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(C) The given differential equation is $(1+x^{2}) \frac{dy}{dx} + 2xy = 4x^{2}$.Dividing by $(1+x^{2})$, we get $\frac{dy}{dx} + \left(\frac{2x}{1+x^{

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$\frac{1}{6}$

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(C) The given differential equation is $(1+x^{2}) \frac{dy}{dx} + 2xy = 4x^{2}$.Dividing by $(1+x^{2})$, we get $\frac{dy}{dx} + \left(\frac{2x}{1+x^{2}}\right)y = \frac{4x^{2}}{1+x^{2}}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{2x}{1+x^{2}}$ and $Q(x) = \frac{4x^{2}}{1+x^{2}}$.The integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{2x}{1+x^{2}} dx} = e^{\ln(1+x^{2})} = 1+x^{2}$.The general solution is given by $y(IF) = \int Q(x)(IF) dx + C$.$y(1+x^{2}) = \int \left(\frac{4x^{2}}{1+x^{2}}\right)(1+x^{2}) dx + C$.$y(1+x^{2}) = \int 4x^{2} dx + C = \frac{4x^{3}}{3} + C$.Using the initial condition $y(0) = -1$, we substitute $x=0$ and $y=-1$:$-1(1+0^{2}) = \frac{4(0)^{3}}{3} + C \Rightarrow C = -1$.Thus, $y(1+x^{2}) = \frac{4x^{3}}{3} - 1$.For $x=1$, $y(1+1^{2}) = \frac{4(1)^{3}}{3} - 1$.$2y = \frac{4}{3} - 1 = \frac{1}{3}$.Therefore, $y(1) = \frac{1}{6}$.

Let $y(x)$ be a solution of $(1+x^{2}) \frac{dy}{dx} + 2xy - 4x^{2} = 0$ and $y(0) = -1$. Then $y(1)$ is equal to

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