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) + n y
(n−1)
(1) + 0 = 0
i.e.
y
(n+2)
+ y
(n+1)
+ x y
(n)
+ n y
(n−1)
= 0
(15)
(ii) At x = 0, equation (15) becomes:
y
(n+2)
+ y
(n+1)
+ n y
(n−1)
= 0
from which, y (n+2) = −{y (n+1) + n y (n−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 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, at x = 0,
y = 0, thus ( y) 0 = 0, and at x = 0,
d y
dx
= 1, thus
( y ) 0 = 1
From the given differential equation,
y + y + x y = 0, and, at x = 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 +
d y
dx
+ x y = 0 is
given by:
y = x−
x 2
2!
+
x 3
3!
−
3x 4
4!
+
6x 5
5!
−
10x 6
6!
+
25x 7
7!
−
61x 8
8!
+· · ·
Now try the following exercise
Exercise 195 Further problems on power
series solutions by the Leibniz–Maclaurin
method
1. Determine the power series solution of the differential equation:
d 2 y
dx 2 + 2x
d y
dx
+ y = 0 using
the Leibniz–Maclaurin method, given that at
x = 0, y = 1 and
d y
dx
= 2.
⎡
⎢
⎢
⎢
⎢
⎢
⎢
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⎢
⎢
⎣
y =
1 −
x 2
2!
+
5x 4
4!
−
5 × 9x 6
6!
+
5 × 9 × 13x 8
8!
− · · ·
+ 2
x −
3x 3
3!
+
3 × 7x 5
5!
−
3 × 7 × 11x 7
7!
+ · · ·
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