If y = y ( x ) is the solution of the differential equation d y d x + 4 x ( x 2 - 1 ) y = x + 2 ? ( x 2 - 1 ) 5 / 2 , x 1 such that ( 2 ) = 2 9 log e ( 2 + 3 ) and ( 2 ) = log e , then is equal to ______ . [2023]

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Solution

Q.
If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{4 x}{(x^{2} - 1)} y = \frac{x + 2}{{(x^{2} - 1)}^{5 / 2}}, x > 1$ such that $y (2) = \frac{2}{9} \log_{e} (2 + \sqrt{3})$ and $y (\sqrt{2}) = α \log_{e} (\sqrt{α} + β) + β - \sqrt{γ}, α, β, γ \in ℕ,$ then $α β γ$ is equal to ______ . [2023]

Ans.
(6)

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

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

$I.F. = e^{2 \ln (x^{2} - 1)} = {(x^{2} - 1)}^{2}$

$\Rightarrow y \times {(x^{2} - 1)}^{2} = \int \frac{x + 2}{{(x^{2} - 1)}^{5 / 2}} \times {(x^{2} - 1)}^{2} d x$

$\Rightarrow$ $y \times {(x^{2} - 1)}^{2} = \int \frac{x + 2}{\sqrt{x^{2} - 1}} d x$

$\Rightarrow y \times {(x^{2} - 1)}^{2} = \int \frac{x d x}{\sqrt{x^{2} - 1}} + 2 \int \frac{d x}{\sqrt{x^{2} - 1}}$

$\Rightarrow y \times {(x^{2} - 1)}^{2} = \sqrt{x^{2} - 1} + 2 \ln$ $| x + \sqrt{x^{2} - 1} |$ $+ C$

$\Rightarrow y (2) = {(3)}^{- 3 / 2} + \frac{2 \ln | 2 + \sqrt{3} |}{9} + \frac{C}{9}$

$\Rightarrow \frac{2}{9} \log_{e} (2 + \sqrt{3}) = {(3)}^{- 3 / 2} + \frac{2}{9} \log_{e}$ $| 2 + \sqrt{3} |$ $+ \frac{C}{9}$

$\Rightarrow \frac{C}{9} = - {(3)}^{- 3 / 2} \Rightarrow C = - \sqrt{3}$

$∴$ $y = {(x^{2} - 1)}^{- \frac{3}{2}} + \frac{2 \log | x + \sqrt{x^{2} - 1} |}{{(x^{2} - 1)}^{2}} - \frac{\sqrt{3}}{{(x^{2} - 1)}^{2}}$

$Now, y (\sqrt{2}) = {(2 - 1)}^{- \frac{3}{2}} + \frac{2 \log | \sqrt{2} + \sqrt{1} |}{1} - \frac{\sqrt{3}}{1}$

$\Rightarrow α \log_{e} (\sqrt{α} + β) + β - \sqrt{γ} = 1 + \log (\sqrt{2} + 1) - \sqrt{3}$

$∴$ $α = 2, β = 1, γ = 3 \Rightarrow α β γ = 6$

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