$$\cos x{{dy} \over {dx}} - y\sin x = 6x$$

$$ \Rightarrow $$ $${{dy} \over {dx}} - y\tan x = 6x.\sec x$$

This is a linear differential equation.

$$ \therefore $$ I.F = $${e^{\int { - \tan xdx} }}$$ = $$\left| {\cos x} \right|$$

As 0 < x < $${\pi \over 2}$$, so I.F = $$\cos x$$

So solution is

y(I.F) = $$\int {6x.\sec x\left( {I.F} \right)} dx$$

$$ \Rightarrow $$ y cosx = $$\int {6x.\sec x\cos x} dx$$

$$ \Rightarrow $$ y cosx = $$6\int x dx$$

$$ \Rightarrow $$ y cosx = 3x² + C

As $$y\left( {{\pi \over 3}} \right) = 0 $$

$$ \therefore $$ $$ \left( 0 \right) \times \left( {{1 \over 2}} \right) = {{3{\pi ^2}} \over 9} + C $$

$$\Rightarrow C = {{ - {\pi ^2}} \over 3}$$

$$ \Rightarrow y\cos x = 3{x^2} - {{{\pi ^2}} \over 3}$$

For $$y\left( {{\pi \over 6}} \right)$$

$$y{{\sqrt 3 } \over 2} = {{3{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 3}$$

$$ \Rightarrow $$ $${{\sqrt 3 y} \over 2} = {{ - 3{\pi ^2}} \over {12}} $$

$$\Rightarrow y = {{ - {\pi ^2}} \over {2\sqrt 3 }}$$