let y = y(x) be the solution of the differential equation $\sin x \frac{d y}{d x} + y \cos x = 4 x, x \in (0, π)$ . If $y = (\frac{π}{2}) = 0, t h e n y (\frac{π}{6}) i s e q u a l t o :$

$- \frac{4}{9} π^{2}$

$\frac{4}{9 \sqrt{3}} π^{2}$

$\frac{- 8}{9 \sqrt{3}} π^{2}$

$- \frac{8}{9} π^{2}$

Solution

$\frac{- 8}{9 \sqrt{3}} π^{2}$

$\sin x \frac{d y}{d x} + y \cos x = 4 x, x \in (0, π) \frac{d y}{d x} + y c o t x = \frac{4 x}{\sin x} ∴ I . F = e^{\int c o t x d x} = \sin x ∴ s o l u t i o n i s g i v e n b y y \sin x = \int \frac{4 x}{\sin x} . \sin x d x y = \sin x = 2 x^{2} + C W h e n x = \frac{π}{2}, y = 0 \Rightarrow C = - \frac{π^{2}}{2} ∴ E q u a t i o n i s : y \sin x = 2 x^{2} - \frac{π^{2}}{2} w h e n x = \frac{π}{6} t h e n y . \frac{1}{2} = 2 . \frac{π^{2}}{36} - \frac{π^{2}}{2} t h e r e f o r e, y = - \frac{8 π^{2}}{9}$

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