Problem:
$ y'' - y = \cos x $
Solution:
Let's first solve $ z'' - z = 0 $, using the characteristic polynomial method, we know the answer is $ z = Ae^x + Be^{-x} $.
It remains to find just one function that fits $ w'' - w = \cos x $, then we know $ y = w + z $ and have the degree of freedom we need.
At this point, the so called "method of undetermined coefficients" is really just about guessing the solution, let's guess our solution is $ w = \cos x $ and see
$ (\cos x)'' - \cos x = -2 \cos x $.
So now the solution is obviously $ w = \frac{-1}{2}\cos x $
The full solution is then $ Ae^x + Be^{-x} - \frac{1}{2}\cos x $.
$ y'' - y = \cos x $
Solution:
Let's first solve $ z'' - z = 0 $, using the characteristic polynomial method, we know the answer is $ z = Ae^x + Be^{-x} $.
It remains to find just one function that fits $ w'' - w = \cos x $, then we know $ y = w + z $ and have the degree of freedom we need.
At this point, the so called "method of undetermined coefficients" is really just about guessing the solution, let's guess our solution is $ w = \cos x $ and see
$ (\cos x)'' - \cos x = -2 \cos x $.
So now the solution is obviously $ w = \frac{-1}{2}\cos x $
The full solution is then $ Ae^x + Be^{-x} - \frac{1}{2}\cos x $.
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