Higher Engineering Mathematics

498 DIFFERENTIAL EQUATIONS

52.5 Power series solution by the

Frobenius method

A differential equation of the form y′′+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 proceduremay be used in
the 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=0 gives:

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)

(iv) The sum of these three terms forms the left-

hand side of the equation. Since the right-hand

side is zero, the coefficients of each power of x

can be equated to zero.

For example, the coefficient of xc−^1 is

equated to zero giving: 3a 0 c(c−1)+a 0 c= 0

or a 0 c[3c− 3 +1]=a 0 c(3c−2)= 0 (18)

The coefficient ofxcis equated to zero giving:

3 a 1 c(c+1)+a 1 (c+1)−a 0 = 0

i.e. a 1 (3c^2 + 3 c+c+1)−a 0

=a 1 (3c^2 + 4 c+1)−a 0 = 0

or a 1 (3c+1)(c+1)−a 0 = 0 (19)

In each of series (a), (b) and (c) anxcterm

is involved, after which, a general relationship

can be obtained forxc+r, wherer≥0.

In series (a) and (b), terms inxc+r−^1 are

present; replacingrby (r+1) will give the cor-

responding terms inxc+r, which occurs in all

three equations, i.e.

in series (a), 3ar+ 1 (c+r)(c+r+1)xc+r

in series (b), ar+ 1 (c+r+1)xc+r

in series (c), −arxc+r

Equating the total coefficients ofxc+rto zero

gives:

3 ar+ 1 (c+r)(c+r+1)+ar+ 1 (c+r+1)

−ar= 0

which simplifies to:

ar+ 1 {(c+r+ 1 )( 3 c+ 3 r+ 1 )}−ar= 0 (20)

Equation (18), which was formed from the

coefficients of the lowest power ofx, i.e.xc−^1 ,

is called theindicial equation, from which,