The solution of the equation dy dx = y x ( y x + 1 ) is - Vidyadip

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
$y = xy + c$

✓

Answer

Correct option: A.

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

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

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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