Series Solutions Near an Ordinary Point
We are considering methods of solving second order linear equations when the
coefficients are functions of the independent variable.
We consider the second order linear homogeneous equation
d2y
dy
P(x) 2 + Q(x) + R(x)y = 0 (1)
dx
dx
since the procedure for the non-homogeneous equation is similar.
Many problems in mathematical physics lead to equations of this form having polynomial
coefficients; examples include the Bessel equation
x2y’’ + xy’ + (x2 – a2) y = 0
where a is a constant, and the Legendre equation
(1 – x2) y’’ – 2xy’ + c(c + 1) y = 0
where c is a constant.
For this reason, as well as to simplify the algebraic computations, we primarily consider
the case in which P, Q, and R are polynomials. However, we will see that the method can
be applied when P, Q, and R are analytic functions.
For the time being, we assume P, Q, and R are polynomials and they do not have
common factors. Suppose that we wish to solve equation (1) in a neighborhood of a point
a. The solution of equation (1) in an interval containing point a is closely related with the
behavior of P(x) in the interval.
Definition: Given the equation
P(x)
the equation
d2y
dy
+ Q(x) + R(x)y = 0
2
dx
dx
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)
is called the equivalent normalized form of equation(1).
!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Definition: The point a is called an ordinary point of equation (1) if both of the
functions p(x) and q(x) in the equivalent normalized form, are analytic functions at the
point a.
If either or both of these functions are not analytic at a, then the point a is a singular
point of equation (1).
Examples:
1) Given y’’ + xy’ + (x2 + 2) = 0
since p(x) = x and q(x) = x2 + 2 are polynomials, then they are analytic functions
everywhere, then every real number is an ordinary point.
1
y = 0 (x2 – 1)
x
the equivalent normalized equation is
x
1
y ''+ 2
y '+
y=0
x !1
x x2 ! 1
2) Given (x 2 ! 1)y ''+ xy '+
(
where!p(x) =
)
x
1
!and!q(x)
=
x2 ! 1
x x2 ! 1
(
)
then the points 1, -1, and 0 are singular points of the equation, any other real number is an
ordinary point of the equation.
Theorem: If a is an ordinary point of the differential equation (1),
then p(x) = Q(x)/P(x) and q(x) = R(x)/P(x) are analytic at a, and there are two nontrivial
linearly independent power series solutions of equation (1) of the form
"
"
n=0
n=0
# cn (x ! a)n ,! # dn (x ! a)n !
Furthermore, the radius of convergence for each of the series solutions is at least as large
as the minimum of the radii of convergence of the series for p(x) and q(x).
Example:
1) The point a = 0 is an ordinary point of the equation
y’’ + (x3 + 1) y’ – xy = 0
then, the differential equation has a solution in the form
!
" cnxn
whose radius of
n=0
convergence is ρ = ∞.
2) The differential equation
(x2 - 4)y’’ + 3xy’ + y = 0
has two singular points 2 and – 2.
Then the differential equation has a solution in the form
!
" cnxn
whose radius of
n=0
convergence is at least ρ = 2.
The Method of Solution
We assume that a is an ordinary point of the equation (1), so it has a series solution near a
of the form
"
# cn (x ! a)n
that converges for |x –a| < ρ .
n=0
We want to determine the coefficients c0, c1, c2, … in the series solution.
Let’s differentiate the series termwise twice.
y(x) =
"
# cn (x ! a)n = c0 + c1 (x ! a) + c2 (x ! a)2 + ....
n=0
"
y '(x) = # c n n(x ! a)n!1 = c1 + 2c 2 (x ! a) + 3c 3 (x ! a)2 + ....
y ''(x) =
n=1
"
# cn n(n ! 1)(x ! a)n!2 = 2c2 + 6c 3 (x ! a) + 12c 4 (x ! a)2 + ....
n=2
Substituting in the differential equation (1), we get
K0 + K1(x - a) + K2(x – a)2 + ….. + Kn(x –a)n + …. = 0 (4)
where Kn is function of certain coefficients ci.
In order that series (4) be valid for all x in some interval |x – a| < ρ, we must have
K0 = K1 = …. = Kn = …. = 0
This leads to a set of conditions that the coefficients cn must satisfy.
