Higher Engineering Mathematics

504 DIFFERENTIAL EQUATIONS

2. Use the Frobenius method to determine the
   general power series solution of the differen-

tial equation:

d^2 y

dx^2

+y= 0

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

y=A

(

1 −

x^2

2!

+

x^4

4!

− ···

)

+B

(

x−

x^3

3!

+

x^5

5!

− ···

)

=Pcosx+Qsinx

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

3. Determine the power series solution of the

differential equation: 3x

d^2 y

dx^2

+ 4

dy

dx

−y= 0

using the Frobenius method.

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

y=A

{

1 +

x

(1×4)

+

x^2

(1×2)(4×7)

+

x^3

(1× 2 ×3)(4× 7 ×10)

+···

}

+Bx−

1

3

{

1 +

x

(1×2)

+

x^2

(1×2)(2×5)

+

x^3

(1× 2 ×3)(2× 5 ×8)

+ ···

}

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

4. Show, using the Frobenius method, that
   the power series solution of the differential

equation:

d^2 y

dx^2

−y=0 may be expressed as

y=Pcoshx+Qsinhx, wherePandQare

constants. [Hint: check the series expansions

for coshxand sinhxon page 48]

52.6 Bessel’s equation and Bessel’s
functions

One of the most important differential equations in

applied mathematics isBessel’s equationand is of

the form:

x^2

d^2 y

dx^2

+x

dy

dx

+(x^2 −v^2 )y= 0

wherevis a real constant. The equation, which has

applications in electric fields, vibrations and heat

conduction, may be solved using Frobenius’ method

of the previous section.

Problem 10. Determine the general power

series solution of Bessels equation.

Bessel’s equationx^2

d^2 y

dx^2

+x

dy

dx

+(x^2 −v^2 )y= 0

may be rewritten as:x^2 y′′+xy′+(x^2 −v^2 )y= 0

Using the Frobenius method from page 498:

(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+···} (34)

wherea 0 =0,

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

+ ···+arxc+r+··· (35)

(ii) Differentiating equation (35) gives:

y′=a 0 cxc−^1 +a 1 (c+1)xc

+a 2 (c+2)xc+^1 +···

+ar(c+r)xc+r−^1 +···

andy′′=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:x^2 y′′+xy′+(x^2 −v^2 )y= 0

gives:

a 0 c(c−1)xc+a 1 c(c+1)xc+^1

+a 2 (c+1)(c+2)xc+^2 +···

+ar(c+r−1)(c+r)xc+r+···+a 0 cxc

+a 1 (c+1)xc+^1 +a 2 (c+2)xc+^2 +···

+ar(c+r)xc+r+···+a 0 xc+^2 +a 1 xc+^3

+a 2 xc+^4 +···+arxc+r+^2 +···−a 0 v^2 xc

−a 1 v^2 xc+^1 −···−arv^2 xc+r+··· = 0

(36)

(iv) Theindicial equationis obtained by equating

the coefficient of the lowest power ofxto zero.

Hence, a 0 c(c−1)+a 0 c−a 0 v^2 = 0

from which, a 0 [c^2 −c+c−v^2 ]= 0

i.e. a 0 [c^2 −v^2 ]= 0

from which, c=+vorc=−vsincea 0 = 0