Question

If $x\frac{dy}{dx}=y\left(logy-logx+1\right),$ then the solution of the equation is:

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

Answer 1

Answer: (B)

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