Question

The general solution of the differential equation, $x\left(\frac{dy}{dx}\right)=y\ln \left(\frac{y}{x}\right),$ where $c$ is an arbitrary constant, is :

• A. $\frac{y}{x}=e^{1+cx}$
• B. $\frac{x}{y}=e^{1-cx}$
• C. $\frac{y}{x}=e^{1-cx}$
• D. $\frac{x}{y}=e^{1+cx}$

Answer 1

Answer: (A), (C)

$\frac{y}{x}=e^{1-cx}$

Given: $x\left(\frac{dy}{dx}\right)=y\ln \left(\frac{y}{x}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}\ln \left(\frac{y}{x}\right)\cdots \left(i\right)$

Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$

The equation $\left(i\right)$ transforms to:
$v+x\frac{dv}{dx}=v\ln v$
$\Rightarrow x\frac{dv}{dx}=v(\ln v-1)$
$\Rightarrow \int \frac{dv}{v(\ln v-1)}=\int \frac{1}{x}dx$
$\Rightarrow \ln |\ln v-1|=\ln x+\ln c\Rightarrow |\ln v-1|=cx$
$\Rightarrow |\ln \frac{y}{x}-1|=cx$
$\Rightarrow \frac{y}{x}=e^{1\pm cx}$