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(C) The given differential equation is $\frac{dy}{dx} = (1+x^2)(1-y+y^2)$.Separating the variables, we get $\int \frac{dy}{y^2-y+1} = \int (1+x^2) dx$

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$\sqrt{3}	an \left(\frac{11\sqrt{3}}{12}ight)$

Vedclass · Answer

(C) The given differential equation is $\frac{dy}{dx} = (1+x^2)(1-y+y^2)$.Separating the variables, we get $\int \frac{dy}{y^2-y+1} = \int (1+x^2) dx$.Completing the square in the denominator: $y^2-y+1 = (y-1/2)^2 + 3/4$.Thus, $\int \frac{dy}{(y-1/2)^2 + (\sqrt{3}/2)^2} = x + \frac{x^3}{3} + C$.Using the formula $\int \frac{du}{u^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{u}{a})$, we get $\frac{2}{\sqrt{3}} 	an^{-1} \left(\frac{2y-1}{\sqrt{3}}ight) = x + \frac{x^3}{3} + C$.Given $y(0) = 1/2$, we have $\frac{2}{\sqrt{3}} 	an^{-1}(0) = 0 + 0 + C$, which implies $C = 0$.So, $\frac{2}{\sqrt{3}} 	an^{-1} \left(\frac{2y-1}{\sqrt{3}}ight) = x + \frac{x^3}{3}$.At $x = 1$, $\frac{2}{\sqrt{3}} 	an^{-1} \left(\frac{2y(1)-1}{\sqrt{3}}ight) = 1 + 1/3 = 4/3$.$	an^{-1} \left(\frac{2y(1)-1}{\sqrt{3}}ight) = \frac{4}{3} \cdot \frac{\sqrt{3}}{2} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$.Therefore, $\frac{2y(1)-1}{\sqrt{3}} = 	an \left(\frac{2}{\sqrt{3}}ight)$, which gives $2y(1)-1 = \sqrt{3} 	an \left(\frac{2}{\sqrt{3}}ight)$.Note: The provided options appear to have a calculation discrepancy in the argument of the tangent function. Based on the standard derivation, the correct value is $\sqrt{3} 	an \left(\frac{2}{\sqrt{3}}ight)$.

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} = (1+x^2)(1-y+y^2)$, with the initial condition $y(0) = \frac{1}{2}$. Then $(2y(1) - 1)$ is equal to:

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