The solution curve of the differential equation, 1 + e - x 1 + y 2 d y d x = y 2 which passes through the…

The solution curve of the differential equation, 1+e-x1+y2dydx=y2 which passes through the point 0, 1, is
  1. y2+1=yloge1+e-x2+2
  2. y2+1=yloge1+ex2+2
  3. y2=1+yloge1+ex2
  4. y2=1+yloge1+e-x2

Solution

1+e-x1+y2dydx=y2

Separate variables

⇒1+y2dyy2=dx1+e-x

Integrate both side

⇒∫1+y2dyy2=∫dx1+e-x

⇒∫1+1y2dy=∫exdxex+1

⇒y-1y=lnex+1+C

Since curve pass through 0, 1.

⇒1-11=lne0+1+C ⇒C=-ln2

⇒y-1y=lnex+1-ln2

⇒y2-1y=lnex+12

⇒y2=1+yloge1+ex2

Asked in: JEE Main 2020 (03 Sep Shift 1)

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