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(D) Given differential equation is $(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$.Dividing by $(1 + x^2)$,we get $\frac{dy}{dx} + \frac{2x}{1 + x^2}y = \frac{

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

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(D) Given differential equation is $(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$.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 $\frac{dy}{dx} + Py = Q$,where $P = \frac{2x}{1 + x^2}$ and $Q = \frac{4x^2}{1 + x^2}$.Integrating factor $(I.F.) = e^{\int P dx} = e^{\int \frac{2x}{1 + x^2} dx} = e^{\ln(1 + x^2)} = 1 + x^2$.The general solution is $y 	imes (I.F.) = \int Q 	imes (I.F.) dx + C$.$y(1 + x^2) = \int \frac{4x^2}{1 + x^2} 	imes (1 + x^2) dx + C$.$y(1 + x^2) = \int 4x^2 dx + C$.$y(1 + x^2) = \frac{4x^3}{3} + C$.Since the curve passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ to find $C$.$0(1 + 0) = \frac{4(0)^3}{3} + C \implies C = 0$.Thus,the equation of the curve is $y(1 + x^2) = \frac{4x^3}{3}$,which simplifies to $3(1 + x^2)y = 4x^3$.

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

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