Example:
1) Solve (x2 – 4)y’’ + 3xy’ + y = 0 in powers of x.
The only singular points of the equation are 2 and -2, so we want to find the series
solution of the equation near zero. The series will have radius of convergence at least 2.
So y(x) =
"
"
n=0
n=1
# cn xn ,!!y '(x) = # ncn xn!1,!y ''(x) =
Substituting the series in the equation
"
"
"
# n(n ! 1)cn xn!2
n=2
"
"
x 2 # n(n ! 1)c n x n!2 ! 4 # n(n ! 1)c n x n!2 + 3x # nc n x n!1 + # c n x n =
n=2
n=2
n=1
"
"
"
n=2
"
n=2
"
n=1
n=0
"
# n(n ! 1)cn xn ! # 4n(n ! 1)cn xn!2 + # 3ncn xn + # cn xn =
n=0
"
# n(n ! 1)cn x ! # 4 ( n + 2 ) (n + 1)cn+2 x + # 3ncn x
n
n=2
n
n=0
n
n=1
"
+ # cnxn = 0
n=0
we shift the index of summation in the second series by 2, replacing n with n + 2 and
using the initial value n = 0.
Since we want to express everything in only one summation sign, we have to start the
summation at n = 2 in every series,
"
"
"
# n(n ! 1)cn x ! # 4 ( n + 2 ) (n + 1)cn+2 x + # 3ncn x
n
n=2
"
n
n=0
n=1
n
"
+ # cnxn =
n=0
"
"
n=2
n=2
# n(n ! 1)cn xn ! 4(2)(1)c2 ! 4(3)(2)c 3x ! # 4(n + 2)(n + 1)cn xn + 3c1x + # 3ncn xn +
n=2
"
c 0 + c1x + # c n x n =! 0
n=2
"
(c 0 ! 8c 2 ) + (4c1 ! 24c 3 )x ! # [ !4(n + 2)(n + 1)c n+2 + (n(n ! 1) + 3n + 1)c n ] x n =
n=2
"
(c 0 ! 8c 2 ) + (4c1 ! 24c 3 )x ! # $%(n + 1)2 c n ! 4(n + 1)(n + 2)c n+2 &' x n = !0
n=2
Now
c0 - 8c2 = 0, then c2 = c0/8
4c1 – 24c3 = 0, then c3 = 4c1/24 = c1/6
cn
,!n " 2
4(n + 2)
This last condition is called a recurrence formula, we can express each cn+2 in terms of a
previous coefficient cn.
3c
3c
3c
n = 2,!c 4 = 2 = 2 0 = 0
4 ! 4 4 ! 2 ! 4 128
4c
2 ! 4c
c
n = 3,!c 5 = 3 = 2 1 = 1
4 ! 5 4 ! 3 ! 5 30
5c
3 ! 5c 0
5c 0
n = 4,!c 6 = 4 = 3
=
4 ! 6 4 ! 2 ! 4 ! 6 1024
6c
2 ! 4 ! 6c1
c
n = 5,!c 7 = 5 = 3
= 1
4 ! 7 4 ! 3 ! 5 ! 7 140
(n + 1)2 c n ! 4(n + 1)(n + 2)c n+2 = 0,!then!c n+2 =
Notice that each even coefficient is expressed in terms of c0 and each odd coefficient is
expressed in terms of c1.
Then, the general solution is:
1
3 4
5 6
1
1 5
1 7
!
$
!
$
y(x) = c 0 # 1 + x 2 +
x +
x + ...& + c1 # x + x 3 +
x +
x + ...&
"
%
"
%
8
128
1024
6
30
140
2) Solve y’’ + xy’ + (x2 + 2)y = 0 in powers of x.
There are no singular points of the equation, so we want to find the series solution of the
equation near zero. The series converges for all x.
