From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is
[tex]3r^2-6r+6=0\iff r^2-2r+2=0[/tex]
which has roots at [tex]r=1\pm i[/tex]. This admits the two fundamental solutions
[tex]y_1=e^x\cos x[/tex]
[tex]y_2=e^x\sin x[/tex]
The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form
[tex]y_p=u_1y_1+u_2y_2[/tex]
where
[tex]u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx[/tex]
[tex]u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx[/tex]
and [tex]W(y_1,y_2)[/tex] is the Wronskian of the fundamental solutions. We have
[tex]W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}[/tex]
and so
[tex]u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx[/tex]
[tex]u_1=\dfrac13\ln|\cos x|[/tex]
[tex]u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx[/tex]
[tex]u_2=\dfrac13x[/tex]
Therefore the particular solution is
[tex]y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x[/tex]
so that the general solution to the ODE is
[tex]y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x[/tex]
