Solve the differential equation (dy/dx)- 3y cot x =sin 2x ; y=2 when…

The given differential equation is (dy/dx) -3ycotx = sin 2x

which is of the form dy/dx + py =Q (Where p= -3cot x and Q= sin 2x)

Now, I.F. = e^∫pdx

I.F. = e^-3∫cotdx

So. I.F. = 1/(sin ³x)

The solution is given by: y.(I.F.) = ∫Q x(I.F.)dx +C

⇒ y. 1/(sin ³x) = ∫{sin2x.  1/(sin ³x)} dx +C

⇒ y. cosec³x = 2∫(cot x.cosecx dx) +C

⇒ y= 2sin²x+ C sin ³x   ......................(1)

now, y =2 at x= π/2

Therefore, we get:

2 =-2 +C

C= 4

So from equation (1)

y= 4sin ³x -2sin²x