# If y=y(x) is the solution of the differential equation,

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:

1. (1) $1+\log _{e} 2$

2. (2) $2+\log _{e} 2$

3. (3) $2 e$

4. (4) $\log _{e} 2$

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 $x=1 \quad \therefore \quad y=1+\log _{e} 2$

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