Assignment 4 – Solutions 1. Note that 4x =0 when x =0, so x =0 is a singular point. Dividing 4xy ′′ +2 y ′ + y =0 (1) by the coefficient of y ′′ gives y ′′ + 1 2x y ′ + 1 4x y =0 (2) and by examining the denominators of (2) we see that x =0 is a regular singular point of the differ- ential equation (1). We can therefore assume that (1) has a Frobenius series solution about x 0 =0 y(x)= m=0 ∞ a m x m+c (3) where a 0 0. There is no singular point of (1) other than x =0, therefore the distance R from x = 0 to the nearest singular point is R = ∞ and so the Frobenius solution obtained will converge for 0 < |x | < ∞. Substituting (3) into (1) gives 4x m=0 ∞ (m + c - 1)(m + c)a m x m+c−2 +2 m=0 ∞ (m + c)a m x m+c−1 + m=0 ∞ a m x m+c ≡ 0 ⇒ m=0 ∞ 4(m + c - 1)(m + c)a m x m+c −1 + m=0 ∞ 2(m + c)a m x m+c−1 + m=0 ∞ a m x m+c ≡ 0 and grouping like powers of x gives m=0 ∞ (4(m + c - 1)(m + c) + 2(m + c))a m x m+c −1 + m=0 ∞ a m x m+c ≡ 0 ⇒ m=0 ∞ 2(m + c)(2m +2c - 1)a m x m+c −1 + m=0 ∞ a m x m+c ≡ 0 (4) The indicial equation is obtained from the coefficient of the lowest power of x. In this case, this lowest power of x is the first term of the first series of (4), so we isolate this term 2c(2c - 1)a 0 x c −1 + m=1 ∞ 2(m + c)(2m +2c - 1)a m x m+c −1 + m=0 ∞ a m x m+c ≡ 0 (5) and the indicial equation is 2c(2c - 1) = 0; the indicial roots are c =0 and c = 1 2 . 1
Transcript
Assignment 4 – Solutions
1. Note that 4x= 0 when x =0, so x =0 is a singular point. Dividing
4xy ′′+ 2y ′+ y =0 (1)
by the coefficient of y ′′ gives
y ′′+1
2xy ′+
1
4xy = 0 (2)
and by examining the denominators of (2) we see that x = 0 is a regular singular point of the differ-ential equation (1). We can therefore assume that (1) has a Frobenius series solution about x0 = 0
y(x)=∑
m=0
∞
amxm+c (3)
where a0 � 0. There is no singular point of (1) other than x = 0, therefore the distance R from x =0 to the nearest singular point is R = ∞ and so the Frobenius solution obtained will converge for0 < |x|<∞.
Substituting (3) into (1) gives
4x
∑
m=0
∞
(m + c− 1)(m + c)am
xm+c−2 + 2
∑
m=0
∞
(m + c)am
xm+c−1 +
∑
m=0
∞
am
xm+c
≡0
⇒
∑
m=0
∞
4(m + c − 1)(m + c)am
xm+c−1 +
∑
m=0
∞
2(m + c)am
xm+c−1 +
∑
m=0
∞
am
xm+c
≡ 0
and grouping like powers of x gives
∑
m=0
∞
(4(m + c− 1)(m + c)+ 2(m + c))amxm+c−1 +∑
m=0
∞
amxm+c≡ 0
⇒∑
m=0
∞
2(m + c)(2m + 2c− 1)amxm+c−1 +∑
m=0
∞
amxm+c≡ 0 (4)
The indicial equation is obtained from the coefficient of the lowest power of x. In this case, thislowest power of x is the first term of the first series of (4), so we isolate this term
2c(2c− 1)a0xc−1 +
∑
m=1
∞
2(m + c)(2m + 2c − 1)amxm+c−1 +
∑
m=0
∞
amxm+c≡ 0 (5)
and the indicial equation is
2c(2c− 1)= 0;
the indicial roots are
c =0 and c =1
2.
