Question

Solve the differential equation$x\frac{dy}{dx}=y(\log y-\log x+1)$

• A. $\log y-\log x=x+c$
• B. $\log y+\log x=cx.$
• C. $\log y-\log x=cx.$
• D. $\log y-\log x+cx=0.$

Answer 1

The correct option is C $\log y-\log x=cx.$

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

$\Rightarrow \frac{1}{y}\frac{dy}{dx}-\frac{1}{x}\log y=\frac{1-\log x}{x.}$
Put $\log y=v\Rightarrow \frac{1}{y}dy=dv$
$\therefore \frac{dv}{dx}-\frac{v}{x}=\frac{1-\log x}{x}$ ...(1)
Here $P=\frac{-1}{x}\Rightarrow \int Pdx=\int \frac{-1}{x}dx=-\log x$
$\therefore I.F=e^{-\log x}=e^{\log x-1}=\frac{1}{x}$
Multiplying (1) by $I.F,$ we get
$\frac{1}{x}\frac{dv}{dx}-\frac{v}{x^2}=\frac{1-\log x}{x^2}$
Integrating both sides we get
$\frac{1}{x}v=\frac{\log x}{x}+c$

$\Rightarrow \log y-\log x=cx$