What to do when particular integral is part of complementary function? I am considering the equation $\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y=e^{2x}$. The complementary function is found to be $Ae^{2x}+Be^{3x}$. Ordinarily I would let $y=\lambda e^{2x}$ to find the particular integral, but as this I a part of the complementary function it cannot satisfy the whole equation. My text book then says to let $y=\lambda xe^{2x}$ without justification. I was wondering why we need the x here and do not need it otherwise. Also, in what cases can we simply add an x for the solution to work? When is adding an x necessary, and when is it allowed?
Thank you in advance :)
 A: I will present two ways to arrive at the term $xe^{2x}$. I hope they would help you understand the matter better.
Change of Basis
For any function $y$ and constant $a$, observe that
$$
(D - a)y = e^{ax}D(e^{-ax}y)
$$
where $D$ is the differential operator $\frac{d}{dx}$.
In other words, the operator $D - a$ is similar to $D$, via the change of basis $e^{ax}$.
To use this to solve the equation $(D - 2)(D - 3)y = e^{2x}$, rewrite the equation as
\begin{align}
(D - 2)(D - 3)y & = e^{2x} \\
e^{2x}D(e^{-2x}(D - 3)y) & = e^{2x} \\
D(e^{x}D(e^{-3x}y)) & = 1 && \text{The right-hand side is a non-zero constant}\\
e^{x}D(e^{-3x}y) & = x + c \\
D(e^{-3x}y) & = xe^{-x} + ce^{-x} \\
e^{-3x}y & = -xe^{-x} + Ae^{-x} + B \\
y & = -xe^{2x} + Ae^{2x} + Be^{3x}.
\end{align}
Annihilator
By recognizing that $e^{2x}$ is in the null space of $(D - 2)$, we can apply $(D - 2)$ to the original equation to obtain
$$
(D - 2)^2(D - 3)y = 0.
$$
Now, the method to find the homogeneous solution should give you the form
$y = Ae^{2x} + Be^{3x} + Cxe^{2x}$.
Substitute back into the original equation and solve for $C$. (You will get $C = -1$.)