1
Notice that the difference of these indicial roots is
1
2− 0=
1
2
which is NOT an integer, so from Theorem 2 of Assignment 4, we will obtain two Frobenius seriessolutions
y1(x)=∑
m=0
∞
amxm+0 , y2(x) =∑
m=0
∞
amxm+
1
2 (6)
which correspond to the indicial roots c = 0 and c =1
2respectively. The series solutions (6) are
determined by their coefficients am and we shall obtain these coefficients by using recurrence rela-tions. For the series solution corresponding to c = 0 we shall need one recurrence relation and for
the solution corresponding to c =1
2we will need a different recurrence relation.
However, we can avoid some work by deriving both recurrence relations from a single general recur-rence:
let c =0 or c =1
2. For these values of c the first term of (5) vanishes
0 +∑
m=1
∞
2(m + c)(2m + 2c− 1)amxm+c−1 +∑
m=0
∞
amxm+c≡ 0.
Now reindex the second summation so that powers of x match
0 +∑
m=1
∞
2(m + c)(2m + 2c− 1)amxm+c−1 +∑
m=1
∞
am−1xm+c−1≡ 0.
Notice that both series have the same starting index ‘m = 1’ and have matching powers of x, hencewe can group them
∑
m=1
∞(
2(m + c)(2m + 2c− 1)am + am−1)xm+c−1≡ 0
and for this identity to hold, the coefficient of each power of x must vanish and we get
2(m + c)(2m + 2c− 1)am + am−1 = 0 for m =1, 2, 3�and solving for am gives the desired general recurrence relation
am =− am−1
2(m + c)(2m + 2c− 1)for m = 1, 2, 3� (7)
We now substitute c = 0 and c =1
2into (7) to get the different solutions y1(x) and y2(x) of (1).
c = 0
Substituting c =0 into (7) gives the recurrence relation
am =− am−1
2m(2m− 1)for m = 1, 2, 3� (8)
2
Substituting a few values of m into the recurrence (8) gives
a1 =(− 1)a0
(2)(1)
a2 =(− 1)a1
(4)(3)
a3 =(− 1)a2
(6)(5)
a4 =(− 1)a3
(8)(7)
and by ‘back substituting’ we have
a1 =(− 1)a0
2.1
a2 =(− 1)2a0
4.3.2.1
a3 =(− 1)3a0
6.5.4.3.2.1
a4 =(− 1)4a0
8.7.6.5.4.3.2.1The general formula for am is
am =(− 1)ma0
(2m)!for m = 0, 1, 2, 3�
Note that the above formula holds for m = 0. Substituting the coefficients am into the Frobeniussolution for c = 0 we get
y1(x)=∑
m=0
∞
amxm+0
= a0
(
∑
m=0
∞
(− 1)mxm
(2m)!
)
.
c =1
2
Substituting c =1
2into (7) gives the recurrence relation
am =− am−1
2(
m +1
2
)(
2m + 2(
1
2
)
− 1) for m = 1, 2, 3�
which simplifies to
am =− am−1
(2m + 1)(2m)for m = 1, 2, 3� (9)
Substituting a few values of m into the recurrence (9) gives
a1 =(− 1)a0
(3)(2)
a2 =(− 1)a1
(5)(4)
a3 =(− 1)a2
(7)(6)
a4 =(− 1)a3
(9)(8)
3
and by ‘back substituting’ we have
a1 =(− 1)a0
3.2
a2 =(− 1)2a0
5.4.3.2
a3 =(− 1)3a0
7.6.5.4.3.2
a4 =(− 1)4a0
9.8.7.6.5.4.3.2
The general formula for am is
am =(− 1)ma0
(2m + 1)!for m = 0, 1, 2, 3�
Note that this formula holds for m = 0. Substituting the coefficients am into the Frobenius solutionfor c =
1
2we get
y2(x)=∑
m=0
∞
amxm+
1
2
= a0
∑
m=0
∞
(− 1)mxm+
1
2
(2m + 1)!
.
We have found two linearly independent solutions
∑
m=0
∞
(− 1)mxm
(2m)!,∑
m=0
∞
(− 1)mxm+
1
2
(2m +1)!
corresponding to the indicial roots c = 0 and c =1
2respectively. Therefore the general solution of
the given differential equation is
y =A
∑
m=0
∞
(− 1)mxm
(2m)!+ B
∑
m=0
∞
(− 1)mxm+
1
2
(2m + 1)!
which, from above, is valid for 0< |x|<∞.
