Q. If $x\ \frac{dy}{dx} = y(\log y - \log x + 1)$, then the solution of the equation is
A) $\log\left( \frac{x}{y} \right) = cy$
B) $\log\left( \frac{y}{x} \right) = cx$
C) $x\log\left( \frac{y}{x} \right) = cy$
D) $y\log\left( \frac{x}{y} \right) = cx$
Explanation
(b)
Given, $\frac{dy}{dx} = \left( \frac{y}{x} \right)\lbrack\log{\left( \frac{y}{x} \right) + 1\rbrack\ }$
Put $\frac{y}{x} = t$
$\Rightarrow$ $y = xt$
$\Rightarrow$ $\frac{dy}{dx} = t + x\ \frac{dt}{dx}$
$\therefore$ $t + x\ \frac{dt}{dx} = t(\log{t + 1})$
$\Rightarrow$ $\frac{1}{t\log t}dt = \frac{dx}{x}$
$\Rightarrow$ $\log{(\log t)} = \log x + \log c$ [integrating]
$\Rightarrow$ $\log{\left( \frac{y}{x} \right) = cx}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.