Hello everyone. I have been struggling with these variation of parameters problems.
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Any help would be appreciated. I am getting tangled up in the trig I think.
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 $\displaystyle u' sin^2(6t)+ v'cos(6t)sin(6t)= 0$ while multiplying both sides of the second equation by cos(6t) gives $\displaystyle u' cos^2(6t)- v'sin(6t)cos(6t)= -2 sec(6t)cos(6t)= -2$ 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 $\displaystyle v'= \frac{1}{2} tan(6t)$ so $\displaystyle v= -\frac{1}{12}ln(cos(6t))$.