If y=yx is the solution of the differential equation ,e^ydy/dx-1=e^x such that y0=0, then y1 is equal to

Correct answer: (A) 1+log_e2. e^y=t e^ydy/dx=dt/dx dt/dx-t=e^x IF=e^(-∫dx) ⇒t·e^(-x)=x+1 e^(y-x)=x+1 y=x+ln⁡x+1 at x=1,y=1+ln⁡2

Q41 / 75 · JEE Main 2020 (07 Jan Shift 1) · Mathematics

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

$e^{y} = t$
$e^{y} \frac{d y}{d x} = \frac{d t}{d x}$
$\frac{d t}{d x} - t = e^{x}$
$I F = e^{- \int d x}$

$\Rightarrow t \cdot e^{- x} = x + 1$
$e^{y - x} = x + 1$
$y = x + \ln (x + 1)$
at $x = 1, y = 1 + \ln (2)$

Hence, the correct answer is option: 'A'.