4
2. Note that 2x(x+ 3)= 0 when x= 0, so x= 0 is a singular point. Dividing
2x(x +3)y ′′− 3(x + 1)y ′+ 2y = 0 (10)
by the coefficient of y ′′ gives
y ′′−3(x + 1)
2x(x +3)y ′+
1
x(x+ 3)y = 0 (11)
and by examining the denominators of (11) we see that x = 0 is a regular singular point of the dif-ferential equation (10). We can therefore assume that that (10) has a Frobenius series solutionabout x0 =0
y(x) =∑
m=0
∞
amxm+c (12)
where a0 � 0. We obtain the singular points of the differential equation (10) by setting the coeffi-cient of y ′′ equal to zero
2x(x+ 3) =0 ⇒ x= 0, x =− 3
so x = − 3 is also a singular point of the differential equation (10). So the distance R from x = 0 tothe nearest singular point is R = 3
and therefore the Frobenius solution (12) will converge for 0 < |x| < 3. Substituting (12) into (10)gives
(2x2 + 6x)
∑
m=0
∞
(m + c− 1)(m + c)am
xm+c−2
− 3(x + 1)∑
m=0
∞
(m + c)am
xm+c−1 +2
∑
m=0
∞
am
xm+c
≡0
⇒
∑
m=0
∞
2(m + c− 1)(m + c)am
xm+c +
∑
m=0
∞
6(m + c− 1)(m + c)am
xm+c−1
−
∑
m=0
∞
3(m + c)am
xm+c
−∑
m=0
∞
3(m + c)am
xm+c−1 +
∑
m=0
∞
2am
xm+c
≡ 0
and grouping like powers of x gives
+∑
m=0
∞
(6(m + c− 1)(m + c)− 3(m + c))am
xm+c−1 +
∑
m=0
∞
(2(m + c − 1)(m + c)− 3(m + c) + 2)am
xm+c
≡ 0
⇒∑
m=0
∞
(6(m + c)2− 9(m + c))amxm+c−1 +∑
m=0
∞
(2(m + c)2− 5(m + c)+ 2)amxm+c≡ 0
⇒∑
m=0
∞
3(m + c)(2(m + c)− 3)amxm+c−1 +∑
m=0
∞
(2(m + c)− 1)((m + c)− 2)amxm+c +≡ 0 (13)
The indicial equation is obtained from the coefficient of the lowest power of x. In this case, thislowest power of x is the first term of the first series of (13), so we isolate this term
3c(2c − 3)a0xc−1 +
∑
m=1
∞
3(m + c)(2(m + c)− 3)amxm+c−1 +
∑
m=0
∞
(2(m + c)− 1)((m + c)− 2)amxm+c≡ 0
5
and so the indicial equation is
3c(2c− 3)= 0
and the indicial roots are
c =0 and c =3
2.
Notice that the difference of these indicial roots is
3
2− 0=
3
2
which is NOT an integer, so from Theorem 2 of Assignment 4, we will obtain two Frobenius seriessolutions
y1(x)=∑
m=0
∞
amxm+0 , y2(x)=∑
m=0
∞
amxm+
3
2 (14)
which correspond to the indicial roots c = 0 and c =3
2respectively. The series solutions (14) are
determined by their coefficients am and we shall obtain these coefficients by using recurrence rela-tions. For the series solution corresponding to c = 0 we shall need one recurrence relation and for
the solution corresponding to c =3
2we will need a different recurrence relation.
