Question

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

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

Answer 1

Answer: (C)

$\frac{dy}{dx}=\frac{e^x}{e^y}(e^x-e^y)$
$\Rightarrow$ $e^y\frac{dy}{dx}+e^x\cdot e^y=e^x\cdot e^x$
Let $e^y=t$ $\Rightarrow$ $e^y\frac{dy}{dx}\cdot \frac{y}{x}=\frac{dt}{dx}$
Then, given equation reduces to
$\frac{dt}{dx}+e^{x}t=e^{2x}$
Here, $P=e^x$ and $Q=e^{2x}$
$\therefore$ $IF=e^{\int Pdx}=e^{\int e^{x}dx}=e^{e^x}$
Required solution is $t\cdot e^{e^x}=\int e^{2x}\cdot e^{e^x}dx+C$
$\Rightarrow$ $e^y=(e^x-1)+C\cdot e^{-e^x}$