Question

The solution of $ \frac{dy}{dx} - 1 = e^{x-y} $ is

• A. $e^{x +y} + x = c$
• B. $e^{- x +y} + x = c$
• C. $e^{-(x +y)} = x + c$
• D. $e^{- x +y} = x + c$

Answer 1

Answer: (D)

$\frac{dy}{dx} -1=e^{x-y} $ can be written as
$\frac{dy}{dx} -1=\frac{e^{x}}{e^{y}} $
$ \Rightarrow e^{y}dy -e^{y }dx =e^{x} dx$
$ \Rightarrow e^{y-x}\left(dy -dx\right) =dx $
Integrating both sides,
$\int e^{y-x}d\left(y-x\right)=x+c $
Put $y -x = t \Rightarrow d\left(y -x\right) =dt$
$\therefore \:\:\: \int e^{t }dt =x+c \Rightarrow e^{y-x} =x+c$