Let y = y(x) be the solution of the differential equation $\sin x{{dy} \over {dx}} + y\cos x = 4x$, $x \in \left( {0,\pi } \right)$. If $y\left( {{\pi \over 2}} \right) = 0$, then $y\left( {{\pi \over 6}} \right)$ is equal to :

Question

Let y = y(x) be the solution of the differential equation

$\sin x{{dy} \over {dx}} + y\cos x = 4x$, $x \in \left( {0,\pi } \right)$.

If $y\left( {{\pi \over 2}} \right) = 0$, then $y\left( {{\pi \over 6}} \right)$ is equal to :

$ - {4 \over 9}{\pi ^2}$

${4 \over {9\sqrt 3 }}{\pi ^2}$

$ - {8 \over {9\sqrt 3 }}{\pi ^2}$

$ - {8 \over 9}{\pi ^2}$

Solution

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