Find all Solutions of the Following Differential Equation I was asked to find all of the solutions of the following equation: 
$$-6y(x) + 11y'(x) - 12y''(x) + 12y^{(3)}(x) - 6y^{(4)}(x) + y^{(5)}(x) = xe^x \\$$
I'll start with dividing the above polynomial by the factors $(x-1), (x-2), (x-3)$.
BUT, how do I go about finding the solution basis for the homogeneous equation (and then go further)?
 A: Undetermined Coefficients: Superposition Principle
$\newcommand{\l}{\lambda}$
The solution to your question can be defined using a simple procedure that involves using the associated homogeneous solution, and then the solving for the particular. To answer your problem one can do the following:
\begin{align*}\l^5-6\l^4+12\l^3-12\l^2+11\l-6=0\end{align*}
After that one has to find the zeroes for the auxiliary equation. So using Possible Rational Zeroes one arrives at the conclusion that the possible rational zeroes are the following:\begin{align*}\pm1,\pm2,\pm3,\pm6\end{align*}
Luckily one of the zeroes is $\l=1$. Then setting up our synthetic division one arrives at the following polynomials.\begin{array}{c|cccccc}1&1&-6&12&-12&11&-6\\&&1&-5&7&-5&6\\&\hline1&-5&7&-5&6&0\end{array}
Thus our polynomial to reduces to this factorized form:
\begin{align*}(\l-1)(\l^4-5\l^3+7\l^2-5\l+6)&=0\end{align*}
Then we would have to rerun the Possible Rational Zeroes to decompose that polynomial, where these would be our possible factors:
\begin{align*}\pm1,\pm2,\pm3\pm6\end{align*}
Luckily another factor that works is $\l=2$. One can begin to setup the following synthetic division.\begin{array}{c|ccccc}2&1&-5&7&-5&6\\ &&2&-6&2&-6\\ &\hline1&-3&1&-3&0\end{array} This becomes a even easier term to achieve complete complete factorization of the auxiliary equation. Then the factorized form that we achieve at is the following:$$(\l-1)(\l-2)(\l^3-3\l^2+\l-3)=0$$
Then we can do the following: \begin{align*}(\l-1)(\l-2)(\l^2+1)(\l-3)&=0\end{align*}
Now we can set each term equal to $0$, and we get the following answers: $$\l=1,2,3,\pm i$$Giving the following homogeneous solution: $$y_h=c_1e^x+c_2e^{2x}+c_3e^{3x}+c_4\sin(x)+c_5\cos(x)$$
The particular solution can then be solved by using this substitution:$$y_p=(Ax^2+Bx+C)e^x$$
The way to solve it is to plug in the $y_p$ into the DE for $y(x)$.\begin{align}-6y(x) + 11y'(x) - 12y''(x) + 12y^{(3)}(x) - 6y^{(4)}(x) + y^{(5)}(x) = xe^x\end{align}
After that you get a system of equations after plugging in and combining like terms then using system of equations one gets that the particular solution is
$$y_p=\frac{1}8x^2e^x+\frac{1}8xe^x$$
So the total solution is that $$\displaystyle{y(x)=c_1e^x+c_2e^{2x}+c_3e^{3x}+c_4\sin(x)+c_5\cos(x)+\frac{1}8x^2e^x+\frac{1}8xe^x}$$