So y(x) =
"
"
n=0
n=1
# cn xn ,!!y '(x) = # ncn xn!1,!y ''(x) =
"
# n(n ! 1)cn xn!2
n=2
Substituting the series in the equation
"
# n(n ! 1)cn x
n=2
"
n!2
"
+ x # nc n x
n=1
"
n!1
+x
2
"
# cnx
n=0
"
n
"
+ 2 # cnxn =
n=0
"
# n(n ! 1)cn xn!2 + # ncn xn + # cn xn+2 + # 2cn xn =
n=2
"
n=1
n=0
n=0
"
"
"
n=1
n=2
n=0
# ( n + 2 ) (n + 1)cn+2 xn + # ncn xn + # cn!2 xn + # 2cn xn = 0
n=0
We shift the index of summation in the first series by 2, replacing n with n + 2 and using
the initial value n = 0. We shift the index of summation in the third series by -2, replacing
n by n – 2 and using the initial value n = 2
Since we want to express everything in only one summation sign, we have to start the
summation at n = 2 in every series,
!
!
!
n=1
n=2
!
" ( n + 2 ) (n + 1)cn+2 xn + " ncn xn + " cn#2 xn + " 2cn xn =
n=0
!
n=0
!
!
2c 2 + 6c 3 + " ( n + 2 ) (n + 1)c n+2 x +!c1x + " nc n x + 3c1x + " c n#2 x n +
n
n=2
!
n
n=2
n=2
2c 0 + 2c1x + " c n x n =! 0
n=2
"
(2c 0 + 2c 2 ) + (3c1 + 6c 3 )x ! # [ (n + 2)(n + 1)c n+2 + (n + 2)c n + c n!1 ] x n = !0
n=2
Now
2c0 + 2c2 = 0, then c2 = -c0
3c1 + 6 c3 = 0, then c3 = -1/2 c1
(n + 2)c n + c n!2
,!n " 2
( n + 1) (n + 2)
This last condition is called a recurrence formula, we can express each cn+2 in terms of a
previous coefficient cn-2 and cn.
4c + c 0
!4c 0 + c 0 3c 0 1
n = 2,!c 4 = ! 2
=!
=
= c0
12
12
12 4
5 ! 12 c1 + c1 3c1
5c + c
n = 3,!c 5 = ! 3 1 = !
=
20
20
40
1
6 c0 ! c0
6c + c 2
c
n = 4,!c 6 = ! 4
=! 4
=! 0
5"6
30
60
3
1
7
c1 ! 2 c1
7c + c 3
c
n = 5, c 7 = ! 5
= ! 40
=! 1
6"7
42
1680
(n + 2)(n + 1)c n+2 + (n + 2)c n + c n!2 = 0,!then!c n+2 = !
( )
( )
( )
Notice that each even coefficient is expressed in terms of c0 and each odd coefficient is
expressed in terms of c1.
Then, the general solution is:
1
1
1
3
1 7
"
%
"
%
y(x) = c 0 $ 1 ! x 2 + x 4 ! x 6 + ...' + c1 $ x ! x 3 + x 5 !
x + ...'
#
&
#
&
4
60
2
4
1680
3) Solve y’’ - xy’ - x2y = 0 in powers of x.
There are no singular points of the equation, so we want to find the series solution of the
equation near zero. The series converges for all x.
So y(x) =
"
"
n=0
n=1
# cn xn ,!!y '(x) = # ncn xn!1,!y ''(x) =
"
# n(n ! 1)cn xn!2
n=2
Substituting the series in the equation
"
"
n=2
"
n=1
"
"
# n(n ! 1)cn xn!2 ! x # ncn xn!1 ! x2 # cn xn =
n=0
"
# n(n ! 1)cn xn!2 ! # ncn xn ! # cn xn+2 =
n=2
"
n=1
n=0
"
# ( n + 2 ) (n + 1)cn+2 x ! # ncn x
n
n=0
n
n=1
"
! # c n!2 x n =! 0
n=2
We shift the index of summation in the first series by 2, replacing n with n + 2 and using
the initial value n = 0. We shift the index of summation in the third series by -2, replacing
n by n – 2 and using the initial value n = 2
Since we want to express everything in only one summation, we have to start the
summation at n = 2 in every series,
"
"
"
n=1
n=2
# ( n + 2 ) (n + 1)cn+2 xn ! # ncn xn ! # cn!2 xn =
n=0
"
"
"
2c 2 + 6c 3x + # ( n + 2 ) (n + 1)c n+2 x !!c1x ! # nc n x + # c n!2 x n =
n
n=2
n=2
n
n=2
"
2c 2 + (6c 3 ! c1 )x + # [ (n + 2)(n + 1)c n+2 ! nc n ! c n!2 ]x n = 0
n=2
Now
2c2 = 0, then c2 = 0
6c3 – c1 = 0, then c3 = -1/6 c1
nc n + c n!2
,!n " 2
(n + 2)(n + 1)
This last condition is called a recurrence formula, we can express each cn+2 in terms of a
previous coefficient cn-2 and cn.
