If $y=y(x)$ is the solution of the differential equation

$$\frac{d y}{d x}+\frac{4 x}{\left(x^{2}-1\right)} y=\frac{x+2}{\left(x^{2}-1\right)^{\frac{5}{2}}}, x > 1$$ such that

$$y(2)=\frac{2}{9} \log _{e}(2+\sqrt{3}) \text { and } y(\sqrt{2})=\alpha \log _{e}(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}, \alpha, \beta, \gamma \in \mathbb{N} \text {, then } \alpha \beta \gamma \text { is equal to }$$ :

<p>We can solve the given differential equation using an integrating factor. <br/><br/>The integrating factor is given by : <br/><br/>$\mu(x) = e^{\int \frac{4x}{x^2 - 1} dx} = e^{2\ln(x^2 - 1)} = (x^2 - 1)^2$ <br/><br/>Multiplying both sides of the differential equation by $\mu(x)$, we get :</p> <p>$$(x^2-1)^2 \frac{d y}{d x} + 4x y = \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}}$$</p> <p>We can rewrite the left-hand side using the product rule:</p> <p>$$\frac{d}{dx} \left((x^2-1)^2 y\right) = \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}}$$</p> <p>Integrating both sides with respect to $x$, we get:</p> <p>$$(x^2-1)^2 y = \int \frac{x+2}{\left(x^{2}-1\right)^{\frac{1}{2}}} dx = \sqrt{x^2-1}+2 \ln\left|x+\sqrt{x^2-1}\right|+C$$</p> <p>where $C$ is the constant of integration. Using the initial condition $y(2) = \frac{2}{9} \log_e(2+\sqrt{3})$, we can solve for $C$:</p> At $x=2$, <br/><br/>$9 \cdot \frac{2}{9} \ln (2+\sqrt{3})=2 \ln (2+\sqrt{3})+\sqrt{3}+C$ <br/><br/>$C=-\sqrt{3}$ <br/><br/>At $x=\sqrt{2}$ <br/><br/>$y(\sqrt{2})=2 \ln (1+\sqrt{2})+1-\sqrt{3}$ <br/><br/>$\beta=1, \alpha=2, \gamma=3$ <br/><br/>$\Rightarrow \alpha \beta \gamma=6$