If the solution curve y = y(x) of the differential equation , x > 0 passes through the point (1, 1) and then is __________. [2024]

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Solution

Q.
If the solution curve y = y(x) of the differential equation $(1 + y^{2}) (1 + \log_{e} x) d x + x d y = 0$ , x > 0 passes through the point (1, 1) and $y (e) = \frac{α - \tan (\frac{3}{2})}{β + \tan (\frac{3}{2})}$ then $α + 2 β$ is __________. [2024]

Ans.
(3)

We have, $(1 + y^{2}) (1 + \log_{e} x) d x + x d y = 0$

$\Rightarrow \frac{d y}{1 + y^{2}} = - \frac{(1 + \log_{e} x) d x}{x}$

Integrating both sides, we get

$\tan^{- 1} y = - \frac{1}{2} {(1 + \log_{e} x)}^{2} + C$ ...(i)

At (1, 1), $\frac{π}{4} = \frac{- 1}{2} + C \Rightarrow C = \frac{π}{4} + \frac{1}{2}$

$∴ y = \tan {\frac{π}{4} + \frac{1}{2} [1 - {(1 + \log_{e} x)}^{2}]}$

$\Rightarrow y (e) = \tan (\frac{π}{4} + \frac{1}{2} - 2)$

$= \tan [\frac{π}{4} - \frac{3}{2}] = \frac{\tan \frac{π}{4} - \tan \frac{3}{2}}{1 + \tan \frac{π}{4} \tan \frac{3}{2}} = \frac{1 - \tan \frac{3}{2}}{1 + \tan \frac{3}{2}}$

So, $α$ = 1 and $β$ = 1

$∴ α + 2 β = 3$ .

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