Question

Is the pair
x1(t) = sin(3t)
x2(t) = cos(3t+4)
linearly independent solutions of x'' = -9x?

How do I show this using the Wronskian? I know I have to show that the Wronskian determinate is 0 or never 0, but when I find the determinate, I get
W(t) = (sin(3t))(-3sin(3t+4) - (3cos(3t))(cos(3t+4)).

I don't know where to go from there. Also, I don't understand how this relates to the given DE x'' = -9x.

Help would be appreciated. :) Thank you.