Question

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

• A. $\log \frac{y}{x}=cx$
• B. $\frac{y}{x}=\log y+c$
• C. $y=\log y+1$
• D. $y=xy+c$

Answer 1

The correct option is A $\log \frac{y}{x}=cx$

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

$y=Vx$

$\frac{dy}{dx}=V+x\frac{dV}{dx}$

$V+x\frac{dV}{dx}=V(\ln V+1)$

$V+x\frac{dV}{dx}=V\ln V+V$

$\int \frac{dV}{V\ln v}=\int \frac{dx}{x}$

$\ln V=t$

$\frac{1}{V}dV=dt$

$\int \frac{dt}{t}=\ln x+c$

$\ln t=\ln x+c\Rightarrow \ln \left(\frac{t}{x}\right)=c$

$\frac{t}{x}=c$

$\ln V=cx$

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