Let y = y(x) be the solution of the differential equation d y d x = ( tan x ) + y sin x ( sec x - sin x tan x ) ,satisfying the condition y( 4 ) = 2 . Then, y ( 3 ) is [2024]

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Solution

Q.
Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} = \frac{(\tan x) + y}{\sin x (\sec x - \sin x \tan x)}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is [2024]

1 $\sqrt{3} (2 + \log_{e} 3)$

2 $\sqrt{3} (2 + \log_{e} \sqrt{3})$

3 $\frac{\sqrt{3}}{2} (2 + \log_{e} 3)$

4 $\sqrt{3} (1 + 2 \log_{e} 3)$

Ans.
(2)

We have, $\frac{d y}{d x} = \frac{\tan x + y}{\sin x (\sec x - \sin x \cdot \tan x)}$

$= \frac{\frac{\sin x}{\cos x} + y}{\sin x (\frac{1}{\cos x} - \sin x \cdot \frac{\sin x}{\cos x})}$

$= \frac{\sin x + y \cos x}{\sin x (1 - \sin^{2} x)} = \frac{\sin x + y \cos x}{\sin x \cdot \cos^{2} x}$

$\frac{d y}{d x} = \sec^{2} x + y \cdot \frac{1}{\sin x \cdot \cos x}$

Now, $IF = e^{- \int \frac{1}{\sin x \cdot \cos x} d x} = e^{- 2 \int cosec 2 x d x}$

$= e^{- 2} (\frac{\log | cosec 2 x - \cot 2 x |}{2}) = e^{- \log | cosec 2 x - \cot 2 x |}$

$= \frac{1}{| cosec 2 x - \cot 2 x |} = \frac{1}{| \tan x |}$

The solution is

$y \cdot \frac{1}{| \tan x |} = \int \frac{1}{| \tan x |} \cdot \sec^{2} x d x = \log | \tan x | + c$

$y = | \tan x | \cdot \log | \tan x | + c | \tan x |$

At $x = \frac{π}{4}, y = 2 \Rightarrow 2 = 0 + c \Rightarrow c = 2$

At $x = \frac{π}{3}$

$y = \sqrt{3} \cdot \log \sqrt{3} + 2 \sqrt{3} = \sqrt{3} (\log \sqrt{3} + 2)$ .

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