Regular Singular Points
We now investigate the solution of the homogeneous second-order linear equation
d2y
dy
P(x) 2 + Q(x) + R(x)y = 0 (1)
dx
dx
near a singular point. Recall that if the functions P(x), Q(x), and R(x) are polynomials
having no common factors, then the singular points of the equation (1) are simply those
points where P(x) = 0.
Example:
1) The point x = 0 is the only singular point of the Bessel equation of order n,
x2y’’ + xy’ +(x2 – n2)y = 0
2) The points x =1 and x = -1are the two singular points of the Legendre equation
(1 –x2)y’’ -2xy’ + n(n+1)y = 0
It turns out that some important features of the solutions of such equations are largely
determined by the behavior near their singular points.
Remark: We will restrict our attention to the case in which x = 0 is a singular point of
the equation. A differential equation having x = a as a singular point is easily transform
by the substitution t = x – c into one having the corresponding singular point at 0.
Types of Singular Points
A differential equation having a singular point at 0, in general, will not have power series
solution of the form y(x) =
!
" cn xn , so the straightforward method of the previous
n=0
section fails in this case.
To investigate the form that the solution of such equation might take, we assume that
equation (1) has analytic coefficient functions and we rewrite it in standard form,
d 2 y Q(x) dy R(x)
d2y
dy
!!!!!!!!!!!!!!!!!!!!!!!!!!!! 2 +
+
y = 0!or! 2 + p(x) + q(x)y = 0
P(x) dx P(x)
dx
dx
dx
Q(x)
R(x)
where!p(x) =
!and!q(x) =
P(x)
P(x)
Recall that x = 0 is an ordinary point (as opposed to a singular point) of equation if both
p(x) and q(x) are analytic functions at x = 0, this means that they both have a power
series expansion around 0.
It can be proved that each of the functions p(x) and q(x) either is analytic or approach ±∞
as x approaches 0. Consequently, x = 0 is a singular point of the equation provided that
either p(x) or q(x) (or both) approaches ±∞ as x approaches 0.
Definition: Given the normalized equation
d 2 y Q(x) dy R(x)
d2y
dy
+
+
y
=
0!or!
+ p(x) + q(x)y = 0
2
2
P(x) dx P(x)
dx
dx
dx
Q(x)
R(x)
where!p(x) =
!and!q(x) =
P(x)
P(x)
let x0 be a singular point.
If the functions
A(x) = (x – x0) p(x), and B(x) = (x – x0)2 q(x)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!
are both analytic at x0, then x0 is called a regular singular point of the equation.
If either (or both) of the functions A(x) or B(x) is not analytic at x0, then x0 is called an
irregular singular point.
Example:
1) Given (1 + x)y’’ + 2xy’ – 3y = 0,
the normalized equation is
2x
3
y ''+
y '!
y=0
1+ x
1+ x
Then, x = -1 is a singular point of the equation since p(x) and q(x) are not analytic at -1.
Since
A(x) = (x +1) p(x) = 2x, and B(x) = (x +1)2 q(x) = 3(x + 1)
are both analytic functions at x = -1, then x = -1 is a regular singular point.
2) Given x2(1- x)2y’’ + x(4 – x2)y’ + (2 + 3x)y = 0,
the normalized equation is
x(4 ! x 2 )
2 + 3x
y ''+ 2
y '+ 2
=0
2
x (1 ! x)
x (1 ! x)2
Then, x = 0 and x = 1 are singular points since p(x) and q(x) are not analytic functions at
0 and 1.
4 ! x2
2 + 3x
2
Since A(x) = xp(x) =
and
B(x)
=
x
q(x)=
(1 ! x )2
(1 ! x )2
are both analytic functions at x = 0, then x = 0 is a regular singular point of the equation.
2 + 3x
4 ! x2
But, since C(x) = (1 –x) p(x) = 2
and D(x) = (1 – x)2 q(x) =
x2
x (1 ! x )
Are not both analytic at x =1, then x =1 is an irregular singular point.
3) Given x3y’’ + x2y’ + y = 0,
the normalized equation is:
1
1
y '+ 3 y = 0
x
x
Then, x = 0 is a singular point, since p(x) and q(x) are not analytic at x = 0.
