Question

If $y=y\left(x\right)$ is the solution of the differential equation, $e^y(\frac{dy}{dx}-1)=e^x$ such that $y\left(0\right)=0$, then $y\left(1\right)$ is equal to

• A. ${\log }_{e}2$
• B. $2e$
• C. $2+{\log }_{e}2$
• D. $1+{\log }_{e}2$

Answer 1

The correct option is D $1+{\log }_{e}2$

$e^y(\frac{dy}{dx}-1)=e^x$
$\Rightarrow \frac{dy}{dx}=e^{x-y}+1$

Let $x-y=t$
$\Rightarrow 1-\frac{dy}{dx}=\frac{dt}{dx}$
So, we can write
$1-\frac{dt}{dx}=e^t+1$
$\Rightarrow -e^{-t}dt=dx$
$\Rightarrow e^{-t}=x+c$
$\Rightarrow e^{y-x}=x+c$

$y\left(0\right)=0$
$\Rightarrow 1=0+c\Rightarrow c=1$
$\Rightarrow e^{y-x}=x+1$

At $x=1$,
$e^{y-1}=2$
$\Rightarrow y=1+{\log }_{e}2$