Question

The family of curves passing through $(0,0)$ and satisfying the differential equation $\frac{y^{\prime \prime }}{y^{\prime }}=1$ $(\text{where}{y}^{\prime }=\frac{dy}{dx}\text{and}{y}^{\prime \prime }=\frac{d^{2}y}{d{x}^2})$ is

• A. $y=k$
• B. $y=kx$
• C. $y=k(e^x+1)$
• D. $y=k(e^x-1)$

Answer 1

The correct option is D $y=k(e^x-1)$

$\frac{dp}{dx}=p(\text{where}p=\frac{dy}{dx})$
$\ln p=x+c\Rightarrow p=e^{x+c}$
$\frac{dy}{dx}=k{e}^x\Rightarrow y=k{e}^x+\lambda$
Satisfying $(0,0)$, So $\lambda =-k$
$y=k(e^x-1)$