Higher Engineering Mathematics, Sixth Edition

500 Higher Engineering Mathematics

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,

isgivenby:y= 1 +x^2 +exgiventheboundary

conditions that atx=0,y=

dy

dx

=1.

3. Find the particular solution of the differ-
   ential 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-
   minethepower series solution for thedifferen-
   tial equation:x

d^2 y

dx^2

+

dy

dx

+xy=1 given that

atx=0,y=1and

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

+···

}

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⎥⎥

⎦

52.5 Power series solution by the

Frobenius method

A differential equation of the formy′′+Py′+Qy=0,

wherePandQare both functions ofx, such that the

equation can be represented by a power series, may be

solved by theFrobenius method.

The following4-step proceduremaybeusedinthe

Frobenius method:

(i) Assume a trial solution of the form y=

xc

{

a 0 +a 1 x+a 2 x^2 +a 3 x^3 +···+arxr+···

}

(ii) differentiate the trial series,

(iii) substitute the results in the given differential

equation,

(iv) equate coefficients of corresponding powers of

the variable on each side of the equation;

this enables index c and coefficientsa 1 , a 2 ,

a 3 , ... from the trial solution, to be determined.

This introductory treatment of the Frobenius method

covering the simplest cases is demonstrated, using the

above procedure, in the following worked problems.

Problem 7. Determine, using the Frobenius

method, the general power series solution of the

differential equation: 3x

d^2 y

dx^2

+

dy

dx

−y=0.

The differential equation may be rewritten as:

3 xy′′+y′−y=0.

(i) Let a trial solution be of the form

y=xc

{

a 0 +a 1 x+a 2 x^2 +a 3 x^3 +···

+arxr+···

}

(16)

wherea 0 =0,

i.e. y=a 0 xc+a 1 xc+^1 +a 2 xc+^2 +a 3 xc+^3

+ ···+arxc+r+··· (17)

(ii) Differentiating equation (17) gives:

y′=a 0 cxc−^1 +a 1 (c+ 1 )xc

+a 2 (c+ 2 )xc+^1 +···

+ar(c+r)xc+r−^1 +···

and y′′=a 0 c(c− 1 )xc−^2 +a 1 c(c+ 1 )xc−^1

+a 2 (c+ 1 )(c+ 2 )xc+···

+ar(c+r− 1 )(c+r)xc+r−^2 + ···

(iii) Substitutingy,y′andy′′into each term of the

given equation 3xy′′+y′−y=0gives:

3 xy′′= 3 a 0 c(c− 1 )xc−^1 + 3 a 1 c(c+ 1 )xc

+ 3 a 2 (c+ 1 )(c+ 2 )xc+^1 +···

+ 3 ar(c+r− 1 )(c+r)xc+r−^1 +···(a)

y′=a 0 cxc−^1 +a 1 (c+ 1 )xc+a 2 (c+ 2 )xc+^1

+···+ar(c+r)xc+r−^1 +··· (b)

−y=−a 0 xc−a 1 xc+^1 −a 2 xc+^2 −a 3 xc+^3

−···−arxc+r−··· (c)