Higher Engineering Mathematics

Ess-For-H8152.tex 19/7/2006 18: 2 Page 717

ESSENTIAL FORMULAE 717

Leibniz’s theorem

To find then’th derivative of a producty=uv:

y(n)=(uv)(n)=u(n)v+nu(n−1)v(1)

+

n(n−1)

2!

u(n−2)v(2)

+

n(n−1)(n−2)

3!

u(n−3)v(3)+···

Power series solutions of second order differential

equations.

(a)Leibniz-Maclaurin method

(i) Differentiate the given equationntimes,

using the Leibniz theorem,

(ii) rearrange the result to obtain the recurrence

relation atx=0,

(iii) determine the values of the derivatives at

x=0, i.e. find (y) 0 and (y′) 0 ,

(iv) substitute in the Maclaurin expansion for

y=f(x),

(v) simplify the result where possible and apply

boundary condition (if given).

(b)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+ ···} a 0 =0,

(ii) differentiate the trial series to find y′

andy′′,

(iii) substitute the results in the given differential

equation,

(iv) equate coefficients of corresponding pow-

ers of the variable on each side of the

equation: this enables indexcand coeffi-

cientsa 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= 0

is:

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 of Bessel functions and gamma

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

−···

}