Official solution
\[\cos x{{dy} \over {dx}} - y\sin x = 6x\]
\[ \Rightarrow \] \[{{dy} \over {dx}} - y\tan x = 6x.\sec x\]
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 \[y\left( {{\pi \over 6}} \right)\]
\[y{{\sqrt 3 } \over 2} = {{3{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 3}\]
\[ \Rightarrow \] \[{{\sqrt 3 y} \over 2} = {{ - 3{\pi ^2}} \over {12}} \]
\[\Rightarrow y = {{ - {\pi ^2}} \over {2\sqrt 3 }}\]