Question

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

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

Answer 1

Answer: (D)

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

$x\frac{dy}{dx}=y(\log y-\log x+1)$
$\therefore$ $\frac{dy}{dx}=\frac{y}{x}(\log \left(\frac{y}{x}\right)+1)$

Now substitute $\frac{y}{x}=v$
$\therefore$ $v\log vdx=xdy$ $\Rightarrow$ $\frac{dy}{v\log v}=\frac{dx}{x}$ $\Rightarrow$ $\log \left(\frac{y}{x}\right)=cx$