Question:
If $y=y(x)$ is the solution of the differential equation, $e^{y}\left(\frac{d y}{d x}-1\right)=e^{x}$ such that $y(0)=0$, then $y(1)$ is equal to:
Correct Option: 1
Solution:
Let $e^{y}=t$
$e^{y} \frac{d y}{d x}=\frac{d t}{d x}$
$\therefore \quad \frac{d t}{d x}-t=e^{x}$ $\left[\because e^{y} \frac{d y}{d x}-e^{y}=e^{x}\right]$
I.F. $=e^{\int-1 . d x}=e^{-x}$
$t\left(e^{-x}\right)=\int e^{x} \cdot e^{-x} d x \Rightarrow e^{y-x}=x+c$
Put $x=0, y=0$, then we get $c=1$
$e^{y-x}=x+1$
$y=x+\log _{e}(x+1)$
Put $x=1 \quad \therefore \quad y=1+\log _{e} 2$