Higher Engineering Mathematics

POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS 497

I

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 196 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

+ 2 x

dy

dx

+y= 0

using the Leibniz–Maclaurin method, given

that atx=0,y=1 and

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!

+···

)

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

2. Show that the power series solution of the dif-

ferential equation:(x+ 1 )

d^2 y

dx^2

+(x−1)

dy

dx

−

2 y=0, using the Leibniz–Maclaurin method,

is given by:y= 1 +x^2 +exgiven the bound-

ary conditions that atx=0,y=

dy

dx

=1.

3. Find the particular solution of the differen-

tial equation: (x^2 +1)

d^2 y

dx^2

+x

dy

dx

− 4 y= 0

using the Leibniz–Maclaurin method, given

the boundary conditions that atx=0,y= 1

and

dy

dx

=1.

[

y= 1 +x+ 2 x^2 +

x^3

2

−

x^5

8

+

x^7

16

+···

]

4. Use the Leibniz–Maclaurin method to deter-
   mine the power series solution for the differ-

ential equation:x

d^2 y

dx^2

+

dy

dx

+xy=1given

that atx=0,y=1 and

dy

dx

=2.

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

y=

{

1 −

x^2

22

+

x^4

22 × 42

−

x^6

22 × 42 × 62

+ ···

}

+ 2

{

x−

x^3

32

+

x^5

32 × 52

−

x^7

32 × 52 × 72

+···

}

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