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 $y-x=t\Rightarrow d\left(y-x\right)=dt$
$\therefore \int e^{t}dt=x+c\Rightarrow e^{y-x}=x+c$