Did you get sin(6t) and cos(6t) as independent solutions to the corresponding homogeneous equation?

Then try a solution of the form y= u(t)sin(6t)+ v(t)cos(6t). y'= u' sin(6t)+ 6u cos(6t)+ v' cos(6t)- 6v sin(6t). We simplify the calculation by restricting u and v to such that u' sin(6t)+ v' cos(6t)= 0.

Now y'= 6u cos(6t)- 6v sin(6t) and y"= 6u' cos(6t)- 36u sin(6t)- 6v' sin(6t)- 36 v cos(6t). Putting that into the differential equation, 6u' cos(6t)- 36 u sin(6t)- 6v' sin(6t)- 36 v cos(6t)+ 36(u sin(6t)+ v cos(6t)= 6u' cos(6t)- 6v' sin(6t)= -12 sec(6t).

Now you have two equations, u' sin(6t)+ v' cos(6t)= 0 and u' cos(6t)- v' sin(6t)= -2 sec(6t) to solve for u' and v'.

For example, if you multiply both sides of the first equation by sin(6t) you get while multiplying both sides of the second equation by cos(6t) gives because sec(6t)= 1/cos(6t).

Adding those two equations eliminates v' and gives u'= -2 so u= -2t (since we are only looking for a particular solution we can take the constant of integration to be 0) and u sin(6t)= -2t sin(6t).

Setting u'= -2 makes the first equation -2 sin(6t)+ v' cos(6t)= 0 so so .