Let α x = exp x β y γ be the solution of the differential equation 2 x 2 y d y - 1 - x y 2 d…

Given the solution of differential equation,

$\alpha x=e^{x^{\beta }·y^{\gamma }}......\left(1\right)$

Now calculating the solution of differential equation, 

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

Let $y^2=t\Rightarrow 2\frac{dy}{dx}=\frac{dt}{dx}$

$\therefore x^2\frac{dt}{dx}=1-xt$

$\frac{\mathrm{dt}}{\mathrm{dx}}+\frac{t}{x}=\frac{1}{x^2} .....\left(2\right)$

Above differential equation is linear in $t$

We know solutions of differential equation of the form $\frac{dy}{\mathrm{dx}}+Py=Q$ is given by,

$y.e^{\int Pdx}=\int Q.e^{\int Pdx}+C (e^{\int Pdx}=I.F.)$

Now calculating, $I.F.=e^{\int \frac{1}{x}dx}=e^{\ln x}=x$

Therefore, the solution of equation $\left(2\right)$ is given by,

$t.x=\int \frac{1}{x^2}·xdx+C$

$\Rightarrow y^2·x=\ln x+\mathrm{C}$

$\because y\left(2\right)=\sqrt{\ln 2}$

$\therefore 2.\ln 2=\ln 2+C\Rightarrow C=\ln 2$

Hence, solution is $x{y}^2=\ln x+\ln 2$

 $\Rightarrow x{y}^2=\ln 2x$

$\Rightarrow 2x=e^{x·y^2}......\left(3\right)$

On comparing equations $\left(1\right)$ and $\left(3\right)$ we get,

 $\alpha =2,\beta =1,\gamma =2$

$\therefore \alpha +\beta -\gamma =1$