MCQ
The solution of the equation $\frac{{dy}}{{dx}} = \frac{y}{x}\left( {\log \frac{y}{x} + 1} \right)$ is
- ✓$\log \left( {\frac{y}{x}} \right) = cx$
- B$\frac{y}{x} = \log y + c$
- C$y = \log y + 1$
- D$y = xy + c$
Put $y = vx$ ==> $\frac{{dy}}{{dx}} = v + x.\frac{{dv}}{{dx}}$
$\therefore v + x.\frac{{dv}}{{dx}} = v(\log v + 1)$
$v + x\frac{{dv}}{{dx}} = v\log v + v$ ==> $x\frac{{dv}}{{dx}} = v\log v$==> $\frac{{dv}}{{v\log v}} = \frac{{dx}}{x}$
Integrating both sides, $\int_{}^{} {\frac{{dv}}{{v\log v}}} = \int_{}^{} {\frac{{dx}}{x}} $
$\log \log v = \log x + \log c$$ \Rightarrow $$\log v = xc$ ==> $\log (y/x) = \,x\,c$.
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