Question

If $x \frac{dy}{dx}=y \left(log\,y-log\,x+1\right),$ 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. $y\,log \left(\frac{x}{y}\right)=cx$

Answer 1

Answer: (B)

$x \frac{dy}{dx}=y\left(log\,y-log\,y+1\right)$
$\therefore \frac{dy}{dx}=\left(\frac{y}{x}\right)\left(log\left(\frac{y}{x}\right)+1\right)$
Now put $\frac{y}{x}=t$
$\Rightarrow y=t\,x \Rightarrow \frac{dy}{dx}=t+x \frac{dt}{dx}$
$\therefore t+x \frac{dt}{dx}=t\,log\,t+t$
$\Rightarrow t\,log\,t\, dx=x\,dt$
$\Rightarrow \frac{dt}{t\,log\,t}=\frac{dx}{x}$
$\Rightarrow log\left(\frac{y}{x}\right)=cx$