But, A(x) = xp(x) = 1 and B(x) = x2 q(x) = 1/x are not both analytic at 0, then x = 0 is a
irregular singular point of the equation.
y ''+
The Method of Frobenius
We need a new approach to solve a second-order linear differential equation near a
regular singular point x = 0. The simplest of such equation is the constant-coefficient
equidimensional equation
x2y’’ + p0xy’ + q0y = 0
where A(x) = p0 and B(x) = q0 are constants.
We can see that the simple power function y(x) = xr is a solution of the equation if and
only if r is a root of the quadratic equation
r(r – 1) + p0r + q0 = 0
In the general case, in which p(x) and q(x) are functions rather than constants, it is
reasonable to conjecture that the differential equation might have a solution of the form
!
y(x) = x r " c n x n =
n=0
!
" cn xn+r = c0 xr + c1xr+1 + ..... (3)
n=0
It is a fruitful conjecture since the method of Frobenius states that every equation of the
form in (1) having x = 0 as regular singular point, has at least one such solution.
An infinite series in the form (3) is called a Frobenius series. Note that in general the
Frobenius series is not a power series, since r could be negative integer or rational
number.
Theorem:
Let x0 be a regular singular point of the second-order linear differential equation (1).
Equation (1) has at least one non-trivial solution of the form
y(x) = x ! x 0
r
#
$ cn (x ! x0 )n !!with!c0 " 0
n=0
where r is a definite (real or complex) constant which may be determined, and this
solution is valid in some deleted interval 0 < |x – x0| < ρ, ρ > 0 (the center x0 is not part of
the interval)
Method
1) Let x0 be a regular singular point of the second-order linear differential equation (1).
We seek a non-trivial solution of the form
y(x) = ( x ! x 0 )
r
#
$ cn (x ! x0 )n !!with!c0 " 0
n=0
valid on some interval 0 < x – x0 < ρ.
To obtain the solution on -ρ < x – x0 < 0, replace x – x0 by –(x – x0).
Write the solution in the form
y(x) =
"
# cn (x ! x0 )n+r !!
n=0
2) Take the first and second derivative
y '(x) =
y ''(x) =
"
# (n + r)cn (x ! x0 )n+r!1 !!
n=0
"
# (n + r)(n + r ! 1)cn (x ! x0 )n+r!2 !!
n=0
3) Substitute into equation (1) and we get
K0(x –x0)r+k + K1(x – x0)r+k+1 + K2(x – x0)r+k+2 + …….. = 0 (4)
where k is a certain integer, and Ki is a function of r and certain coefficients cn.
4) Expression (4) is valid for all x in the deleted interval 0 < x – x0 < ρ if
K0 = K1 = ……. = Kn = ……. = 0
5) Since K0 contain c0 and c0 ≠ 0, then K0 = 0 produces a quadratic equation in r. this
equation is called the indicial equation. The two roots of the indicial equation are called
the exponents of the equation (1).
Let’s called the exponent of the equation r1 and r2 where Re(r1) > Re(r2).
If r1 and r2 are real numbers then r1 = Re(r1) and r2 = Re(r2), so r1 > r2
6) Equate to zero the remaining coefficients K1, K2, …. . This will lead to a set of
conditions, involving the constant r and some coefficients cn.
7) Substitute the root r1 for r into the conditions obtained in step 6 and then choose the cn
that satisfy these conditions.
If the cn are chosen, the resulting series with r = r1 is a solution of the equation (1).
8) If r1 ≠ r2, we may repeat step 7 using the root r2 instead r1. In this way a second
solution is obtained.
Remark: In some cases the second solution obtained in step 8 is not linearly independent
of the solution obtained in step 7. Also if r1 = r2 (real numbers), then both solutions are
equal. So we will see how we proceed in these cases later.
Example:
1) Use the method of Frobenius to find the solution of the equation
2xy’’ + (x +1)y’ + 3y = 0
in some interval 0 < x < ρ.
Note that x = 0 is a regular singular point of the equation, then we can assume there is a
solution in the form
y(x) = x
in some interval 0 < x < ρ.
