B :aBb oe " and B# ; aBb oe #B , which are both analytic everywhere. We set C oe B< a+! +" B +# B# +8 B8 b. Substitution into the ODE results in
8oe!
" a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8 #" +8 B<8" oe !
_ _ _ 8oe! 8oe!
After adjusting the indices in the last series, we obtain +! c<a< "b <dB< " ca< 8ba< 8 "b+8 a< 8b+8 # +8" dB<8 oe !
_ 8oe"
Assuming +! ! , the indicial equation is <# oe ! , with double root < oe ! . Setting the remaining coefficients equal to zero, we have for 8 " , + 8 a< b oe It follows that + 8 a< b oe a8 < b # # +8" a<b .
ca8 <ba8 < "ba" <bd# C" aBb oe "
_
a "b 8 # 8
+! , 8 " .
Since < oe ! , one solution is given by
a "b 8 # 8 a8xb#
B8 .
8oe!
For a second linearly independent solution, we follow the discussion in Section &( . First note that
w + 8 a< b " " " oe #" . + 8 a< b 8< 8<" "< w + 8 a !b
Setting < oe ! , oe # L8 +8 a!b oe # L8
a "b 8 # 8 a8xb#
Therefore, C# aBb oe C" aBb 68 B #"
_
a "b 8 # 8 L 8 a8xb#
B8 .
8oe!
4. Here B :aBb oe % and B# ; aBb oe # B , which are both analytic everywhere. We set C oe B< a+! +" B +# B# +8 B8 b. Substitution into the ODE results in ________________________________________________________________________
page 245
---------------------------------------------------- CHAPTER 5. ---- " a< 8ba< 8 "b+8 B<8 %" a< 8b+8 B<8
_ _ 8oe! _
8oe!
" +8 B<8" #" +8 B<8 oe !
_ 8oe! 8oe!
After adjusting the indices in the second-to-last series, we obtain +! c<a< "b %< #dB< " ca< 8ba< 8 "b+8 %a< 8b+8 # +8 +8" dB<8 oe !
_ 8oe"
Assuming +! ! , the indicial equation is <# $< # oe ! , with roots <" oe " and <# oe # . Setting the remaining coefficients equal to zero, we have for 8 " , + 8 a< b oe It follows that + 8 a< b oe " +8" a<b . a8 < "ba8 < #b
a "b 8 +! , 8 " . ca8 < "ba8 <ba# <bdca8 < #ba8 <ba$ <bd C" aBb oe B" "
_
Since <" oe " , one solution is given by
a "b 8 B8 . a8bxa8 "bx 8oe!
For a second linearly independent solution, we follow the discussion in Section &( . Since <" <# oe R oe " , we find that + " a< b oe " , a< #ba< $b
with +! oe " . Hence the leading coefficient in the solution is + oe lim a< #b +" a<b oe " .
<#
Further, a < #b + 8 a < b oe
Let E8 a<b oe a< #b +8 a<b . It follows that
a8 < #b ca8 < "ba8 <ba$ <bd#
a "b 8
w E8 a<b " " " " oe #" . E8 a<b 8<# 8<" 8< $<
Setting < oe <# oe # ,
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---------------------------------------------------- CHAPTER 5. ----
w E8 a #b " " " oe #" " E8 a #b 8 8" 8# oe L8 L8"
Hence
-8 a #b oe aL8 L8" b E8 a #b a "b 8 oe aL8 L8" b . 8xa8 "bx C# aBb oe C" aBb 68 B B " "
_ #
Therefore,
a "b8 aL8 L8" b 8 B --. 8xa8 "bx 8oe"
6. Let CaBb oe @aBbB . Then C w oe B"# @ w B$# @# and C ww oe B"# @ ww B$# @ w $ B&# @% . Substitution into the ODE results in " B$# @ ww B"# @ w $ B"# @%` B"# @ w B"# @#` OEB# B"# @ oe ! . % Simplifying, we find that with general solution @aBb oe -" -9= B -# =38 B . Hence @ ww @ oe ! ,
CaBb oe -" B"# -9= B -# B"# =38 B .
