Let $$y = y(x)$$, $$y > 0$$, be a solution curve of the differential equation $$(1 + x^2)dy = y(x - y)dx$$. If $$y(0) = 1$$ and $$y(2\sqrt{2}) = \beta$$, then

We solve the differential equation $$(1 + x^2)dy = y(x - y)dx$$ with $$y(0) = 1$$, and find $$\beta = y(2\sqrt{2})$$.

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

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

This is a Bernoulli equation with $$n = 2$$.

Since $$\frac{dv}{dx} = -\frac{1}{y^2}\frac{dy}{dx}$$, dividing the equation by $$-y^2$$:

$$\frac{dv}{dx} + \frac{x}{1+x^2} \cdot v = \frac{1}{1+x^2}$$

$$\mu = e^{\int \frac{x}{1+x^2} dx} = e^{\frac{1}{2}\ln(1+x^2)} = \sqrt{1+x^2}$$

$$\frac{d}{dx}\left[v\sqrt{1+x^2}\right] = \frac{1}{\sqrt{1+x^2}}$$

$$v\sqrt{1+x^2} = \int \frac{1}{\sqrt{1+x^2}} \, dx = \ln\left(x + \sqrt{1+x^2}\right) + C$$

At $$x = 0$$: $$v(0) = 1$$, so $$1 \cdot 1 = \ln(0 + 1) + C = 0 + C$$, giving $$C = 1$$.

$$\frac{\sqrt{1+x^2}}{y} = \ln\left(x + \sqrt{1+x^2}\right) + 1$$

$$\sqrt{1 + 8} = 3$$, so:

$$\frac{3}{\beta} = \ln(2\sqrt{2} + 3) + 1 = 1 + \ln(3 + 2\sqrt{2})$$

$$\beta = \frac{3}{1 + \ln(3 + 2\sqrt{2})}$$

From $$\frac{3}{\beta} = 1 + \ln(3 + 2\sqrt{2})$$, we get:

$$e^{3/\beta} = e^{1 + \ln(3+2\sqrt{2})} = e \cdot (3 + 2\sqrt{2})$$

This matches Option A: $$e^{3\beta^{-1}} = e(3 + 2\sqrt{2})$$.

The correct answer is Option A.