okay. You know that all the s^(4x) and e^(-4x) stuff will zero out. So we need some quadratic stuff that will leave a particular solution.
So, let's say that y_p is
yp = Ax^2+Bx+C
yp" = 2A
yp" - 16yp = 2A - 16(Ax^2+Bx+C)
So, we need
-16Ax^2 - 16B + 2A-16C = 256x^2
A = -16
B = 0
C = -2
y = c1*e^(4x) + c2*e^(-4x) - 16x^2 - 2
Check to be sure y"-16y = 256x^2
We consider the non-homogeneous problem {y''-16y} = {256x^{2}}
First we consider the homogeneous problem {y''-16y} = 0 :
1) the auxiliary equation is a r^2+br+c = =0 . ________________
2) The roots of the auxiliary equation are (enter answers as a comma separated list). ____________________ (PLEASE SHOW YOUR WORK IN DETAIL)
3) A fundamental set of solutions is (enter answers as a comma separated list). Using these we obtain the the complementary solution y_c=c_1 y_1+ c_2 y_2 for arbitrary constants c_1 and c_2 .
____________________ (PLEASE SHOW YOUR WORK IN DETAIL)
Next we seek a particular solution y_p of the non-homogeneous problem {y''-16y} = {256x^{2}} using the method of undetermined coefficients (See the link below for a help sheet)
4) Apply the method of undetermined coefficients to find y_p = ____________________ (PLEASE SHOW YOUR WORK IN DETAIL)
I have figured out the auxilirary equation r^2-16, the roots 4 and -4, and the fundamental set of solutions e^(4x),e^(-4x), but cannot find y_p. I know that A=-16 but that is as far as I can get. Please help
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