Solve $x\frac{{dy}}{{dx}} = y(\log y - \log x + 1)$

Given, $x\frac{{dy}}{{dx}} = y(\log y - \log x + 1)$

$\Rightarrow$ $x\frac{{dy}}{{dx}} = y\log \left( {\frac{y}{x} + 1} \right)$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{y}{x}\left( {\log \frac{y}{x} + 1} \right)$

…….(i)
which is a homogeneous equation.

Put $\frac{y}{x} = v$ or $y = vx$

$\therefore \quad \frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$

On substituting these values in Eq.(i),

we get$v + x\frac{{dv}}{{dx}} = v(\log v + 1)$

$\Rightarrow$ $x\frac{{dv}}{{dx}} = v(\log v + 1 - 1)$

$\Rightarrow$ $x\frac{{dv}}{{dx}} = v(\log v)$

$\Rightarrow$ $\frac{{dv}}{{v\log v}} = \frac{{dx}}{x}$

On integrating both sides,

we get $\int {\frac{{dv}}{{v\log v}}} = \int {\frac{{dx}}{x}}$

On putting $\log v = u$ in LHS integral,

we get$\frac{1}{v} \cdot dv = du$

$\int {\frac{{du}}{u}} = \int {\frac{{dx}}{x}}$

$\Rightarrow$ $\log u = \log x + \log C$

$\Rightarrow$ $\log u = \log Cx$

$\Rightarrow$ $u = Cx$

$\Rightarrow$ $\log v = Cx$

$\Rightarrow$ $\log \left( {\frac{y}{x}} \right) = Cx$

OBJECTIVE TYPE