Let y=y(x) be the solution of the differential equation sin⁡xdydx+ycos⁡x=4x,x∈(0,π). if yπ2=0 then yπ6 is equal to

$\frac{d y}{d x} + (\cot x) y = 4 x cosec x$

I.F. $= e^{\int \cot x d x} = e^{\log (\sin x)} = \sin x$

Then the solution is given by

$\begin{array}{l} y \cdot \sin x = \int 4 x cosec (x) \sin x d x + C \end{array}$

i.e., $y \sin x = 2 x^{2} + C$

As $y (π / 2) = 0$ we have $C = - π^{2} / 2$

So, $y \sin x = 2 x^{2} - π^{2} / 2$

$∴ y (π / 6) = 2 \{\frac{2 π^{2}}{36} - \frac{π^{2}}{2}\} = 2 π^{2} \{\frac{1}{18} - \frac{1}{2}\} = - \frac{8}{9} π^{2}$

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
$y (\frac{π}{2}) = 0$ then $y (\frac{π}{6})$ is equal to