Question

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

• 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

The correct option is C $e^y=(e^x-1)+C{e}^{-e^x}$

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 $\frac{dt}{dx}+Pt=Q$, 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
$t\cdot e^{e^x}=\int e^{2x}{e}^{e^x}dx+C$
$\Rightarrow e^y{e}^{e^x}=(e^x-1)e^{e^x}+C$
$\Rightarrow e^y=(e^x-1)+C{e}^{-e^x}$