Question

The differential equation representing the family of curves $y={xe}^{cx},$ where $c$ is a constant, is

• A. $\frac{dy}{dx}=\frac{y}{x}(1-\ln \frac{y}{x})$
• B. $\frac{dy}{dx}=\frac{y}{x}\ln \left(\frac{y}{x}\right)+1$
• C. $\frac{dy}{dx}=\frac{y}{x}(1+\ln \frac{y}{x})$
• D. $\frac{dy}{dx}+1=\frac{y}{x}\ln \frac{y}{x}$

Answer 1

The correct option is C $\frac{dy}{dx}=\frac{y}{x}(1+\ln \frac{y}{x})$

$y=x{e}^{cx}$
$\Rightarrow \frac{y}{x}=e^{cx}$
Taking $\ln$ both sides, we get
$\ln \frac{y}{x}=\ln e^{cx}$
$\Rightarrow \ln \frac{y}{x}=cx$
$\Rightarrow \frac{1}{x}\ln \frac{y}{x}=c$
Differentiating both sides w.r.t. $x$ :
$\frac{1}{x}(\frac{x}{y}\times \frac{1}{x^2}(x\frac{dy}{dx}-y))+\ln \frac{y}{x}(-\frac{1}{x^2})=0$
$\Rightarrow \frac{1}{xy}\left(\frac{dy}{dx}\right)=\frac{1}{x^2}(\ln \frac{y}{x}+1)$
$\Rightarrow \frac{dy}{dx}=\frac{y}{x}(\ln \frac{y}{x}+1)$