If the differential equation representing the family of all circles touching $$x$$-axis at the origin is $$(x^2 - y^2)\frac{dy}{dx} = g(x)y$$, then $$g(x)$$ equals:

The problem involves finding the function $$g(x)$$ in the differential equation $$(x^2 - y^2)\frac{dy}{dx} = g(x)y$$ that represents the family of all circles touching the x-axis at the origin.

First, consider the family of circles. Since each circle touches the x-axis at the origin $$(0,0)$$, the center must lie on the y-axis. This is because the radius to the point of tangency is perpendicular to the tangent (x-axis), so the center has coordinates $$(0, k)$$ for some constant $$k$$. The radius is the distance from $$(0,k)$$ to $$(0,0)$$, which is $$|k|$$. Thus, the equation of the circle is:

$$x^2 + (y - k)^2 = k^2$$

Expanding this:

$$x^2 + y^2 - 2ky + k^2 = k^2$$

Simplifying by subtracting $$k^2$$ from both sides:

$$x^2 + y^2 - 2ky = 0 \quad \text{(1)}$$

To form the differential equation, eliminate the arbitrary constant $$k$$. Differentiate equation (1) with respect to $$x$$:

$$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2ky) = \frac{d}{dx}(0)$$

Applying the chain rule and noting that $$k$$ is constant:

$$2x + 2y \frac{dy}{dx} - 2k \frac{dy}{dx} = 0$$

Divide the entire equation by 2:

$$x + y \frac{dy}{dx} - k \frac{dy}{dx} = 0$$

Rearrange the terms:

$$x + (y - k) \frac{dy}{dx} = 0 \quad \text{(2)}$$

Solve for $$k$$ from equation (1):

$$x^2 + y^2 - 2ky = 0 \implies 2ky = x^2 + y^2 \implies k = \frac{x^2 + y^2}{2y} \quad \text{(3)}$$

Substitute this expression for $$k$$ into equation (2):

$$x + \left(y - \frac{x^2 + y^2}{2y}\right) \frac{dy}{dx} = 0$$

Simplify the expression inside the parentheses:

$$y - \frac{x^2 + y^2}{2y} = \frac{2y \cdot y}{2y} - \frac{x^2 + y^2}{2y} = \frac{2y^2 - (x^2 + y^2)}{2y} = \frac{2y^2 - x^2 - y^2}{2y} = \frac{y^2 - x^2}{2y}$$

So the equation becomes:

$$x + \left(\frac{y^2 - x^2}{2y}\right) \frac{dy}{dx} = 0$$

Multiply every term by $$2y$$ to clear the denominator:

$$2y \cdot x + (y^2 - x^2) \frac{dy}{dx} = 0$$

Rearrange:

$$(y^2 - x^2) \frac{dy}{dx} = -2xy$$

Note that $$y^2 - x^2 = -(x^2 - y^2)$$, so:

$$-(x^2 - y^2) \frac{dy}{dx} = -2xy$$

Multiply both sides by $$-1$$:

$$(x^2 - y^2) \frac{dy}{dx} = 2xy$$

Compare this with the given form $$(x^2 - y^2)\frac{dy}{dx} = g(x)y$$. Here, $$2xy = 2x \cdot y$$, so $$g(x)y = 2x y$$, which implies $$g(x) = 2x$$.

Now, check the options:

A. $$\frac{1}{2}x^2$$

B. $$2x$$

C. $$\frac{1}{2}x$$

D. $$2x^2$$

Thus, $$g(x) = 2x$$ corresponds to option B.

Hence, the correct answer is Option B.