As in question 1, we can avoid some work by deriving both recurrence relations from a single gen-eral recurrence:
let c =0 or c =3
2. For these values of c the first term of (14) vanishes
0 +∑
m=1
∞
3(m + c)(2(m + c)− 3)amxm+c−1 +∑
m=0
∞
(2(m + c)− 1)((m + c)− 2)amxm+c≡0
Now reindex the second summation so that powers of x match
∑
m=1
∞
3(m + c)(2(m + c)− 3)amxm+c−1 +∑
m=1
∞
(2((m− 1)+ c))− 1)((m− 1)+ c)− 2)am−1xm+c−1
≡0
⇒∑
m=1
∞
3(m + c)(2(m + c)− 3)amxm+c−1 +∑
m=1
∞
(2(m + c)− 3)((m + c)− 3)am−1xm+c−1
≡0
Notice that both series have the same starting index ‘m = 1’ and have matching powers of x, hencewe can group them
Notice that this formula for am does not hold for m = 0. Substituting the coefficients am into the
Frobenius solution for c =3
2we get
y2(x)=∑
m=0
∞
amxm+
3
2
= a0x3
2 +∑
m=1
∞
amxm+
3
2
= a0
x3
2 +∑
m=1
∞
(− 1)m+1xm+
3
2
3m−1(2m− 1)(2m +1)(2m + 3))
We have found two linearly independent solutions
1 +2x
3+
x2
9, x
3
2 +∑
m=1
∞
(− 1)m+1xm+
3
2
3m−1(2m− 1)(2m +1)(2m +3))
8
corresponding to the indicial roots c = 0 and c =3
2respectively. Therefore the general solution of
the given differential equation is
y =A
(
1+2x
3+
x2
9
)
+ B
x3
2 +∑
m=1
∞
(− 1)m+1xm+
3
2
3m−1(2m− 1)(2m + 1)(2m +3))
which, from above, is valid for 0< |x|< 3.
9
3. Note that x(x− 1)= 0 when x= 0, so x= 0 is a singular point. Dividing
x(x− 1)y ′′+(3x− 1)y ′+ y = 0 (19)
by the coefficient of y ′′ gives
y ′′ +(3x− 1)
x(x− 1)y ′+
1
x(x− 1)y = 0 (20)
and by examining the denominators of (20) we see that x = 0 is a regular singular point of the dif-ferential equation (19). We can therefore assume that that (19) has a Frobenius series solutionabout x0 =0
y(x) =∑
m=0
∞
amxm+c (21)
where a0 � 0. We obtain the singular points of the differential equation (19) by setting the coeffi-cient of y ′′ equal to zero
x(x− 1)= 0 ⇒ x =0, x= 1
so x = 1 is also a singular point of the differential equation (19). So the distance R from x = 0 tothe nearest singular point is R = 1
x-axis0 1−1
R = 1
and therefore the Frobenius solution (21) will converge for 0< |x|< 1.
Substituting (21) into (19) gives
(x2− x)
∑
m=0
∞
(m + c − 1)(m + c)am
xm+c−2 + (3x− 1)
∑
m=0
∞
(m + c)am
xm+c−1 +
∑
m=0
∞
am
xm+c
≡0
⇒
∑
m=0
∞
(m + c− 1)(m + c)am
xm+c
−
∑
m=0
∞
(m + c− 1)(m + c)am
xm+c−1 +
∑
m=0
∞
3(m + c)am
xm+c
−∑
m=0
∞
(m + c)am
xm+c−1 +
∑
m=0
∞
am
xm+c
≡ 0
and grouping like powers of x gives
−
∑
m=0
∞
((m + c− 1)(m + c) + (m + c))am
xm+c−1 +
∑
m=0
∞
((m + c − 1)(m + c)+ 3(m + c) + 1)am
xm+c
≡ 0
⇒ −∑
m=0
∞
(m + c)2amxm+c−1 +∑
m=0
∞
((m + c)2 + 2(m + c)+ 1)amxm+c≡ 0
⇒ −∑
m=0
∞
(m + c)2amxm+c−1 +∑
m=0
∞
(m + c + 1)2amxm+c≡ 0 (22)
The indicial equation is obtained from the coefficient of the lowest power of x. In this case, thislowest power of x is the first term of the first series of (22), so we isolate this term
− c2a0xc−1−
∑
m=1
∞
(m + c)2amxm+c−1 +∑
m=0
∞
(m + c + 1)2amxm+c≡ 0 (23)
10
and so the indicial equation is
c2 = 0
and we only have one (repeated) indicial root
c =0.
