Solve the differential equation `(y)/(dx)=y sin 2x, " given that " y(0) =1.`

Answer

`y=e^(sin^(2)x)`

Explanation

`int (1)/(y)dy = int sin2x dx +C rArr log |y| = -(1)/(2)cos2x +C." " `...(i)
Putting x = 0 and y = 1 in (i), we get C = `(1)/(2).`
`therefore log |y|=(1)/(2)(1-cos2x)=((1)/(2) xx 2 sin^(2)x)=sin^(2)x.`
Hence, `y = e^(sin^(2)x).`