If e^y + xy = e,the ordered pair left(frac{dy | vedclass

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(D) Given equation is $e^y + xy = e$. At $x = 0$,$e^y + 0 = e \Rightarrow e^y = e \Rightarrow y = 1$. Differentiating $e^y + xy = e$ with respect to $

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given equation is $e^y + xy = e$. At $x = 0$,$e^y + 0 = e \Rightarrow e^y = e \Rightarrow y = 1$. Differentiating $e^y + xy = e$ with respect to $x$: $e^y \frac{dy}{dx} + x \frac{dy}{dx} + y = 0$ $\Rightarrow \frac{dy}{dx} (e^y + x) = -y$ $\Rightarrow \frac{dy}{dx} = -\frac{y}{e^y + x}$. At $x = 0$ and $y = 1$: $\frac{dy}{dx} = -\frac{1}{e^1 + 0} = -\frac{1}{e}$. Now,differentiate $\frac{dy}{dx} (e^y + x) = -y$ with respect to $x$: $\frac{d^2y}{dx^2} (e^y + x) + \frac{dy}{dx} (e^y \frac{dy}{dx} + 1) = -\frac{dy}{dx}$ $\Rightarrow \frac{d^2y}{dx^2} (e^y + x) + e^y \left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} = -\frac{dy}{dx}$ $\Rightarrow \frac{d^2y}{dx^2} (e^y + x) + e^y \left(\frac{dy}{dx}\right)^2 + 2 \frac{dy}{dx} = 0$. Substituting $x = 0, y = 1, \frac{dy}{dx} = -\frac{1}{e}$: $\frac{d^2y}{dx^2} (e + 0) + e \left(-\frac{1}{e}\right)^2 + 2 \left(-\frac{1}{e}\right) = 0$ $e \frac{d^2y}{dx^2} + e \left(\frac{1}{e^2}\right) - \frac{2}{e} = 0$ $e \frac{d^2y}{dx^2} + \frac{1}{e} - \frac{2}{e} = 0$ $e \frac{d^2y}{dx^2} = \frac{1}{e}$ $\frac{d^2y}{dx^2} = \frac{1}{e^2}$. Thus,the ordered pair is $\left(-\frac{1}{e}, \frac{1}{e^2}\right)$.

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

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