Question

The solution of the differential equation $\frac{dy}{dx}=e^{x-y}(e^x-e^y)$

• A. $e^y=(e^x+1)+C{e}^{-e^x}$
• B. $e^y=(e^x-1)+C$
• C. $e^y=(e^x-1)+C{e}^{-e^x}$
• D. None of the above

Answer 1

Answer: (C)

Given equation can be rewritten as $\frac{dy}{dx}=\frac{e^x}{e^y}(e^x-e^y)$ $\Rightarrow$ $e^y\frac{dy}{dx}=e^{2x}-e^x{e}^y$ $\Rightarrow$ $e^y\frac{dy}{dx}+e^x{e}^y=e^{2x}$ Put $e^y=t\Rightarrow e^y\frac{dy}{dx}=\frac{dt}{dx}$ $\therefore$ $\frac{dt}{dx}+e^{x}t=e^{2x}$ On comparing with $\frac{dx}{dy}+Pt=Q,$ we get $P=e^x$ and $Q=e^{2x}$ $\therefore$ If $=e^{{\int }_{}^{}pdx}=e^{{\int }_{}^{}{e}^{x}dx}=e^{e^x}$ Required solution is $t.e^{e^x}={\int }_{}^{}{e}^{2x}{e}^{e^x}dx+C$ $\Rightarrow$ $e^y{e}^{e^x}=(e^x-1)e^{e^x}+C$ $\Rightarrow$ $e^y=(e^x-1)+C{e}^{-e^x}$