1.

Let y=y(x) be the solution of the differential equation dydx=(y+1)⎛⎜⎜⎝(y+1)ex22−x⎞⎟⎟⎠,0<x<2, such that y(2)=0. Then the value of dydx at x=1 is

Answer»

Let y=y(x) be the solution of the differential equation dydx=(y+1)
(y+1)ex22x
,0<x<2,
such that y(2)=0. Then the value of dydx at x=1 is



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