Question

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

• A. $\log y=\mathrm{cx}+logx$
• B. $x\log y=cx$
• C. $x\log y=cy$
• D. $y=x{e}^{\log x}$

Answer 1

Answer: (A)

$\log y=\mathrm{cx}+logx$

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

$\Rightarrow y=vx$

$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$v+x\frac{dv}{dx}=\frac{y}{x}(\log \frac{y}{x}+1)$

$v+x\frac{dv}{dx}=v(\log v+1)$

$v+x\frac{dv}{dx}=v\log v+v$

$x\frac{dv}{dx}=v\log v$

$\frac{dv}{v\log v}=\frac{dx}{x}$

Integrating both sides

$\Rightarrow \int \frac{dv}{v\log v}=\int \frac{dx}{x}$

$\log \left(v\right)=\log x+\log c$

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

$\log y=cx+\log x$

Hence, the answer is $\log y=cx+\log x.$