Power series methods of solving ordinary differential equations 505
Substitutingyandy′′into each term of the given
equationy′′− 2 y=0gives:
y′′− 2 y=a 0 c(c− 1 )xc−^2 +a 1 c(c+ 1 )xc−^1
+[a 2 (c+ 1 )(c+ 2 )− 2 a 0 ]xc+···
+[ar+ 2 (c+r+ 1 )(c+r+ 2 )
− 2 ar]xc+r+···= 0 (30)
(iv) Theindicial equationis obtained by equating
the coefficient of the lowest power ofxto zero.
Hence, a 0 c(c− 1 )=0 from which,c= 0 or
c= 1 sincea 0 = 0
For the term inxc−^1 ,i.e.a 1 c(c+ 1 )= 0
Withc= 1 ,a 1 = 0 ;however,whenc= 0 ,a 1 is
indeterminate, since any value ofa 1 combined
with the zero value ofcwould make the product
zero.
For the term inxc,
a 2 (c+ 1 )(c+ 2 )− 2 a 0 =0 from which,
a 2 =
2 a 0
(c+ 1 )(c+ 2 )
(31)
For the term inxc+r,
ar+ 2 (c+r+ 1 )(c+r+ 2 )− 2 ar= 0
from which,
ar+ 2 =
2 ar
(c+r+ 1 )(c+r+ 2 )
(32)
(a)Whenc=0:a 1 is indeterminate, and from
equation (31)
a 2 =
2 a 0
( 1 × 2 )
=
2 a 0
2!
In general,ar+ 2 =
2 ar
(r+ 1 )(r+ 2 )
and
whenr=1,a 3 =
2 a 1
( 2 × 3 )
=
2 a 1
( 1 × 2 × 3 )
=
2 a 1
3!
whenr=2,a 4 =
2 a 2
3 × 4
=
4 a 0
4!
Hence,y=x^0
{
a 0 +a 1 x+
2 a 0
2!
x^2 +
2 a 1
3!
x^3
+
4 a 0
4!
x^4 + ···
}
from equation (28)
=a 0
{
1 +
2 x^2
2!
+
4 x^4
4!
+···
}
+a 1
{
x+
2 x^3
3!
+
4 x^5
5!
+···
}
Since a 0 and a 1 are arbitrary constants
depending on boundary conditions, leta 0 =P
anda 1 =Q,then:
y=P
{
1 +
2 x^2
2!
+
4 x^4
4!
+···
}
+Q
{
x+
2 x^3
3!
+
4 x^5
5!
+···
}
(33)
(b)Whenc=1: a 1 =0,andfromequation(31),
a 2 =
2 a 0
( 2 × 3 )
=
2 a 0
3!
Since c=1, ar+ 2 =
2 ar
(c+r+ 1 )(c+r+ 2 )
=
2 ar
(r+ 2 )(r+ 3 )
from equation (32) and whenr=1,
a 3 =
2 a 1
( 3 × 4 )
=0sincea 1 = 0
whenr=2,
a 4 =
2 a 2
( 4 × 5 )
=
2
( 4 × 5 )
×
2 a 0
3!
=
4 a 0
5!
whenr=3,
a 5 =
2 a 3
( 5 × 6 )
= 0
Hence, whenc= 1 ,
y=x^1
{
a 0 +
2 a 0
3!
x^2 +
4 a 0
5!
x^4 +···
}
from equation (28)
i.e. y=a 0
{
x+
2 x^3
3!
+
4 x^5
5!
+...
}
Again,a 0 is an arbitrary constant; leta 0 =K,
then y=K
{
x+
2 x^3
3!
+
4 x^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 con-
stantsPandQgives the general solution.Thisis always