Power series methods of solving ordinary differential equations 499
equation (13), each term is differentiated n times,
which gives:
y(n+^2 )+y(n+^1 )+y(n)(x)+ny(n−^1 )( 1 )+ 0 = 0
i.e. y(n+^2 )+y(n+^1 )+xy(n)+ny(n−^1 )= 0
(15)
(ii) Atx=0, equation (15) becomes:
y(n+^2 )+y(n+^1 )+ny(n−^1 )= 0
from which, y(n+^2 )=−{y(n+^1 )+ny(n−^1 )}
This is therecurrence relationand applies for
n≥ 1
(iii) Substitutingn= 1 , 2 , 3 ,...will produce a set of
relationships between the various coefficients.
For n=1, (y′′′) 0 =−{(y′′) 0 +(y) 0 }
n=2, (y(^4 )) 0 =−{(y′′′) 0 + 2 (y′) 0 }
n=3, (y(^5 )) 0 =−{(y(^4 )) 0 + 3 (y′′) 0 }
n=4, (y(^6 )) 0 =−{(y(^5 )) 0 + 4 (y′′′) 0 }
n=5, (y(^7 )) 0 =−{(y(^6 )) 0 + 5 (y(^4 )) 0 }
n=6, (y(^8 )) 0 =−{(y(^7 )) 0 + 6 (y(^5 )) 0 }
From the given boundary conditions, atx=0,
y=0, thus(y) 0 =0, and atx=0,
dy
dx
=1, thus
(y′) 0 = 1
From the given differential equation,
y′′+y′+xy=0, and, atx=0,
(y′′) 0 +(y′) 0 +( 0 )y=0 from which,
(y′′) 0 =−(y′) 0 =− 1
Thus,(y) 0 = 0 ,(y′) 0 = 1 ,(y′′) 0 =− 1 ,
(y′′′) 0 =−{(y′′) 0 +(y) 0 }=−(− 1 + 0 )= 1
(y(^4 )) 0 =−{(y′′′) 0 + 2 (y′) 0 }
=−[1+ 2 ( 1 )]=− 3
(y(^5 )) 0 =−{(y(^4 )) 0 + 3 (y′′) 0 }
=−[− 3 + 3 (− 1 )]= 6
(y(^6 )) 0 =−{(y(^5 )) 0 + 4 (y′′′) 0 }
=−[6+ 4 ( 1 )]=− 10
(y(^7 )) 0 =−{(y(^6 )) 0 + 5 (y(^4 )) 0 }
=−[− 10 + 5 (− 3 )]= 25
(y(^8 )) 0 =−{(y(^7 )) 0 + 6 (y(^5 )) 0 }
=−[25+ 6 ( 6 )]=− 61
(iv) Maclaurin’s theorem states:
y=(y) 0 +x(y′) 0 +
x^2
2!
(y′′) 0 +
x^3
3!
(y′′′) 0
+
x^4
4!
(y(^4 )) 0 + ···
and substituting the above values into
Maclaurin’s theorem gives:
y= 0 +x( 1 )+
x^2
2!
{− 1 }+
x^3
3!
{ 1 }+
x^4
4!
{− 3 }
+
x^5
5!
{ 6 }+
x^6
6!
{− 10 }+
x^7
7!
{ 25 }
+
x^8
8!
{− 61 }+···
(v) Simplifying, the power series solution of
the differential equation:
d^2 y
dx^2
+
dy
dx
+xy=0is
given by:
y=x−
x^2
2!
+
x^3
3!
−
3 x^4
4!
+
6 x^5
5!
−
10 x^6
6!
+
25 x^7
7!
−
61 x^8
8!
+···
Now try the following exercise
Exercise 195 Further problemson power
series solutions by the Leibniz–Maclaurin
method
- Determine the power series solutionof the dif-
ferential equation:
d^2 y
dx^2
+ 2 x
dy
dx
+y=0using
the Leibniz–Maclaurin method, given that at
x=0,y=1and
dy
dx
=2.
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
y=
(
1 −
x^2
2!
+
5 x^4
4!
−
5 × 9 x^6
6!
+
5 × 9 × 13 x^8
8!
−···
)
+ 2
(
x−
3 x^3
3!
+
3 × 7 x^5
5!
−
3 × 7 × 11 x^7
7!
+···
)
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