Solve the initial value problem: `(dy)/(dx)-3ycotx=sin2x ; y=2\ w h e n\ x=\ pi/2` .

Explanation

``
`frac{d y}{d x}-3 cot x cdot y =sin 2 x`


`text { I.F. } =e^{int-3 cot x d x} `


`=e^{-3 log (sin x)}=(sin x)^{-3}`


`={cosec}^{3} x`


``
`therefore` Solution is :


``
...