Method 1: Because 1+i is not the root of the characteristic equation of homogeneous linear equation, let the special solution of non-homogeneous linear equation y * = e x (acosx+bsinx) be substituted.

(-A-2B)cosx+(2A-B)sinx=cosx

Therefore, -A-2B= 1, 2A-B=0, while A=- 1/5 and B=-2/5.

So y * =- 1/5e x (cosx+2sinx).

Method 2: e^xcosx is the real part of e (( 1+I) x), so first find the special solution of y'-4y'+3y = e (( 1+I) x) and set it as y * = AE ((65438)).

So y * = (-1/5+2i/5) e ((1+i) x) = (-1/5+2i/5) e x (cosx+isinx), which is actually -65438+.

So a special solution of y''-4y'+3y = e x cosx y * =-1/5ex cosx-2/5exsinx.