If y=y(x) is the solution of the differential

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 $\mathrm{y}(1)$ is equal to :

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

  2. $2 \mathrm{e}$

  3. $\log _{e} 2$

  4. $1+\log _{\mathrm{e}} 2$


Correct Option: , 4

Solution:

$e^{y} \frac{d y}{d x}-e^{y}=e^{x}$, Let $e^{y}=t$

$\Rightarrow \mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}$

$\frac{d t}{d x}-t=e^{x}$

I.F. $=e^{\int-d x}=e^{-x}$

$t e^{-x}=x+c \Rightarrow e^{y-x}=x+c$

$y(0)=0 \Rightarrow c=1$

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

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