Official solution
Given,
sin x \[{{dy} \over {dx}} + y\cos y = 4x\]
\[ \Rightarrow \,\,\,\,{{dy} \over {dx}}\,\] + y cot x = 4x cosec x
This is a linear differential equation of form,
\[{{dy} \over {dx}}\,\] + py = Q
Where p = cot x and Q = 4x cosec x
So, Integrating factor (I. F)
= \[{e^{\int {pdx} }}\]
= \[{e^{\int {\cot dx} }}\]
= \[{e^{\ln \left| {\sin \,x} \right|}}\]
= sin x as \[x \in \left( {0,\pi } \right)\]
Solution of the differential equation is
y sin x = \[\int {} \] 4x cosecx sinx dx + c
\[ \Rightarrow \,\,\,\,\] y sinx = \[\int {4x\,dx\, + c} \]
\[ \Rightarrow \,\,\,\,\] y sinx = 4.\[{{{x^2}} \over 2} + c\]
\[ \Rightarrow \,\,\,\,\] y sinx = 2x2 + c . . . . . (1)
Given that, \[y\left( {{\pi \over 2}} \right) = 0\]
\[\therefore\,\,\,\] x = \[y\left( {{\pi \over 2}} \right) = 0\] and y = 0
Put this x = \[{\pi \over 2}\] and y = 0 at equation (1)
0.1 = 2. \[\left( {{\pi \over 2}} \right)\]2 + c
\[ \Rightarrow \,\,\,\] c =\[ - {{{\pi ^2}} \over 2}\]
So, differential equation is
y sin x = 2x2 - \[{{{\pi ^2}} \over 2}\,\,.....(2)\]
Now we have to find y \[\left( {{\pi \over 6}} \right).\]
So, put x = \[{{\pi \over 6}}\] at equation (2)
y . sin \[{{\pi \over 6}}\] = 2 \[{\left( {{\pi \over 6}} \right)^2} - {{{\pi ^2}} \over 2}\]
\[ \Rightarrow \,\,\,\,\,\,\,y.{1 \over 2}\] = 2. \[{{{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 2}\]
\[ \Rightarrow \,\,\,\,\,{y \over 2} = {{{\pi ^2}} \over {18}} - {{{\pi ^2}} \over 2}\]
\[ \Rightarrow \,\,\,\,{y \over 2} = {{{\pi ^2} - 9{\pi ^2}} \over {18}}\]
\[ \Rightarrow \,\,\,\,\,y = - {{8{\pi ^2}} \over 9}\]