text: I do not know if this is the 'annihilator method' but here goes:\n\ny'' + 4y' + 4y = 2x + 6\n\nFirst solve y'' + 4y' + 4y = 0\nK^2 + 4K + 4 = 0 \nSo (K + 2)^2 = 0\nie K = -2, -2\nSolution is y = Ae^(-2t) + Bte^(-2t)\n\nNow for particular solution\n\nyp = Ax + B\nyp' = A\nyp'' = 0\ny'' + 4y' + 4y = Ax + (4A + B)\n= 2x + 6\nSo A = 2 and 4A + B = 6 \nSo B = -2\n\nThus the general solution is y = Ae^(-2t) + Bte^(-2t) + 2x - 2