Then
r
"
# cnx
n=0
n
=
"
# cn xn+r ,!c0 ! 0
n=0
y '(x) =
y ''(x) =
"
# (n + r)cn xn+r!1
n=0
"
# (n + r)(n + r ! 1)cn xn+r!2
n=0
Substitute into the equation
"
"
n=0
"
n=0
"
"
n=0
n=0
# 2(n + r)(n + r ! 1)cn xn+r!1 + # ( n + r )cn xn+r + # (n + r)cn xn+r!1 + # 3cn xn+r =
"
# $% 2 ( n + r )( n + r ! 1) + ( n + r )&'cn xn+r!1 + # $%( n + r ) + 3&'cn xn+r =
n=0
"
n=0
"
n=0
n=1
# $% 2 ( n + r )( n + r ! 1) + ( n + r )&'cn xn+r!1 + # [ n ! 1 + r + 3]cn!1xn+r!1 = 0
we shift the index of summation in the second series by -1, replacing n with n -1 and
using the initial value n = 1.
Since we want to express everything in only one summation, we have to start the
summation at n = 1 in every series,
&
&
' "# 2 ( n + r )( n + r ! 1) + ( n + r )$%cn xn+r!1 + ' [ n ! 1 + r + 3]cn!1xn+r!1 =
n=0
n=1
&
r(2r ! 1)c 0 x r!1 + ' "#( n + r ) ( 2n + 2r ! 1) c n + ( n + r + 2 ) c n!1 $%x n+r!1 = 0
n=0
Since c0 ≠ 0, r(2r – 1) = 0 is the indicial equation
Then r1 = ½ and r2 = 0 are the exponents of the equation.
They are real numbers and not equal.
Now, for n ≥ 1,
cn = !
r+n+2
c
( n + r ) ( 2n + 2r ! 1) n!1
Letting r1 = ½ ,
cn = !
5
2
+n
( n + 12 ) 2n
c n!1
n =1, c1 = -7/2 c0
n = 2, c2 = 21/40 c0
n = 3, c3 = - 11/80 c0
1
(
y(x) = x 2 c 0 ! 67 c 0 x +
21
c x2
40 0
!
11
c x3
80 0
)
(
1
3
+ ...... = c 0 x 2 ! 67 x 2 +
Letting r2 = 0,
cn = !
n+2
c n!1
n ( 2n ! 1)
21 52
x
40
!
11 27
x
80
)
+ .......
n =1, c1 = -3c0
n = 2, c2 = 2c0
n = 3, c3 = -2/3 c0
y(x) = x 0 c 0 ! 3c 0 x + 2c 0 x 2 ! 23 c 0 x 3 + ...... = c 0 1 ! 3x + 2x 2 ! 23 x 3 + .......
(
)
(
The two solutions are linearly independent, then the general solution is:
(
1
3
y(x) = c1 x 2 ! 67 x 2 +
21 52
x
40
!
11 27
x
80
)
(
)
)
+ ....... + c 2 1 ! 3x + 2x 2 ! 23 x 3 + ......
Theorem:
Let x0 be a regular singular point of the second-order linear differential equation (1).
Let r1 and r2 (Re(r1) ≥ Re(r2)) be the roots of the indicial equation associated with x0.
Conclusion 1:
If r1 – r2 ≠ 0 or r1 - r2 ≠ N (positive integer), then the differential equation (1) has two
non-trivial linearly independent solutions of the form
!!!!!!!!!!!!!!!!!!!!!!!!!!!y1 (x) = x ! x 0
r1
#
$ cn (x ! x0 )n !!with!c0 " 0
n=0
and
!!!!!!!!!!!!!!!!!!!!!!!!!!y 2 (x) = x ! x 0
r2
#
$ dn (x ! x0 )n !!with!d0 " 0
n=0
Conclusion 2:
If r1 - r2 = N (positive integer), then the differential equation (1) has two non-trivial
linearly independent solutions of the form
!!!!!!!!!!!!!!!!!!!!!!!!!!!y1 (x) = x ! x 0
r1
#
$ cn (x ! x0 )n !!with!c0 " 0
n=0
and
!!!!!!!!!!!!!!!!!!!!!!!!!!y 2 (x) = x ! x 0
r2
#
$ dn (x ! x0 )n + cy1 (x)ln x ! x0 !!with!d0 " 0
n=0
where c is a constant that may or may not be zero.
