Question

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

Answer 1

$$\begin{gathered} e^x+e^y=e^{x+y} \\ \Rightarrow e^x+e^y\frac{dy}{dx}=e^{x+y}\left(1+\frac{dy}{dx}\right) \\ \Rightarrow e^x+e^y\frac{dy}{dx}=e^{x+y}+e^{x+y}\frac{dy}{dx} \\ \Rightarrow e^y\frac{dy}{dx}-e^{x+y}\frac{dy}{dx}=e^{x+y}-e^x \\ \Rightarrow \frac{dy}{dx}\left(e^y-e^{x+y}\right)=e^{x+y}-e^x \\ \Rightarrow \frac{dy}{dx}=\frac{e^{x+y}-e^x}{e^y-e^{x+y}} \\ =\frac{e^x\left(e^y-1\right)}{e^y\left(1-e^x\right)} \\ =-\frac{e^x\left(e^y-1\right)}{e^y\left(e^x-1\right)} \end{gathered}$$