Solution of the equation `cos^2x(dy/dx)-(tan2x)y=cos^4x, when `y(pi/6)=(3sqrt(3))/8` is

Options

A. `y=tan2xcos^(2)x`
B. `y=cot2xcos^(2)x`
C. `y=1/2tan2xcos^(2)x`
D. `y=1/2cot2xcos^(2)x`

Answer

C

Explanation

The given differential equation can be written as
`(dy)/(dx) -(tan2x)/(cos^(2)x)y`
`=cos^(2)x` which is linear different equation of first order.
`intPdx=int(-sin2x)/(cos2xcos^(2)x)dx`
`=-int(2sin2xdx)/(cos2x(1+cos2x))`
`=int(dt)/(t(1+t))`
`=int(1/t-1/(1+t))dt`
`=logt/(1+t)` where `t=cos2x`
`therefore e^(int Pdx) = e^(log(cos2x)/(1+cos2x))=(cos2x)/(1+cos2x) = (cos2x)/(2cos^(2)x)`
Thus, the solution is
`y(cos2x)/(2cos^(2)x)=int(cosx^(2)xcos2x)/(2cos^(2)x)dx+C`
`=1/4sin2x+C`
When `x=pi/6, y=(3sqrt(3))/(8)`
`therefore (3sqrt(3))/(8)(4)/(2 xx 2 xx 3)=1/4(sqrt(3)/2)+C` or C=0
`therefore y=1/2tan2xcos^(2)x`