Passage: Consider the differential equation e^ x+y (dy)/(dx)=e^ x-y...

Given the differential equation:

$e^{x+y}\dfrac{dy}{dx} = e^{x-y}$

$e^x e^y \dfrac{dy}{dx} = e^x e^{-y}$

Dividing both sides by $e^x$:

$\dfrac{dy}{dx} = \dfrac{e^{-y}}{e^y} = e^{-2y}$

Differentiating both sides with respect to $x$:

$\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}(e^{-2y}) = -2e^{-2y} \dfrac{dy}{dx}$

Substituting $\dfrac{dy}{dx} = e^{-2y}$:

$\dfrac{d^2y}{dx^2} = -2e^{-2y}(e^{-2y}) = -2e^{-4y}$

From $\dfrac{dy}{dx} = e^{-2y}$, we get:

$\dfrac{dx}{dy} = \dfrac{1}{e^{-2y}} = e^{2y}$

Squaring both sides:

\left(\dfrac{dx}{dy}\right)^2 = (e^{2y})^2 = e^{4y}$<br><br>Now, multiplying the two results:<br><br>$\dfrac{d^2y}{dx^2}\left(\dfrac{dx}{dy}\right)^2 = (-2e^{-4y})(e^{4y}) = -2e^0 = -2$<br><br>Answer:$-2