Chapter 52
Power series methods of solving ordinary
differential equations
Why it is important to understand: Power series methods of solving ordinary differential equations
The differential equations studied so far have all had closed form solutions, that is, their solutions
could be expressed in terms of elementary functions, such as exponential, trigonometric, polynomial
and logarithmic functions, and most such elementary functions have expansions in terms of power
series. However, there are a whole class of functions which are not elementary functions and which
occur frequently in mathematical physics and engineering. These equations can sometimes be solved
by discovering a power series that satis es the differential equation, but the solution series may not be
summable to an elementary function. In this chapter the methods of solution to such equations are
explained.
At the end of this chapter, you should be able to:
appreciate the reason for using power series methods to solve differential equations
determine higher order differential coef cients as a series
use Leibniz’s theorem to obtain the nth derivative of a given function
obtain a power series solution of a differential equation by the Leibniz-Maclaurin method
obtain a power series solution of a differential equation by the Frobenius method
determine the general power series solution of Bessel’s equation
express Bessel’s equation in terms of gamma functions
determine the general power series solution of Legendre’s equation
determine Legendre polynomials
determine Legendre polynomials using Rodrigues’ formula
52.1
Introduction
Second-order ordinary differential equations that cannot be solved by analytical methods (as shown in
Chapters 50 and 51), i.e. those involving variable coef cients, can often be solved in the form of an in nite
series of powers of the variable. This chapter looks at some of the methods that make this possible – by the
Leibniz–Maclaurin and Frobenius methods, involving Bessel’s and Legendre’s equations, Bessel and
gamma functions and Legendre’s polynomials. Before introducing Leibniz’s theorem, some trends with
higher differential coef cients are considered. To better understand this chapter it is necessary to be able to:
(a) differentiate standard functions (as explained in Chapters 25 and 30),
(b) appreciate the binomial theorem (as explained in Chapter 5), and
(c) use Maclaurin’s theorem (as explained in Chapter 37).
52.2
Higher order differential coefficients as series
The following is an extension of successive differentiation (see page 337), but looking for trends, or series,
as the differential coef cient of common functions rises.
(a) If y
ax
= e
, then
If we abbreviate
dy
dx
dy
dx
ax
= ae
2
as y′,
d y
2
dx
,
2
d y
2
dx
2
ax
= a e
, and so on.
n
as y′′, … and
d y
n
dx
as y
(n)
, then y
′
(n)
emerging pattern gives:
For example,
(b) If y
A
A
A
A
if y
= sin ax
2x
= 3e
, then
y
ax
= ae
n
= a
′
′′
y
7
=
=
=
′′′
y
′′
=
=
In general,
(c) If y
= cosax
, and the
(1)
(7)
= y
=
7
2x
3(2 ) e
2x
= 384e
acos ax = a sin (ax +
2
−a
2
a
−a
a
sin ax = a
sin (ax +
3
3
2
2π
2
π
2
)
Soufyan Jamal
sin(ax + π)
)
cos ax
3π
sin (ax +
(n)
2
)and so on.
n
y
= a
sin(ax +
nπ
2
)
(2)
5
=
ax
ax
e
For example, if
y
2
= a e
7
d y
dx
,
y
,y
sin 3x, then
5
=
3
sin (3x +
=
243 cos 3x
d y
5
dx
5π
2
(5)
= y
5
) = 3
sin (3x +
,
(
)
π
2
)
′
y
=
′′
y
′′′
y
−a sin ax = a cos (ax +
2
=
−a
=
a
3
2
cos ax = a
3
sin ax = a
(n)
For example, if y
d y
cos (ax +
= a
2π
3π
2
cos(ax +
2
)
)and so on.
nπ
2
)
(3)
,
= 4(2
6
(6)
= y
6
dx
)
= 4 cos 2x
6
then
2
cos (ax +
n
In general,y
π
) cos (2x +
6π
2
)
= 4(2 ) cos (2x + 3π)
= 4(2 ) cos (2x + π)
= −256 cos 2x
6
6
(d) If y =
and y
x
(n)
a
,y
′
a−1
= ax
,y
′′
a−2
= a(a − 1)x
,y
′′′
a−3
= a(a − 1)(a − 2)x
,
a−n
= a(a − 1)(a − 2) … (a − n + 1) x
(n)
ory
=
a!
a−n
(a − n)!
Soufyan Jamal
where a is a positive integer.
For example, if y
x
6
= 2x
, then
(4)
4
d y
4
dx
(4)
= y
= (2)
6!
6−4
(6−4)!
= (2)
= 720x
x
6×5×4×3×2×1
2×1
2
x
2
(e) (v) If y
= sinh ax
,
′
y
= acoshax
′′
I f y = si y
′′′
y
=
=
2
a
3
a
sinh ax
cosh ax, and so on
Since sinh ax is not periodic (see graph on page 154), it is more dif cult to nd a general statement
for y . However, this is achieved with the following general series:
(n)
(n)
y
=
a
n
2
n
{[1 + (−1)
] sinh ax
n
+ [1 − (−1)
For example, if
] cosh ax}
(5)
5
y
=
d y
sinh 2x, then
5
2
=
2
5
dx
(5)
= y
5
{[1 + ( − 1) ] sinh 2x
5
+ [1 − ( − 1) ] cosh 2x}
5
2
=
2
=
(f) If y
= cosh ax
{[0] sinh 2x + [2] cosh 2x}
Soufyan Jamal
32 cosh 2x
,
′
y
′′
y
′′′
y
=
a sinh ax
=
a
=
2
3
a
cosh ax
sinh ax, and so on
Since cosh ax is not periodic (see graph on page 154), again it is more dif cult to nd a general
statement for y . However, this is achieved with the following general series:
(n)
(n)
y
a
=
n
2
{[1 − ( − 1)
n
] sinh ax
+ [1 + ( − 1)
For example, if y
then
1
=
cosh 3x
9
7
d y
(7)
= y
7
dx
1
= (
9
n
] cosh ax}
(6)
,
7
)
3
2
(2 sinh 3x)
= 243 sinh3x
(g) If y
,
′
= ln ax y
,y
1
=
′′
x
,y
1
= −
′′′
2
x
=
2
3
x
, and so on.
(n)
In general,y
For example, if y
= ln 5x
6
d y
6
dx
(6)
= y
6−1
= (−1)
Note that if y = ln x, y
( − 1) = 1 and if y =
0
x
n
(7)
, then
5!
(
′
n−1 (n − 1)!
= (−1)
6
x
=
′
1
x
) = −
120
x
6
; if in equation (7), n = 1 then y = ( − 1)
then (0)! = 1 (check that ( − 1) = 1 and (0)! = 1 on a calculator).
1
′
0 (0)!
1
x
x
0
Now try the following Practice Exercise
Exercise 236 Higher-order differential coefficients as series (Answers on page 896)
Determine the following derivatives:
1. y
when y = e
2. y
when y = sin 3t (b) y
3. y
when y
= cos 2x
4. y
when y
= 2x
(4)
(4)
(8)
(7)
(b) y
2x
when y = 8 e
(5)
(7)
9
when y =
(b) y
(9)
(b) y
(6)
t
2
1
sin 5θ
50
when y
= 3 cos
when y
=
7
t
8
2
3
t
5. y
when y
6. y
when y
=
7. y
when y
= 2ln 3θ
(7)
(7)
(4)
52.3
=
1
4
sinh 2x
cosh 2x
(b) y
(6)
(b) y
(8)
(b) y
when y
when y
when y
(7)
= 2 sinh 3x
1
=
1
=
cosh 3x
9
3
ln 2t
Leibniz’s theorem
If
y = uv
(8)
where u and v are each functions of x, then by using the product rule,
′
′
y
′
= uv + vu
(9)
′′
y
′′
= uv
′
′
′′
+ v u + vu
′
′
+ u v
(10)
′′
′
′
′′
= u v + 2u v + uv
′′′
′′
y
′
′′′
′
= u v + vu
′′
+ 2u v
′
′′
+ 2v u
′′′
′′
+ uv
′
+ v u
(11)
′′′
= u
(4)
(4)
y
= u
′′
′
′
′′
v + 3u v + 3u v
(3)
v + 4u
(1)
v
′′′
+ uv
(2)
+ 6u
(2)
v
(1)
+ 4u
(3)
v
(4)
+ uv
(12)
From equations (8) to (12) it is seen that
(a) the nth derivative of u decreases by 1 moving from left to right,
(b) the nth derivative of v increases by 1 moving from left to right,
(c) the coef cients 1, 4, 6, 4, 1 are the normal binomial coef cients (see page 49).
In fact, (uv) may be obtained by expanding (u + v) using the binomial theorem (see page 50), where
the ‘powers’ are interpreted as derivatives. Thus, expanding (u + v) gives:
(n)
(n)
(n)
(n)
y
=
(n)
(uv)
+
+
(n)
= u
n(n − 1)
2!
(n−2)
u
n(n − 1)(n − 2)
3!
(n−1)
v + nu
(1)
v
(2)
v
(n−3)
u
(3)
v
+
⋯
(13)
Equation (13) is a statement of Leibniz’s theorem 1, which can be used to differentiate a product n times.
