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

The given differential equation is $$(1 + x^2)\frac{dy}{dx} + 2xy = 4x^2$$, and the curve passes through the origin, so when $$x = 0$$, $$y = 0$$.

First, rewrite the equation in standard linear form $$\frac{dy}{dx} + P(x)y = Q(x)$$. Divide both sides by $$(1 + x^2)$$:

$$\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{4x^2}{1 + x^2}$$

Here, $$P(x) = \frac{2x}{1 + x^2}$$ and $$Q(x) = \frac{4x^2}{1 + x^2}$$. The integrating factor is $$e^{\int P(x) dx} = e^{\int \frac{2x}{1 + x^2} dx}$$.

Compute the integral $$\int \frac{2x}{1 + x^2} dx$$. Notice that the derivative of $$1 + x^2$$ is $$2x$$, so:

$$\int \frac{2x}{1 + x^2} dx = \ln|1 + x^2|$$

Since $$1 + x^2 > 0$$ for all real $$x$$, this simplifies to $$\ln(1 + x^2)$$. Thus, the integrating factor is $$e^{\ln(1 + x^2)} = 1 + x^2$$.

Multiply both sides of the differential equation by the integrating factor:

$$(1 + x^2) \frac{dy}{dx} + (1 + x^2) \cdot \frac{2x}{1 + x^2} y = (1 + x^2) \cdot \frac{4x^2}{1 + x^2}$$

Simplify:

$$(1 + x^2) \frac{dy}{dx} + 2x y = 4x^2$$

The left side is the derivative of $$(1 + x^2)y$$:

$$\frac{d}{dx} \left[ (1 + x^2)y \right] = 4x^2$$

Integrate both sides with respect to $$x$$:

$$\int \frac{d}{dx} \left[ (1 + x^2)y \right] dx = \int 4x^2 dx$$

$$(1 + x^2)y = \frac{4}{3}x^3 + C$$

where $$C$$ is the constant of integration.

Use the condition that the curve passes through the origin $$(0, 0)$$. Substitute $$x = 0$$ and $$y = 0$$:

$$(1 + 0^2) \cdot 0 = \frac{4}{3}(0)^3 + C$$

$$0 = 0 + C$$

Thus, $$C = 0$$. The equation becomes:

$$(1 + x^2)y = \frac{4}{3}x^3$$

Multiply both sides by 3 to eliminate the fraction:

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

Comparing with the options, this matches option D.

Hence, the correct answer is Option D.