Let y = y ( x ) be a solution of the differential equation ( x cos x ) d y + ( x y sin x + y cos x - 1 ) d x = 0 , 0 x 2 . If 3 y ( 3 ) = 3 , then | 6 ? y ' ' ( 6 ) + 2 ? y ' ( 6 ) | is equal to ______ . [2023]

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Solution

Q.
Let $y = y (x)$ be a solution of the differential equation $(x \cos x) d y + (x y \sin x + y \cos x - 1) d x = 0, 0 < x < \frac{π}{2} .$ If $\frac{π}{3} y (\frac{π}{3}) = \sqrt{3},$ then $| \frac{π}{6} y'' (\frac{π}{6}) + 2 y' (\frac{π}{6}) |$ is equal to ______ . [2023]

Ans.
(2)

$(x \cos x) d y + (x y \sin x + y \cos x - 1) d x = 0,$

$0 < x < \frac{π}{2}$ $\Rightarrow \frac{d y}{d x} + \frac{(x \sin x + \cos x)}{x \cos x} y = \frac{1}{x \cos x}$

$I.F. = e^{\int (\frac{x \sin x + \cos x}{x \cos x}) d x} = e^{\int (\tan x + \frac{1}{x}) d x} = e^{\ln | \sec x | + \ln | x |} = e^{\ln | x \sec x |}$

$I.F. = x \sec x$

$y \times x \sec x = \int \frac{x \sec x}{x \cos x} d x \Rightarrow y \times x$ $s e c x = \tan x + C$

Given, $y (\frac{π}{3}) = \frac{3 \sqrt{3}}{π}$

$So, \frac{3 \sqrt{3}}{π} \times \frac{π}{3} \cdot 2 = \sqrt{3} + C$ $\Rightarrow 2 \sqrt{3} = \sqrt{3} + C \Rightarrow C = \sqrt{3}$

$∴$ $y x \sec x = \tan x + \sqrt{3} \Rightarrow x y = \sin x + \sqrt{3} \cos x$

$\Rightarrow x y' + y = \cos x - \sqrt{3} \sin x$

$\Rightarrow x y'' + 2 y' = - \sin x - \sqrt{3} \cos x$

$∴$ $| \frac{π}{6} \cdot y'' (\frac{π}{6}) + 2 y' (\frac{π}{6}) |$ $=$ $| - \frac{1}{2} - \frac{3}{2} |$ $= 2$

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