Conclusion 3:
If r1 - r2 = 0, then the differential equation (1) has two non-trivial linearly independent
solutions of the form
!!!!!!!!!!!!!!!!!!!!!!!!!!!y1 (x) = x ! x 0
r1
#
$ cn (x ! x0 )n !!with!c0 " 0
n=0
and
!!!!!!!!!!!!!!!!!!!!!!!!!!y 2 (x) = x ! x 0
r1 +1
#
$ dn (x ! x0 )n + y1 (x)ln x ! x0 !!with!d0 " 0
n=0
All these solutions are valid in some deleted interval 0 < |x – x0| < ρ, ρ > 0.
Example:
1) Solving the equation using the method of Frobenius
xy’’ + (x - 1)y’ – y = 0
Note that x = 0 is a regular singular point of the equation, then we can assume there is a
solution in the form
"
y(x) = x r # c n x n =
n=0
"
# cn xn+r ,!c0 ! 0
n=0
in some interval 0 < x < ρ.
Then
y '(x) =
y ''(x) =
"
# (n + r)cn xn+r!1
n=0
"
# (n + r)(n + r ! 1)cn xn+r!2
n=0
Substitute into the equation
"
"
n=0
"
n=0
"
"
n=0
n=0
# (n + r)(n + r ! 1)cn xn+r!1 + # ( n + r )cn xn+r ! # (n + r)cn xn+r!1 ! # cn xn+r =
"
# $%( n + r )( n + r ! 1) ! ( n + r )&'cn xn+r!1 + # $%( n + r ) ! 1&'cn xn+r =
n=0
"
"
n=0
n=0
"
n=0
"
n=0
n=1
# $%( n + r )( n + r ! 1 ! 1)&'cn xn+r!1 + # [ n + r ! 1]cn xn+r =
# $%( n + r )( n + r ! 1 ! 1)&'cn xn+r!1 + # [ n ! 1 + r ! 1]cn!1xn+r!1 = 0
we shift the index of summation in the second series by -1, replacing n with n -1 and
using the initial value n = 1.
Since we want to express everything in only one summation, we have to start the
summation at n = 1 in every series,
&
' "#( n + r )( n + r ! 2 )$%cn x
n+r!1
n=0
&
+ ' [ n + r ! 2 ]c n!1x n+r!1 =
n=1
&
r(r ! 2)c 0 x r!1 + ' "#( n + r ) ( n + r ! 2 ) c n + ( n + r ! 2 ) c n!1 $%x n+r!1 = 0
n=0
Since c0 ≠ 0, r(r – 2) = 0 is the indicial equation
Then r1 = 2 and r2 = 0 are the exponents of the equation.
Notice 2 – 0 = 0 is an integer
Now, for n ≥ 1,
cn = !
r+n!2
c
c n!1 = ! n!1
n+r
( n + r )( n + r ! 2 )
Letting r1 = 2,
cn = !
c n!1
n+2
c0
3
c
c
n = 2,!c 2 = ! 1 = 0
4 3" 4
c
c0
n = 3,!c 3 = ! 2 = !
5
3" 4 " 5
n = 1,!c1 = !
(
y(x) = x 2 c 0 ! 13 c 0 x +
d
(
1 2
x
2!
!
1 3
x
3!
+
1 4
x
4!
!
1
3"4
c0x2 !
1 5
x
5!
1
c x3
3"4"5 0
) (
)
(
+ ...... = c 0 x 2 ! 13 x 3 +
+ ....... = d 1 ! x + e !x
)
1
3"4
x4 !
1
x5
3"4"5
taking d = 2c0.
When we consider the smallest root r2 = 0, conclusion 2 says there is a linearly
independent solution of the
form !!!!!!!!!!!!!!!!!!!!!!!!!!y 2 (x) =
#
$ dn (x ! x0 )n+r
2
+ cy1 (x)ln x!!with!d 0 " 0
n=0
where c may or may not be zero.
We tentatively assume we have a solution of the form
!
" dn xn+r
2
n=0
Letting r2 = 0,
cn = !
n =1, c1 = -c0
n = 2, c2 = -½ c1 = ½ c0
n = 3, c3 = -1/3 c2 = - 1/3!c0
1
y(x) = x 0 c 0 ! c 0 x + 2!
c0x2 !
(
1
c x3
3! 0
c n!1
, n≥1
n
)
(
+ ...... = c 0 1 ! x +
1 2
x
2!
