y'' - y' - 2y = 0
let y = e^(kt)
y' = k.e^(kt)
y'' = k^2.e^(kt)
k^2.e^(kt) - k.e^(kt) - 2e^(kt) = 0
e^(kt) (k^2 - k -2) = 0
e^(kt) cant = 0
therefore k^2 - k -2 = 0
k = 2 or -1
therfore Yh = Ae^2t + Be^-t
f(t) = 6t^2 - 1
guess Yp = Ct^2 + Dt + E
Yp' = 2Ct + D
Yp'' = 2C
sub back into original equation
(2C) - (2Ct + D) - 2(Ct^2 + Dt + E) = 6t^2 - 1
2C(1-t-t^2) + D(-1-2t) - 2E = 6t^2 - 1
dont know how to find C D or E