# Solve the following quadratic equations

Question:

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

1. $\left(-\frac{1}{\mathrm{e}}, \frac{1}{\mathrm{e}^{2}}\right)$

2. $\left(\frac{1}{\mathrm{e}}, \frac{1}{\mathrm{e}^{2}}\right)$

3. $\left(\frac{1}{\mathrm{e}},-\frac{1}{\mathrm{e}^{2}}\right)$

4. $\left(-\frac{1}{\mathrm{e}},-\frac{1}{\mathrm{e}^{2}}\right)$

Correct Option: 1

Solution:

$e^{y}=x y=e$

differentiate w.r.t. $\mathrm{x}$

$e^{y} \frac{d y}{d x}+x \frac{d y}{d x}+y=0$

$\frac{d y}{d x}\left(x+e^{y}\right)=-y,\left.\frac{d y}{d x}\right|_{(0,1)}=-\frac{1}{e}$

again differentiate w.r.t. $\mathrm{x}$

$e^{y} \cdot \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \cdot e^{y} \cdot \frac{d y}{d x}+x \cdot \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+\frac{d y}{d x}=0$

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

$e \frac{d^{2} y}{d x^{2}}+\frac{1}{e^{2}} e+2\left(-\frac{1}{e}\right)=0$

$\therefore \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{1}{\mathrm{e}^{2}}$