I am having some trouble justifying to myself something. Please see my other question in which I talked about solving if the functions f and g are linearly independent or not where
$$f(x)=\cos(3x)$$
and
$$g(x)=5\cos^3 x−10\cos(x)$$
So , I was able to do this following the hint provided in Linear independence of trig functions and I solved that it was linearly independent, and that was apparently the correct answer.
Here is my issue, when I try to do the same problem using the Wronskian, (noting) $$f′(x)=−3\sin (3x)$$ and $$g′(x)=−15\cos^2 x\sin x+10\sin x$$
but I just can't seem to see how it is always non zero. There seems to be roots
For some reason, I keep getting the Wronskian as $W=-50\sin x \cos^3 x$ which indeed does have real roots. So this is confusing my because I was under the impression that if the Wronskian was ever zero than this would imply via Abels theorem that it was zero everywhere as well in that interval.
So is it that I must only evaluate exactly zero when dealing with functions in regard to the Wronskian? Doe plugging in x into the final answer or form on wronskian not tell us anything ?
Thank you all
here is what i know about the wronskians. suppose $f, g$ are solutions of a linear differential equation $$Ly = y'' + by' + cy = 0$$ and $w = fg' - f'g.$ then $w$ satisfies the ables equation $$w' = aw$$
from these you can conclude that if the wronskian of $f$ and $g$ are either identically zero or is never zero.
this does not mean the wronskian, which is defined for any two smooth functions, has the same property as two solutions of the second order linear differential equation $Ly = 0.$ what this implies is that $$f(x) = \cos 3x, g(x) = 5 \cos^3 x - 10 \cos x$$ cannot be the solutions of $$y'' + by' + cy = 0$$ for any $b, c.$