The theorem is demonstrated in the following worked problems.
∗
Problem 1. Determine y
(n)
when y = x
2
3x
e
For a product y = uv, the function taken as
1. u is the one whose nth derivative can readily be determined (from equations (1) to (7)),
2. v is the one whose derivative reduces to zero after a few stages of differentiation.
Thus, when y = x e , v = x , since its third derivative is zero, and
known from equation (1), i.e. 3 e
Using Leinbiz’s theorem (equation (13)),
2
3x
2
n
(n)
y
(n)
=
u
Hence, y
(n)
n
3x
(3 e
2
= x
,v
(1)
,
(2)
= 2x v
(n−1)
v + nu
n(n−1)(n−2)
+
where in this case v
3x
u = e
since the nth derivative is
ax
and v
(3)
n(n−1)
+
(n−3)
u
3!
= 2
(1)
v
2!
(3)
v
+
(n−2)
u
(2)
v
⋯
Soufyan Jamal
= 0
=
2
n−1
)(x ) + n(3
3x
e
)(2x)
+
+
=3
n−2
3x
e
2
2
(3 x
n(n−1)
2!
n−2
(3
n(n−1)(n−2)
3!
3x
e
)(2)
n−3
(3
3x
e
)(0)
+ n(3)(2x)
+ n(n − 1) + 0)
i.e.
(n)
y
3x
e
Problem 2. If x
2
′′
y
′
+ 2xy + y = 0
=
n−2
3
show that: xy
(n+2)
Differentiating each term of x
2
′′
y
2
(9x
(n+1)
+ 2(n + 1)xy
(n+2)
2
x
(n+1)
+ {y
i. e.
(n+1)
+ ny
(2x) +
(n)
(2x) + n y
2
(n+2)
(n+1)
x y
(n+2)
x y
2
+ (n
{
(n)
+ n(n − 1)y
(n)
+ 2n y
(n)
+ y
(n)
− n + 2n + 1)y
= 0
= 0
Soufyan Jamal
(n+1)
+ 2(n + 1) x y
(n)
+ n + 1)y
Problem 3. Differentiate the following differential equation n times: (1
By Leibniz’s equation, equation (13),
} = 0
+ 2(n + 1)xy
+ (n
2
(n)
(n+1)
2
or
= 0
(2) + 0}
(2) + 0} + {y
+ 2xy
i. e.
(n)
y
2!
+ 2n xy
(n+2)
(n)
+ n + 1)y
n(n−1)
(n+1)
x y
2
2
+ (n
n times, using Leibniz’s theorem of equation (13), gives:
′
+ 2xy + y = 0
{y
+ 6nx + n(n − 1))
2
′′
+ x )y
+
= 0
′
2xy − 3y = 0
(n+2)
2
{y
(n+1)
(1 + x ) + ny
(n+1)
+ 2{y
(n)
(x) + n y
2
i. e.
(n+2)
(n+1)
2
or
Problem 4. Find the fth derivative of y
4
= x
sin x
4
= x
(n)
(n+2)
(n+1)
(n)
+ n − 3)y
=
nπ
[sin (x +
+
+
+
+
(5)
and y
=
n(n−1)
2
=
3!
4
x
+
π
2
sin(x + 2π) ≡ sin x, sin (x +
and
) ≡
3π
2
(5)
4
y
3
= x
cos x + 20x
cos x
(5)(4)
2
(5)
4
= (x
3
+ (20x
2
− 120x
− 240x) sin x
Now try the following Practice Exercise
5π
2
2
(5)(4)(3)
(3)(2)
(n−2)π
2
)12x ]
2
(n−3)π
2
)24x]
(n−4)π
[sin (x
2
3
) + 20x
sin(x + 2π)
3π
2
)
(24x) sin (x + π)
(5)(4)(3)(2)
(4)(3)(2)
,
sin x + 120x (− cos x)
+ 120)cos x
3
)4x ]
(12x ) sin (x +
,
2
gives:
)24]
sin (x +
) ≡ − cos x
+ 240x(− sin x) + 120 cos x
y
2
[ sin (x +
sin (x + π) ≡ − sin x,
then
i.e.
(n−1)π
4!
2
4
= x
4
n(n−1)(n−2)(n−3)
(n−4)π
and v
)x ]
n(n−1)(n−2)
+
) ≡ sin (x +
sin x
[sin (x +
2!
+
2
= 0
sin x
Soufyan Jamal
5π
= 0
+ 2(n + 1)xy
+ n[sin (x +
Since sin (x +
= 0
+ 2(n + 1)xy
, then using Leibniz’s equation with u
y
(n)
(n+1)
(1 + x )y
2
} = 0
(n)
− 3y
(n)
+ (n
If y
(n)
(2) + 0}
+ n(n − 1)y
− n + 2n − 3)y
2
i. e.
(n)
+ 2 ny
(1 + x )y
+ (n
(n)
y
(1) + 0} − 3{y
+ 2n xy
(n+2)
2
2!
(n+1)
(1 + x )y
+ 2xy
n(n−1)
(2x) +
(24) sin (x +
π
2
)
Practice Exercise 237 Leibniz’s theorem (Answers on page 896)
Use the theorem of Leibniz in the following problems:
1. Obtain the nth derivative of: x
2
2. If y
3
nd y
2x
(n)
= x e
y
and hence y
(3)
3. Determine the fourth derivative of: y = 2x
3
4. If y = x
3
cos x
determine the fth derivative
5. Find an expression for y
(4)
6. If y = x
5
2
8. If y = (x
3
52.4
′′
y
if y = e
−t
sin t
nd y
(3)
ln 2x
7. Given 2x
−x
e
′
+ xy
2
+ 3y = 0
2x
+ 2x )e
show that 2x
2
(n+2)
y
(n+1)
+ (4n + 1)xy
2
+ (2n
(n)
− n + 3)y
= 0
determine an expansion for y
(5)
Power series solution by the Leibniz–Maclaurin method
For second-order differential equations that cannot be solved by algebraic methods, the Leibniz–Maclaurin*
method produces a solution in the form of in nite series of powers of the unknown variable. The following
simple ve-step procedure may be used in the Leibniz–Maclaurin method:
(a) Differentiate the given equation n times, using the Leibniz theorem of equation (13),
(b) rearrange the result to obtain the recurrence relation at x = 0,
(c) determine the values of the derivatives at x
= 0
, i.e. nd (y) and (y ) ,
(d) substitute in the Maclaurin expansion for y
= f (x)
′
0
0
(see page 452, equation (5)),
(e) simplify the result where possible and apply boundary conditions, (if given).
The Leibniz–Maclaurin method is demonstrated, using the above procedure, in the following worked
problems.
Problem 5. Determine the power series solution of the differential equation:
method, given the boundary conditions that at x
,
= 0 y = 1
and
dy
dx
2
d y
2
dx
+ x
dy
dx
+ 2y = 0
using the Leibniz–Maclaurin
= 2
Following the above procedure:
(a) The differential equation is rewritten as: y + xy + 2y =
equation (13), each term is differentiated n times, which gives:
′′
(n+2)
y
i.e.
(n+1)
+ {y
(n+2)
y
(b) At x = 0, equation (14) becomes:
′
(n)
(x) + n y
(n+1)
+ xy
0
and from the Leibniz theorem of
(n)
(1) + 0} + 2 y
(n)
+ (n + 2) y
=
0
=
0
Soufyan Jamal
(n+2)
(n)
y
+ (n + 2) y
= 0
from which, y
= − (n + 2) y
This equation is called a recurrence relation or recurrence formula, because each recurring term
depends on a previous term.
(n+2)
(c) Substituting n
For
= 0
(n)
, 1, 2, 3,…will produce a set of relationships between the various coef cients.
,
1, (y
2, (y
= −2(y)
= −3(y )
= −4(y ) =
= 2 × 4(y)
= −5(y ) =
= 3 × 5(y )
′′
n = 0 (y )0
n =
n =
′′′
0
′
)0
(4)
0
′′
)
− 4{−2(y)0 }
0
0
0
,
(5)
n = 3 (y
′′′
)
′
− 5{−3(y )0 }
0
0
′
0
,
(6)
n = 4 (y
= −6(y ) = −6{2 × 4(y) }
= −2 × 4 × 6(y)
= −7(y ) = −7{3 × 5(y ) }
= −3 × 5 × 7(y )
= −8(y ) =
(4)
)
0
0
0
0
,
(7)
n = 5 (y
(5)
)
0
′
0
0
′
0
,
(8)
n = 6 (y
(6)
)
0
0
−8{−2 × 4 × 6(y)0 } = 2 × 4 × 6 × 8(y)0
(d) Maclaurin’s theorem from page 452 may be written as:
y
=
0
+ x(y )
x
4!
(4)
(y
)
0
x
+
0
4
+
2
′
(y)
+
3
′′
(y )
0
2!
+
x
3!
′′′
(y
)
0
⋯
Substituting the above values into Maclaurin’s theorem gives:
y
=
0
+ x(y )
0
3
Soufyan Jamal
+
x
3!
x
5!
x
7!
x
2!