2c + c 0 c 0
n = 2,!c 4 = 2
=
12
12
1
c1 + c1 3c1
3c + c
n = 3,!c 5 = 3 1 = 2
=
20
20
40
1
4c + c 2 4 12 c 0 c 0
n = 4,!c 6 = 4
=
=
30
30
90
3
1
5c + c 3 40 c1 + 6 c1 13c1
n = 5,!c 7 = 5
=
=
42
42
1080
c n+2 =
( )
Notice that each even coefficient is expressed in terms of c0 and each odd coefficient is
expressed in terms of c1.
Then, the general solution is:
1
1
1
3 5
13 7
!
$
!
$
y(x) = c 0 # 1 + x 4 + x 6 + ...& + c1 # x + x 3 +
x +
x + ....&
" 12
%
"
%
90
6
40
1080
4) Solve (x2 + 1)y’’ - y’ + y = 0 in powers of x.
There are no singular points of the equation, so we want to find the series solution of the
equation near zero. The series converges for all x.
So y(x) =
"
"
n=0
n=1
# cn xn ,!!y '(x) = # ncn xn!1,!y ''(x) =
Substituting the series in the equation
"
"
"
# n(n ! 1)cn xn!2
n=2
"
"
x 2 # n(n ! 1)c n x n!2 + # n(n ! 1)c n x n!2 ! # nc n x n!1 + # c n x n =
n=2
n=2
n=1
"
"
"
n=2
"
n=2
"
n=1
n=2
n=0
n=0
"
# n(n ! 1)cn xn + # n(n ! 1)cn xn!2 ! # ncn xn!1 + # cn xn =
n=0
"
"
n=0
n=0
# n(n ! 1)cn xn + # ( n + 2 ) (n + 1)cn+2 xn ! # ( n + 1) cn xn + # cn xn = 0
we shift the index of summation in the second series by 2, replacing n with n + 2 and
using the initial value n = 0, and we shift the index of summation in the third series by 1,
replacing n with n + 1 and using the initial value n = 0.
Since we want to express everything in only one summation sign, we have to start the
summation at n = 2 in every series,
"
"
# n(n ! 1)cn x + # ( n + 2 ) (n + 1)cn+2 x
n
n=2
n=0
"
n
"
"
! # ( n + 1) c n+1x + # c n x n =
n
n=0
"
n=0
"
2c 2 + (3)(2)c 3x + # (n + 2)(n + 1)c n+2 x n + # n(n ! 1)c n x n ! c1 ! 2c 2 x + # ( n + 1) c n+1x n +
n=2
n=2
n=2
"
c 0 + c1x + # c n x n =! 0
n=2
"
(2c 2 ! c1 + c 0 ) + (6c 3 ! 2c 2 + c1 )x + # [ (n + 2)(n + 1)c n+2 + n(n ! 1)c n ! (n + 1)c n+1 + c n ] x n =
n=2
"
(
)
(2c 2 ! c1 + c 0 ) + (6c 3 ! 2c 2 + c1 )x + # $%(n + 2)(n + 1)c n+2 + n 2 ! n + 1 c n ! (n + 1)c n+1 &' x n = !0
n=2
Now
2c2 –c1 = c0 = 0, then c2 = (c1 - c0)/2
6c3 -2c2 + c1= 0, then c3 = (2c2 - c1)/6 = -c0/6
!c n+2 =
(
)
(n + 1)c n+1 ! n 2 ! n + 1 c n
(n + 2)(n + 1)
,!n " 2
This last condition is called a recurrence formula, we can express each cn+2 in terms of a
previous coefficient cn+1 and cn.