From Theorem 2 of Assignment 4
“If c1 = c2 then there is only one Frobenius series solution about the point x = x0. The other solu-tion may be obtained by using the reduction of order method.” Therefore in this case we will obtainonly one Frobenius solution that corresponds to c = 0 and this solution will take the form
y(x)=∑
m=0
∞
amxm+0. (24)
We substitute c =0 into (23) to determine the coefficients am of (24)
0−∑
m=1
∞
m2amxm−1 +∑
m=0
∞
(m +1)2amxm≡ 0.
Reindex the second summation so that powers of x match
−∑
m=1
∞
m2amxm−1 +∑
m=1
∞
((m− 1)+ 1)2am−1xm−1≡ 0
⇒−∑
m=1
∞
m2amxm−1 +∑
m=1
∞
m2am−1xm−1≡ 0.
Notice that both series have the same starting index ‘m = 1’ and have matching powers of x, hencewe can group them
∑
m=1
∞(
−m2am + m2am−1
)
xm+c−1≡ 0
and for this identity to hold, the coefficient of each power of x must vanish and we get
−m2am + m2am−1 = 0 when m =1, 2, 3,� (25)
and as m2� 0 when m =1, 2, 3,� we can divide (25) by m2 to get
− am + am−1 = 0 when m =1, 2, 3,�⇒ am = am−1 when m = 1, 2, 3,� (26)
Substituting a few values of m into the recurrence (26) gives
a1 = a0
a2 = a1
a3 = a2
11
and by ‘back substituting’ we have
a1 = a0
a2 = a0
a3 = a0.
The general formula for am is
am = a0 for m =0, 1, 2, 3�Notice this general formula holds when m = 0. Substituting this general formula for am into theFrobenius solution (24) gives
y(x) =∑
m=0
∞
amxm
=∑
m=0
∞
a0xm
= a0
(
∑
m=0
∞
xm
)
(27)
which, from above, is valid for 0< |x|< 1. Recall that
∑
m=0
∞
xm = 1 +x + x2 +�is a geometric series with first term a = 1 and common ratio r = x. From the formula for the sum ofa geometric series we have
∑
m=0
∞
xm =1
1− x
and so we may write the Frobenius solution (27) as
y = a0
(
1
1− x
)
.
We therefore have that
y1(x)=1
1− x=(1− x)−1 (28)
is a solution of the differential equation (19). We obtain a second linearly independent solution bythe reduction of order method, that is, we assume our second solution y2(x) of the differentialequation (19) takes the form
and as y2 is assumed, by the reduction of order method, to be a solution of the differential equation(19) we substitute y2 into (19) to determine v(x) –
Now use the substitution w = v ′ to convert the second order differential equation (33) to a firstorder differential equation
xw ′+ w = 0 (34)
1. Substituting equation (32) into the differential equation (31) removes the terms in v. (31) then takes the form
( )v ′+ ( )v ′′= 0.
We then reduce the order of this second order differential equation by using the substitution w = v′ to obtain the first order
differential equation
( )w + ( )w′= 0
which, in this case, can be solved by seperation of variables.
13
Write (34) as
xdw
dx+ w =0
and seperating variables gives∫
dw
w=−
∫
dx
x(35)
By integrating (35) we have
lnw =− lnAx (36)
where A is a constant. From (36) we obtain
w =B
x
and as we only seek ONE second solution we may assume B = 1, that is
w =1
x. (37)
Recall that w = v ′, so by integrating (37) we have that
v(x) = ln x
(again we only seek ONE second solution so we may assume the constant of integration is equal tozero in this case). Therefore, from (29)
y2 = (1− x)−1lnx
is a second solution to the differential equation (19). Since
y1 =(1− x)−1 and y2 = (1− x)−1lnx
are linearly independent, it follows that the general solution to the differential equation (29) is
y = A(1− x)−1 + B (1− x)−1lnx
which clearly may be written as
y = A
(
1
1−x
)
+ B
(
lnx
1−x
)
.
From above, this solution is valid for 0< |x|< 1.