{−2(y) }
0
4
′
0
x
4!
{2 × 4(y) }
0
6
′
{3 × 5(y ) } +
0
7
+
+
{−3(y ) } +
5
+
2
′
(y)
x
{−2 × 4 × 6(y) }
0
6!
′
{−3 × 5 × 7(y ) }
0
8
+
x
8!
{2 × 4 × 6 × 8(y) } + ⋯ .
0
(e) (v) Collecting similar terms together gives:
{
2
y
=
4
2x
(y) {1 −
0
6
1
2×4×6×8x
+
6!
− ⋯
4!
8
2×4×6x
−
2×4x
+
2!
8!
3
5
3x
′
} + (y ) {x −
0
2
+
3!
3×5x
5!
7
3×5×7x
−
+
7!
⋯}
2
i. e. y
=
4
x
(y) {1 −
0
6
x
+
1
x
−
1×3
3×5
8
x
+
−
3×5×7
′
+ (y )
0
3
x
× {
The boundary conditions are that at x
5
x
−
1
1
+ ⋯
2×4×6
=
= 0, y = 1
x
{1 −
2
and
x
+
1
x
+
dy
= 2
dx
4
−
,
x = 0 y = 0
and
dx
= 1
2
x
5
−
2×4
= 1
0
+ x
dy
and (y )
+ 2y = 0
dx
′
0
= 2
is:
6
3×5
⋯} + 2{
x
x
1
−
x
3
1×2
7
+
2×4×6
Problem 6. Determine the power series solution of the differential equation:
dy
d y
dx
x
−
1×3
, i.e. (y)
2
8
3×5×7
+
x
2×4
}
2
Hence, the power series solution of the differential equation:
y
+
1×2
2
1
7
x
−
Soufyan Jamal
⋯}
2
d y
dy
+
2
dx
⋯}
+ xy = 0
dx
given the boundary conditions that at
, using the Leibniz–Maclaurin method.
Following the above procedure:
(i) The differential equation is rewritten as: y + y + xy
(13), each term is differentiated n times, which gives:
′′
(n+2)
y
i.e.
(ii) At x
= 0
(n+1)
+ y
(n+2)
y
′
(n)
+ y
(n+1)
+ y
= 0
and from the Leibniz theorem of equation
(n−1)
(x) + n y
(n)
+ xy
(1) + 0
(n−1)
+ ny
=
0
=
0
, equation (15) becomes:
(n+2)
(n+1)
y
(n−1)
+ y
+ ny
= 0
from which, y
= −{y
+ ny
}
(1)
This is the recurrence relation and applies for n ≥ 1
(iii) Substituting n = 1, 2, 3, … will produce a set of relationships between the various coef cients.
(n+2)
For
,
n = 2,
n = 1
(n+1)
(n−1)
′′′
(y
(4)
(y
= −{(y ) + (y) }
= −{(y ) + 2(y )
′′
)
0
)
0
0
0
′′′
′
0
0
}
n = 3
,
(6)
n = 4 (y
,
(7)
n = 5 (y
,
(8)
n = 6 (y
,
(5)
(y
)
0
= −{(y
)0
= −{(y
)
= −{(y
0
(5)
)
0
(6)
(7)
′′′
+ 4(y
(5)
+ 6(y
′
0
,
= 0 y = 0
′′
′′
, thus (y)
= 0
0
, and at x
= 0
,
dy
dx
= 1
, thus
(4)
(y
0
)
0
(y
)
0
= −{(y
= −{(y
(6)
)
0
(y
)
0
(8)
)
0
,
′′
(y )
′
0
+ (y )
0
+ (0)y = 0
′
+ 2(y ) }
0
0
− [1 + 2(1)] =
)
0
− 3
Soufyan Jamal
′′
+ 3(y ) }
0
− [−3 + 3(−1)] = 6
= −{(y
(5)
)
0
′′′
+ 4(y
) }
0
− [6 + 4(1)] =
= −{(y
(6)
)
0
(4)
+ 5(y
− 10
) }
0
− [−10 + 5(−3)] = 25
= −{(y
(7)
=
− (−1 + 0) = 1
0
)
= −{(y
=
+ (y) } =
0
(4)
=
(7)
)
′′′
=
(y
x = 0
0
=
(5)
, and, at
′′
0
′′
)
+ xy = 0
0
′
0
′
′
0
(y
0
= 1
0
′′′
+ 3(y ) }
) }
From the given differential equation, y + y
from which, (y ) = − (y ) = − 1
Thus, (y) = 0, (y ) = 1, (y ) = − 1,
(y
0
)0 }
From the given boundary conditions, at x
(y )
′′
)
0
)0 + 5(y
0
(4)
0
) }
(4)
)
= −{(y
)
)
0
(5)
+ 6(y
) }
0
− [25 + 6(6)] =
− 61
(iv) Maclaurin’s theorem states:
y
0
+ x(y )
0
4
+
2
′
(y)
=
x
4!
(4)
(y
+
)0 +
x
2!
3
′′
(y )
0
+
x
′′′
(y
3!
)
0
⋯
and substituting the above values into Maclaurin’s theorem gives:
2
y
=
x
5!
{−1} +
2!
5
+
3
x
0 + x(1) +
x
3!
6
{6} +
x
6!
4
x
{1} +
4!
{−3}
7
{−10} +
x
7!
{25}
8
+
x
8!
{−61} + ⋯
(v) Simplifying, the power series solution of the differential equation:
y
=
x −
+
Now try the following Practice Exercise
x
25x
7!
2
2!
+
7
−
x
3
3!
61x
8!
−
3x
4
4!
+
8
+
⋯
6x
2
d y
2
dx
5
5!
−
+
10x
6!
6
dy
dx
+ xy = 0
is given by:
Practice Exercise 238 Power series solutions by the Leibniz-Maclaurin method
(Answers on page 896)
1. Determine the power series solution of the differential equation:
that at x
,
= 0 y = 1
and
dy
dx
2
d y
dy
dx
+ y = 0
using the Leibniz–Maclaurin method, given
= 2
2. Show that the power series solution of the differential equation:
method, is given by: y
+ 2x
2
dx
2
= 1 + x
−x
+ e
d y
dx
2
+ (x − 1)
given the boundary conditions that at x
3. Find the particular solution of the differential equation: (x
2
the boundary conditions that at x
2
(x + 1)
,
= 0 y = 1
and
dy
2
+ 1)
d y
dx
2
+ x
dy
dx
dy
dx
,
− 2y = 0
= 0 y = 2and
− 4y = 0
dy
dx
, using the Leibniz–Maclaurin
=
using the Leibniz–Maclaurin method, given
= 1
dx
4. Use the Leibniz–Maclaurin method to determine the power series solution for the differential equation: x
that at x
52.5
,
= 0 y = 1
and
dy
dx
− 1
2
d y
2
dx
+
dy
dx
+ xy = 0
given
= 2
Power series solution by the Frobenius method
A differential equation of the form y + P y + Qy = 0, where P and Q are both functions of x, such that
the equation can be represented by a power series, may be solved by the Frobenius method.*
′′
′
The following four-step procedure may be used in the Frobenius method:
(i) Assume a trial solution of the form y = x {a + a x + a x + a x + ⋯ + a x + ⋯}
(ii) differentiate the trial series,
(iii) substitute the results in the given differential equation,
(iv) equate coef cients of corresponding powers of the variable on each side of the equation; this enables
index c and coef cients a1, a2, a3, … from the trial solution, to be determined.
c
2
0
1
3
2
r
3
r
This introductory treatment of the Frobenius method covering the simplest cases is demonstrated, using the
above procedure, in the following worked problems.
Problem 7. Determine, using the Frobenius method, the general power series solution of the differential equation:
2
3x
d y
2
dx
+
dy
dx
− y = 0
The differential equation may be rewritten as:
′′
3xy
′
+ y
− y = 0
(i) Let a trial solution be of the form
c
y
2
3
= x {a0 + a1 x + a2 x
+ a3 x
+ ⋯
r
+ ar x
where a
0
≠ 0
+ ⋯}
(16)
,
i.e.
c
y
c+1
= a0 x
+ a1 x
c+2
+ a2 x
c+3
+ a3 x
c+r
+ ⋯ + ar x
+ ⋯
(17)
(ii) Differentiating equation (17) gives:
′
y
c−1
=
a0 cx
c
+ a1 (c + 1)x
c+1
+ a2 (c + 2)x
+ ar (c + r)x
′′
and
y
c−2
=
a0 c(c − 1)x
Soufyan Jamal
+ ⋯
c+r−1
+ ⋯
c−1
+ a1 c(c + 1)x
c
+ a2 (c + 1)(c + 2)x
+ ⋯
c+r−2
+ ar (c + r − 1)(c + r)x
(iii) Substituting y, y′ and y′′ into each term of the given equation 3xy
′′
c−1
′′
= 3a0 c(c − 1)x
3xy
+
⋯
′
+ y − y = 0
gives:
c
+ 3a1 c(c + 1)x
c+1
+ 3a2 (c + 1)(c + 2)x
+ ⋯
c+r−1
+ 3ar (c + r − 1)(c + r)x
′
y
c−1
= a0 cx
c
+ a1 (c + 1)x
+ ⋯
(a)
c+1
+ a2 (c + 2)x
c+r−1
+ ⋯ + ar (c + r)x
+ ⋯
(b)
−y
c
= −a0 x
c+1
− a1 x
c+2
− a2 x
c+3
− a3 x
c+r
− ⋯ − ar x
− ⋯
(c)
(iv) The sum of these three terms forms the left-hand side of the equation. Since the right-hand side is
zero, the coef cients of each power of x can be equated to zero.