8. The absolute value of the ratio of consecutive terms is
+#7# B#7# kBk#7# ##7 a7 "bx 7x kBk# oe . oe +#7 B#7 %a7 #ba7 "b kBk#7 ##7# a7 #bxa7 "bx lim 7_ +#7# B#7# kBk# oe lim oe !. 7_ +#7 B#7 %a7 #ba7 "b
Applying the ratio test,
Hence the series for N" aBb converges absolutely for all values of B . Furthermore, since the series for N! aBb also converges absolutely for all B, term-by-term differentiation results in
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---------------------------------------------------- CHAPTER 5. ---- N! aBb oe "
_ w _
a "b7" B#7" oe " #7" # a7 "bx 7x 7oe! oe
a "b7 B#7" ##7" 7xa7 "bx 7oe"
Therefore, N!w aBb oe N" aBb .
B _ a "b7 B#7 " #7 . # 7 oe ! # a7 "bx 7x
9a+b. Note that B :aBb oe " and B# ; aBb oe B# / # , which are both analytic at B oe ! . Thus B oe ! is a regular singular point. Furthermore, :! oe " and ;! oe / # . Hence the indicial equation is <# / # oe ! , with roots <" oe / and <# oe / . a,b. Set C oe B< a+! +" B +# B# +8 B8 b. Substitution into the ODE results in " a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8
_ _ 8oe! _
8oe!
" +8 B<8# / # " +8 B<8 oe !
_ 8oe! 8oe!
After adjusting the indices in the second-to-last series, we obtain
_
+! <a< "b < / # `B< +" a< "b< a< "b / # `
8oe#
" a< 8ba< 8 "b+8 a< 8b+8 / # +8 +8# `B<8 oe !
Setting the coefficients equal to zero, we find that +" oe ! , and +8 oe a < 8b # / # " +8# ,
for 8 # . It follows that +$ oe +& oe oe +#7" oe oe ! . Furthermore, with <oe/, +8 oe So for 7 oe " # , +#7 oe " +#7# #7a#7 #/ b a "b 7 oe #7 +! . # 7xa" / ba# / ba7 " / ba7 / b " +8# . 8a8 #/ b
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page 248
---------------------------------------------------- CHAPTER 5. ---- Hence one solution is
/
C" aBb oe B " "
_
a "b 7 B #7 S < --. 7xa" / ba# / ba7 " / ba7 / b # 7oe"
a- b. Assuming that <" <# oe #/ is not an integer, simply setting < oe / in the above results in a second linearly independent solution C# aBb oe B/ " "
_
a "b 7 B #7 S < -- 7xa" / ba# / ba7 " / ba7 / b # 7oe"
a. b. The absolute value of the ratio of consecutive terms in C" aBb is
+#7# B#7# kBk#7# ##7 7xa" @ba7 / b oe +#7 B#7 kBk#7 ##7# a7 "bxa" @ba7 " / b kBk# oe . %a7 "ba7 " / b lim 7_ +#7# B#7# kBk# oe lim oe !. 7_ +#7 B#7 %a7 "ba7 " / b
Applying the ratio test,
Hence the series for C" aBb converges absolutely for all values of B . The same can be shown for C# aBb . Note also, that if / is a positive integer, then the coefficients in the series for C# aBb are undefined. 10a+b. It suffices to calculate PcN! aBb 68 Bd. Indeed, cN! aBb 68 Bd w oe N!w aBb 68 B N! aBb B
and
N!w aBb N! aBb cN! aBb 68 Bd oe N! aBb 68 B # . B B#
ww ww
Hence
Since B# N!ww aBb B N!w aBb B# N! aBb oe ! ,
PcN! aBb 68 Bd oe B# N!ww aBb 68 B #B N!w aBb N! aBb B N!w aBb 68 B N! aBb B# N! aBb 68 B PcN! aBb 68 Bd oe #B N!w aBb .
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page 249
---------------------------------------------------- CHAPTER 5. ---- a,b. Given that PcC# aBbd oe ! , after adjusting the indices in Part a+b, we have ," B ## ,# B# " ^8# ,8 ,8# B8 oe #B N!w aBb .
_ 8oe$
Using the series representation of N!w aBb in Problem ) ,
_ # # 8oe$
," B # ,# B " ^8# ,8 ,8# B8 oe # "
_
a "b8 a#8bB#8 ##8 a8xb#
.
