Question

If $x\frac{dy}{dx}=y(logy-logx+1)$, then the solution of the equation is

• A. $ylog\left(\frac{x}{y}\right)=cx$
• B. $xlog\left(\frac{y}{x}\right)=cy$
• C. $log\left(\frac{y}{x}\right)=cx$
• D. $log\left(\frac{x}{y}\right)=cx$

Answer 1

The correct option is C $log\left(\frac{y}{x}\right)=cx$

$\frac{dy}{dx}=\frac{y}{x}(log\frac{y}{x}+1)$
Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\therefore v+x\frac{dv}{dx}=vlogv+v\Rightarrow \frac{1}{vlogv}dv=\frac{1}{x}dx\Rightarrow \int \frac{1}{vlogv}dv=\int \frac{1}{x}dx\Rightarrow log\left(logv\right)=logx+logc$
$\Rightarrow log\frac{y}{x}=cx$