For example, the coef cient of x
is equated to zero giving: 3a
c−1
or
0 c(c
− 1) + a0 c = 0
a0 c[3c − 3 + 1] = a0 c(3c − 2) = 0
The coef cient of xc is equated to zero giving: 3a
1 c(c
2
i.e.
a1 (3c
+ 1) + a1 (c + 1) − a0 = 0
+ 3c + c + 1) − a0
2
= a1 (3c
or
(18)
+ 4c + 1) − a0 = 0
a1 (3c + 1)(c + 1) − a0 = 0
(19)
In each of series (a), (b) and (c) an xc term is involved, after which, a general relationship can be
obtained for x , where r ≥ 0
c+r
In series (a) and (b), terms in x
are present; replacing r by (r
terms in x , which occurs in all three equa-tions, i.e.
in series (a), 3a (c + r)(c + r + 1)x
c+r−1
+ 1)
will give the corresponding
c+r
c+r
r+1
in series (b),
in series (c),
Soufyan Jamal
c+r
ar+1 (c + r + 1)x
c+r
−ar x
Equating the total coef cients of x
c+r
to zero gives:
3ar+1 (c + r)(c + r + 1) + ar+1 (c + r + 1)
− ar = 0
which simpli es to:
ar+1 {(c + r + 1)(3c + 3r + 1)} − ar = 0
(20)
Equation (18), which was formed from the coef cients of the lowest power of x, i.e. x , is called
the indicial equation, from which the value of c is obtained. From equation (18), since a ≠ 0, then
c = 0 or c =
c−1
0
2
3
(a) When c
= 0
From equation (19), if c
= 0 a1 (1 × 1) − a0 = 0
From equation (20), if c
= 0 ar+1 (r + 1)(3r + 1) − ar = 0
Thus, when r
= 1
, a2
,
, i.e. a
1
= a0
,
=
a1
(2×4)
, i.e. a
r+1
=
since a
1
a0
(2×4)
= a0
=
ar
(r + 1)(3r + 1)
r ≥ 0
when r
= 2
, a3
=
a2
or
when r
= 3
, a4
a0
=
(3×7)
(2×4)(3×7)
a0
(2×3)(4×7)
=
(4×10)
=
(2×3×4)(4×7×10)
a3
a0
Soufyan Jamal
and so on.
From equation (16), the trial solution was:
y
Substituting c
= 0
c
=
2
x {a0 + a1 x + a2 x
3
r
+ a3 x
+ ⋯ + ar x
+ ⋯}
and the above values of a1, a2, a3,… into the trial solution gives:
y
a0
0
=
x {a0 + a0 x + (
+(
+(
a0
2
(2×4)
)x
3
(2×3)(4×7)
)x
a0
4
(2×3×4)(4×7×10)
)x
+ ⋯}
2
i.e.
y
=
3
x
a0 {1 + x +
+
(2×4)
x
(2×3)(4×7)
4
+
(b) When c
=
3
=
From equation (20), if c
=
i.e. a
i.e. a
2
r+1 (r
r+1
2
3
,a
1 (3)(
5
3
) − a0 = 0
5
+
=
3
2
= 1
a0
=
5
,
+ 8r + 5) − ar = 0
, a2
=
a1
when r
= 2
since a
=
= 3
, a3
a0
=
(2×8)
1
when r
,
r ≥ 0
(r + 1)(3r + 5)
Thus, when r
1
3
)(3r + 3) − ar = ar+1 (3r
ar
, i.e. a
2
+ r + 1)(2 + 3r + 1) − ar = 0
3
+ ⋯}
2
From equation (19), if c
ar+1 (
x
(2×3×4)(4×7×10)
(2×5×8)
=
a0
5
a2
(3×11)
, a4
=
(2×3)(5×8×11)
=
(4×14)
=
(2×3×4)(5×8×11×14)
a0
a3
a0
and so on.
From equation (16), the trial solution was:
y
Substituting c
=
2
3
=
c
2
x {a0 + a1 x + a2 x
3
+ a3 x
r
+ ⋯ + ar x
+ ⋯}
and the above values of a1, a2, a3, … into the trial solution gives:
2
y
=
x
{a0 + (
3
2
{1 +
3
2
2×5×8
)x
3
)x
(2×3)(5×8×11)
a0
+(
= a0 x
a0
)x + (
5
a0
+(
i.e.y
a0
4
(2×3×4)(5×8×11×14)
)x
+ ⋯}
2
x
x
+
5
(2×5×8)
3
+
x
(2×3)(5×8×11)
4
+
x
(2×3×4)(5×8×11×14)
+
⋯}
(22)
Since a0 is an arbitrary (non-zero) constant in each solution, its value could well be different.
Let a = A in equation (21), and a = B in equation (22). Also, if the rst solution is denoted by u(x)
and the second by v(x), then the general solution of the given differential equation is y = u(x) + v(x).
Hence,
0
0
y
=
x
A{1 + x +
+
Soufyan Jamal
x
+
2
x
{1 +
3
x
x
3
(2 × 3)(4 × 7)
4
+
(2 × 3 × 4)(4 × 7 × 10)
+Bx
+
2
(2 × 4)
x
+
5
⋯}
2
(2 × 5 × 8)
3
(2 × 3)(5 × 8 × 11)
x
+
4
+
(2 × 3 × 4)(5 × 8 × 11 × 14)
⋯}
Problem 8. Use the Frobenius method to determine the general power series solution of the differential equation:
2
2
d y
2x
2
dy
− x
+ (1 − x)y = 0
dx
dx
The differential equation may be rewritten as: 2x
2
′′
y
′
− xy
+ (1 − x)y = 0
(i) Let a trial solution be of the form
y
=
c
2
x {a0 + a1 x + a2 x
r
+ ar x
3
+ a3 x
+ ⋯
+ ⋯}
where a0 ≠ 0,
i.e.
y
c
= a0 x
c+1
+ a1 x
c+2
+ a2 x
c+3
+ a3 x
c+r
+ ⋯ + ar x
(ii) Differentiating equation (24) gives:
+ ⋯
(24)
′
y
=
c−1
a0 cx
c
+ a1 (c + 1)x
c+1
+ a2 (c + 2)x
c+r−1
+ ⋯ + ar (c + r)x
′′
and y
c−2
a0 c(c − 1)x
=
+ ⋯
c−1
+ a1 c(c + 1)x
c
+ a2 (c + 1)(c + 2)x
+ ⋯
c+r−2
+ ar (c + r − 1)(c + r)x
(iii) Substituting y, y′ and y′′ into each term of the given equation 2x
2
2
′′
2x y
c
=
2a0 c(c − 1)x
′′
y
′
− xy
+ ⋯
c+r
+ 2ar (c + r − 1)(c + r)x
c
= −a0 cx
+ ⋯
c+1
− a1 (c + 1)x
Soufyan Jamal
c+2
− a2 (c + 2)x
c+r
− ar (c + r)x
(1 − x)y
gives:
c+1
c+2
′
+ (1 − x)y = 0
+ 2a1 c(c + 1)x
+ 2a2 (c + 1)(c + 2)x
−xy
+ ⋯
=
c
c+3
+ a3 x
=
c+1
(1 − x)(a0 x
c
a0 x
+ a1 x
c+1
c+4
− a3 x
c+2
− a1 x
c+2
+ a2 x
c+r
+ ⋯ + ar x
− a0 x
c+2
c+r
+ a1 x
(b)
+ a2 x
+ ⋯ + ar x
c+1
− ⋯
− ⋯
+ ⋯)
c+3
+ a3 x
+ ⋯
c+3
− a2 x
c+r+1
− ⋯ − ar x
− ⋯
(c)
(iv) The indicial equation, which is obtained by equating the coef cient of the lowest power of x to zero,
gives the value(s) of c. Equating the total coef cients of xc (from equations (a) to (c)) to zero gives:
=0
[2c(c − 1) − c + 1] = 0
a [2c − 2c − c + 1] = 0
a [2c − 3c + 1] = 0
a [(2c − 1)(c − 1)] = 0
2a0 c(c − 1) − a0 c + a0
i.e.
i.e.
i.e.
i.e.
a0
2
0
2
0
0
from which,
c = 1 or c =
Equating the coef cient of the general term, i.e. x
1
2
c+r
, (from equations (a) to (c)) to zero gives:
2ar (c + r − 1)(c + r) − ar (c + r)
+ ar − ar−1 = 0
from which,
ar [2(c + r − 1)(c + r) − (c + r) + 1] = ar−1
and ar =
ar−1
2(c+r−1)(c+r)−(c+r)+1
(25)
(a) When c = 1, ar
ar−1
=
2(r)(1+r)−(1+r)+1
ar−1
=
2
2r+2r −1−r+1
ar−1
=
Thus, when r
= 1
ar−1
=
2
2r +r
r (2r+1)
,
a0
a1 =
a0
=
1(2+1)
1×3
when r = 2,
a2
a1
=
a1
=
2(4+1)
(2×5)
a0
=
a0
or
(1×3)(2×5)
(1×2)×(3×5)
when r = 3,
a3
a2
=
a2
=
3(6+1)
(3×7)
a0
=
(1×2×3)×(3×5×7)
Soufyan Jamal
when r = 4,
a4
=
=
a3
a3
=
4(8+1)
4×9
a0
(1×2×3×4)×(3×5×7×9)
and so on.
