Nonhomogeneous L.D.E. - y'' - 2y' - 8y= 3e^-2x + 5x^2 - 4x

[forza1nter]

Ok guys, this problem is driving me crazy,

y'' - 2y' - 8y = 3e^-2x + 5x^2 - 4x (method of undetermined coefficients)

the LHS, of course is easy to find(obviously), but to me honest im stumped on even how to setup the RHS. Can anyone set up the RHS with instructions on how to compute it?

its weird, but im ok with the cos/ sin type of probs, but this polynomial is really kicking my butt (3 hrs and still no soln!!!!)

thx in advance

I/m

 

[Unco]

The method of undetermined coefficients handles such a problem fine. If the RHS was just \(\displaystyle 3e^{-2x}\) or just \(\displaystyle 5x^2 - 4x\) I suspect you would know to choose \(\displaystyle Ae^{-2x}\) or \(\displaystyle Ax^2+Bx+C\) for particular solutions, respectively. Well, by linearity we can just as well combine the two to consider a particular solution of the form \(\displaystyle y_p = Ae^{-2x} + Bx^2+Cx+D\) (A, B, C, D constants).

Show us your complete work if you get stuck.