Let $\alpha x=\exp \left(x^{\beta} y^{\gamma}\right)$ be the solution of the differential equation $$2 x^{2} y \mathrm{~d} y-\left(1-x y^{2}\right) \mathrm{d} x=0, x > 0,y(2)=\sqrt{\log _{e} 2}$$. Then $\alpha+\beta-\gamma$ equals :

$\begin{aligned} & 2 x^2 y d y-\left(1-x y^2\right) d x=0 \\\\ & \Rightarrow 2 x^2 y \frac{d y}{d x}-1+x y^2=0 \\\\ & \Rightarrow 2 y \frac{d y}{d x}-\frac{1}{x^2}+\frac{y^2}{x}=0 \\\\ & \Rightarrow 2 y \frac{d y}{d x}+\frac{y^2}{x}=\frac{1}{x^2} \text { (L.D.E) } \\\\ & Let, y^2=t \Rightarrow 2 y \frac{d y}{d x}=\frac{d t}{d x} \\\\ & \frac{d t}{d x}+\frac{t}{x}=\frac{1}{x^2} \\\\ & \text { I.F }=e^{\int \frac{1}{x} d x}=x\end{aligned}$ <br/><br/>So, the general solution is <br/><br/>$$ \begin{aligned} & \Rightarrow t \times x=\int x \cdot \frac{1}{x^2} d x \\\\ & \Rightarrow y^2 \cdot x=\ln x+c \\\\ & \text { Also,y(2) }=\sqrt{\log _e 2} \\\\ & \log _e^2 2=\log _e 2+c \Rightarrow c=\log _e 2 \\\\ & \Rightarrow y^2 x=\ln x+\ln 2 \\\\ & \Rightarrow y^2 x=\ln 2 x \\\\ & 2 x=\exp \left(x^1 y^2\right) \end{aligned} $$ <br/><br/>By compare it with given solution we get, <br/><br/>$$ \begin{aligned} & \alpha=2,\beta=1, \gamma=2 \\\\ & \Rightarrow \alpha+\beta-\gamma=2+1-2=1 \end{aligned} $$