Higher Engineering Mathematics, Sixth Edition

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

1. 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!

+···

)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