Let `y=g(x)` be the solution of the differential equation `sin (dy)/(dx)+y cos x=4x, x in (0,pi)` If y(pi/2)=0`, then `y(pi/6)` is equal to

Options

A. `-4/9pi^(2)`
B. `4/(9sqrt(3))pi^(2)`
C. `-8/(9sqrt(3))pi^(2)`
D. `-8/9pi^(2)`

Answer

D

Explanation

`sinx(dy)/(dx)+ycosx=4x`
`therefore (dy)/(dx) +y.cotx=(4x)/(sinx)`
This is a linear differential equation.
I.F. `=e^(intPdx)=e^(intcotx.dx) = e^(log(sinx))=sinx`
Therefore, solution is
`ysin=int(4xdx+C)`
`therefore ysinx=2x^(2)+C`
Given that `y(pi/2)=0`.
`therefore 0 xx sin pi/2 =2 xx (pi/2)^(2)+C`
or `C=-pi^(2)/2`
`therefore ysinx=2x^(2)-pi^(2)/2`
`therefore ysinx=2x^(2)-pi^(2)/2`
Putting `x=pi/6`, we get
`y/2=2(pi/6)^(2)-pi^(2)/2`
`therefore y=(-8pi^(2))/9`