Question

Solve the differential equation: $\frac{dy}{dx}=e^{x-y}(e^x-e^y).$

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

Answer 1

Answer: (B)

$e^y=(e^x-1)+c{e}^{-e^x}.$

Given, $\frac{dy}{dx}=e^{x-y}(e^x-e^y)$

$\Rightarrow \frac{dy}{dx}=\frac{e^x}{e^y}(e^x-e^y)\Rightarrow e^y\frac{dy}{dx}+e^x.e^y=e^{2x}$
Put $e^y=v\Rightarrow e^y\frac{dy}{dx}=\frac{dv}{dx}$
$\therefore \frac{dv}{dx}+v{e}^x=e^{2x}$ ...(1)
Here $P=e^x\Rightarrow \int Pdx=\int e^{x}dx=e^x$
$\therefore I.F.=e^{e^x}$
Multiplying (1) by $I.F.$ we get
$e^{e^x}\frac{dv}{dx}+v{e}^{e^x}{e}^x=e^{e^x}{e}^{2x}$
Integrating both sides , we get
$v.e^{e^x}=\int e^x.e^x.e^{e^x}dx+c$

Putting $e^x=t\Rightarrow e^{x}dx=dt$

$\therefore v{e}^t=\int t{e}^{t}dt+c=e^t(t-1)+c$

$\Rightarrow v=(t-1)+c{e}^{-t}$

$\Rightarrow e^y=(e^x-1)+c{e}^{-e^x}$