Solution of First Order Differential Equation with Initial Condition

To find the value of $(2y(1) - 1)$, we start by solving the given first-order differential equation using the method of separation of variables.

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

Separating the variables $x$ and $y$, we get: $\frac{dy}{y^2 - y + 1} = (1 + x + x^2) dx$

Integrating both sides: $\int \frac{dy}{y^2 - y + 1} = \int (1 + x + x^2) dx$

Completing the square in the denominator of the left-hand side: $y^2 - y + 1 = \left(y - \frac{1}{2}\right)^2 + \frac{3}{4} = \left(y - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$

Now, using the integration formula $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right)$, we integrate the left side: $\int \frac{dy}{\left(y - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{\sqrt{3}}{2}} \arctan\left(\frac{y - \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \frac{2}{\sqrt{3}} \arctan\left(\frac{2y - 1}{\sqrt{3}}\right)$

Integrating the right side: $\int (1 + x + x^2) dx = x + \frac{x^2}{2} + \frac{x^3}{3} + C$

Thus, the general solution is: $\frac{2}{\sqrt{3}} \arctan\left(\frac{2y - 1}{\sqrt{3}}\right) = x + \frac{x^2}{2} + \frac{x^3}{3} + C$

We are given the initial condition $y(0) = \frac{1}{2}$. Substituting $x = 0$ and $y = \frac{1}{2}$: $\frac{2}{\sqrt{3}} \arctan\left(\frac{2\left(\frac{1}{2}\right) - 1}{\sqrt{3}}\right) = 0 + 0 + 0 + C$ $\frac{2}{\sqrt{3}} \arctan(0) = C \implies C = 0$

Substituting $C = 0$ back into the solution gives: $\frac{2}{\sqrt{3}} \arctan\left(\frac{2y - 1}{\sqrt{3}}\right) = x + \frac{x^2}{2} + \frac{x^3}{3}$

Now, we need to find $y(1)$. Substitute $x = 1$: $\frac{2}{\sqrt{3}} \arctan\left(\frac{2y(1) - 1}{\sqrt{3}}\right) = 1 + \frac{1}{2} + \frac{1}{3}$ $\frac{2}{\sqrt{3}} \arctan\left(\frac{2y(1) - 1}{\sqrt{3}}\right) = \frac{11}{6}$

Rearranging the equation to solve for the tangent argument: $\arctan\left(\frac{2y(1) - 1}{\sqrt{3}}\right) = \frac{11}{6} \cdot \frac{\sqrt{3}}{2} = \frac{11\sqrt{3}}{12}$

Taking the tangent of both sides: $\frac{2y(1) - 1}{\sqrt{3}} = \tan\left(\frac{11\sqrt{3}}{12}\right)$

Multiplying both sides by $\sqrt{3}$: $2y(1) - 1 = \sqrt{3} \tan\left(\frac{11\sqrt{3}}{12}\right)$

This matches Option C.