!
1 3
x
3!
(
)
+ ....... = c 0 e !x
The two solutions are linearly independent, then the general solution is:
y(x) = c1 x ! 1 + e !x + c 2 e !x = d1 (x ! 1) + d 2 e !x
)
2) Solving the equation using the method of Frobenius
3xy’’ + 2y’ + x2y = 0
Note that x = 0 is a regular singular point of the equation, then we can assume there is a
solution in the form
y(x) = x
in some interval 0 < x < ρ.
Then
r
"
# cnx
n=0
n
=
"
# cn xn+r ,!c0 ! 0
n=0
)
+ ....... =
"
# (n + r)cn xn+r!1
y '(x) =
y ''(x) =
n=0
"
# (n + r)(n + r ! 1)cn xn+r!2
n=0
Substitute into the equation
"
# 3(n + r)(n + r ! 1)cn x
n+r!1
n=0
"
"
+ # 2(n + r)c n x
n+r!1
n=0
"
# $%( n + r )( n + r ! 1) ! 2 ( n + r )&'cn xn+r!1 + #
n=0
"
"
+ # c n x n+r+2 =
n=0
c n x n+r+2 =
n=0
"
# $%( n + r )( 3n + 3r ! 1)&'cn xn+r!1 + #
n=0
"
"
n=0
n= 3
c n x n+r+2 =
n=0
# $%( n + r )( n + r ! 1)&'cn xn+r!1 + #
c n! 3x n+r!1 = 0
we shift the index of summation in the second series by -3, replacing n with n -3 and
using the initial value n = 3.
Since we want to express everything in only one summation, we have to start the
summation at n = 3 in every series,
&
&
n=0
n= 3
' "#( n + r )( 3n + 3r ! 1)$%cn xn+r!1 + '
r(3r ! 1)c 0 x
r!1
c n! 3x n+r!1 =
+ (r + 1)(3r + 2)c1x + (r + 2)(3r + 5)c 2 x
r
r+1
&
+ ' "#( n + r ) ( 3n + 3r ! 1) c n + c n! 3 $%x n+r!1 = 0
n= 3
Since c0 ≠ 0, r(3r – 1) = 0 is the indicial equation
Then r1 = 1/3 and r2 = 0 are the exponents of the equation.
Notice that neither r1 = 1/3 nor r2 = 0 makes the second and the third term vanish,
therefore we have to assume that c1 = 0 and c2 = 0.
Now, for n ≥ 3,
cn = !
c n! 3
,!
( n + r ) ( 3n + 3r ! 1)
then c1 = c4 = c7 = ….. = 0
and c2 = c5 = c8 = …… = 0
Letting r1 = 1/3,
cn = !
n = 3, c3 = -7/30 c0
(
c n! 3
n + 13 3n
)
n = 6, c6 = 1/3420 c0
n = 9, c9 = - 1/861840 c0
1
(
y(x) = x 3 c 0 !
1
c x3
30 0
+
1
c x6
3420 0
!
1
c x9
861840 0
(
)
1
+ ...... = c 0 x 3 !
1
1
x + 3420
x2
30
)
!
1
x3
861840
+ .......
!
1
x9
572832
+ .......
Letting r2 = 0,
cn = !
n = 3, c3 = -1/24 c0
n = 6, c6 = 1/2448 c0
n = 9, c9 = -1/572832 c0
1
y(x) = x 0 c 0 ! 24
c0x 3 +
(
1
c x6
2448 0
!
c n! 3
,n≥3
n ( 3n ! 1)
1
c x9
572832 0
)
(
+ ...... = c 0 1 !
1
24
x3 +
1
x6
2448
The two solutions are linearly independent, then the general solution is:
(
1
y(x) = c1 x 3 !
1
21
x + 3420
x2
30
!
11
x3
861840
)
(
+ ....... + c 2 1 !
1
24
x3 +
1
x6
2448
!
1
x9
572832
)
+ .......
4) Solving the equation using the method of Frobenius
x2y’’ + (x2 – 3x)y’ + 3y = 0
Note that x = 0 is a regular singular point of the equation, then we can assume there is a
solution in the form
"
y(x) = x r # c n x n =
in some interval 0 < x < ρ.
