checking the Solution of Bessel differential equation I want to check the first part of the solution and help in the second part.

Obtain the solution
  $$y_{1}(x)=J_{0}(x)=1-\frac{x^{2}}{2^{2}}+\frac{x^{4}}{2^{2}\cdot 4^{2}}-\ldots+\frac{(-1)^{n} x^{2 n}}{2^{2 n}(n !)^{2}}+\ldots$$
  of the differential equation $$x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x y=0$$
  Show that $$y_{2}(x)=u(x) J_{0}(x)$$ is a second solution if $$u(x)=\int \frac{d x}{x J_{0}^{2}}$$

Solution :
Firstly
Let $y(x)=\sum_{\lambda=0}^{\infty} a_{\lambda} x^{k+\lambda}$.
We differentiate and substitute. The result is
$$\sum_{\lambda=0}^{\infty} a_{\lambda}(k+\lambda)(k+\lambda-1) x^{k+\lambda-1}+\sum_{\lambda=0}^{\infty} a_{\lambda}(k+\lambda) x^{k+\lambda-1}+\sum_{\lambda=0}^{\infty} a_{\lambda} x^{k+\lambda+1}=0$$
By setting $\lambda=0,$ we get the coefficient of $x^{k-1},$ the lowest power of $x$ appearing on the
left-hand side,
$$\quad a_{0}\left[k(k-1)+k\right]=0$$
and $a_{0} \neq 0$ by definition. Equation therefore yields the indicial equation
with solutions $k=0$
It is of some interest to examine the coefficient of $x$ also. Here we obtain
$a_{1}=0$.
Proceeding to the coefficient of $x^{J-1}$ for $k=0,$ we set $\lambda=j$ in the first, and second terms and $\lambda=j-2$ in the third term. By requiring the resultant
coefficient of $x$ to vanish, we obtain
$$a_{j}\left[j(j-1)+j\right]+a_{j-2}=0$$
When $j$ is replaced by $j+2,$ this can be rewritten for $j \geq 0$ as
$a_{j+2}=-a_{j} \frac{1}{(j+2)(j+2)}$
which is the desired recurrence relation.
Repeated application of this recurrence relation
leads to
$$a_{2}=-a_{0} \frac{1}{2(2)}=-\frac{a_{0} }{2^{2}}$$
$$a_{4}=-a_{2} \frac{1}{4(4)}=\frac{a_{0} }{2^{4} (2!)^{2}}$$
$$a_{6}=-a_{4} \frac{1}{6(6)}=-\frac{a_{0} }{2^{6} (3!)^{2}}, \quad$$ and so on
and in general,
$$a_{2 p}=(-1)^{p} \frac{a_{0}}{2^{2 p} (p!)^{2}}$$
Inserting these coefficients in our assumed series solution and take $a_0=1$, we have
$$y_1(x)=J_0(x)= \left[1-\frac{ x^{2}}{2^{2} (1!)^{2}}+\frac{ x^{4}}{2^{4} (2 !)^{2}}-\cdots\right]$$
In summation form
$$y_1(x)=J_0(x)=\sum_{j=0}^{\infty}(-1)^{j} \frac{ x^{2 j}}{2^{2j} (j!)^{2} }= \sum_{j=0}^{\infty}(-1)^{j} \frac{1}{(j !)^{2}}\left(\frac{x}{2}\right)^{2 j}$$
Secondly 
If $$y_{2}(x)=u(x) J_{0}(x)$$ where $$u(x)=\int \frac{d x}{x J_{0}^{2}}$$ then $$y'_{2}(x)=u(x) J'_{0}(x)+u'(x) J_{0}(x)$$ and $$y''_{2}(x)=u'(x) J'_{0}(x)+u(x) J''_{0}(x)+u''(x) J_{0}(x)+u'(x) J'_{0}(x)$$ We get 
$$x(u'(x) J'_{0}(x)+u(x) J''_{0}(x)+u''(x) J_{0}(x)+u'(x) J'_{0}(x))+u(x) J'_{0}(x)+u'(x) J_{0}(x)+x(u(x) J_{0}(x))$$
$$xu'(x) J'_{0}(x)+xu(x) J''_{0}(x)+xu''(x) J_{0}(x)+xu'(x) J'_{0}(x)+u(x) J'_{0}(x)+u'(x) J_{0}(x)+xu(x) J_{0}(x)$$
$$2xu'(x) J'_{0}(x)+xu(x) J''_{0}(x)+xu''(x) J_{0}(x)+u(x) J'_{0}(x)+u'(x) J_{0}(x)+xu(x) J_{0}(x)$$
$$u(x)(x J''_{0}(x)+ J'_{0}(x)+x J_{0}(x))+xu''(x) J_{0}(x)+u'(x) J_{0}(x)+2xu'(x) J'_{0}(x)$$
I can not continue.. 

Your series solution looks good.
As for finding the second solution... it would help a great deal if we actually used the definition of $u$.
\begin{align*}y_2(x) &= u(x)J_0(x)\\
y_2'(x) &= u(x)J_0'(x) + u'(x)J_0(x) = u(x)J_0'(x) + \frac{1}{xJ_0(x)}\\
y_2''(x) &= u(x)J_0''(x) + u'(x)J_0'(x) - \frac{J_0(x)+xJ_0'(x)}{x^2J_0^2(x)}\\
&= u(x)J_0''(x) + \frac{J_0'(x)}{xJ_0^2(x)} -\frac{1}{x^2J_0(x)}-\frac{J_0'(x)}{xJ_0^2(x)}\\
y_2''(x) &= u(x)J_0''(x) - \frac{1}{x^2J_0(x)}\end{align*}
We have of course used the Fundamental Theorem to differentiate $u$. And now, we put those calculated derivatives into the differential equation:
\begin{align*}xy_2(x) + y_2'(x) + xy_2(x) &= xu(x)J_0(x) + u(x)J_0'(x) + \frac{1}{xJ_0(x)} + xu(x)J_0''(x) - \frac{1}{xJ_0(x)}\\
&= u(x)\left(xJ_0(x)+J_0'(x)+xJ_0''(x)\right) = u(x)\cdot 0 = 0\end{align*}
We use that $J_0$ is a solution of the differential equation, and there it is.
