Let y = y(x) be the solution of the differential equation $$\left( {(x + 2){e^{\left( {{{y + 1} \over {x + 2}}} \right)}} + (y + 1)} \right)dx = (x + 2)dy$$, y(1) = 1. If the domain of y = y(x) is an open interval ($\alpha$, $\beta$), then | $\alpha$ + $\beta$| is equal to ______________.

Let y + 1 = Y and x + 2 = X<br><br>dy = dY<br><br>dx = dX<br><br>$\left( {X{e^{{X \over X}}} + Y} \right)dX = XdY$<br><br>$\Rightarrow {{XdY - YdX} \over {{X^2}}} = {{{e^{{Y \over X}}}} \over X}dX$<br><br>$\Rightarrow {e^{ - {Y \over X}}}d\left( {{Y \over X}} \right) = {{dX} \over X}$<br><br>$\Rightarrow - {e^{ - {Y \over X}}} = \ln |X| + c$<br><br>$$ \Rightarrow - {e^{ - \left( {{{y + 1} \over {x + 2}}} \right)}} = \ln |x + 2| + c$$<br><br>$\because$ (1, 1) satisfy this equation<br><br>So, $c = - {e^{ - {2 \over 3}}} - \ln 3$<br><br>Now, $$y = - 1 - (x + 2)\ln \left( {\ln \left( {\left| {{3 \over {x + 2}}} \right|} \right) + {e^{ - {2 \over 3}}}} \right)$$<br><br>Domain :<br><br>$\ln \left| {{3 \over {x + 2}}} \right| &gt; {e^{ - {e^{ - {2 \over 3}}}}}$<br><br>$$ \Rightarrow {3 \over {\left| {x + 2} \right|}} &gt; {e^{ - {e^{ - {2 \over 3}}}}}$$<br><br>$\Rightarrow \left| {x + 2} \right| &lt; 3{e^{{e^{ - {2 \over 3}}}}}$<br><br>$$ \Rightarrow - 3{e^{{e^{ - {2 \over 3}}}}} - 2 &lt; x &lt; 3{e^{{e^{ - {2 \over 3}}}}} - 2$$<br><br>So, $\alpha + \beta = - 4$<br><br>$\Rightarrow \left| {\alpha + \beta } \right| = 4$