14
4. Note that 4x= 0 when x =0, so x =0 is a singular point. Dividing
xy ′′− (4+ x)y ′+ 2y = 0 (38)
by the coefficient of y ′′ gives
y ′′−(4+ x)
xy ′ +
2
xy = 0 (39)
and by examining the denominators of (39) we see that x = 0 is a regular singular point of the dif-ferential equation (38). We can therefore assume that (38) has a Frobenius series solution aboutx0 = 0
y(x) =∑
m=0
∞
amxm+c (40)
where a0� 0. There is no singular point of (38) other than x = 0, therefore the distance R from x =0 to the nearest singular point is R = ∞ and so the Frobenius solution obtained will converge for0 < |x|<∞.
Substituting (40) into (38) gives
x
∑
m=0
∞
(m + c− 1)(m + c)am
xm+c−2
− (4+ x)∑
m=0
∞
(m + c)am
xm+c−1 + 2
∑
m=0
∞
am
xm+c
≡0
⇒
∑
m=0
∞
(m + c − 1)(m + c)am
xm+c−1
−
∑
m=0
∞
4(m + c)am
xm+c−1
−
∑
m=0
∞
(m + c)am
xm+c +
∑
m=0
∞
2am
xm+c
≡ 0
and grouping like powers of x gives
∑
m=0
∞
((m + c− 1)(m + c)− 4(m + c))amxm+c−1−
∑
m=0
∞
((m + c)− 2)amxm+c≡ 0
⇒∑
m=0
∞
(m + c)(m + c− 5)amxm+c−1−
∑
m=0
∞
((m + c)− 2)amxm+c≡ 0 (41)
The indicial equation is obtained from the coefficient of the lowest power of x. In this case, thislowest power of x is the first term of the first series of (41), so we isolate this term
c(c− 5)a0xc−1 +
∑
m=1
∞
(m + c)(m + c− 5)amxm+c−1
−
∑
m=0
∞
((m + c)− 2)amxm+c≡ 0 (42)
and the indicial equation is
c(c− 5)= 0;
and the indicial roots are
c = 5 and c = 0.
15
Notice that the difference of these indicial roots is
5− (0) =5
which IS an integer, so from Theorem 2 of Assignment 4
“If c1− c2 is an integer then one of the following alternatives occur
a) A Frobenius series solution about the point x = x0 can be obtained from the larger indicialroot, but a second linearly independent Frobenius series solution cannot be obtained fromthe smaller root.
b) A Frobenius series solution about the point x = x0 can be obtained from the larger indicialroot and two linearly independent Frobenius series solutions about the point x = x0 can beobtained from the smaller root”
As in questions 1, 2 & 3 we obtain the general recurrence relation. It is perhaps best to first substi-tute the smaller root into this general recurrence in case the smaller root does yield both Frobeniussolutions.
Let c =5 or c = 0. For these values of c the first term of (42) vanishes
0+∑
m=1
∞
(m + c)(m + c− 5)amxm+c−1−
∑
m=0
∞
(m + c− 2)amxm+c≡ 0
Now reindex the second summation so that powers of x match
∑
m=1
∞
(m + c)(m + c− 5)amxm+c−1−
∑
m=1
∞
(m + c− 3)am−1xm+c−1≡ 0.
Notice that both series have the same starting index ‘m = 1’ and have matching powers of x, hencewe can group them
∑
m=1
∞
((m + c)(m + c− 5)am − (m + c− 3)am−1)xm+c−1≡ 0
and for this identity to hold, the coefficient of each power of x must vanish and we get
(m + c)(m + c− 5)am − (m + c− 3)am−1 =0 for m = 1, 2, 3� (43)
However, at this point, we do not solve for am immediately as this will eventually lead to divisionby zero. We substitute the indicial roots, starting with the smaller root c = 0 in the hope that thissmaller root will yield both solutions.
