Question

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

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

Answer 1

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

$x\frac{dy}{dx}=y(\log y-\log x+1)$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}(\log \frac{y}{x}+1)$
Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=v(\log v+1)$
$\Rightarrow \frac{dv}{v\log v}=\frac{dx}{x}$
Integrating both sides,
$\Rightarrow \log \left(\log v\right)=\log x+logc=\log cx$
$\Rightarrow \log v=cx\Rightarrow \log \frac{y}{x}=cx$