3c ! 3c 2 c 3 ! c 2 ! 16 c 0 ! 12 ( c1 ! c 0 ) 2c 0 ! 3c1
n = 2,!c 4 = 3
=
=
=
4"3
4
128
24
c
# 2c ! 3c1 &
4% 0
+7 0
(
$
4c ! 7c 3
3c ! c
24 '
6
n = 3,!c 5 = 4
=
= 0 1
5"4
5"4
40
36c1 ! 17c 0
n = 4,!c 6 ==
720
Then, the general solution is:
c !c
c
2c ! 3c1 4 3c 0 ! c1 5 36c1 ! 17c 0
y(x) = c 0 + c1x + 1 0 x 2 ! 0 x 3 + 0
x +
x +
+ .... =
2
6
24
40
720
1
1
1
3 5
1
1
1 5
"
%
"
%
c0 $ 1 ! x2 ! x 3 + x 4 +
x + ...' + c1 $ x + x 2 ! x 4 !
x + ...'
#
&
#
&
2
6
12
40
2
8
40
5) Solve the non-homogeneous equation
y’’ – 2x2y’ + 4xy = x2 + 2x + 2
in powers of x.
There are no singular points of the equation, so we want to find the series solution of the
equation near zero. The series converges for all x.
So y(x) =
"
"
n=0
n=1
# cn xn ,!!y '(x) = # ncn xn!1,!y ''(x) =
Substituting the series in the equation
"
# n(n ! 1)cn x
n=2
"
n!2
! 2x
2
"
# ncn x
n=1
"
n!1
"
# n(n ! 1)cn xn!2
n=2
"
+ 4x # c n x n ! x 2 ! 2x ! 2 = 0
n=0
"
# n(n ! 1)cn xn!2 ! # 2ncn xn+1 + # 4cn xn+1 ! x2 ! 2x ! 2 = 0
n=2
"
n=1
n=0
"
"
n=2
n=1
# ( n + 2 ) (n + 1)cn+2 xn ! # 2(n ! 1)cn!1xn + # 4cn!1xn ! x2 ! 2x ! 2 = 0
n=0
We shift the index of summation in the first series by 2, replacing n with n + 2 and using
the initial value n = 0. We shift the index of summation in the second series by -1,
replacing n by n – 1 and using the initial value n = 2. We shift the index of summation in
the third series by -1, replacing n by n – 1 and using the initial value n = 1.
Since we want to express everything in only one summation sign, we have to start the
summation at n = 3 in every series (notice the term x2 by itself),
"
"
"
# ( n + 2 ) (n + 1)cn+2 xn ! # 2(n ! 1)cn!1xn + # 4cn!1xn ! x2 ! 2x ! 2 =
n=0
n=2
n=1
"
"
n=2
n=2
"
2c 2 + 6c 3x + # ( n + 2 ) (n + 1)c n+2 x n ! # 2(n ! 1)c n!1x n + 4c 0 x + # 4c n!1x n ! x 2 ! 2x ! 2 =
n=2
"
( 2c2 ! 2 ) + ( 6c 3 + 4c0 ! 2 ) x + ( !1 + 12c 4 + 2c1 ) x 2 + # [(n + 2)(n + 1)cn+2 ! 2(n ! 1)cn!1 + 4cn!1 ] x n = 0
n= 3
Now
2c2 – 2 = 0, then c2 = 1
6c3 + 4 c0 – 2 = 0, then c3 = 1/3 – 2/3 c0
-1 + 12 c4 + 2c1 = 0, then c4 = 1/12 – 1/6 c1
(n + 2)(n + 1) cn+2 – 2(n -3) cn-1 = 0, n ≥ 3
2(n ! 3)
c n+2 =
c n!1,!n " 3
(n + 1)(n + 2)
n = 3, c5 = 0
n = 4, c6 = 2/30 c4 = 1/15(1/3 – 2/3c0) = 1/45 – 2/45c0
n = 5, c7 = 4/42 c4 = 1/126 – 1/63 c1
y(x) = c 0 + c1x + x 2 + ( 13 ! 23 c 0 )x 3 +
(
c 0 1 ! 23 x 3 !
