$${{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{{{x^2}} \over 2}}} - x} \right)$$

$$ \Rightarrow {{ - 1} \over {{{(y + 1)}^2}}}{{dy} \over {dx}} - x\left( {{1 \over {y + 1}}} \right) = - {e^{{{{x^2}} \over 2}}}$$

Put, $${1 \over {y + 1}} = z$$

$$ - {1 \over {{{(y + 1)}^2}}}.{{dy} \over {dx}} = {{dz} \over {dx}}$$

$$ \therefore $$ $${{dz} \over {dx}} + z( - x) = - {e^{{{{x^2}} \over 2}}}$$

$$I.F = {e^{\int { - xdx} }} = {e^{{{ - {x^2}} \over 2}}}$$

$$z.\left( {{e^{ - {{{x^2}} \over 2}}}} \right) = - \int {{e^{ - {{{x^2}} \over 2}}}.{e^{{{{x^2}} \over 2}}}dx} = - \int {1.dx = - x + C} $$

$$ \Rightarrow {{{e^{ - {{{x^2}} \over 2}}}} \over {y + 1}} = - x + C$$ .... (1)

Given y = 0 at x = 2 Put in (1)

$${{{e^{ - 2}}} \over {0 + 1}} = - 2 + C$$

$$C = {e^{ - 2}} + 2$$ .... (2)

From (1) and (2)

$$y + 1 = {{{e^{ - {{{x^2}} \over 2}}}} \over {{e^{ - 2}} + 2 - x}}$$

Again, at x = 1

$$ \Rightarrow y + 1 = {{{e^{3/2}}} \over {{e^2} + 1}}$$

$$ \Rightarrow y + 1 = {{{e^{3/2}}} \over {{e^2} + 1}}$$

$$ \therefore $$ $${\left. {{{dy} \over {dx}}} \right|_{x = 1}} = {{{e^{3/2}}} \over {{e^2} + 1}}\left( {{{{e^{3/2}}} \over {{e^2} + 1}} \times {e^{1/2}} - 1} \right)$$

$$ = - {{{e^{3/2}}} \over {{{({e^2} + 1)}^2}}}$$