Question

If $xdy=y(logy-logx+1)dx$, then the solution of the equation is

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

Answer 1

Answer: (C)

$\log \left(y/x\right)=cx$

$xdy=y(logy-logx+1)dx$

or, $xdy-ydx=y\left(\log \frac{y}{x}\right)dx$

or, $\frac{xdy-ydx}{x^2}=\frac{y}{x}\left(\log \frac{y}{x}\right)\frac{dx}{x}$

or, $d\left(\frac{y}{x}\right)=\frac{y}{x}\left(\log \frac{y}{x}\right)\frac{dx}{x}$

or, $\frac{d\left(\frac{y}{x}\right)}{\frac{y}{x}\log \frac{y}{x}}=\frac{dx}{x}$

or, $d\left\{\log \left(\log \frac{y}{x}\right)\right\}=\frac{dx}{x}$

Now integrating we get,

$\log \left(\log \frac{y}{x}\right)=\log cx$ [ $c$ is integrating constant]

or, $\log \frac{y}{x}=cx$.