We have $y=b\cos \left [ n\log \left ( \dfrac {x}{n} \right ) \right ]$

$y_1 = -b \sin \left [ \left ( \dfrac {x}{n} \right ) \right ]n\cdot \dfrac {1}{x}\cdot \dfrac {1}{n} = -b \sin \left [ n\log \left ( \dfrac {x}{n} \right ) \right ] \dfrac {n}{x}$

$ \therefore xy_1 = - nb \sin \left [ n\log \left ( \dfrac {x}{n} \right ) \right ]$

Differentiating once again,

$xy_2 + y_1 = -nb \cos \left [ n \log \left ( \dfrac {x}{n} \right ) \right ]n\cdot \dfrac {1}{\frac{x}{n}} \cdot \dfrac {1}{n} = \dfrac {-yn^2}{x}$

$\therefore x^2y_2+ xy_1 + yn^2 =0$

Applying Leibnitz's theorem

$x^2y_{n+2}+n(2x)y_{n+1}+ \dfrac {n(n-1)}{2!} 2y_n + xy_{n+1}+ ny_n + n^2 y_n =0$

$\text{Hence, } x^2 y_{n+2} + (2n+1)xy_{n+1}+2n^2y_n=0$