Answer to Question #193696 in Differential Equations for Sherlock

Question

Find a solution of the Bessel’s equation of order zero. 

5. Find the series solutions of the Laguerre equation (1 ) 0

2

x y   x y  ny  of the form 

( ) ( ) 0 y x c L x  n

, where 

r r

n

r

n x

n r r

n

L x

2

0 ( )![( )!]

( )! ( ) ( 1)



  



. Hence find first four polynomials.

Answer 1

Given equation,

$xy''+(1-x)y'+py=0$

Let the solution of equation be-

$y=\sum_{r=0}^{\infty} a_rx^{k+r}, a_o\neq 0 ~~~~~~-(1)$

Differentiating w.r.t. x-

$\dfrac{dy}{dx}=\sum_{r=0}^{\infty} a_r (k+r) x^{k+r-1}~~~~~~-(2) \\[9pt] \dfrac{d^2y}{dx^2}=\sum_{r=0}^{\infty} a_r (k+r)(k+r+1) x^{k+r-2}~~~~-(3)$

Putting the values of y,y' and y'' we get-

$x[\sum_{r=0}^{\infty} a_r (k+r)(k+r+1) x^{k+r-2}]+(1-x)[\sum_{r=0}^{\infty} a_r (k+r) x^{k+r-1}]+p[\sum_{r=0}^{\infty} a_rx^{k+r}]=0$

$\Rightarrow \sum_{r=0}^{\infty} a_r [(k+r)^2x^{k+r-2}-(k+r-p)x^{k+r}]=0~~~~~~-(4)$

The recurrence relation is-

$a_{r+1}=\dfrac{k+r-p}{(k+r+1)^2}a_r~~~~~~~-(5)$

When $k=0, a_{r+1}=\dfrac{r-p}{(r+1)^2} a_r ~~~~~~~~~-(6)$

putting r=0,1,2...

$a_1=-pa_o\\[9pt]a_2=(-1)^2\dfrac{p(p-1)}{(2!)^2}a_o \\[9pt]a_3=\dfrac{(-1)^3p(p-1)(p-2)}{(3!)^2} a_o$

On equation the coefficient 

From eqn (1)-

$y=\sum_{r=0}^{\infty} a_r x^r=a_o+a_1x+a_2x^2+...+a_r x^r+...$

   $=a_o\sum_{r=0}^{\infty} \dfrac{(-1)^r p(p-1)...(p-r+1)}{(r!)^2}x^r$