If zeroes of the equation of DE are $\lambda=1$ and $\lambda=\pm i$ then how to know the particular solution? Given DE: $$x'''-x''+x'-x=e^t~.$$ First we find a solution to homogeneous DE rewritten as equation: $\lambda^3-\lambda^2+\lambda-1=0 \; \lambda_1=1 \; \lambda_{2,3}=\pm i \Rightarrow x_h=C_1e^t+C_2\cos t +C_3\sin t$.
It is said that particular solution is $Ate^t ( \text{ because } \alpha =1)$. 
I don't understand why such particular solution? 
Could we take any one of those three components in general solution and substitute $C$ with $A$ and that would give us the right particular solution or is there some other reasoning as to how to find a particular solution if general solution is given?
 A: The particular solution should be since$e^t$ is already part of the homogeneous solution:
$$x_p(t)=Ate^t$$
You can also try this:
$$x'''-x''+x'-x=e^t$$
$$e^{⁻t}(x'''-x'')+e^{⁻t}(x'-x)=1$$
$$(x''e^{⁻t})'+(xe^{⁻t})'=1$$
$$x''e^{⁻t}+xe^{⁻t}=c_1+t$$
$$x''+x=(c_1+t)e^{t}$$
$$R^2+1=0 \implies R=\pm i$$
$$x(t)=C\cos(t)+D\sin(t)$$
And the particular solution is now :
$$x_p(t)=(A+Bt)e^t$$
A: $$x'''-x''+x'-x=e^t$$
$$\implies (D^3-D^2+D-1)x=e^t$$where $~~D\equiv \dfrac{d}{dt}~$.
Roots of the auxiliary equation are  $~1,~~i,~~-i~$.
Here $~~f(D)=D^3-D^2+D-1~$.
Clearly $~f(1)=0~$, but $~f'(1)=2\ne 0~$
For particular integral (P.I.),
$\text{P.I.  =}~~\dfrac{1}{D^3-D^2+D-1}~~e^t$
$~~~~~~=~t~\dfrac{1}{2}~~e^t~~~~~$ (by rule $2$)
$~~~~~~=~~\dfrac{1}{2}~t~e^t~~~~~~$


Consider a differential equation of the form $f(D)y=X$
If $X=e^{ax}$, then
$1.$ P.I.$\quad = \frac{1}{f(D)}e^{ax}=\frac{e^{ax}}{f(a)}$, if $f(a)\neq 0$
$2.$ P.I.$\quad =\frac{1}{f(D)}e^{ax}=x~\frac{e^{ax}}{f'(a)}$, if $~f(a)=0~$but$~f'(a)\ne 0$
$3.$ P.I.$\quad =\frac{1}{f(D)}e^{ax}=x^2~\frac{e^{ax}}{f''(a)}$, if $~f(a)=0,~f'(a)= 0~$but$~f''(a)\ne 0$
and so on.

