Given, $\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}(\log{\text{y}-\log\text{x}+1})$
$\Rightarrow \text{x}\frac{\text{dy}}{\text{dx}}=\text{y}\log\Big(\frac{\text{y}}{\text{x}} + 1\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\Big(\log\frac{\text{y}}{\text{x}} + 1\Big)\ .....(\text{i})$
Which is a homogeneous equation.
Put $\frac{\text{y}}{\text{x}}=\text{v}$ or $\text{y}=\text{vx}$
$\therefore\frac{\text{dy}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}$
On substituting these values in Eq. (i), we get
$\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}(\log\text{v}+1)$
$\Rightarrow\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}(\log\text{v}+1-1)$
$\Rightarrow\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}(\log\text{v})$
$\Rightarrow\frac{\text{dv}}{\text{v}\log\text{v}}=\frac{\text{dx}}{\text{x}}$
On integrating both sides, we get
$\Rightarrow\int\frac{\text{dv}}{\text{v}\log\text{v}}=\int\frac{\text{dx}}{\text{x}}$
On putting $\log\text{v}=\text{u}$ in LHS integral, we get
$\frac{1}{\text{v}}.\text{dv}=\text{du}$
$\int\frac{\text{du}}{\text{u}}=\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\log\text{u}=\log\text{x}+\log\text{C}$
$\Rightarrow\log\text{u}=\log\text{C}\text{x}$
$\Rightarrow\text{u}=\text{C}\text{x}$
$\Rightarrow\log\text{v}=\text{C}\text{x}$
$\Rightarrow\log\Big(\frac{\text{y}}{\text{x}}\Big)=\text{C}\text{x}$
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