Then
y '(x) =
y ''(x) =
n=0
"
# cn xn+r ,!c0 ! 0
n=0
"
# (n + r)cn xn+r!1
n=0
"
# (n + r)(n + r ! 1)cn xn+r!2
n=0
Substitute into the equation
"
"
n=0
"
n=0
"
"
n=0
n=0
# (n + r)(n + r ! 1)cn xn+r + # ( n + r )cn xn+r+1 ! # 3(n + r)cn xn+r + # 3cn xn+r =
"
# $%( n + r )( n + r ! 1 ! 3) + 3&'cn xn+r + # ( n + r )cn xn+r+1 =
n=0
"
"
n=0
n=0
"
n=0
"
n=0
n=1
# $%( n + r )( n + r ! 4 ) + 3&'cn xn+r + # (n + r)cn xn+r+1 =
# $%( n + r )( n + r ! 4 ) + 3&'cn xn+r + # (n + r ! 1)cn!1xn+r = 0
)
we shift the index of summation in the second series by -1, replacing n with n -1 and
using the initial value n = 1.
Since we want to express everything in only one summation, we have to start the
summation at n = 1 in every series,
&
&
' "#( n + r )( n + r ! 4 ) + 3$%cn xn+r + ' (n + r ! 1)cn!1xn+r =
n=0
n=1
&
[ r(r ! 4) + 3] c0 x r + ' "# "#( n + r ) ( n + r ! 4 ) + 3$% cn + ( n + r ! 1) cn!1 $%x n+r = 0
n=1
Since c0 ≠ 0, r(r – 4) +3 = r2 – 4r + 3 = 0 is the indicial equation
Then r1 = 3 and r2 = 1 are the exponents of the equation.
Notice 3 – 1 = 2 is an integer
Now, for n ≥ 1,
cn = !
n + r !1
c
( n + r ) ( n + r ! 4 ) + 3 n!1
Letting r1 = 3,
n + 3!1
n+2
n+2
c
c n!1 = ! 2
c n!1 = !
c n!1 = ! n!1
(n + 3)(n + 3 ! 4) + 3
n(n + 2)
n
n + 2n ! 3 + 3
n = 1,!c1 = !c 0
c
c
n = 2,!c 2 = ! 1 = 0
2 2!
c2
c
n = 3,!c 3 = ! = ! 0
3
3!
cn = !
(
y1 (x) = x 3 c 0 ! c 0 x +
1
c x2
2! 0
!
1
c x3
3! 0
)
(
+ ...... = c 0 x 3 1 ! x +
1 2
x
2!
!
1 3
x
3!
+
1 4
x
4!
)
+ ....... = c 0 x 3e !x
When we consider the smallest root r2 = 1,
n(n-2) cn + ncn-1 = 0, n ≥ 1
c
c n = ! n!1 ,!n " 1,!n # 2
n!2
n =1, c1 = -c0
n = 2, the expression is undefined, or 0c2 + 2c1 = 0 that implies c1 = 0 then c0 = 0 that is a
contradiction.
Therefore, there is no solution of the form
!
x r2 " c n x n
n=0
conclusion 2 says there is a linearly independent solution of the form
!!!!!!!!!!!!!!!!!!!!!!!!!!y 2 (x) =
#
$ dn (x ! x0 )n+r
2
+ cy1 (x)ln x!!with!d 0 " 0
n=0
where c may or may not be zero.
We will use the method of reduction of order to get such solution.
Then the second solution has the form y(x) = x3e-x v(x).
y’(x) = x3e-x v’ + (3x2e-x –x3e-x)v
y’’(x) = x3e-x v’’ + 2(3x2e-x –x3e-x)v’ + (x3e-x – 6x2e-x + 6xe-x)v.
