# 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?

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 $C = -1$.)

• Thank you for your reply! I was just wondering if you could explain the first equation under the change of basis further. I am actually in high school so have no formal knowledge of operators, although I am really interested in quantum mechanics so know enough about them from there to understand the majority of your post (which has been very enlightening!). I just need some help with that first step? Thank you!
– Meep
Commented Feb 23, 2015 at 7:06
• There was nothing magical about the first equation. You can derive it by using the product rule of differentiation on the right-hand side. The remark about change of basis has nothing to do with the derivation. Commented Feb 23, 2015 at 8:09