Question

If $e^y+xy=e,$ the ordered pair $(\frac{dy}{dx},\frac{d^{2}y}{d{x}^2})$ at $x=0$ is equal to :

• A. $(-\frac{1}{e},\frac{1}{e^2})$
• B. $(\frac{1}{e},-\frac{1}{e^2})$
  
• C. $(\frac{1}{e},\frac{1}{e^2})$
• D. $(-\frac{1}{e},-\frac{1}{e^2})$

Answer 1

The correct option is A $(-\frac{1}{e},\frac{1}{e^2})$

Given equation, $e^y+xy=e,$
As $x=0\Rightarrow y=1$
$\Rightarrow e^y\frac{dy}{dx}+x\frac{dy}{dx}+y=0\cdots \left(1\right)$
$\Rightarrow \frac{dy}{dx}=-\frac{y}{e^y+x}$
At $x=0,y=1\frac{dy}{dx}=-\frac{1}{e}$
On differentiating $\left(1\right)$ again, we have -
$e^y{\left(\frac{dy}{dx}\right)}^2+e^y\frac{d^{2}y}{d{x}^2}+x\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}=0$

At $x=0,y=1$
$\Rightarrow e{(-\frac{1}{e})}^2+e\cdot \frac{d^{2}y}{d{x}^2}-\frac{2}{e}=0$

$\Rightarrow e\cdot \frac{d^{2}y}{d{x}^2}=\frac{1}{e}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{e^2}$