2 6
x
45
)
(
( 121 ! 16 c1 ) x 4 + ( 451 ! 452 c0 ) x6 + .( 1261 ! 631 ) x 7 ..... =
)
1
+ ..... + c1 x ! 16 x 4 ! 126
x 7 ..... + (x 2 + 13 x 3 + 121 x 4 +
1 6
x
45
1
+ 126
x 7 ......)
6) Solve the non-homogeneous equation
y’’ – xy’ + 2y = e-x
in powers of x.
There are no singular points of the equation, so we want to find the series solution of the
equation near zero. The series converges for all x.
"
"
"
"
xn
So y(x) = # c n x n ,!!y '(x) = # nc n x n!1,!y ''(x) = # n(n ! 1)c n x n!2 ,!e !x = # (!1)n
n!
n=0
n=1
n=2
n=0
Substituting the series in the equation
n
"
"
"
"
n!2
n!1
n
n x
n(n
!
1)c
x
!
x
nc
x
+
2
c
x
!
(!1)
=
#
# n
# n #
n
n!
n=2
n=1
n=0
n=0
"
# n(n ! 1)cn x
n=2
"
n!2
"
"
"
xn
! # nc n x + # 2c n x ! # (!1)
=
n!
n=1
n=0
n=0
n
n
n
"
"
"
n=1
n=0
n=0
# ( n + 2 ) (n + 1)cn+2 xn ! # ncn xn + # 2cn xn ! # (!1)n
n=0
xn
= !0
n!
We shift the index of summation in the first series by 2, replacing n with n + 2 and using
the initial value n = 0. Since we want to express everything in only one summation sign,
we have to start the summation at n = 1 in every series,
"
"
"
"
n=0
n=1
n=0
n=0
# ( n + 2 ) (n + 1)cn+2 xn ! # ncn xn + # 2cn xn ! # (!1)n
xn
=!
n!
$
(!1)n ' n
( 2c2 + 2c0 ! 1) + # &( n + 2 )( n + 1) cn+2 ! (n ! 2)cn !
)x = 0
n! (
%
Now,
2c2 + 2c0 – 1 = 0, then c2 = ½ - ½ c0
(!1)n
(n + 2)(n + 1)c n+2 ! (n ! 2)c n !
= 0,!n " 1
n!
(!1)n
(n ! 2)c n +
n! ,!n " 1
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!c n+2 =
(n + 2)(n + 1)
!c ! 1
c 1
n = 1,!c 3 = 1
=! 1!
3# 2
6 6
1
1
n = 2,!c 4 = 2 =
4 # 3 24
c 3 ! 16 c 3
1
1'
1
c
1
1 $ c1
n = 3,!c 5 =
=
!
= 20
=! 1 !
&% ! ! )( !
5#4
20 120
6 6
120
120 60
n = 4,!c 6 =
n = 5,!c 7 =
2c 4 +
1
24
6#5
1
3c 5 ! 120
7#6
c4
1
1
1
3
1
+
=
+
=
=
15 720 15 # 24 720 720 240
c
1
1'
1
c
7
c
1
$ c
= 5!
= 141 & ! 1 ! ) !
=! 1 !
=! 1 !
% 120 60 ( 5040
14 5040
1680 5040
1680 720
=
1 1
c 1
1 4
c
1
1 6
y(x) = c 0 + c1x + ( ! c 0 )x 2 + (! 1 ! )x 3 +
x + (! 1 ! )x 5 +
x +
2 2
6 6
24
120 60
240
c
1
(! 1 !
)x 7 + ........ =
1680 720
1
1
1 5
1 7
1
1
1 4 1 5
1 7
c 0 (1 ! x 2 ) + c1 (x ! x 3 !
x !
x + .....) + ( x 2 ! x 3 +
x !
x !
x + ...)
2
6
120
1680
2
6
24
60
720
Translated Series Solutions
If we look for a solution of the equation P(x)
y=
"
d2y
dy
+
Q(x)
+ R(x)y = 0 of the form
dx
dx 2
# cn (x ! a)n , we can make a change of variable x – a = t, obtaining a new
n=0
differential equation for y as a function of t, and then look for solutions of this new
equation of the form y(t) =
!
" cn t n .
