Question

consider the differential equation y"+xy=0 the recurrence formula for the coefficient is​

Answer 1

$a_{n+2}$ = - $a_{n-1}\frac{(n+1)}{(n+2)}$

where, n = 1, 2, 3, 4, 5....

Given:

Airy's differential equation = y" + xy = 0

In this equation, the $x_{0}$ = 0.

Suppose,

y = $Sum \sum_{n=0}$ $a_{n} x^{n}$ be the sequence solution for the Airy's differential equation

∴ $y^{'}$ = $Sum \sum_{n=1}$ n $a_{n} x^{n-1}$ and $y^{''}$ = $Sum \sum_{n=0}$ $a_{n} x^{n+1}$

Substituting the values of $y^{'}$ and $y^{''}$, we get,

$Sum \sum_{n=1}$ n(n-1) $a_{n} x^{n-2}$ + x $Sum \sum_{n=0}$ $a_{n} x^{n}$ = 0,

$Sum \sum_{n=1}$ n(n+1) (n+2) $a_{n+2} x^{n}$ + $Sum \sum_{n=0}$ $a_{n} x^{n+1}$ = 0.

$Sum \sum_{n=1}$ n(n+1) (n+2) $a_{n+2} x^{n}$ + $Sum \sum_{n=0}$ $a_{n} x^{n}$+ 2$a_{2}$ = 0

$Sum \sum_{n=1}$ [(n+1) (n+2)( $a_{n+2}$ + $a_{n-1}$)]$x^{n}$ + 2$a_{2}$ = 0

(n+1) (n+2)( $a_{n+2}$ + $a_{n-1}$) = 0,

The co-efficients,

2$a_{2}$ = 0, 3.2 $a_{3}$ + $a_{0}$ =0, 4.3. $a_{4}$+$a_{1}$ = 0

∴ $a_{n+2}$ = - $a_{n-1}\frac{(n+1)}{(n+2)}$