From equation (23), the trial solution was:
y
=
c
2
x {a0 + a1 x + a2 x
r
+ ar x
Substituting c
= 1
+ ⋯
+ ⋯}
and the above values of a1, a2, a3,…into the trial solution gives:
y
=
1
x {a0 +
+
+
+
i.e. y
3
+ a3 x
1
= a0 x {1 +
x
(1×3)
a0
(1×3)
x +
a0
a0
x
a0
4
(1×2×3×4)×(3×5×7×9)
1
x
3
(1×2×3)×(3×5×7)
⋯
2
(1×2)×(3×5)
x
}
2
2
+
x
(1×2)×(3×5)
3
+
x
(1×2×3)×(3×5×7)
4
+
+
x
(1×2×3×4)×(3×5×7×9)
⋯
1
2
}
(26)
1
(b) When c =
2
ar
ar−1
=
2(c+r−1)(c+r)−(c+r)+1
f rom equation (25)
i.e. ar
when r
when r
= 2
= 3
= 4
1
+r−1)(
1
2(r−
2
1
+r)−(
+r)+1
2
2
)(r+
1
2
)−
1
−r+1
2
ar−1
2
2(r −
1
4
)−
1
2
ar−1
1
2
2r −
2
−
1
2
−r+1
−r+1
=
ar−1
2
2r −r
ar−1
=
when r
2
ar−1
=
=
= 1
1
2(
=
Soufyan Jamal
Thus, when r
ar−1
=
r(2r − 1)
, a1
, a2
, a3
, a4
=
1(2−1)
=
2(4−1)
=
(2×3)
=
3(6−1)
=
(2×3)×(3×5)
=
4(8−1)
=
(2×3×4)×(3×5×7)
a0
=
a1
=
a0
1×1
a1
(2×3)
a0
a2
=
a2
3×5
a0
a3
=
a3
4×7
a0
and so on.
From equation (23), the trial solution was:
y
c
=
2
x {a0 + a1 x + a2 x
r
+ ar x
Substituting c
=
1
2
3
+ a3 x
+ ⋯
+ ⋯}
and the above values of a1, a2, a3,…into the trial solution gives:
1
y
=
x
2
+
= a0 x
2
(2×3)
a0
2
x
4
(2×3×4)×(3×5×7)
1
i.e.y
a0
{a0 + a0 x +
x
+
+
a0
3
(2×3)×(3×5)
x
⋯}
2
{1 + x +
x
(2×3)
3
+
x
(2×3)×(3×5)
4
+
+
x
(2×3×4)×(3×5×7)
⋯
1
2
}
(27)
Since a0 is an arbitrary (non-zero) constant in each solution, its value could well be different.
Let a = A in equation (26), and a = B in equation (27). Also, if the rst solution is denoted by
u(x) and the second by v(x), then the general solution of the given differential equation is
y = u(x) + v(x),
0
0
{
i.e. y
=
x
Ax{1 +
+
+
Soufyan Jamal
1
2
+
+
(1 × 3)
x
x
+
2
(1 × 2) × (3 × 5)
3
(1 × 2 × 3) × (3 × 5 × 7)
x
4
(1 × 2 × 3 × 4) ×(3 × 5 × 7 × 9)
1
+
⋯} + B x
x
{1 + x +
2
x
2
(2 × 3)
3
(2 × 3) × (3 × 5)
x
4
+
(2 × 3 × 4) ×(3 × 5 × 7)
⋯}
Problem 9. Use the Frobenius method to determine the general power series solution of the differential equation:
2
d y
dx
− 2y = 0
2
The differential equation may be rewritten as:
′′
y
− 2y = 0
(i) Let a trial solution be of the form
y
c
2
3
= x {a0 + a1 x + a2 x
+ a3 x
+ ⋯
r
+ ar x
where a
≠ 0
0
+ ⋯}
(28)
,
i.e.y
c
= a0 x
c+1
+ a1 x
c+2
c+3
+ a2 x
+ a3 x
c+r
+ ⋯ + ar x
+ ⋯
(29)
(ii) Differentiating equation (29) gives:
′
c−1
=
y
a0 cx
c
+ a1 (c + 1)x
c+1
+ a2 (c + 2)x
c+r−1
+ ⋯ + ar (c + r)x
′′
andy
=
c−2
a0 c(c − 1)x
+ ⋯
c−1
+ a1 c(c + 1)x
c
+ a2 (c + 1)(c + 2)x
+ ⋯
c+r−2
+ ar (c + r − 1)(c + r)x
(iii) Replacing r by (r
+ 2)
in a
c+r−2
r (c
+ r − 1)(c + r) x
+ ⋯
gives:
c+r
ar+2 (c + r + 1)(c + r + 2)x
Substituting y and y′′ into each term of the given equation y
′′
′′
y
− 2y
c−2
= a0 c(c − 1)x
− 2y = 0
gives:
c−1
+ a1 c(c + 1)x
c
+ [a2 (c + 1)(c + 2) − 2a0 ]x
+ ⋯
+ [ar+2 (c + r + 1)(c + r + 2)
c+r
− 2ar ] x
+ ⋯ = 0
(30)
(iv) The indicial equation is obtained by equating the coef cient of the lowest power of x to zero.
Hence,
from which, c
a0 c(c − 1) = 0
For the term in x
c−1
, i.e. a
1 c(c
= 0
or c
since a
= 1
0
≠ 0
+ 1) = 0
With c = 1, a = 0; however, when c = 0, a is indeterminate, since any value of a1 combined
with the zero value of c would make the product zero.
1
1
For the term in xc,
a2 (c + 1)(c + 2) − 2a0 = 0 f rom which,
2a0
a2 =
For the term in x
c+r
(c+1)(c+2)
(31)
,
ar+2 (c + r + 1)(c + r + 2) − 2ar = 0
from which,
ar+2 =
(a) When c
= 0 :
a1
r+2
when r
when r
=
,
= 1 a3 =
,
= 2 a4 =
(32)
is indeterminate, and from equation (31)
a2 =
In general, a
2ar
(c+r+1)(c+r+2)
(2 × 3)
2a2
(3×4)
2a1
=
=
2a0
=
2!
and
2ar
(r + 1)(r + 2)
2a1
2a0
(1×2)
(1 × 2 × 3)
=
Soufyan Jamal
2a1
3!
4a0
4!
Hence, y
=
2a0
0
x {a0 + a1 x +
+
4a0
4!
4
x
+
2
x
2!
2a1
+
3!
3
x
⋯ }
f romequation(28)
2
=
a0 {1 +
4
2x
4x
+
2!
3
5
2x
+a1 {x +
+ ⋯ }
4!
+
3!
4x
5!
+ ⋯ }
Since a0 and a1 are arbitrary constants depending on boundary conditions, let
then:
y
=
P {1 +
2x
2
2!
+ Q{x +
+
2x
4x
+
and a
1
= Q,
4
4!
3
3!
a0 = P
+
4x
⋯}
5
5!
+
⋯}
(33)
(b) When c
= 1 :
a1 = 0
, and from equation (31),
a2 =
Since c
,
2ar
= 1 ar+2 =
(c+r+1)(c+r+2)
from equation (32) and when r
= 1
= 2
= 3
3!
(r+2)(r+3)
,
2a1
(3×4)
= 0 since a1 = 0
,
2a2
a4 =
when r
2a0
=
2ar
=
a3 =
when r
2a0
(2×3)
(4×5)
=
2
(4×5)
2a0
×
=
3!
4a0
5!
,
a5 =
Hence, when c
2a3
(5×6)
Soufyan Jamal
= 0
= 1,
1
y = x {a0 +
2a0
3!
2
x
+
4a0
5!
4
x
+ ⋯
1
2
}
from equation (28)
i.e.
3
y = a0 {x +
2x
3!
5
+
4x
5!
+
…}
Again, a0 is an arbitrary constant; let a
0
then
= K
,
y = K{x +
2x
3
3!
+
4x
5
5!
