Let = exp( x ^ β y ^ γ ) be the solution of the differential...

Given the solution of differential equation,

$\alphax = {e}^{{x}^{\beta} \cdot {y}^{\gamma}} . . . . . . ( 1 )$

Now calculating the solution of differential equation,

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

Let ${y}^{2} = t ⇒ 2\dfrac{d y}{d x} = \dfrac{d t}{d x}$

$∴ {x}^{2}\dfrac{dt}{dx} = 1 - xt$

$\dfrac{dt}{dx} + \dfrac{t}{x} = \dfrac{1}{{x}^{2}} . . . . . ( 2 )$

Above differential equation is linear in $t$

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

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

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

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

$t . x = ∫ \dfrac{1}{{x}^{2}} \cdot xdx + C$

$⇒ {y}^{2} \cdot x = \ln x + C$

$∵ y ( 2 ) = \sqrt{\ln 2}$

$∴ 2 . \ln 2 = \ln 2 + C ⇒ C = \ln 2$

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

$⇒ x{y}^{2} = \ln 2x$

$⇒ 2x = {e}^{x \cdot {y}^{2}} . . . . . . ( 3 )$

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

$\alpha = 2, \beta = 1, \gamma = 2$

$∴ \alpha + \beta - \gamma = 1$