Finding the general solution of $xy''+y'+xy=0$ using reduction of order. 
Use reduction of order to show that the general solution of $$xy''+y'+xy=0$$ can be written as $$y=c_1J_0(x)+c_2J_0(x) \int^x_1 \frac{1}{s[J_0(x)]^2} ds.$$ Show that if $c_2 \neq 0, y \approx c_2 \ln(x)$ as $x \rightarrow 0^+.$

Notes: First we need to transform the original ODE into Bessel's ODE $x^2y''+xy'+x^2y=0$. As $J_0(x)$ (a 0 order Bessel function of the first kind) is a solution to Bessel's ODE we can write it as the second solution as 
\begin{align*}
y_2&=J_0(x)u(x) \\
y'_2&=J_0'(x)u(x)+J_0(x)u'(x) \\
y''_2 &=J_0^{\prime\prime}u+2J_0^\prime u^\prime+J_0u^{\prime\prime}
\end{align*}
We can be substituted into $x^2y''+xy'+x^2y=0$ to find the value of $u(x)$
From there my main issues are finding $J_0'(x)$ and $J_0''(x)$, as well as finding the definite integral  $\int^x_1 \frac{1}{s[J_0(x)]^2} ds$. How does reduction of order give a definite integral for $u(x)$?
 A: Note that you do not need to find $J_0^\prime$ or $J_0^{\prime\prime}$ to find $u$ and the form $\int_0^x\dfrac{1}{sJ_0(s)}\,ds$ is merely a way of writing the indefinite integral explicitly as a function of $x$. It essentially means the same thing as $\int\dfrac{1}{xJ_0}\,dx$.
\begin{eqnarray}
y_2&=&J_0u\\
y_2^\prime&=&J_0^\prime u+J_0u^\prime\\
y_2^{\prime\prime}&=&J_0^{\prime\prime}u+2J_0^\prime u^\prime+J_0u^{\prime\prime}
\end{eqnarray}
\begin{equation}
x^2y_2^{\prime\prime}+xy_2^\prime+x^2y_2=0
\end{equation}
\begin{equation}
x^2\left(J_0^{\prime\prime}u+2J_0^\prime u^\prime+J_0u^{\prime\prime}\right)+x\left(J_0^\prime u+J_0u^\prime\right)+x^2u=0
\end{equation}
\begin{equation}
(x^2J_0^{\prime\prime}+xJ_0^\prime+x^2J_0)u+x^2(2J_0^\prime u^\prime+J_0u^{\prime\prime})+xJ_0u^\prime=0
\end{equation}
But
\begin{equation}
x^2J_0^{\prime\prime}+xJ_0^\prime+x^2J_0=0
\end{equation}
So
\begin{equation}
x^2(2J_0^\prime u^\prime+J_0u^{\prime\prime})+xJ_0u^\prime=0 \tag{1}
\end{equation}
Substitute $v=J_0u^\prime$. Then $v^\prime=J_0^\prime u^\prime+J_0y^{\prime\prime}$ and $u^\prime=\dfrac{V}{J_0}$.
So equation ($1$) can be re-written as
\begin{eqnarray}
x^2\left(\dfrac{J_0^\prime}{J_0}v+v^\prime\right)+xV&=&0\\
x^2v^\prime+\left(x+x^2\cdot\dfrac{J_0^\prime}{J_0}\right)v&=&0\\
v^\prime+\left(\dfrac{1}{x}+\dfrac{J_0^\prime}{J_0}\right)v&=&0
\end{eqnarray}
which has an integrating factor $\mu=xJ_0$ giving $u^\prime J_0=\dfrac{1}{xJ_0}$ so that
\begin{equation}
u=\int\dfrac{1}{xJ_0^2}\,dx
\end{equation}