+
⋯}
However, this latter solution is not a separate solution, for it is the same form as the second series in
equation (33). Hence, equation (33) with its two arbitrary constants P and Q gives the general solution. This
is always the case when the two values of c differ by an integer (i.e. whole number). From the above three
worked problems, the following can be deduced, and in future assumed:
(i) if two solutions of the indicial equation differ by a quantity not an integer, then two independent
solutions y = u(x) + v(x) result, the general solution of which is y = Au + Bv (note: Problem 7
had c = 0 and and Problem 8 had c = 1 and ; in neither case did c differ by an integer)
2
1
3
2
(ii) if two solutions of the indicial equation do differ by an integer, as in Problem 9 where c = 0 and 1,
and if one coef cient is indeterminate, as with when c = 0, then the complete solution is always
given by using this value of c. Using the second value of c, i.e. c = 1 in Problem 9, always gives a
series which is one of the series in the rst solution.
Now try the following Practice Exercise
Practice Exercise 239 Power series solutions by the Frobenius method (Answers on
page 896)
1. Produce, using the Frobenius method, a power series solution for the differential equation:
2
d y
2x
+
2
dx
dy
2. Use the Frobenius method to determine the general power series solution of the differential equation:
3. Determine the power series solution of the differential equation: 3x
2
d y
dx
dy
+ 4
2
4. Show, using the Frobenius method, that the power series solution of the differential equation:
y = P
52.6
cosh x
+ Q
2
d y
2
dx
+ y = 0
using the Frobenius method.
− y = 0
dx
− y = 0
dx
2
d y
dx
2
may be expressed as
− y = 0
sinh x, where P and Q are constants. [Hint: check the series expansions for cosh x and sinh x on page 159].
Bessel’s equation and Bessel’s functions
One of the most important differential equations in applied mathematics is Bessel’s4 equation and is of the
form:
2
2 d y
x
2
dx
+ x
dy
2
+ (x
dx
2
− v )y = 0
where v is a real constant. The equation, which has applications in electric elds, vibrations and heat
conduction, may be solved using Frobenius’ method of the previous section.
Problem 10. Determine the general power series solution of Bessel’s equation.
2
Bessel’s equation x
+ x
+ (x − v )y = 0 may be rewritten as: x
Using the Frobenius method from page 586:
2 d y
dy
2
dx
dx
2
2
2
′′
y
′
2
+ xy + (x
2
− v )y = 0
(i) Let a trial solution be of the form
y
c
=
2
x {a0 + a1 x + a2 x
r
+ ar x
where a
0
≠ 0
3
+ a3 x
Soufyan Jamal
+ ⋯
+ ⋯}
,
i.e.
y
c
= a0 x
c+1
+ a1 x
c+2
+ a2 x
c+3
+ a3 x
c+r
+ ⋯ + ar x
(ii) Differentiating equation (35) gives:
′
y
=
c−1
c
a0 cx
+ a1 (c + 1)x
c+1
+ a2 (c + 2)x
+ ⋯
c+r−1
+ ar (c + r)x
′′
andy
=
c−2
a0 c(c − 1)x
+ ⋯
c−1
+ a1 c(c + 1)x
c
+ a2 (c + 1)(c + 2)x
+ ⋯
c+r−2
+ ar (c + r − 1)(c + r)x
+ ⋯
+ ⋯
(35)
(iii) Substituting y, y′ and y′′ into each term of the given equation: x
2
c
a0 c(c − 1)x
c+r
c+1
+ a1 (c + 1)x
c+r
+ ar (c + r)x
2
gives:
c
+ ⋯ + a0 cx
c+2
+ a2 (c + 2)x
c+2
+ ⋯ + a0 x
c+r+2
+ ⋯ + ar x
c+1
− a1 v x
2
− v )y = 0
+ ⋯
+ ar (c + r − 1)(c + r)x
c+4
2
c+1
c+2
+ a2 x
′
+ xy + (x
+ a1 c(c + 1)x
+ a2 (c + 1)(c + 2)x
Soufyan Jamal
′′
y
2
c+3
+ a1 x
2
c
+ ⋯ − a0 v x
c+r
− ⋯ − ar v x
+ ⋯
− ⋯ = 0
(iv) The indicial equation is obtained by equating the coef cient of the lowest power of x to zero.
Hence,
2
a0 c(c − 1) + a0 c − a0 v
from which, a
2
0 [c
i.e. a [c − v
from which,
2
0
2
= 0
2
− c + c − v ] = 0
] = 0
c =
For the term in x
c+r
+ v
or c
=
− v
since a
0
≠ 0
,
(
)(
)
(
)
ar (c + r − 1)(c + r) + ar (c + r) + ar−2
2
− ar v
= 0
2
ar [(c + r − 1)(c + r) + (c + r) − v ] =
i.e.
2
ar [(c + r)(c + r − 1 + 1) − v ] =
i.e.
a [(c + r)
− v ] =
i.e. the recurrence relation is:
2
2
ar−2
ar =
c+1
− ar−2
− ar−2
r
For the term in x
− ar−2
f or
2
2
v
− (c + r)
r ≥ 2
(37)
,
2
a1 [c(c + 1) + (c + 1) − v ]
=0
i.e.
a1 [(c + 1)
but if c = v
2
a1 [(v + 1)
i.e.
Similarly, if c = −v a
1 [1
2
2
− v ]
− v ]
a1 [2v + 1]
− 2v]
2
=0
=0
=0
=0
The terms (2v + 1) and (1 − 2v) cannot both be zero since v is a real constant, hence a
Since a = 0, then from equation (37) a = a = a = … = 0 and
1
1
3
Soufyan Jamal
5
7
whenr
=
2, a2 =
whenr
=
4, a4
whenr
=
a6
=
a0
2
2
v −(c+2)
a0
2
2
2
2
[v −(c+2) ][v −(c+4) ]
6,
a0
2
2
2
2
2
2
[v −(c+2) ][v −(c + 4) ][v −(c + 6) ]
and so on.
When c
=
+ v
,
a2
=
=
a4
a0
a0
=
2
2
2
−a0
4+4v
=
2
v −v −4v−4
v −(v+2)
−a0
2
2 (v+1)
a0
=
2
2
2
2
[v −(v+2) ][v −(v+4) ]
=
=
=
a6
=
=
=
=
a0
2
3
[−2 (v+1)][−2 (v+2)]
a0
5
2 (v+1)(v+2)
a0
4
2 ×2(v+1)(v+2)
a0
2
2
2
2
2
2
[v −(v+2) ][v −(v+4) ][v −(v+6) ]
a0
4
[2 ×2(v+1)(v+2)][−12(v+3)]
−a0
4
2
2 ×2(v+1)(v+2)×2 ×3(v+3)
−a0
6
2 ×3!(v+1)(v+2)(v+3)
and so on.
= 0
The resulting solution for c
=
+ v
y
is given by:
=
u =
2
v
4
x
A x {1 −
+
2
2 (v+1)
x
4
2 ×2!(v+1)(v+2)
6
x
−
6
2 ×3!(v+1)(v+2)(v+3)
+ ⋯
1
2
}
which is valid provided v is not a negative integer and where A is an arbitrary constant.
When c = − v,
a2 =
a0
2
2
v −(−v+2)
=
a0
4v−4
a4 =
=
=
=
Soufyan Jamal
a0
2
2
v −(v −4v+4)
a0
=
2
2 (v−1)
a0
2
2
2
[2 (v−1)][v −(−v+4) ]
a0
2
3
[2 (v−1)][2 (v−2)]
a0
4
2 ×2(v−1)(v−2)
Similarly,
a0
a6 =
6
2 ×3!(v−1)(v−2)(v−3)
Hence,
y
=
w =
−v
Bx
2
{1 +
4
x
2
2 (v−1)
x
+
4
2 ×2!(v−1)(v−2)
6
x
+
6
2 ×3!(v−1)(v−2)(v−3)
+ ⋯}
which is valid provided v is not a positive integer and where B is an arbitrary constant.
The complete solution of Bessel’s equation:
2
2 d y
x
2
dx
+ x
dy
dx
2
+ (x
2
− v )y = 0
y
=
is:
u + w =
x
v
A x {1 −
−
x
6
2
−v
+
+
{1 +
x
x
+
⋯}
+
⋯}
2
2
4
× 2!(v − 1)(v − 2)
6
× 3!(v − 1)(v − 2)(v − 3)
x
2
4
× 2!(v + 1)(v + 2)
2 (v − 1)
4
2
x
4
2
6
× 3!(v + 1)(v + 2)(v + 3)
+Bx
+
2
2
2 (v + 1)
6
(39)
The gamma function
The solution of the Bessel equation of Problem 10 may be expressed in terms of gamma functions. Γ is the
upper case Greek letter gamma, and the gamma function Γ (x) is de ned by the integral
∞
∞
x−1
Γ (x) = ∫
t
−t
e
dt
0
and is convergent for x
(40)
> 0
∞
Fromequation(40),
x
Γ (x+1)= ∫
-t
t e
dt
0
and by using integration by parts (see page 491):
Γ (x + 1)
x
=
[(t )(
e
Soufyan Jamal
)]
−1
∞
−∫
∞
−t
0
−t
e
x−1
(
)x t
dt
−1
0
∞
=
−t
(0 − 0) + x ∫
e
x−1
t
dt
0
=
xΓ (x)
f romequation(40)
This is an important recurrence relation for gamma functions.
