Question

If \[e^x + e^y = e^{x + y} , \text{ prove that } \frac{dy}{dx} = - \frac{e^x \left( e^y - 1 \right)}{e^y \left( e^x - 1 \right)} or \frac{dy}{dx} + e^{y - x} = 0\] ?

Answer 1

\[e^x + e^y = e^{x + y} \]

\[ \Rightarrow e^x + e^y \frac{dy}{dx} = e^{x + y} \left( 1 + \frac{dy}{dx} \right)\]

\[ \Rightarrow e^x + e^y \frac{dy}{dx} = e^{x + y} + e^{x + y} \frac{dy}{dx}\]

\[ \Rightarrow e^y \frac{dy}{dx} - e^{x + y} \frac{dy}{dx} = e^{x + y} - e^x \]

\[ \Rightarrow \frac{dy}{dx}\left( e^y - e^{x + y} \right) = e^{x + y} - e^x \]

\[ \Rightarrow \frac{dy}{dx} = \frac{e^{x + y} - e^x}{e^y - e^{x + y}}\]

\[ = \frac{e^x \left( e^y - 1 \right)}{e^y \left( 1 - e^x \right)}\]

\[ = - \frac{e^x \left( e^y - 1 \right)}{e^y \left( e^x - 1 \right)}\]