Differential Equation: Modifying Particular Solution For a particular solution of $$ y^{(4)} - y''' - y'' + y' = t^2 + 4 + te^t$$
The solution to the homogeneous solution is represented by the characteristic equation:
$$ r(r-1)^2 (r+1)$$
so the solutions of the homogeneous solution are:  $ 0,-1$ and $1$ with multiplicity 2.
So the homogeneous solution will have the form:
$$ y_h = Ae^t + Bte^t + Ce^{-t} + D$$
As for the particular solution, I think it will have the form of
$$ y_p = A + Bt + Ct^2 + D + Ee^t + Ete^t$$
and then modify it so that it doesn't overlap with $y_h$. 
I am not sure how to go from here. (The question is merely looking for the "form" of the particular solution)
 A: $y_p$ can have the form
$$ y_p = A_1 t + A_2 t^2 + A_3 t^3 +  B_1 t^2 e^{t} + B_2 t^3 e^{t}. $$
Here is a general technique tells you how to find $y_p$.
Added: We will apply the general technique to find $y_p$. To annihilate the right hand side of the ode, we apply the operator $D^3(D-1)^2$, where $D=\frac{d}{dt}$, to both sides of the ode which results in 
$$ D^4(D-1)^4(D+1) y(t)=0, $$
since $D^3 (t^4+4)=0$ and $(D-1)^2 te^{t} = 0 $.
Now, the above is a new homogenous ode with the the solution,
$$ y_h = c_0 + c_1 t + c_2 t^2 + c_3 t^3 + c_5 e^{t}+ c_6 te^{t}+ c_7 t^2 e^{t} + c_8 t^3e^{t} + c_9 e^{-t} $$
$$ = (c_0 + c_5e^{t}+c_6te^{t} + c_9e^{-t})+ c_1 t + c_2 t^2 + c_3 t^3 + c_7 t^2 e^{t} + c_8 t^3e^{t}.$$ 
You can see in the above equation what is between the brackets corresponds to $y_h$ of the old ode
$$ y_{h_{o}} = Ae^t + Bte^t + Ce^{-t} + D. $$
So, we can conclude our $y_p$ to be 
$$ y_p = A_1 t + A_2 t^2 + A_3 t^3 +  B_1 t^2 e^{t} + B_2 t^3 e^{t}.$$
Note that, the above form will do, but still you can choose the more general form which is based on the above form
$$ y_p = A_0+A_1 t + A_2 t^2 + A_3 t^3 +  (B_0+ B_1  t + B_2 t^2 + B_3 t^3) e^{t} $$
A: For each source term of the form $P(t)\,e^{\lambda t}$, where $P$ is a polynomial, the form of the particular solution will include $t^kQ(t)\,e^{\lambda t}$ where 


*

*$Q$ is a generic polynomial (with undetermined coefficients) of the same degree as $P$

*$k$ is the multiplicity of $\lambda$ as a root of the characteristic equation. In particular, $k=0$ if $\lambda$ is not a root of the characteristic equation. 


You have (second degree polynomial) $e^{0\cdot t}$ + (first degree polynomial)$e^{1\cdot t}$. And you already noticed that 


*

*$0$ is a root of multiplicity $k=1$ 

*$1$ is a root of multiplicity $k=2$


Is the path clear now?
