If \(x \frac{\mathrm{~d} y}{\mathrm{~d} x}=y(\log y-\log x+1)\), then general solution of this equation is

\(\begin{aligned}
& x \frac{\mathrm{~d} y}{\mathrm{~d} x}=y(\log y-\log x+1) . \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x}\left[\log \left(\frac{y}{x}\right)+1\right] \\
& \text { Put } \mathrm{v}=\frac{y}{x} \\
& \therefore \quad y=\mathrm{v} x \\
& \frac{\mathrm{~d} y}{\mathrm{~d} x}=\mathrm{v}+x \frac{\mathrm{~d} v}{\mathrm{~d} x} \\
& \text { From (i) }
\end{aligned}\)
\(\begin{aligned}
& \therefore \quad \mathrm{v}+x \frac{\mathrm{~d} v}{\mathrm{~d} x}=\mathrm{v}(\log \mathrm{v}+1) \\
& \mathrm{v}+x \frac{\mathrm{~d} v}{\mathrm{~d} x}=\operatorname{vlog} \mathrm{v}+\mathrm{v} \\
& x \frac{\mathrm{~d} v}{\mathrm{~d} x}=\mathrm{v} \log \mathrm{v} \\
& \frac{1}{v \log v} d v=\frac{d x}{x}
\end{aligned}\)
Integrating on both sides, we get
\(\int \frac{1}{\operatorname{vlog} v} \mathrm{dv}=\int \frac{\mathrm{d} x}{x}\)
\(\begin{aligned} & \log (\log \mathrm{v})=\log x+\mathrm{c}_1 \\ & \log (\log \mathrm{v})=\log x+\log \mathrm{c} \text { where, } \mathrm{c}_1=\log \mathrm{c} \\ & \Rightarrow \log (\log \mathrm{v})=\log (x \mathrm{c}) \\ & \Rightarrow \log \mathrm{v}=x \mathrm{c} \\ & \Rightarrow \log \left(\frac{y}{x}\right)=\mathrm{c} x\end{aligned}\)