Solve the differential equation $(D^3-3D^2+4D-2)y = (e^x +cos x)$, where $D = frac{mathrm{d}}{mathrm{d}x}$.

As an alternative method, you could find the particular integral through the method of undetermined coefficients.

$$(D^3-3D^2+4D-2)y = e^x +cos x$$ $$y'''-3y''+4y'-2y = e^x +cos x$$

Complementary Solution:

$$y'''-3y''+4y'-2y=0$$ The characteristic equation for this differential equation and its roots are: begin{align}r^3-3r^2+4r-2&=(r - 1) (r^2 - 2 r + 2)\&= (r-1)(r-(1+i))(r-(1-i))\&=0 end{align} which implies that $$r_1=1,~r_2=1+i,~r_3=1-i$$

The complementary solution is then $$y_c=c_1e^x+c_2e^{x}sin x+c_3e^xcos x$$

Particular Solution:

We apply the method of undetermined coefficients. As the right-hand side is of the form $e^x +cos x$, a reasonable guess would be $y_p=Ae^x+Bsin x + Ccos x$. However, $e^x$ is already part of the complementary solution, so we instead make the guess

$$y_p=Axe^x+Bsin x + Ccos x$$

Differentiating thrice we find

$$y'_p=Ae^x(x+1)-Csin x+Bcos x$$ $$y''_p=Ae^x(x+2)-Bsin x-Ccos x$$ $$y'''_p=Ae^x(x+3)+Csin x-Bcos x$$

Therefore

$$y'''_p-3y''_p+4y'_p-2y_p =Ae^x+(B-3C)sin x+(3B+C)cos x=e^x+cos x$$

$$begin{cases} A=1\ B-3C=0 \ 3B+C=1 end{cases}implies B=frac{3}{10},~ C=frac{1}{10}$$

Thus, the particular solution takes the form

$$y_p=xe^x+frac{3}{10}sin x + frac{1}{10}cos x$$

$$(D^3-3D^2+4D-2)y = cos x$$ For this particular integral: $$y_p =dfrac 1 {(D^3-3D^2+4D-2)} cos x$$ $$y_p =dfrac 1 {3D+1} cos x$$ $$y_p =dfrac {3D-1} {9D^2-1} cos x$$ $$-10y_p = ({3D-1}) cos x$$ $$-10y_p =-3sin x -cos x$$ $$y_p =dfrac {3sin x +cos x}{10}$$

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$$(D^3-3D^2+4D-2)y = e^x$$ For this particular integral: $$y_p =dfrac 1 {(D-1)(D^2-2D+2)}e^x$$ $$y_p =dfrac 1 {(D-1)}e^x=e^xdfrac 1 {D}1$$ $$ implies y_p =xe^x$$

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Finally the particular integral is: $$boxed {y_p =dfrac {3sin x +cos x}{10}+xe^x}$$