Question

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

• A. $e^y=e^x-1+c.e^{-ex}$
• B. $e^y=e^x+1+c.e^{-ex}$
• C. $e^y=e^x-1-c.e^{-ex}$
• D. $e^y=e^x-2+c.e^{-ex}$

Answer 1

The correct option is A $e^y=e^x-1+c.e^{-ex}$

$\begin{array}{c}\frac{dy}{dx}=e^{x-y}\left(e^x-e^y\right)\\ Rewaiting,\\ \frac{dy}{dx}=e^{2x}\cdot e^{-y}-e^x\\ \frac{dy}{dx}+w^x=e^{2x}\cdot e^{-y}\\ Dividingby{e}^{-y},e^y\left(\frac{dy}{dx}\right)+e^x\cdot e^y=e^{2x}\\ put{e}^y+zsothat,e^y\left(\frac{dy}{dx}\right)=\frac{dz}{dx}\\ weget\\ \frac{dz}{dx}+e^x\cdot z=e^{2x}\\ I.F.=e^{\int exdx}\\ then,\\ z\cdot e^{ex}=\int e^{2x}\cdot e^{ex}+c=\int e^x\cdot e^{ex}\cdot e^{x}dx+c\\ \Rightarrow \int t{e}^{t}dt+c\left(puuting{e}^x=tsothat{e}^{x}dx=dt\right)\\ \Rightarrow t\cdot e^t-\left(1-e^t\right)dt+c\left(byparts\right)\\ \Rightarrow t{e}^t-e^t=e^t\left(t-1\right)+c\Rightarrow e^y\cdot e^x=e^{ex}\left(e^x-1\right)+c\\ \Rightarrow e^y=e^x-1+c{e}^{-ex}\\ z=e^x\&t=e^x\end{array}$