Question

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

• A. $x=y{e}^{Cy}$
• B. $y=x{e}^{Cy}$
• C. $x=y{e}^{Cx}$
• D. None of these

Answer 1

Answer: (D)

None of these

We have, $\frac{dy}{dx}=\frac{y}{x}(\log \left(\frac{y}{x}\right)+1)$
Put $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=v(\log v+1)$
$\Rightarrow v+x\frac{dv}{dx}=v\log v+v$
$\Rightarrow \frac{dv}{v\log v}=\frac{dx}{x}$
$\Rightarrow \log \left(\log v\right)=\log x+\log C$
$\Rightarrow \log v=Cx$
$\Rightarrow v=e^{Cx}$
$\Rightarrow \frac{y}{x}=e^{Cx}$
$\Rightarrow y=x{e}^{Cx}$