Question

The family of curves passing through (0,0) and satisfying the differential equation $\frac{y_2}{y_1}=1$ (where $y_n=\frac{d^{n}y}{d{x}^n})$ 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 $P=x+c\Rightarrow p=e^{x+c}$
$\frac{dy}{dx}=k{e}^{x}y=k{e}^x+\lambda$
Satisfying (0,0), So $\lambda =-k$
$y=k(e^x-1)$