Let $$\alpha x = \exp(x^{\beta}y^{\gamma})$$ be the solution of the differential equation $$2x^2 y dy - (1 - xy^2)dx = 0$$, $$x \gt 0$$, $$y(2) = \sqrt{\log_e 2}$$. Then $$\alpha + \beta - \gamma$$ equals:

We need to solve the differential equation $$2x^2 y \, dy - (1 - xy^2) dx = 0$$ with $$y(2) = \sqrt{\ln 2}$$.

Rearranging the given equation $$2x^2 y \, dy - (1 - xy^2) dx = 0$$ leads to $$2x^2 y \, dy = (1 - xy^2) dx$$ and hence $$\frac{dy}{dx} = \frac{1 - xy^2}{2x^2 y}$$. This can be rewritten as $$2x^2 y \frac{dy}{dx} + xy^2 = 1$$.

Let $$v = y^2$$, so that $$\frac{dv}{dx} = 2y\frac{dy}{dx}$$. Substituting into the previous equation gives $$x^2 \frac{dv}{dx} + xv = 1$$ or equivalently $$\frac{dv}{dx} + \frac{v}{x} = \frac{1}{x^2}$$.

The integrating factor is $$e^{\int \frac{1}{x} dx} = x$$, which yields $$\frac{d}{dx}(xv) = \frac{1}{x}$$ and hence $$xv = \ln x + C$$. Substituting back $$v = y^2$$ gives $$xy^2 = \ln x + C$$.

Applying the initial condition $$y(2) = \sqrt{\ln 2}$$ results in $$2 \ln 2 = \ln 2 + C$$ so that $$C = \ln 2$$. Therefore $$xy^2 = \ln x + \ln 2 = \ln(2x)$$.

Exponentiating both sides gives $$e^{xy^2} = 2x$$, which can be written in the form $$\alpha x = \exp(x^\beta y^\gamma)$$ with $$\alpha = 2\,,\;\beta = 1\,,\;\gamma = 2$$. Thus $$\alpha + \beta - \gamma = 2 + 1 - 2 = 1$$. The correct answer is Option A: $$1$$.