1. Rewrite the differential equation

Given $2x^2y\,dy-(1-xy^2)\,dx=0.$

Rearrange in derivative form: $2x^2y\frac{dy}{dx}-(1-xy^2)=0$ $2x^2y\frac{dy}{dx}=1-xy^2.$

So, $\frac{dy}{dx}=\frac{1-xy^2}{2x^2y}.$

2. Make a substitution

Let $u=y^2.$ Then $\frac{du}{dx}=2y\frac{dy}{dx}.$

Using the differential equation, $2y\frac{dy}{dx}=\frac{1-xy^2}{x^2}=\frac{1-xu}{x^2}.$

Hence, $\frac{du}{dx}=\frac{1}{x^2}-\frac{u}{x}.$

So we get the linear differential equation $\frac{du}{dx}+\frac{1}{x}u=\frac{1}{x^2}.$

3. Solve the linear equation

Integrating factor: $\text{I.F.}=e^{\int \frac1x\,dx}=x \qquad (x>0).$

Multiplying throughout by $x$: $x\frac{du}{dx}+u=\frac{1}{x}.$

Thus, $\frac{d}{dx}(xu)=\frac{1}{x}.$

Integrating, $xu=\ln x + C.$

Therefore, $u=\frac{\ln x + C}{x}.$

Since $u=y^2$, $y^2=\frac{\ln x + C}{x}.$

4. Use the initial condition

Given $y(2)=\sqrt{\ln 2}.$

So, $y^2(2)=\ln 2.$

Substitute in the general solution: $\ln 2=\frac{\ln 2 + C}{2}.$

Therefore, $2\ln 2=\ln 2 + C$ $C=\ln 2.$

Hence, $y^2=\frac{\ln x + \ln 2}{x}=\frac{\ln(2x)}{x}.$

So, $xy^2=\ln(2x).$

5. Convert to the required form

Exponentiating both sides, $e^{xy^2}=2x.$

This matches $\alpha x=\exp\left(x^{\beta}y^{\gamma}\right).$

Comparing, $\alpha=2,\quad \beta=1,\quad \gamma=2.$

Thus, $\alpha+\beta-\gamma=2+1-2=1.$

6. Check options

The value is $1$, which corresponds to Option A.