If y = y x is the solution of the differential equation , e y d y d x - 1 = e x such that y 0 = 0 , then y 1…

If y=yx is the solution of the differential equation ,eydydx-1=ex such that y0=0, then y1 is equal to
  1. 1+loge2
  2. 2+loge2
  3. 2e
  4. loge2

Solution

ey=t
eydydx=dtdx
dtdx-t=ex
IF=e-∫dx

⇒t·e-x=x+1
ey-x=x+1
y=x+ln⁡x+1
at x=1,y=1+ln⁡2

Asked in: JEE Main 2020 (07 Jan Shift 1)

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