Question

The differential equation representing the family of curves given by $y=a{e}^{-3x}+b,$ where $a$ and $b$ are arbitrary constants, is :
$\frac{d^{2}y}{d{x}^2}+3\frac{dy}{dx}-2y=0$

• A. $\frac{d^{2}y}{d{x}^2}+3\frac{dy}{dx}-2y=0$
• B. $\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}=0$
• C. $\frac{d^{2}y}{d{x}^2}-3\frac{dy}{dx}-2y=0$
• D. $\frac{d^{2}y}{d{x}^2}+3\frac{dy}{dx}+2y=0$
• E. $\frac{d^{2}y}{d{x}^2}+3\frac{dy}{dx}=0$

Answer 1

The correct option is E $\frac{d^{2}y}{d{x}^2}+3\frac{dy}{dx}=0$

Given differential equation is
$y=a{e}^{-3x}+b$ ...(i)
On differtiating w.r.t. $x$, we get
$\frac{dy}{dx}=-3a{e}^{-3x}+0$ ...(ii)
Again differentiating, we get
$\frac{d^{2}y}{d{x}^2}=9a{e}^{-3x}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=3(-\frac{dy}{dx})$ ....[From Eq. (i)]
$\Rightarrow \frac{d^{2}y}{d{x}^2}+3\frac{dy}{dx}=0$