Solve $x \frac{d y}{d x}=y(\log y-\log x+1)$
$$\begin{array}{ll} \text { Given, } & x \frac{d y}{d x}=y(\log y-\log x+1) \\ \Rightarrow & x \frac{d y}{d x}=y \log \left(\frac{y}{x}+1\right) \\ \Rightarrow & \frac{d y}{d x}=\frac{y}{x}\left(\log \frac{y}{x}+1\right)\quad\text{.... (i)} \end{array}$$
which is a homogeneous equation.
$$\begin{aligned} \text { Put } & \frac{y}{x} =v \text { or } y=v x \\ \therefore & \frac{d y}{d x} =v+x \frac{d v}{d x} \end{aligned}$$
$$\begin{aligned} &\text { On substituting these values in Eq.(i), we get }\\ &\begin{array}{rlrl} \Rightarrow & v+x \frac{d v}{d x} =v(\log v+1) \\ \Rightarrow & x \frac{d v}{d x} =v(\log v+1- \\ \Rightarrow & x \frac{d v}{d x} =v(\log v) \\ \Rightarrow & \frac{d v}{v \log v}=\frac{d x}{x} \end{array} \end{aligned}$$
On integrating both sides, we get
$$\int \frac{d v}{v \log v}=\int \frac{d x}{x}$$
On putting $\log v=u$ in LHS integral, we get
$$\begin{aligned} & \frac{1}{v} \cdot d v=d u \\ & \int \frac{d u}{u}=\int \frac{d x}{x} \end{aligned}$$
$$\begin{aligned} \Rightarrow \quad & \log u =\log x+\log C \\ \Rightarrow \quad & \log u =\log C x \\ \Rightarrow \quad & u =C x \\ \Rightarrow \quad & \log v =C x \\ \Rightarrow \quad & \log \left(\frac{y}{x}\right) =C x \end{aligned}$$
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