Thus, since
Γ (x + 1) = xΓ (x)
then similarly,
Γ (x + 2)
= (x + 1)Γ (x + 1)
= (x + 1)xΓ (x)
and
Γ (x + 3)
= (x + 2)Γ (x + 2)
= (x + 2)(x + 1)xΓ (x),
and so on.
(41)
These relationships involving gamma functions are used with Bessel functions.
Bessel functions
The power series solution of the Bessel equation may be written in terms of gamma functions as shown in
Problem 11 below.
Problem 11. Show that the power series solution of the Bessel equation of Problem 10 may be written in terms of the Bessel functions
Jv (x)
and J
−v (x)
as:
AJv (x) + BJ−v (x)
= (
x
2
v
) {
2
1
−
Γ (v+1)
x
2
2 (1!)Γ (v+2)
4
+
+ (
x
2
−v
)
x
4
2 (2!)Γ (v+4)
{
1
Γ (1−v)
− ⋯}
2
−
x
2
2 (1!)Γ (2−v)
4
+
x
4
2 (2!)Γ (3−v)
− ⋯}
From Problem 10 above, when c
If we let a
0
,
=
−a0
+ v a2 =
2
2 (v+1)
1
=
v
2 Γ (v+1)
then
a2
=
=
Similarly,
a4 =
a2
2
2
v −(c+4)
−1
2
=
v
2 (v+1) 2 Γ (v+1)
−1
v+2
2
−1
v+2
2
(v+1)Γ (v+1)
f romequation(41)
Γ (v+2)
from equation (37)
=
a2
a2
=
(v−c−4)(v+c+4)
−4(2v+4)
since c = v
−a2
=
−1
=
3
2 (v+2)
Soufyan Jamal
−1
3
v+2
2 (v+2)
2
Γ (v+2)
1
=
v+4
2
(2!)Γ (v+3)
since (v + 2)Γ (v + 2) = Γ (v + 3)
and
a6 =
−1
v+6
2
(3!)Γ (v + 4)
and so on. The recurrence relation is:
r/2
(−1)
ar =
v+r
2
(
r
2
!)Γ (v+
r
+1)
2
and if we let r = 2k, then
k
(−1)
a2k =
v+2k
2
(k!)Γ(v + k + 1)
(42)
for k = 1, 2, 3, …
Hence, it is possible to write the new form for equation (38) as:
y
=
v
2
1
Ax {
v
2 Γ (v+1)
−
x
v+2
2
(1!)Γ (v+2)
4
x
+
v+4
2
− ⋯}
(2!)Γ (v+3)
This is called the Bessel function of the rst-order kind, of order v, and is denoted by J
,
v (x)
i.e.
Jv (x)
=
v
x
(
) {
2
1
−
Γ(v + 1)
+
x
x
2
2
2 (1!)Γ(v + 2)
4
4
2 (2!)Γ(v + 3)
−
⋯}
provided v is not a negative integer.
For the second solution, when c
=
− v
, replacing v by −v in equation (42) above gives:
k
(−1)
a2k =
from which, when k
2k−v
2
(k!)Γ (k−v+1)
0
= 0, a0 =
(−1)
−v
2
(0!)Γ (1−v)
=
1
−v
2
Γ (1−v)
since 0!
= 1
(see page 580)
a2
when k = 1,
=
1
(−1)
2−v
(1!)Γ (1−v+1)
2−v
(1!)Γ (2−v)
4−v
(2!)Γ (2−v+1)
4−v
(2!)Γ (3−v)
2
=
a4
when k = 2,
−1
2
=
2
(−1)
2
=
a6
when k = 3,
1
2
=
3
(−1)
6−v
(3!)Γ (3−v+1)
6−v
(3!)Γ (4−v)
2
=
Hence, y
and so on.
Soufyan Jamal
−1
2
−v
=
Bx
{
2
1
−v
2
x
−
Γ (1−v)
2−v
2
(1!)Γ (2−v)
4
x
+
i.e.
J−v (x)
=
(
x
2
−v
)
4−v
2
{
1
x
−
Γ(1 − v)
x
+
− ⋯}
(2!)Γ (3−v)
2
2
2 (1!)Γ(2 − v)
4
− ⋯}
4
2 (2!)Γ(3 − v)
provided v is not a positive integer.
Jv (x)
and J
−v (x)
are two independent solutions of the Bessel equation; the complete solution is:
y = AJv (x) + BJ−v (x)
where A and B are constants
i.e.
y = AJv (x) + BJ−v (x)
= A(
v
x
) {
2
1
x
+
+ B(
x
2
−v
)
x
−
Γ(v + 1)
4
− ⋯}
4
2 (2!)Γ(v + 3)
{
1
−
Γ(1 − v)
+
2
2
2 (1!)Γ(v + 2)
x
x
2
2
2 (1!)Γ(2 − v)
4
4
2 (2!)Γ(3 − v)
− ⋯}
k
∞
In
general
terms:
Jv (x) = (
x
2
v
)
2k
2
k=0
∞
J−v (x) = (
x
2
−v
)
∑
k=0
k
(k!)Γ (v + k + 1)
2k
(−1) x
2k
2
(k!)Γ (k − v + 1)
Another Bessel function
It may be shown that another series for J
n (x)
Jn (x)
=
(
is given by:
x
2
n
) {
1
n!
+
−
1
(n+1)!
(
1
(2!)(n+2)!
x
2
(
2
)
x
2
2k
(−1) x
∑
4
)
−
⋯}
and
Figure 52.1
Soufyan Jamal
From this series two commonly used functions are derived,
i. e.
J0 (x)
=
1
1
−
1
−
(0!)
(
2
(3!)
6
x
)
2
x
= 1 −
2
x
(
2
(1!)
)
2
x
+
2
=
x
2
{
1
=
x
2
−
(1!)(2!)
1
+
(2!)(3!)
x
−
(
x
2
2
(
4
)
x
2
2
4
)
x
−
6
6
2
+
⋯
2 (3!)
2
)
− ⋯}
3
3
2 (1!)(2!)
x
x
2 (2!)
1
−
(1!)
(
4
4
2 (1!)
and J1 (x)
2
(2!)
+ ⋯
2
2
1
+
x
+
5
5
2 (2!)(3!)
7
+
7
2 (3!)(4!)
⋯
Tables of Bessel functions are available for a range of values of n and x, and in these, J (x) and J (x) are
most commonly used.
Graphs of J (x), which looks similar to a cosine, and J (x), which looks similar to a sine, are shown in
Figure 52.1.
0
0
1
1
Now try the following Practice Exercise
Practice Exercise 240 Bessel’s equation and Bessel’s functions (Answers on page
896)
1. Determine the power series solution of Bes- sel’s equation: x
2
2 d y
+ x
2
dx
dy
dx
2
+ (x
2
− v )y = 0
when v
= 2
, up to and including the
6
term in x .
2. Find the power series solution of the Bessel function:
v = 3
′′
′
+ xy
2
+ (x
2
− v )y = 0
in terms of the Bessel function
7
. Give the answer up to and including the term in x .
3. Evaluate the Bessel functions J
0 (x)
52.7
2
x y
and J
1 (x)
when x
= 1
, correct to 3 decimal places.
Legendre’s equation and Legendre polynomials
J3 (x)
when
Another important differential equation in physics and engineering applications is Legendre’s* equation of
the form: (1 − x
constant.
2
2
)
d y
2
dx
− 2x
dy
dx
+ k(k + 1)y = 0
or (1 − x
2
′′
)y
′
− 2xy + k(k + 1)y = 0
where k is a real
Problem 12. Determine the general power series solution of Legendre’s equation.
