Let y = y(x) be the solution of the differential equation (dy)/(dx)...

The given differential equation is $\dfrac{dy}{dx} = (1 + x + x^2)(1 - y + y^2)$.

Separating the variables:

$\dfrac{dy}{y^2 - y + 1} = (x^2 + x + 1) dx$

Integrating both sides:

$\int \dfrac{dy}{\left(y - \dfrac{1}{2}\right)^2 + \dfrac{3}{4}} = \int (x^2 + x + 1) dx$

$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{y - \dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right) = \dfrac{x^3}{3} + \dfrac{x^2}{2} + x + C$

$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y - 1}{\sqrt{3}}\right) = \dfrac{x^3}{3} + \dfrac{x^2}{2} + x + C$

Given $y(0) = \dfrac{1}{2}$, substituting $x = 0 and$y = \dfrac{1}{2}$:<br><br>$\dfrac{2}{\sqrt{3}} \tan^{-1}(0) = C \Rightarrow C = 0$<br><br>Substituting$x = 1$to find$y(1)$:<br><br>$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{1^3}{3} + \dfrac{1^2}{2} + 1$<br><br>$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{11}{6}$<br><br>$\tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{11\sqrt{3}}{12}$<br><br>$2y(1) - 1 = \sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{12}\right)$<br><br>Answer:$\sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{12}\right)