question: y'' + 2y' - 24y = 16 - (x+2)e^4x
what i've done:
yh : λ^2 + 2λ - 24 = 0
(λ+6)(λ-4) = 0
yh = c1e^(-6x) + c2e^4x
Try: yp = (A1x)e^4x <--- this is ok because it doesn't appear in yh
y'p= (4A1x)e^4x + A1e^4x
y''p= (16A1x)e^4x + 4A1e^4x + 4A1e^4x = (16A1x)e^4x + 8A1e^4x
subbing into the D.E gives...
(16A1x)e^4x + 8A1e^4x + 2[(4A1x)e^4x + A1e^4x] - 24[(A1x)e^4x] = 16 - (x+2)e^4x
so grouping like terms gives: xe^4x: 16A1 + 8A1 - 24A1= 0 which is ok
e^4x: 8A1 + 2A1 = x+2 , A1= (x+2)/10
but where does the constant(16) come from?