To solve Legendre’s equation (1 − x
2
′′
)y
′
− 2xy
+ k(k + 1)y = 0
using the Frobenius method:
(i) Let a trial solution be of the form
c
y
2
= x {a0 + a1 x + a2 x
3
+ a3 x
r
+ ⋯ + ar x
where a
0
≠ 0
+ ⋯}
(43)
,
i.e.
y
c
= a0 x
c+1
+ a1 x
c+2
+ a2 x
c+3
+ a3 x
c+r
+ ⋯ + ar x
(ii) Differentiating equation (44) gives:
+ ⋯
Soufyan Jamal
(44)
′
y
=
c−1
c
a0 cx
+ a1 (c + 1)x
c+1
+ a2 (c + 2)x
+ ⋯
c+r−1
+ ar (c + r)x
′′
andy
=
c−2
a0 c(c − 1)x
+ ⋯
c−1
+ a1 c(c + 1)x
c
+ a2 (c + 1)(c + 2)x
+ ⋯
c+r−2
+ ar (c + r − 1)(c + r)x
(iii) Substituting y, y′ and y′′ into each term of the given equation:
gives:
c−2
a0 c(c − 1)x
c+r−2
c
− a0 c(c − 1)x
c+1
c+2
− ⋯
c+r
− ar (c + r − 1)(c + r)x
c+1
c+r
− 2ar (c + r)x
+ kar x
2
− ⋯
c
− ⋯ + k a0 x
c+2
+ k a2 x
c
c+r
− 2a2 (c + 2)x
2
⋯ + ka0 x
c
− ⋯ − 2a0 cx
c+2
− 2a1 (c + 1)x
+
+ ⋯
− a1 c(c + 1)x
− a2 (c + 1)(c + 2)x
+ k a1 x
′
− 2xy + k(k + 1)y = 0
+ ⋯
+ ar (c + r − 1)(c + r)x
c+1
′′
c−1
c
2
2
(1 − x )y
+ a1 c(c + 1)x
+ a2 (c + 1)(c + 2)x
Soufyan Jamal
+ ⋯
2
c+r
+ ⋯ + k ar x
c+1
+ ka1 x
+ ⋯
+ ⋯ = 0
(45)
(iv) The indicial equation is obtained by equating the coef cient of the lowest power of x (i.e.
zero. Hence, a c(c − 1) = 0 from which, c = 0 or c = 1 since a ≠ 0
0
c−2
x
) to
0
For the term in x , i.e. a c(c + 1) = 0, with c = 1, a = 0; however, when c = 0, a1 is
indeterminate, since any value of a1 combined with the zero value of c would make the product zero.
c−1
1
For
the
1
term
in
c+r
x
,
ar+2 (c + r + 1)(c + r +
2
2) − ar (c + r − 1)(c + r) − 2ar (c + r) + k ar + kar = 0
from which,
2
ar+2
=
=
When c
ar [(c+r−1)(c+r)+2(c+r)−k −k]
(c+r+1)(c+r+2)
ar [(c + r)(c + r + 1)−k(k + 1)]
(c+r+1)(c+r+2)
= 0,
ar+2 =
ar [r(r+1)−k(k+1)]
(r+1)(r+2)
For r = 0,
a2 =
For r = 1,
a0 [−k(k+1)]
(1)(2)
(46)
a3
a1 [(1)(2)−k(k+1)]
=
(2)(3)
2
−a1 [k +k−2]
=
−a1 (k−1)(k+2)
=
3!
3!
For r = 2,
a4
2
a2 [(2)(3)−k(k+1)]
=
−a2 [k +k−6]
=
(3)(4)
(3)(4)
Soufyan Jamal
−a2 (k+3)(k−2)
=
(3)(4)
−(k+3)(k−2)
=
a0 [−k(k+1)]
.
(3)(4)
(1)(2)
a0 k(k+1)(k+3)(k−2)
=
4!
For r = 3,
a5
a3 [(3)(4) − k(k + 1)]
=
(4)(5)
2
=
−a3 [k
+ k − 12]
(4)(5)
−a3 (k + 4)(k − 3)
=
(4)(5)
−(k+4)(k−3)
=
(4)(5)
−a1 (k − 1)(k + 2)
.
(2)(3)
a1 (k − 1)(k − 3)(k + 2)(k + 4)
=
and so on.
5!
Substituting values into equation (43) gives:
y
=
−
+
+
=
2!
a1 (k − 1)(k + 2)
3!
a0 k(k + 1)(k − 2)(k + 3)
4!
3
4
x
5!
⋯
a0 {1 −
+
1
2
5
x
}
k(k + 1)
2!
2
x
k(k + 1)(k − 2)(k + 3)
4!
+ a1 {x −
+
2
x
x
a1 (k − 1)(k − 3)(k + 2)(k + 4)
+
i.e. y
a0 k(k + 1)
0
x {a0 + a1 x −
4
x
(k − 1)(k + 2)
3!
−
3
x
(k − 1)(k − 3)(k + 2)(k + 4)
5!
⋯}
5
x
−
⋯}
From page 591, it was stated that if two solutions of the indicial equation differ by an integer, as in this
case, where c = 0 and 1, and if one coef cient is indeterminate, as with when c = 0, then the complete
solution is always given by using this value of c. Using the second value of c, i.e. c = 1 in this problem,
will give a series which is one of the series in the rst solution. (This may be checked for c = 1 and where
a
= 0; the result will be the second part of equation (47) above.)
1
Legendre’s polynomials
(A polynomial is an expression of the form: f (x) = a + bx + cx + dx + ⋯ . ) When k in equation
(47) above is an integer, say, n, one of the solution series terminates after a nite number of terms. For
example, if k = 2, then the rst series terminates after the term in x2. The resulting polynomial in x,
denoted by P (x), is called a Legendre polynomial. Constants a0 and a1 are chosen so that y = 1 when
x = 1. This is demonstrated in the following worked problems.
2
3
n
Problem 13. Determine the Legendre polynomial P
2
Since in P
,n
2 (x)
= k = 2
(x)
, then from the rst part of equation (47), i.e. the even powers of x:
2(3)
y = a0 {1 −
a0 is chosen to make y
i.e. 1
= 1
when x
2
= a0 {1 − 3(1) } =
Hence,
P2 (x) =
−
1
2
2!
2
2
x
+ 0} = a0 {1 − 3x }
Soufyan Jamal
= 1
− 2a0
, from which, a
=
0
1
2
(1 − 3x ) =
2
2
(3x
1
−
2
− 1)
Problem 14. Determine the Legendre poly- nomial P
3 (x)
Since in P
,n
3 (x)
= k = 3
, then from the second part of equation (47), i.e. the odd powers of x:
y
=
a1 {x −
+
i.e.y
a1 is chosen to make y
i.e. 1
= a1 {1 −
Hence, P
5
3
3 (x) = −
= 1
when x
} = a1 ( −
3
2
(x −
5
3
3
2
3
x )
)
3!
3
x
(k−1)(k−3)(k+2)(k+4)
5!
=
a1 {x −
=
a1 {x −
= 1
(k−1)(k+2)
(2)(5)
3!
5
3
3
x
3
x
5
x
−
⋯}
(2)(0)(5)(7)
+
5!
5
x }
+ 0}
.
from which, a
1
or P
3 (x) =
1
2
=
−
3
(5x
3
2
− 3x)
Rodrigues’ formula
An alternative method of determining Legendre polynomials is by using Rodrigues’* formula, which states:
Pn (x) =
1
n
2
n!
n
d
(x
2
This is demonstrated in the following worked problems.
Problem 15. Determine the Legendre polynomial P
2 (x)
n
− 1)
dx
using Rodrigues’ formula.
n
(48)
n
In Rodrigues’ formula,
and when n
= 2
2
d (x
1
Pn (x) =
n
n
− 1)
n
2 n!
dx
,
2
P2 (x)
2
2
2 2!
dx
2
=
d
dx
2
4
3
dx
4
(x
Hence, P
2 (x)
i.e.
2
=
P2 (x) =
1
4
2
2
1
2
3
+ 1) = 4x
3
=
d(4x −4x)
dx
2
(3x
− 1),
=
1
8
2
(12x
2
= 12x
3 (x)
In Rodrigues’ formula, P
n
n
(x) =
− 4)
the same as in Problem 13.
Problem 16. Determine the Legendre polynomial P
1
n
2 n!
using Rodrigues’ formula.
n
2
d (x −1)
n
dx
and when n
− 4x
− 4
Soufyan Jamal
dx
2
d (x −2x +1)
3
2
− 2x
2
2
2
2
2
dx
4
d (x −2x +1)
1
d (x −2x +1)
and
2
2
d (x −1)
1
=
= 3
,
3
P3 (x)
=
3
3
2 3!
dx
3
Soufyan Jamal
=
2
4
1
3
3
2 (6)
dx
6
4
2
d (x −3x +3x −1)
1
3
(8)(6)
dx
6
4
2
d(x −3x +3x −1)
5
= 6x
dx
5
dx
4
2
d(30x
− 36x
3
= 120x
3
Hence, P
3 (x)
i.e.
+ 6)
dx
=
P3 (x) =
1
2
dx
(5x
4
= 30x
2
− 36x
+ 6
2
3
3
+ 6x
− 72x
4
d (x −3x +3x −1)
(8)(6)
1
6
3
− 12x
3
d(6x −12x +6x)
and
2
d (x −1)(x −2x +1)
3
=
3
2
d (x −1)
1
− 3x),
=
1
(8)(6)
3
(120x
− 72x) =
1
8
3
(20x
− 12x)
the same as in Problem 14.
Now try the following Practice Exercise
Practice Exercise 241 Legendre’s equation and Legendre polynomials (Answers on
page 897)
1. Determine the power series solution of the Legendre equation:
2
′′
(1 − x )y
′
− 2xy + k(k + 1)y = 0
when (a) k
2. Find the following Legendre polynomials: (a) P
= 0
1 (x)
(b) k
(b) P
= 2
4 (x)
, up to and including the term in x5.
(c) P
5 (x)
For fully worked solutions to each of the problems in Practice Exercises 236 to 241 in this
chapter, go to the website: www.routledge.com/cw/bird
* Who was Leibniz? For image and resume, go to www.routledge.com/cw/bird
* Who was Maclaurin? For image and resume of Maclaurin, see page 451. To nd out more go to www.routledge.com/cw/bird