c = 0
Substituting c =0 into (43) gives
m(m− 5)am − (m− 3)am−1 = 0 for m = 1, 2, 3� (44)
16
In the recurrence relation (44), notice that the coefficients m(m − 5) and (m − 3) of am and am−1
vanish at m =5 and m = 3 respectively. Substituting values m = 1, 2, 3, 4, 5 into (44) gives
as a4 = 0. This fifth equation imposes no condition on a5 and implies that a5 can take any value,that is, a5 is an arbitrary constant. We now show that am for m ≥ 6 can be expressed in terms of
a5. Note that the coefficient m(m − 5) of am in (44) is nonzero for m ≥ 6; it follows that solving(44) for am
am =(m− 3)am−1
m(m− 5)m = 6, 7,� (47)
does not lead to division by zero when m = 6, 7,� . Substituting a few values of m into the recur-rence (47) gives
m = 6: a6 =(3)a5
(6)(1)
m = 7: a7 =(4)a6
(7)(2)
m = 8: a8 =(5)a7
(8)(3)
and by ‘back substituting’ we have
a6 =(3)a5
(6)(1)
a7 =(3.4)a5
(6.7)(1.2)
a8 =(3.4.5)a5
(6.7.8)(1.2.3)
and so a general formula for am when m≥ 6 is
am =(3.4� (m− 3))a5
(6.7�m)(m− 5)!m = 6, 7,� (48)
17
and note this formula does not hold when m = 5. The equations (46) and (48) express the coeffi-cients am of the Frobenius solution
y(x) =∑
m=0
∞
amxm+0 (49)
corresponding to c = 0 in terms of a0 and a5 where both a5 and a0 are arbitrary constants. Substi-tuting (46) and (48) into (49) gives
y = a0 + a1x + a2x2 + a3x
3 + a4x4 + a5x
5 +∑
m=6
∞
amxm+0
⇒ y = a0 +1
2a0x +
1
12a0x
2 + 0x3 +0x4 + a5x5 +
∑
m=0
∞
(3.4� (m− 3))a5xm+c
(6.7�m)(m− 5)!
⇒ y = a0
(
1 +1
2x+
1
12x2
)
+ a5
(
x5 +∑
m=6
∞
(3.4� (m− 3))xm
(6.7�m)(m− 5)!
)
(50)
As a5 and a0 are arbitrary constants, (50) is the general solution of the given differential equation;from above this solution is valid for 0< |x|<∞.
18
5.
a) Note that x2 = 0 when x= 0, so x= 0 is a singular point. Dividing
x2y ′′+ xy ′+ (x2− ν2)y = 0 (51)
by the coefficient of y ′′ gives
y ′′+1
xy ′+
(x2− ν2)
x2y = 0 (52)
and by examining the denominators of (52) we see that x = 0 is a regular singular point thedifferential equation (51). We can therefore assume that that (51) has a Frobenius seriessolution about x0 =0
y(x) =∑
m=0
∞
amxm+c (53)
where a0� 0.
Substituting (53) into (51) gives
x2∑
m=0
∞
(m + c− 1)(m + c)am
xm+c−2 + x
∑
m=0
∞
(m + c)am
xm+c−1 + (x2
− ν2)∑
m=0
∞
am
xm+c
≡0
∑
m=0
∞
(m + c− 1)(m + c)am
xm+c +
∑
m=0
∞
(m + c)am
xm+c +
∑
m=0
∞
am
xm+c+2 −
∑
m=0
∞
ν2a
mx
m+c≡ 0
and grouping like powers of x gives
∑
m=0
∞(
(m + c− 1)(m + c)+ (m + c)− ν2)
am
xm+c +
∑
m=0
∞
amxm+c+2≡ 0
⇒∑
m=0
∞(
(m + c)2− ν2)
am
xm+c +
∑
m=0
∞
amxm+c+2≡ 0 (54)
The indicial equation is obtained from the coefficient of the lowest power of x. In this case,this lowest power of x is the first term of the first series of (54), so we isolate this term
(c2− ν2)a0x
c +∑
m=1
∞(
(m + c)2− ν2)
am
xm+c +
∑
m=0
∞
amxm+c+2≡ 0 (55)
and so the indicial equation is
(c2− ν2)= 0
and the indicial roots are
c = ν and c =− ν.
b) We desire only one Frobenius solution – the solution corresponding to c = ν
y(x)=∑
m=0
∞
amxm+ν. (56)
19
The coefficients am of (56) are obtained by substituting c = ν into (55) and obtaining arecurrence relation.
