Higher Engineering Mathematics, Sixth Edition

Essential formulae 671

(ii) differentiate the trial series to find y′

andy′′,

(iii) substitutethe results in thegiven differential

equation,

(iv) equate coefficients of corresponding powers

of the variable on each side of the equa-

tion: this enables indexcand coefficients

a 1 ,a 2 ,a 3 , ...from the trial solution, to be

determined.

Bessel’s equation:

The solution ofx^2

d^2 y

dx^2

+x

dy

dx

+(x^2 −v^2 )y=0is:

y=Axv

{

1 −

x^2

22 (v+ 1 )

+

x^4

24 ×2!(v+ 1 )(v+ 2 )

−

x^6

26 ×3!(v+ 1 )(v+ 2 )(v+ 3 )

+···

}

+Bx−v

{

1 +

x^2

22 (v− 1 )

+

x^4

24 ×2!(v− 1 )(v− 2 )

+

x^6

26 ×3!(v− 1 )(v− 2 )(v− 3 )

+···

}

or, in terms ofBessel functionsandgamma functions:

y=AJv(x)+BJ−v(x)

=A

(x

2

)v{ 1

(v+ 1 )

−

x^2

22 (1!)(v+ 2 )

+

x^4

24 (2!)(v+ 4 )

−···

}

+B

(x

2

)−v{ 1

( 1 −v)

−

x^2

22 (1!)( 2 −v)

+

x^4

24 (2!)( 3 −v)

−···

}

In general terms:

Jv(x)=

(x

2

)v∑∞

k= 0

(− 1 )kx^2 k

22 k(k!)(v+k+ 1 )

and J−v(x)=

(x

2

)−v∑∞

k= 0

(− 1 )kx^2 k

22 k(k!)(k−v+ 1 )

and in particular:

Jn(x)=

(x

2

)n{ 1

n!

−

1

(n+ 1 )!

(x

2

) 2

+

1

(2!)(n+ 2 )!

(x

2

) 4

−···

}

J 0 (x)= 1 −

x^2

22 (1!)^2

+

x^4

24 (2!)^2

−

x^6

26 (3!)^2

+···

and J 1 (x)=

x

2

−

x^3

23 (1!)(2!)

+

x^5

25 (2!)(3!)

−

x^7

27 (3!)(4!)

+···

Legendre’s equation:

The solution of( 1 −x^2 )

d^2 y

dx^2

− 2 x

dy

dx

+k(k+ 1 )y= 0

is:

y=a 0

{

1 −

k(k+ 1 )

2!

x^2

+

k(k+ 1 )(k− 2 )(k+ 3 )

4!

x^4 −···

}

+a 1

{

x−

(k− 1 )(k+ 2 )

3!

x^3

+

(k− 1 )(k− 3 )(k+ 2 )(k+ 4 )

5!

x^5 −···

}

Rodrigue’s formula:

Pn(x)=

1

2 nn!

dn(x^2 − 1 )n

dxn

Statistics and Probability

Mean, median, mode and standard

deviation:

Ifx=variate andf=frequency then:

meanx ̄=

∑

fx

∑

f

Themedianis the middle term of a ranked set of data.