Bessel function of second kind with integer order.

[yungman]

I have a question about deriving the Bessel function of the second kind with integer order. I understand that the Bessel function and the second independent variable is defined as:
L(y)=x^2y''+xy'+(x^{2}-n^{2})y=0
y_{2}(x)=aJ_m(x) ln(x)+\sum_{u=0}^{\infty} C_{u} x^{u+n}
and Bessel first kind for integer order is
J_{n}(x)=\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}

Without going through the series manipulations and factoring out, let me jump to the grouping with terms containing $a ln(x)$
a ln(x)\left[ \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)(2k+n-1)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n+2}}{k!(k+n)!2^{2k+n}} - n^{2}\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}\right]

You can see this is in form of
L(y_{1})=x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1}
Where
y_{1}=\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}
My question is in the next step, the derivation claimed
L(y_{1})=x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1}=0
\Rightarrow\;a ln(x)\left[ \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)(2k+n-1)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n+2}}{k!(k+n)!2^{2k+n}} + n^{2}\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}\right]=0
And all these disappeared!

I understand the definition for the Bessel function is
L(y_{1})=x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1}=0
But that does not imply when you see anything like L(y_{1})=x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1} it is automatically equal to zero. Please explain.

Thanks

 

[yungman]

I resolve it already. It is very simple

It is given already that
L(y_{1})=x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1}=0
Where
y_{1}=\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}

a ln(x)\left[ \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)(2k+n-1)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}(2k+n)x^{2k+n}}{k!(k+n)!2^{2k+n}} + \sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n+2}}{k!(k+n)!2^{2k+n}} - n^{2}\sum_{0}^{\infty}\frac{(-1)^{k}x^{2k+n}}{k!(k+n)!2^{2k+n}}\right]=a ln(x)\left[x^{2} y_{1}'' + xy_{1}'+(x^{2}-n^{2})y_{1}\right]=0