(ν2− ν
2)a0xc +
∑
m=1
∞(
(m + ν)2− ν2)
am
xm+ν +
∑
m=0
∞
amxm+ν+2≡ 0
⇒ 0 +∑
m=1
∞
m(m + 2ν)amxm+ν +∑
m=0
∞
amxm+ν+2≡ 0 (57)
and now re-index the second series of (53) so that the powers match
∑
m=1
∞
m(m +2ν)amxm+ν +∑
m=2
∞
am−2xm+ν ≡ 0 (58)
Note that we cannot immediately group the two series of (54) as the starting indices of theseries do not match. We therefore rewrite the first series of (54)
(1+ 2ν)a1x1+ν +
∑
m=2
∞
m(m +2ν)amxm+ν +∑
m=2
∞
am−2xm+ν ≡ 0
and now we may group the series to obtain
(1 + 2ν)a1x1+ν +
∑
m=2
∞
(m(m +2ν)am + am−2)xm+ν ≡ 0
and for this identity to hold, the coefficient of each power of x must vanish and we get
(1 + 2ν)a1 = 0 (59)
m(m +2ν)am + am−2 = 0 when m =2, 3,� (60)
and, since by assumption ν > 0 we have from (59) that
a1 = 0. (61)
Notice that (60) is a recurrence relation that ‘steps’ by 2. This type of recurrence seperatesinto ‘even’ a2k and ‘odd’ a2k+1, but from (61) it follows that all odd a2k+1 will be zero.Hence we consider only even a2k. First rewrite (60) as
am =− am−2
m(m + 2ν)(62)
and substituting m = 2, 4, 6 into (62) we have
a2 =− a0
2(2+ ν)=
− a0
22(1)(1 + ν)
a4 =− a2
4(4+ ν)=
− a2
22(2)(2 + ν)
a6 =− a4
6(6+ ν)=
− a4
22(3)(3 + ν)
20
and ‘back-substituting’ gives
a2 =− a0
22(1)(1 + ν)
a4 =a0
24(1.2)(1 + ν)(2+ ν)
a6 =− a0
26(1.2.3)(1+ ν)(2 + ν)(3+ ν)
and the general formula for even a is given by
a2k =(− 1)ka0
22kk!(ν +1)(ν + 2)� (ν + k)when k =1, 2, 3,� (63)
and as mentioned above
a2k+1 =0 when k =0, 1, 2, 3,� (64)
Substituting (63) and (64) into the Frobenius solution corresponding to c = ν gives
y(x)=∑
m=0
∞
amxm+ν
=∑
k=0
∞
a2kx2k+ν +
∑
k=0
∞
a2k+1x2k+1+ν
=∑
k=0
∞
a2kx2k+ν + 0
= a0
(
xν +∑
k=1
∞
(− 1)kx2k+ν
22kk!(ν +1)(ν + 2)� (ν + k)
)
.
c) When ν = n where n = 0, 1, 2,� it follows by replacing ν =n in part b) that
y(x) = a0
(
xn +∑
k=1
∞
(− 1)kx2k+n
22kk!(n + 1)(n + 2)� (n + k)
)
(65)
are solutions of Bessel’s equation of index n:
x2y ′′+ xy ′+ (x2−n2)y = 0. (66)
Letting a0 =1
2nn!
in (65) gives a particular solution of (66) which is
y(x)=1
2nn!
(
xn +∑
k=1
∞
(− 1)kx2k+n
22kk!(n + 1)(n + 2)� (n + k)
)
=xn
2nn!+
1
2nn!
∑
k=1
∞
(− 1)kx2k+n
22kk!(n +1)(n +2)� (n + k)
=xn
2nn!+∑
k=1
∞
(− 1)kx2k+n
22k2nk!n!(n +1)(n +2)� (n + k)
21
Now
n!(n +1)(n +2)� (n + k) = (n + k)!
so we have
y(x)=xn
2nn!+∑
k=1
∞
(− 1)kx2k+n
22k+nk!(n + k)!(67)
Notice that
(− 1)kx2(0)+n
22(0)+n0!(n +0)!=
xn
2nn!
that is, the general formula in the series in (67) holds when k = 0. Therefore we can rewrite(67) as
y(x)=∑
k=0
∞
(− 1)kx2k+n
22k+nk!(n + k)!(68)
and so, as required, (68) is a particular solution of Bessel’s equation of index n.