After substituting into the equation, we get
d2v
dv
x 2 + (3 ! x) = 0
dx
dx
Calling w = dv/dx
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!x
dw
+ (3 ! x)w = 0
dx
separating!var iables
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
dw " 3 %
= $ ! + 1' dx
w # x &
whose solution is
ln w = ln x-3 + x
or ln(wx3) = x, or wx3 = ex, or w = x-3ex
then,
(
v(x) = " x !3e x dx = " x !3 1 + x + 12 x 2 + 16 x 3 +
# !3
!2
!1
" %$ x + x + 12 x + 16 +
1
24
1
24
x 4 + ... +
&
x n! 3
1
x + ... +
+ ...( dx = ! 12 x !2 ! + 12 ln x + 16 x +
n!
x
'
1
#
y 2 (x) = x 3e !x % ! 12 x !2 ! + 12 ln x + 16 x +
$
x
1 2
x
48
+
1
x3
360
1 3 !x
x e
2
1
#
ln x + x 3e !x % ! 12 x !2 ! + 16 x +
$
x
1 3 !x
x e
2
ln x + e !x ! 12 x ! x 2 + 16 x 4 +
1 3 !x
x e
2
#
&
x2 x3
ln x + % 1 ! x +
!
+ ....( ! 12 x ! x 2 + 16 x 4 +
2
6
$
'
(
1
y (x)ln x ! 12 x +
2 1
)
+ .... dx =
1 n
x
n!
1 2
x
48
1 5
x
48
+
+
1
x3
360
1
x6
360
(
&
+ .....( =
'
&
+ .....( =
'
)
+ ..... =
1 5
x
48
+
1
x6
360
( !1 ! 12 ) x2 + (1 ! 41 ) x 3 + ( 16 ! 12 + 121 ) x 4 + ........ =
1
y (x)ln x ! 12 x ! 23 x 2
2 1
+ 43 x 3 ! 41 x 4 + ........
The general solution is:
y(x) = c 0 x 3e !x + c1 ! 12 x ! 12 x 2 + 43 x 3 ! 41 x 4 + .... + 12 y1 (x)ln x
(
)
)
+ ..... =
1 2
x
48
+
1
x3
360
+ .....
5) Solving the equation using the method of Frobenius
x2y’’ + xy’ + (x -1)y = 0
Note that x = 0 is a regular singular point of the equation, then we can assume there is a
solution in the form
"
y(x) = x r # c n x n =
n=0
"
# cn xn+r ,!c0 ! 0
n=0
in some interval 0 < x < ρ.
Then
y '(x) =
y ''(x) =
"
# (n + r)cn xn+r!1
n=0
"
# (n + r)(n + r ! 1)cn xn+r!2
n=0
Substitute into the equation
"
"
n=0
"
n=0
"
"
n=0
n=0
# (n + r)(n + r ! 1)cn xn+r + # ( n + r )cn xn+r + # cn xn+r+1 ! # cn xn+r =
"
# $%( n + r )( n + r ! 1) + (n + r) ! 1&'cn xn+r ! #
n=0
"
c n x n+r+1 =
n=0
"
# $%( n + r )( n + r ! 1 + 1) ! 1&'cn xn+r + #
n=0
"
"
n=0
n=1
c n x n+r+1 =
n=0
# $%( n + r )( n + r ) ! 1&'cn xn+r + # cn!1xn+r =
"
"
n=0
n=1
# $%( n + r + 1)( n + r ! 1)&'cn xn+r + # cn!1xn+r = 0
we shift the index of summation in the second series by -1, replacing n with n -1 and
using the initial value n = 1.
Since we want to express everything in only one summation, we have to start the
summation at n = 1 in every series,
&
&
n=0
n=1
' "#( n + r + 1)( n + r ! 1)$%cn xn+r + ' cn!1xn+r =
&
[(r + 1)(r ! 1)] c0 x r + ' "# "#( n + r + 1) ( n + r ! 1)$% cn + cn!1 $%x n+r = 0
n=1
Since c0 ≠ 0, (r + 1)(r – 1) = 0 is the indicial equation
Then r1 = 1 and r2 = -1 are the exponents of the equation.
Notice 1 – (-1) = 2 is an integer
Now, for n ≥ 1,
cn = !
c n!1
( n + r + 1) ( n + r ! 1)
Letting r1 = 1,
cn = !
c n!1
(n + 2)n
c0
2c
= 0
3 "1 3!1!
c
2c 0
2c
n = 2,!c 2 = ! 1 =
= 0
4 " 2 3!1!" 4 " 2 4!2!
c
2c 0
2c
n = 3,!c 3 = ! 2 = !
=! 0
5"3
4!2!" 5 " 3
5!3!
c
2c
n = 5,!c 4 = ! 3 = 0
6 " 4 6!4!
n = 1,!c1 = !
