18. Using the method of variation of parameter, a particular solution to y′′ + 16y = 4 sec(4t) isyp(t) = u1(t) cos(4t) + u2(t) sin(4t). Then u2(t) is equal toA. 1 B. t C. ln | sin 4t| D. ln | cos 4t| E. sec(4t)

Q: the correct answer is C could you please explain

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