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$Let y = y (x) be the solution of the differential equation \sin x \frac{d y}{d x} + y \cos x = 4 x, \in (0, π) . If$ $y (\frac{π}{2}) = 0, then y (\frac{π}{6}) is equal to$

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Correct option is D

Given that sinxdydx+ycosx=4x⇒ddxysinx=4xIntegration on both sides∫dysinx=∫4xdx⇒ysinx=4x22+C⇒ysinx=2x2+C Given yπ2=0C=−π22 Thus ysinx=2x2-π22 Now ysinπ6=2π236-π22⇒y·12=π218−π22⇒y2=−8π218∴y=−8π29

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Let y=y(x) be the solution of the differential equation sinxdydx+ycosx=4x,∈(0,π). If yπ2=0, then yπ6 is equal to

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$Let y = y (x) be the solution of the differential equation \sin x \frac{d y}{d x} + y \cos x = 4 x, \in (0, π) . If$ $y (\frac{π}{2}) = 0, then y (\frac{π}{6}) is equal to$