Question

The solution of $\frac{xdy}{dx}=y\left(ln\,y-ln\,x+1\right)$ is

• A. $x\,ln\, \frac{y}{x}=cy$
• B. $y\,ln\, \frac{x}{y}=cx$
• C. $ln\, \frac{x}{y}=cy$
• D. $ln\, \frac{y}{x}=cx$

Answer 1

Answer: (D)

$\frac{dy}{dx}=\frac{y}{x}\left(1+ln \frac{y}{x}\right)$
Substitute $y=vx$
$\Rightarrow \frac{dy}{dx}=\frac{xdv}{dx}+v$
So, given equation becomes
$\frac{xdv}{dx}+v=v\left(1+ln\,v\right)$
$\Rightarrow \int \frac{dv}{v\,ln\,v}=\int \frac{dx}{x}$
$\Rightarrow ln\left(ln \frac{y}{x}\right)=ln\,x+ln\,c$
$\Rightarrow ln \frac{y}{x}=cx$.