Let y = y ( x ) be the solution of the differential equation d y d x = y + 1 y + 1 e x 2 / 2 - x , 0 < x…

Let y=y(x) be the solution of the differential equation dydx=y+1y+1ex2/2-x,0<x<2.1, with y(2)=0. Then the value of dydx at x=1 is equal to
  1. -e3/2e2+12
  2. -2e21+e22
  3. e5/21+e22
  4. 5e1/2e2+12

Solution

Let y+1=Y

∴dYdx=Y2ex22-xY

Put -1Y=k

⇒dkdx+k-x=ex22

I.F.=e-x22

∴k=x+cex2/2

Put k=-1y+1

∴y+1=-1(x+c)ex2/2   ...i

when x=2, y=0, then c=-2-1e2

Differentiate equation (i) & put x=1 we get, 

dydxx=1=-e3/21+e22

Asked in: JEE Main 2021 (18 Mar Shift 2)

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