When we have finished the calculations, we replace
n=0
t by x –a.
Example:
Solve 2y’’ + xy’ + y = 0 in powers of x - 1.
Notice that x = 1 is an ordinary point of the equation. We look for a solution in the form
y=
"
# cn (x ! 1)n .
n=0
We can simplify the computation of the coefficients cn by translating the center of the
expansion from x = 1 to t = 0.
This is accomplished by the substitution x – 1 = t or t = x + 1, then dt/dx = 1 or dt = dx.
Since y(x) = y(t + 1), using chain rule:
dy dy dt dy
dy
=
=
!1 =
dx dt dx dt
dt
2
d y d " dy % d " dy % d 2 y dt d 2 y
=
=
=
=
dx 2 dx $# dx '& dx $# dt '& dt 2 dx dt 2
d2y
dy
So the equation transforms: 2 2 + (t ! 1) + y = 0 .
dt
dt
We seek a solution in the form y(t) =
!
" cn t n where the coefficients cn are the same that
n=0
appear in solution (2).
So y(x) =
"
"
n=0
n=1
"
# cn t n ,!!y '(x) = # ncn t n!1,!y ''(x) =
# n(n ! 1)cn t n!2
n=2
Substituting the series in the equation
"
2 # n(n ! 1)c n t
n=2
"
n!2
"
+t # nc n t
n!1
n=1
"
"
! # nc n t
n=1
"
n!1
"
+ # cn t n =
n=0
"
# 2n(n ! 1)cn t n!2 + # ncn t n ! # ncn t n!1 + # cn t n =
n=2
"
n=1
n=1
n=0
"
"
"
n=1
n=0
n=0
# 2(n + 2)(n + 1)cn+2 t n + # ncn t n ! # (n + 1)cn+1t n + # cn t n = 0
n=0
We shift the index of summation in the first series by 2, replacing n with n + 2 and using
the initial value n = 0. We shift the index of summation in the third series by 1, replacing
n by n + 1 and using the initial value n = 0
Since we want to express everything in only one summation sign, we have to start the
summation at n = 1 in every series,
!
!
!
!
" 2(n + 2)(n + 1)cn+2 t n + " ncn t n # " (n + 1)cn+1t n + " cn t n =
n=0
n=1
!
n=0
n=0
!
!
!
n=1
n=1
n=0
4c 2 + " 2(n + 2)(n + 1)c n+2 t n + " nc n t n # c1 # " (n + 1)c n+1t n + c 0 + " c n t n =
n=1
!
( 4c2 # c1 + c0 ) + " ![ 2(n + 2)(n + 1)cn+2 + (n + 1)cn # (n + 1)cn+1 ]t n = 0
n=1
Now,
4c2 – c1 + c0 = 0, then c2 = ¼(c1 – c0)
2(n+2)(n+1)cn+2 + (n+1)(cn –cn+1) = 0, n ≥ 1
(n + 1)(c n ! c n+1 )
c ! c n+1
c n+2 = !
=! n
,!n " 1
2(n + 2)(n + 1)
2(n + 2)
c !c
c1 ! 1 0
c1 ! c 2
3c + c
4
n = 1,!c 3 = !
=!
=! 1 0
2#3
6
24
c1 ! c 0 3c1 + c 0
+
c2 ! c 3
9c ! 5c 0
4
24
n = 2,!c 4 = !
=!
=! 1
2#4
8
192
c1 ! c 0 2 3c1 + c 0 3 9c1 ! 5c 0 4
t !
t !
t ! ........ =
4
24
192
1
1 3
5 4
1
1
9 4
"
%
"
%
c0 $ 1 + t 2 !
t +
t + ....' + c1 $ t ! t 2 ! t 3 !
t ! .....'
#
&
#
&
4
24
192
4
8
192
y(t) = c 0 + c1t !
1
1
5
"
%
y(x) = c 0 $ 1 + (x ! 1)2 !
(x ! 1)3 +
(x ! 1)4 + ....' +
#
&
4
24
192
1
1
9
"
%
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!c1 $ (x ! 1) ! (x ! 1)2 ! (x ! 1)3 !
(x ! 1)4 ! .....'
#
&
4
8
192