Problem:
y″−y=cosx
Solution:
Let's first solve z″−z=0, using the characteristic polynomial method, we know the answer is z=Aex+Be−x.
It remains to find just one function that fits w″−w=cosx, 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=cosx and see
(cosx)″−cosx=−2cosx.
So now the solution is obviously w=−12cosx
The full solution is then Aex+Be−x−12cosx.
y″−y=cosx
Solution:
Let's first solve z″−z=0, using the characteristic polynomial method, we know the answer is z=Aex+Be−x.
It remains to find just one function that fits w″−w=cosx, 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=cosx and see
(cosx)″−cosx=−2cosx.
So now the solution is obviously w=−12cosx
The full solution is then Aex+Be−x−12cosx.
No comments:
Post a Comment