First check the characteristic solution:
y'' + 4y' + 4y = 0
has characteristic equation
r ² + 4r + 4 = (r + 2)² = 0
with a double root at r = -2, so the characteristic solution is
[tex]y_c = C_1e^{-2t} + C_2te^{-2t}[/tex]
For the particular solution corresponding to [tex]12te^{-2t}[/tex], we might first try the ansatz
[tex]y_p = (At+B)e^{-2t}[/tex]
but [tex]e^{-2t}[/tex] and [tex]te^{-2t}[/tex] are already accounted for in the characteristic solution. So we instead use
[tex]y_p = (At^3+Bt^2)e^{-2t}[/tex]
which has derivatives
[tex]{y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}[/tex]
[tex]{y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}[/tex]
Substituting these into the left side of the ODE gives
[tex](4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}[/tex]
so that 6A = 12 and 2B = 0, or A = 2 and B = 0.
For the second solution corresponding to [tex]-8t-12[/tex], we use
[tex]y_p = Ct + D[/tex]
with derivative
[tex]{y_p}' = C[/tex]
[tex]{y_p}'' = 0[/tex]
Substituting these gives
[tex]4C + 4(Ct+D) = 4Ct + 4C + 4D = -8t-12[/tex]
so that 4C = -8 and 4C + 4D = -12, or C = -2 and D = -1.
Then the general solution to the ODE is
[tex]y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1[/tex]
Given the initial conditions y (0) = -2 and y' (0) = 1, we have
[tex]-2 = C_1 - 1 \implies C_1 = -1[/tex]
[tex]1 = -2C_1 + C_2 - 2 \implies C_2 = 1[/tex]
and so the particular solution satisfying these conditions is
[tex]y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1[/tex]
or
[tex]\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}[/tex]
