The differential equations, find a particular solution satisfying the given condition: `(dy)/(dx)-3ycotx=sin2x ; y=2`when `x=pi/2`

Explanation

`(dy)/(dx)=3y cot x=sin 2x`
Herem `P=-3cotx`, `Q=sin2x`
`I.F.=e^(intPdx)=e^(-3cotxdx)`
`=e^(-3logsinx)`
`=e^(log(sinx)^(-3))`
`=(sinx)^(-3)=(1)/(sin^(3)x)`
and general solution
`y(I.F.)=intQ*(I.F.)dx+c`
`implies y(1)/(sin^(3)x)=intsin2x*(1)/(sin^(3)x)dx+c`
`implies (y)/(sin^(3)x)=int(2sinxcosx)/(sin^(3)x)dx+c`
`=2int(cosx)/(sin^(2)x)dx+c`
Let `sinx=t`
`implies cosxdx=dt`
`=2int(dt)/(t^(2))+c`
`=-(2)/(t)+c`
`=-(2)/(sinx)+c`
`:. (y)/(sin^(3)x)=-(2)/(sinx)+c`.........`(1)`
Given that, `y=2` at `x=pi//2`
`:. (2)/(sin^(3)((pi)/(2)))=-(2)/(sin((pi)/(2)))+c`
`impliesc=4`
`:.` From equation `(1)`,
`(y)/(sin^(3)x)=(-2)/(sinx)+4`
`implies y=4sin^(3)x-2sin^(2)x`