I was doing a question on finding the general solution of a linear nonhomogeneous function and I was given a solution. But I don't understand the third line of the particular solution bit. Can someone help explain this to me?
$$ \frac{d^6 y}{dx^6} - 2 \frac{d^5y}{dx^5} + \frac{d^4 y}{dx^4} = e^x + e^{2x} $$ We can write it in the form $$ \left(D^6 - 2D^5 +D^4 \right)y = e^x + e^{2x} $$ with $D= \frac{d}{dx}$. Let $y=e^{mx}\neq 0$ be the trivial solution. Therefore, the auxiliary equation is $$ \begin{align} & m^6 - 2m^5 + m^4 = 0 \\ &\Rightarrow m^4 \left(m^2 - 2m +1 \right) = 0 \\ &\Rightarrow m^4 \left(m-1 \right)^2 = 0 \\ &\Rightarrow m = 0, +1, +1 \end{align} $$ Therefore, CF is $$ y = C_1 + (C_2 + c_3 x)e^x $$ where $c_1$, $c_2$ and $c_3$ are arbitrary constants. Now $$ \begin{split} PI &= \frac{1}{D^4 (D-1)^2} \left(e^x + e^{2x} \right) \\ &= \frac{1}{D^4 (D-1)^2} e^x + \frac{1}{D^4 (D-1)^2}e^{2x} \\ &= e^x \frac{x^2}{2} + \frac{e^{2x}}{16} \end{split} $$ Therefore, the required general solution is $$ \begin{split} y &= CF + PI \\ & = C_1 + (C_2 + c_3 x)e^x + e^x \frac{x^2}{2} + \frac{e^{2x}}{16} \end{split} $$ where $c_1$, $c_2$ and $c_3$ are the arbitrary constants.