Question

If $e^y+xy=e$ then at $x=0$,$\frac{d^{2}y}{d{x}^2}=e^{-\lambda ,}$ then numerical quantity $-\lambda$ should be equal to

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

If $e^y+xy=$e then

$e^y+xy=e$

On substituting $x=$ 0, we get $e^y=e$

$y=1$ when $x=0$

On differentiating the relation (i) we get

$e^y\frac{dy}{dx}+1.y+x.\frac{dy}{dx}=0$

On substituting, $x=0$, $y=1$ we get

$e^y\left(\frac{dy}{dx}\right)+1=0\Rightarrow \frac{dy}{dx}=\frac{-1}{e}$ on differentiating solution (ii) we get

$e^y{\left(\frac{dy}{dx}\right)}^2+e^y\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+\frac{dy}{dx}+x\frac{d^{2}y}{d{x}^2}=0$

On substituting
$x=0$, $y=1,\frac{dy}{dx}=\frac{-1}{e}weget$

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

Hence, option 'A' is correct.