Let y = y(x) be the solution of the differential equation

${{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{x^2}/2}} - x} \right)$, 0 < x < 2.1, with y(2) = 0. Then the value of ${{dy} \over {dx}}$ at x = 1 is equal to :

${{dy} \over {dx}} = (y + 1)\left( {(y + 1){e^{{{{x^2}} \over 2}}} - x} \right)$<br><br>$$ \Rightarrow {{ - 1} \over {{{(y + 1)}^2}}}{{dy} \over {dx}} - x\left( {{1 \over {y + 1}}} \right) = - {e^{{{{x^2}} \over 2}}}$$<br><br>Put, ${1 \over {y + 1}} = z$<br><br>$- {1 \over {{{(y + 1)}^2}}}.{{dy} \over {dx}} = {{dz} \over {dx}}$<br><br>$\therefore$ ${{dz} \over {dx}} + z( - x) = - {e^{{{{x^2}} \over 2}}}$<br><br>$I.F = {e^{\int { - xdx} }} = {e^{{{ - {x^2}} \over 2}}}$<br><br>$$z.\left( {{e^{ - {{{x^2}} \over 2}}}} \right) = - \int {{e^{ - {{{x^2}} \over 2}}}.{e^{{{{x^2}} \over 2}}}dx} = - \int {1.dx = - x + C} $$<br><br>$\Rightarrow {{{e^{ - {{{x^2}} \over 2}}}} \over {y + 1}} = - x + C$ .... (1)<br><br>Given y = 0 at x = 2 Put in (1)<br><br>${{{e^{ - 2}}} \over {0 + 1}} = - 2 + C$<br><br>$C = {e^{ - 2}} + 2$ .... (2)<br><br>From (1) and (2)<br><br>$y + 1 = {{{e^{ - {{{x^2}} \over 2}}}} \over {{e^{ - 2}} + 2 - x}}$<br><br>Again, at x = 1<br><br>$\Rightarrow y + 1 = {{{e^{3/2}}} \over {{e^2} + 1}}$<br><br>$\Rightarrow y + 1 = {{{e^{3/2}}} \over {{e^2} + 1}}$<br><br>$\therefore$ $${\left. {{{dy} \over {dx}}} \right|_{x = 1}} = {{{e^{3/2}}} \over {{e^2} + 1}}\left( {{{{e^{3/2}}} \over {{e^2} + 1}} \times {e^{1/2}} - 1} \right)$$<br><br>$= - {{{e^{3/2}}} \over {{{({e^2} + 1)}^2}}}$