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

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$3y(1+x^2) = 4x^3$

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(C) The given differential equation is $\left(1+x^2\right) \frac{dy}{dx} + 2xy = 4x^2$. Dividing by $(1+x^2)$,we get $\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4x^2}{1+x^2}$. 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 $(IF)$ is $e^{\int P(x) dx} = e^{\int \frac{2x}{1+x^2} dx} = e^{\ln(1+x^2)} = 1+x^2$. The solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$. $y(1+x^2) = \int \frac{4x^2}{1+x^2} \cdot (1+x^2) dx + C$. $y(1+x^2) = \int 4x^2 dx + C = \frac{4x^3}{3} + C$. Since the curve passes through the origin $(0,0)$,we substitute $x=0, y=0$: $0(1+0) = 0 + C \implies C = 0$. Thus,$y(1+x^2) = \frac{4x^3}{3}$,which simplifies to $3y(1+x^2) = 4x^3$.

The equation of the curve passing through the origin and satisfying $\left(1+x^2\right) \frac{dy}{dx} + 2xy = 4x^2$ is

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