Part a)
First we make the coefficient of $y''$ one, which results in
\begin{equation*}
y''(x) - \frac{(2\ln x+3)}{x(\ln x+1)} y'(x) + \frac{(2\ln x+3)}{x^2(\ln x+1)}y(x)=0
\end{equation*}
By Abel' Thm,
\begin{equation}
\ln W=\int{\frac{(2\ln x+3)}{x(\ln x+1)}}dx\label{A}
\end{equation}
Using change of variables to evaluate the integral,let
\begin{equation*}
\mu=\ln x + 1
\end{equation*}
The the integral becomes
\begin{equation*}
\int{\frac{2\mu+1}{\mu}}du =\int{2}du+\int{\frac{1}{\mu}}du \label{B}
\end{equation*}
We have \begin{equation*} \ln W=(2\mu+\ln \mu)+C_1 \end{equation*}
Plug in x back, which is \begin{equation} W= C_2x^2(\ln x + 1) \end{equation}
Part b)
To verify y = x is a solution. Just plug in, we have
\begin{equation*} x^3(\ln x+1)\cdot 0 -(2\ln x+3)x^2 \cdot 1 + (2\ln x+3) x^2 = 0 \end{equation*}
Then to find another independent solution, notice that $W > 0$ for $x > 1$
So we can just let $C_2 = 1$. And we know $W = \left| \begin{matrix} x & y \\1 & y’ \end{matrix}\right|$
Now we have a new ode
\begin{equation} y' - \frac{1}{x} y = x(\ln x + 1), \qquad x>1. \end{equation}
Solve the homogeneous ode first,
\begin{equation*} z' - \frac{1}{x}z = 0 \end{equation*}
We get $z=x$. Then we know $y=\mu(x)x$.
And $y'=\mu'x+\mu$, we plug in into (3). We have \begin{equation*} u'=(\ln x + 1) \end{equation*}
Then \begin{equation*} u=\int{\ln x +1}dx \end{equation*}
We use integral by parts to evaluate $\int{\ln x}dx$ first.
\begin{equation*} \int{\ln x} dx = x\ln x - \int{1}dx = x\ln x - 1 \end{equation*}
So we have \begin{equation*} \mu x=\ln x + x + C \end{equation*}
Finally, we have $y_2=(x\ln x+x+C)x$.
