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

Answer 1

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

$\frac{xdy}{dx}=y(logy-logx+1)$
$\frac{dy}{dx}=\frac{y}{x}(log\frac{y}{x}+1)$
Put y=vx
$\frac{dy}{dx}=v+\frac{xdv}{dx}\Rightarrow v+\frac{xdv}{dx}=v(logv+1)$
$\frac{xdv}{dx}=vlogv\Rightarrow \frac{dv}{vlogv}=\frac{dx}{x}$
Put log v=z
$\frac{1}{v}=dz\Rightarrow \frac{dz}{z}=\frac{dx}{x}lnz=lnx+lnc$
$z=cxorlogv=cxorlog\left(\frac{y}{x}\right)=cx$.