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Question: Answered & Verified by Expert
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
MathematicsDifferential EquationsJEE Main
Options:
  • A -e3/2e2+12
  • B -2e21+e22
  • C e5/21+e22
  • D 5e1/2e2+12
Solution:
2436 Upvotes Verified Answer
The correct answer is: -e3/2e2+12

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

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