If x = f(y) is the solution of the differential equation with f(0) = 1, then is equal to : [2025]

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Solution

Q.
If x = f(y) is the solution of the differential equation $(1 + y^{2}) + (x - 2 e^{\tan^{— 1} y}) \frac{d y}{d x} = 0, y \in (- \frac{π}{2}, \frac{π}{2})$ with f(0) = 1, then $f (\frac{1}{\sqrt{3}})$ is equal to : [2025]

1 $e^{π / 6}$

2 $e^{π / 12}$

3 $e^{π / 3}$

4 $e^{π / 4}$

Ans.
(1)

$\frac{d x}{d y} + \frac{x}{1 + y^{2}} = \frac{2 e^{\tan^{- 1} y}}{1 + y^{2}}$ , which is linear differential equation of first order.

$I . F . = e^{\int \frac{d y}{1 + y^{2}}} = e^{\tan^{— 1} y}$

Thus, the required solution is given by

$x e^{\tan^{— 1} y} = \int \frac{2 {(e^{\tan^{— 1} y})}^{2}}{1 + y^{2}} d y$

Put $\tan^{— 1} y = t, \frac{d y}{1 + y^{2}} = d t$ $∴ x e^{\tan^{- 1} y} = e^{2 \tan^{- 1} y} + c$

Given, f(0) = 1, we get 1 = 1 + c $\Rightarrow$ c = 0

Therefore, $x = e^{\tan^{- 1} y}$

$\Rightarrow f (\frac{1}{\sqrt{3}}) = e^{\frac{π}{6}}$

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