(
"
%
*
2x
2x 2
2x 3
(!1)n x n y1 (x) = x $ c 0 ! c 0
+ c0
! c0
+ ......' = c 0 x ,1 + )
/
1!3!
2!4!
3!5!
#
&
+ n=1 n!(n + 2)! .
When we consider the smallest root r2 = -1,
n(n- 2) cn + cn-1 = 0, n ≥ 1
c n!1
cn = !
,!n " 1,!n # 2
n (n ! 2)
n =1, c1 = c0
n = 2, the expression is undefined, or 0c2 + c1 = 0 that implies c1 = 0 then c0 = 0 that is a
contradiction.
Therefore, there is no solution of the form
!
x r2 " c n x n
n=0
conclusion 2 says there is a linearly independent solution of the form
!!!!!!!!!!!!!!!!!!!!!!!!!!y 2 (x) =
#
$ dn (x ! x0 )n+r
2
+ cy1 (x)ln x!!with!d 0 " 0
n=0
where c may or may not be zero.
We will use the method of reduction of order to get such solution.
Then the second solution has the form y(x) = y1(x) v(x).
y’(x) = y1(x) v’ + y1’(x)v
y’’(x) = y1(x) v’’ + 2y1’(x) v’ + y1’’(x) v.
After substituting into the equation, we get
d2v
dv
xy1 2 + (2xy1! + y1 ) = 0
dx
dx
Calling w = dv/dx
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!xy1
dw
+ (2xy1! + y1 )w = 0
dx
separating!var iables
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
# y ! 1&
dw
= " % 2 1 + ( dx
w
$ y1 x '
whose solution is
ln w = -2 ln y1(x) - ln x
w=
1
x![ y1 (x)]
2
Since,
x2 x3 x4
x5
y1 (x) = x !
+
!
+
+ .....
3 24 360 8640
"
% "
%
x2 x3 x4
x5
x2 x3 x4
x5
2
[ y1 (x)] = $ x ! 3 + 24 ! 360 + 8640 + .....' ( $ x ! 3 + 24 ! 360 + 8640 + .....' =
#
& #
&
1
1
1 , 5
) 1 1,
) 1 1 1, 4 ) 1
x2 + + ! ! . x3 + +
+ + . x + +!
!
!
!
x + ..... =
* 3 3* 24 9 24 * 360 72 72 360 .2
7 4 1 5
x2 ! x3 +
x !
x + ......
3
36
45
To get 1/(y1(x))2 we use long division
1 21 1
8
+
+ !
x + ....
2
x
3
x
4
135
2
7 4 1 5
x2 ! x3 +
x !
x + ....
!!1
3
36
45
!
2
7 2 1 3
1! x +
x !
x + ..
3
36
45
2
7 2 1 3
0+ x!
x +
x + ...
3
36
45
!
2
7 2 1 3
!!! x !
x +
x + ...
3
36
45
1
29 3
0 + x2 !
x + ......
4
270
!
1 2 1 3
x ! x + .....
4
6
8 3
0+
x + ......
135
1
x [ y1 (x)]
2
=
1" 1
2 1
8 % 1
2
1
8
+
+ !
x' = 3 + 2 +
!
+ ......
2
$
x #x
3x 4 135 & x
4x 135
3x
2
1
8
1
2 1
8
"1
%
v(x) = ( $ 3 + 2 +
!
+ ......'dx = ! 2 !
+ ln x !
x + ......
4x 135
3x 4
135
3x
2x
#x
&
)
,) 1
x2 x3 x4
2 1
8
,
y 2 (x) = y1 (x)v(x) = + x !
+
!
+ ..... + ! 2 !
+ ln x !
x + ....... =
3 24 360
3x 4
135
*
- * 2x
)
,) 1
1
x2 x3 x4
2
8
,
y1 (x)ln x + + x !
+
!
+ ..... + ! 2 !
!
x + ....... =
4
3 24 360
3x 135
*
- * 2x
1
) 1 5 29
,
y1 (x)ln x + + !
! +
x + ........
* 2x 6 144
4
The general solution is:
"
%
x2 x3 x4
" 1 5 29
%
y(x) = c 0 $ x !
+
!
+ ....' + c1 $ !
! +
x + .......' + 41 y1 (x)ln x
# 2x 6 144
&
3 24 360
#
&