- #1
kostoglotov
- 234
- 6
Homework Statement
Write a trial solution for the method of undetermined coefficients. Do not determine the coefficients.
For: [itex]y'' + 2y' + 10y = x^2e^{-x}\cos{3x}[/itex]
There's a modification performed and I'm not 100% confident as to why.
Homework Equations
The Attempt at a Solution
The complementary equation gives a solution of
[tex]y_c = e^{-x}\left(c_1\cos{3x} + c_2\sin{3x}\right)[/tex]
Now, before considering a modification to the particular solution [itex]y_p[/itex], let's consider the standard trial solution by the method of undetermined coefficients. I would say it's
[tex]y_p = e^{-x}\left[(Ax^2+Bx+C)\cos{3x} + (Dx^2+Ex+F)\sin{3x}\right][/tex]
Now I know that the modification is performed so that "no term in the particular solution can be a solution for the complementary equation"...I do know basically what this means...and I would have thought that for the particular trial solution above, that neither term would suitably solve the complementary equation. Because the particular solution has a polynomial of degree 2 and the complementary solution has a polynomial of degree 0.
So I've thought, maybe we could solve the complementary equation with the trial solution [itex]y_p[/itex] if we set A,B and/or D,E to = 0,0, so that the polynomials in the trial solutions were reduced from 2nd degree to 0 degree.
Is this why we need to modify [itex]y_p[/itex] by multiply it by x? Because we can set A,B,D,E to zero in order to solve for the complementary equation? Thus, it won't really matter what degree a polynomial is in the solutions as far as deciding to modify is concerned?