8oe"
a- b. Equating the coefficients on both sides of the equation, we find that Also, with 8 oe " , ## ,# oe "a"xb# , that is, ,# oe "## a"xb# `. Furthermore, for 7 #, a#7b# ,#7 ,#7# oe # a "b7 a#7b ##7 a7xb# . ," oe ,$ oe oe ,#7" oe oe !
More explicitly, " " " #OE % # " " " ,' oe # # # OE" # % ' # $ ,% oe ## It can be shown, in general, that ,#7 oe a "b7" 11. Bessel's equation of order one is ##7 a7xb# L7 .
B# C ww B C w ^B# "C oe ! .
Based on Problem *, the roots of the indicial equation are <" oe " and <# oe " . Set C oe B< a+! +" B +# B# +8 B8 b. Substitution into the ODE results in
8oe!
" a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8
_ _ 8oe! _
" +8 B<8# " +8 B<8 oe !
_ 8oe! 8oe!
After adjusting the indices in the second-to-last series, we obtain
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page 250
---------------------------------------------------- CHAPTER 5. ---- +! c<a< "b < "dB< +" ca< "b< a< "b "d
_ 8oe#
" ca< 8ba< 8 "b+8 a< 8b+8 +8 +8# dB<8 oe ! + 8 a< b oe +8# a<b a < 8b # " " oe +8# a<b , a8 < "ba8 < "b " a "b 7
Setting the coefficients equal to zero, we find that +" oe ! , and
for 8 # . It follows that +$ oe +& oe oe +#7" oe oe !. Solving the recurrence relation, +#7 a<b oe With < oe <" oe ", a#7 < "ba#7 < "b# a< $b# a< "b a "b 7 +#7 a"b oe #7 +! # a7 "bx 7x +! .
For a second linearly independent solution, we follow the discussion in Section &( . Since <" <# oe R oe # , we find that + # a< b oe " , a< $ba< "b " . #
with +! oe " . Hence the leading coefficient in the solution is + oe lim a< "b +# a<b oe
<"
Further, a< "b +#7 a<b oe
Let E8 a<b oe a< "b +8 a<b . It follows that
a#7 < "b ca#7 < "ba$ <bd#
a "b 7
w E#7 a<b " " " oe #" . E#7 a<b #7 < " #7 < " $<
Setting < oe <# oe " , we calculate
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page 251
---------------------------------------------------- CHAPTER 5. ---- " -#7 a "b oe aL7 L7" bE#7 a "b # " a "b 7 oe aL7 L7" b # #7 ca#7 #b#d# " a "b 7 oe aL7 L7" b #7" # # 7xa7 "bx -#7" a "b oe " . E#7" a<b oe !. .< <oe<#
Note that +#7" a<b oe ! implies that E#7" a<b oe ! , so
Therefore,
Based on the definition of N" aBb,
_ _ " " B #7 a "b 7 a "b7 aL7 L7" b B #7 C# aBb oe B " S < --68 B " " S < -- # 7 oe ! a7 "bx 7x # 7xa7 "bx # B 7oe"
_ " a "b7 aL7 L7" b B #7 C# aBb oe N" aBb 68 B " " S < -- B 7xa7 "bx # 7oe"
12. Consider a solution of the form
CaBb oe B 0 ^!B" . .0 !" B" 0 a0 b . 0 B # B
Then Cw oe in which 0 oe !B" . Hence C ww oe and B# C ww oe !# " # B#" B Substitution into the ODE results in !# " # B#"
. # 0 !# " # B#" .0 !" # B" 0 a0b , . 0# . 0 BB BB %BB .# 0 .0 " !" # B" B B 0 a0b . 0# .0 %
.# 0 .0 " " !" # B" 0 a0b OE!# " # B#" / # " # 0 a0b oe ! # .0 .0 % %
Simplifying, and setting 0 oe !B" , we find that ________________________________________________________________________
page 252
---------------------------------------------------- CHAPTER 5. ---- .# 0 .0 0 ^0# / # 0 a0b oe ! , ab . 0# .0
0#
which is a Bessel equation of order / . Therefore, the general solution of the given ODE is in which 0" a0b and 0# a0b are the linearly independent solutions of ab. CaBb oe B -" 0" ^!B" -# 0# ^!B" `,
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