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

Given differential equation is $\dfrac{dy}{dx} + \dfrac{4x}{\left({x}^{2} - 1\right)}y = \dfrac{x + 2}{\left({x}^{2} - 1\right)\dfrac{5}{2}}, x > 1$

Now $IF = {e}^{∫ \dfrac{4x}{{x}^{2} - 1}dx} = {\left({x}^{2} - 1\right)}^{2}$

The required equation will be $⇒ y \cdot {\left({x}^{2} - 1\right)}^{2} = ∫ \dfrac{x + 2}{{\left({x}^{2} - 1\right)}^{1 / 2}}dx$

$⇒ y \cdot {\left({x}^{2} - 1\right)}^{2} = \dfrac{1}{2} ∫ \dfrac{2x}{{\left({x}^{2} - 1\right)}^{1 / 2}}dx + 2 ∫ \dfrac{dx}{{\left({x}^{2} - 1\right)}^{1 / 2}}$

$⇒ y \cdot {\left({x}^{2} - 1\right)}$^{2} = 2ln$\left(\left|\sqrt{{x}^{2} - 1} + x\right|\right) + \sqrt{{x}^{2} - 1} + C$

Now using, at $x = 2,$ $y ( 2 ) = \dfrac{2}{9}\log_{e}\left(2 + \sqrt{3}\right)$ we get,

$⇒ 9 \cdot \dfrac{2}{9}\ln \left(2 + \sqrt{3}\right) = 2\ln \left(2 + \sqrt{3}\right) + \sqrt{3} + C$

$⇒ C = - \sqrt{3}$

Now finding the value of function at $x = \sqrt{2}$ we get,

$y \times 1 = 2\ln \left(1 + \sqrt{2}\right) + 1 - \sqrt{3}$

Now on comparing we get,

$⇒ \beta = 1, \alpha = 2, \gamma = 3$

$⇒ \alpha\beta\gamma = 1 \times 2 \times 3 = 6$

Hence, this is the required answer.