498 Higher Engineering Mathematics
This equation is called arecurrence relation
orrecurrence formula, because each recurring
term depends on a previous term.
(iii) Substituting n=0, 1, 2, 3,...will produce
a set of relationships between the various
coefficients.
Forn=0, (y′′) 0 =− 2 (y) 0
n=1, (y′′′) 0 =− 3 (y′) 0
n=2,(y(^4 )) 0 =− 4 (y′′) 0 =− 4 {− 2 (y) 0 }
= 2 × 4 (y) 0
n=3,(y(^5 )) 0 =− 5 (y′′′) 0 =− 5 {− 3 (y′) 0 }
= 3 × 5 (y′) 0
n=4,(y(^6 )) 0 =− 6 (y(^4 )) 0 =− 6 { 2 × 4 (y) 0 }
=− 2 × 4 × 6 (y) 0
n=5,(y(^7 )) 0 =− 7 (y(^5 )) 0 =− 7 { 3 × 5 (y′) 0 }
=− 3 × 5 × 7 (y′) 0
n=6,(y(^8 )) 0 =− 8 (y(^6 )) 0 =
− 8 {− 2 × 4 × 6 (y) 0 }= 2 × 4 × 6 × 8 (y) 0
(iv) Maclaurin’s theorem from page 69 may be
written as:
y=(y) 0 +x(y′) 0 +
x^2
2!
(y′′) 0 +
x^3
3!
(y′′′) 0
+
x^4
4!
(y(^4 )) 0 + ···
Substituting the above values into Maclaurin’s
theorem gives:
y=(y) 0 +x(y′) 0 +
x^2
2!
{− 2 (y) 0 }
+
x^3
3!
{− 3 (y′) 0 }+
x^4
4!
{ 2 × 4 (y) 0 }
+
x^5
5!
{ 3 × 5 (y′) 0 }+
x^6
6!
{− 2 × 4 × 6 (y) 0 }
+
x^7
7!
{− 3 × 5 × 7 (y′) 0 }
+
x^8
8!
{ 2 × 4 × 6 × 8 (y) 0 }
(v) Collecting similar terms together gives:
y=(y) 0
{
1 −
2 x^2
2!
+
2 × 4 x^4
4!
−
2 × 4 × 6 x^6
6!
+
2 × 4 × 6 × 8 x^8
8!
− ···
}
+(y′) 0
{
x−
3 x^3
3!
+
3 × 5 x^5
5!
−
3 × 5 × 7 x^7
7!
+ ···
}
i.e.y=(y) 0
{
1 −
x^2
1
+
x^4
1 × 3
−
x^6
3 × 5
+
x^8
3 × 5 × 7
− ···
}
+(y′) 0 ×
{
x
1
−
x^3
1 × 2
+
x^5
2 × 4
−
x^7
2 × 4 × 6
+···
}
The boundary conditions are that atx= 0 ,y= 1
and
dy
dx
=2, i.e.(y) 0 =1and(y′) 0 =2.
Hence, the power series solution of the differen-
tial equation:
d^2 y
dx^2
+x
dy
dx
+ 2 y=0is:
y=
{
1 −
x^2
1
+
x^4
1 × 3
−
x^6
3 × 5
+
x^8
3 × 5 × 7
−···
}
+ 2
{
x
1
−
x^3
1 × 2
+
x^5
2 × 4
−
x^7
2 × 4 × 6
+···
}
Problem 6. Determine the power series solution
of the differential equation:
d^2 y
dx^2
+
dy
dx
+xy=0 given the boundary conditions
that atx=0,y=0and
dy
dx
=1, using
Leibniz–Maclaurin’s method.
Following the above procedure:
(i) The differential equation is rewritten as:
y′′+y′+xy=0and from the Leibniz theorem of