If e^y + xy = e,then the value of frac{d^2y}{ | vedclass

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(B) Given the equation $e^y + xy = e$. Differentiating with respect to $x$,we get: $e^y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$ .....$(i)$ Differenti

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$\frac{1}{e^2}$

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(B) Given the equation $e^y + xy = e$. Differentiating with respect to $x$,we get: $e^y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$ .....$(i)$ Differentiating again with respect to $x$,we get: $e^y \frac{d^2y}{dx^2} + e^y \left( \frac{dy}{dx} \right)^2 + \frac{dy}{dx} + \frac{dy}{dx} + x \frac{d^2y}{dx^2} = 0$ $e^y \frac{d^2y}{dx^2} + e^y \left( \frac{dy}{dx} \right)^2 + 2 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = 0$ .....$(ii)$ Putting $x = 0$ in the original equation $e^y + xy = e$,we get $e^y + 0 = e$,which implies $y = 1$. Substituting $x = 0$ and $y = 1$ in equation $(i)$: $e^1 \frac{dy}{dx} + 1 + 0 = 0$ $e \frac{dy}{dx} = -1$ $\frac{dy}{dx} = -\frac{1}{e}$ Substituting $x = 0$,$y = 1$,and $\frac{dy}{dx} = -\frac{1}{e}$ in equation $(ii)$: $e \frac{d^2y}{dx^2} + e \left( -\frac{1}{e} \right)^2 + 2 \left( -\frac{1}{e} \right) + 0 = 0$ $e \frac{d^2y}{dx^2} + e \left( \frac{1}{e^2} \right) - \frac{2}{e} = 0$ $e \frac{d^2y}{dx^2} + \frac{1}{e} - \frac{2}{e} = 0$ $e \frac{d^2y}{dx^2} - \frac{1}{e} = 0$ $e \frac{d^2y}{dx^2} = \frac{1}{e}$ $\frac{d^2y}{dx^2} = \frac{1}{e^2}$

If $e^y + xy = e$,then the value of $\frac{d^2y}{dx^2}$ for $x = 0$ is

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