$ \text{ The Auxiliary equation is } $

$ D^2 + 5D + 6 = 0 $

$D = -3, -2 $

$ \text{ C.F. is } y_c = c_1e^{-3x} + c_2e^{-2x} $

$ \text{P.I = }y_p = \frac{1}{D^2 + 5D + 6}e^{-2x} sec^2x(1+2tanx)$

$ = \frac{1}{(D+2)(D + 3)}e^{-2x} sec^2x(1+2tanx)$

$ = \frac{1}{(D + 3)}[ e^{-2x} \int e^{2x} e^{-2x}sec^2x(1+2tanx) dx]$

$ = \frac{1}{(D + 3)}[ e^{-2x} \int sec^2x(1+2tanx) dx]$

$ \text {Put tanx = t} $

$ sec^2x dx = dt$

$ = \frac{1}{(D + 3)}[ e^{-2x} \int (1+2t) dt]$

$ = \frac{1}{(D + 3)}[ e^{-2x} (t+t^2) ]$

$ = \frac{1}{(D + 3)}[ e^{-2x} (tanx+tan^2x) ]$

$ = e^{-3x}\int[ e^{3x} e^{-2x} (tanx+tan^2x) dx ]$

$ = e^{-3x}\int[ e^{x} (tanx+sec^2x -1) dx ]$

$ = e^{-3x}[ \int[ e^{x} (tanx+sec^2x) dx - \int e^x dx ]$

$ = e^{-3x}[ e^{x} tanx - e^x ]$

$ = e^{-2x}[ tanx - 1 ] $

$\therefore \text{ The complete solution is y = C.F. + P.I. } $

$ \therefore y = c_1e^{-3x} + c_2e^{-2x} + e^{-2x}[ tanx - 1 ]$