Let y = y(x) be the solution of the differential equation , . If , then is equal to _________. [2025]

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Solution

Q.
Let y = y(x) be the solution of the differential equation $2 \cos x \frac{d y}{d x} = \sin 2 x - 4 y \sin x$ , $x \in (0, \frac{π}{2})$ . If $y (\frac{π}{3}) = 0$ , then $y' (\frac{π}{4}) + y (\frac{π}{4})$ is equal to _________. [2025]

Ans.
(1)

Given $2 \cos x \frac{d y}{d x} = \sin 2 x - 4 y \sin x$ , $x \in (0, \frac{π}{2})$

The simplified is given by $\frac{d y}{d x} = \frac{2 \sin x \cdot \cos x}{2 \cos x} - \frac{4 \sin x}{2 \cos x} y$

$\Rightarrow \frac{d y}{d x} + 2 y \tan x = \sin x$

The integrating factor is given by

$I . F . = e^{\int 2 \tan x d x} = 2 \sec^{2} x$

$∴$ The solution is given by

$\Rightarrow 2 y \sec^{2} x = 2 \int \frac{\sin x}{\cos^{2} x} d x$

$\Rightarrow y \sec^{2} x = \int \tan x \cdot \sec x d x$

$\Rightarrow y \sec^{2} x = \sec x + c$

Since, $y (\frac{π}{3}) = 0 \Rightarrow 0 = 2 + c \Rightarrow c = - 2$

$∴ y = \cos x - 2 \cos^{2} x & y' = - \sin x + 4 \sin x \cos x$

Here, $y (\frac{π}{4}) = \frac{1}{\sqrt{2}} - 1 & y' (\frac{π}{4}) = \frac{- 1}{\sqrt{2}} + 2$

Hence, $y' (\frac{π}{4}) + y (\frac{π}{4}) = \frac{1}{\sqrt{2}} - 1 - \frac{1}{\sqrt{2}} + 2 = 1$ .

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