1. y''' + y'' = 8x^2
2. y'' - 2y' - 3y = 4e^x - 9
Annihilator for any polinomial D^n+1
annihilator for e functions = e^ax = (D-a)^n
The Attempt at a Solution
1. getting the complementary eq.
values of m = 0, 0, -1
yc= c1e^-x +c2 + xc3
then we multiply both sides by the annihilator
D^3<D^2(D+1)> = 8x^2<D^3>
we get y= c1e^-x +c2 + xc3 + x^2c4 + x^3c5
and from the original eq. the x^2 term has a coefficient of 8
therefore c4 = 8
sice there is no x^3 term then c5 should be zero ?
and what about c4?
the answer on the back says the answer is:
c1+c2x+c3e^-x + 2/3 x^4 + 8/3x^3 + 8x^2
2.values of m=3, 2
yc=c1e^3x + c2e^-x
if we multiply by the annihilator alpha = 1 in this case therefore annihilator would be (D-1)
so we have
yc=c1e^3x + c2e^-x + c3e^x
how do I go from there?