Given, $y=e^x(\log x)$
$$
\frac{d y}{d x}=\frac{e^x}{x}+e^x \ln x
$$

$$
\begin{aligned}
& \Rightarrow \quad \frac{d y}{d x}=\frac{e^x}{x}+y \Rightarrow x \frac{d^2 y}{d x^2}=x \frac{x \cdot e^x-e^x}{x^2}+\frac{d y}{d x} x \\
& \Rightarrow x \frac{d^2 y}{d x^2}=\frac{x^2 e^x-x e^x}{x^2}+x \frac{d y}{d x} \\
& \Rightarrow x \frac{d^2 y}{d x^2}+(x-1) y=\frac{x e^x\{x-1\}}{x^2}+x \frac{d y}{d x}+(x-1) y \\
& \Rightarrow \frac{x d^2 y}{d x^2}+(x-1) y=\frac{e^x(x-1)}{x}+(x-1) y+x \frac{d y}{d x} \\
& =e^x-\frac{e^x}{x}+(x-1) e^x \ln x+x \frac{d y}{d x} \\
& =(x-1) e^x\left\{\frac{1}{x}+\ln x\right\}+x \frac{d y}{d x} \\
&
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow \quad x \frac{d^2 y}{d x^2}+(x-1) y=(x-1) \quad\left[\frac{e^x}{x}+e^x \ln x\right]+x \frac{d y}{d x} \\
& \Rightarrow x \frac{d^2 y}{d x^2}+(x-1) y=(x-1+x) \frac{d y}{d x} \\
& x \frac{d^2 y}{d x^2}+(x-1) y=(2 x-1) \frac{d y}{d x} \\
&
\end{aligned}
$$
[by Eq. (i)]