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

Given the differential equation $$2x^2 y\,dy - (1 - xy^2)\,dx = 0$$ with $$x > 0$$ and $$y(2) = \sqrt{\log_e 2}$$. We rearrange the equation as $$2x^2 y\,dy = (1 - xy^2)\,dx$$, which gives $$2x^2 y\,dy + xy^2\,dx = dx$$ and hence $$x(2xy\,dy + y^2\,dx) = dx$$.

Noting that $$d(xy^2) = x \cdot 2y\,dy + y^2\,dx$$, the left side becomes $$x \cdot d(xy^2) = dx$$ so that $$d(xy^2) = \frac{dx}{x}$$. Integrating both sides yields $$xy^2 = \ln x + C$$.

Applying the initial condition $$y(2) = \sqrt{\log_e 2}$$ gives $$2 \cdot (\sqrt{\ln 2})^2 = \ln 2 + C$$ so $$2\ln 2 = \ln 2 + C$$ and hence $$C = \ln 2$$. Thus the particular solution is $$xy^2 = \ln x + \ln 2 = \ln(2x)$$, or equivalently $$2x = e^{xy^2}$$.

Comparing with the given form $$\alpha x = \exp(x^\beta y^\gamma)$$, we see that $$2x = e^{x^1 \cdot y^2}$$ so $$\alpha = 2$$, $$\beta = 1$$, $$\gamma = 2$$ and therefore $$\alpha + \beta - \gamma = 2 + 1 - 2 = 1$$.