MathematicsMedium38×since 2002Q3531Let y = y(x) be the solution of the differential equation dydx=(y+1)((y+1)ex2/2−x){{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{x^2}/2}} - x} \right)dxdy=(y+1)((y+1)ex2/2−x), 0 < x < 2.1, with y(2) = 0. Then the value of dydx{{dy} \over {dx}}dxdy at x = 1 is equal to :Ae5/2(1+e2)2{{{e^{5/2}}} \over {{{(1 + {e^2})}^2}}}(1+e2)2e5/2B5e1/2(e2+1)2{{5{e^{1/2}}} \over {{{({e^2} + 1)}^2}}}(e2+1)25e1/2C−2e2(1+e2)2- {{2{e^2}} \over {{{(1 + {e^2})}^2}}}−(1+e2)22e2D−e3/2(e2+1)2{{ - {e^{3/2}}} \over {{{({e^2} + 1)}^2}}}(e2+1)2−e3/2Check answerSkip