# Solving a Second-Order Linear Equation (Non-zero RHS)

I am reading Mary L. Boas' Mathematical methods in the Physical sciences. I am considering an auxiliary equation of form: $$(D-a)(D-b)y=ke^{x},$$ where k and c are constant.

I know that a general solution to a second order DE can be given by the sum of a complementary function with a particular solution.

This is my problem:

In the book, it states that a particular solution of the above auxiliary equation can be assumed to be: $$Ce^{cx},$$ if c is not equal to a or b (a $\neq$ b). OR $$Cxe^{cx},$$ if c is equal to a or b (a $\neq$ b). OR $$Cx^2e^{cx},$$ if c=a=b.

I am doing Problem 8.6.3 in the book, which asks me to solve: $$y''+y'-2y=e^{2x},$$ I have simplified this to: $$(D-1)(D+2)y=e^{2x}.$$ In this situation, it initially made sense for me to assume a solution of: $$Ce^{2x},$$ This gave me a particular solution of:$$\frac{1}{4}e^{2x}.$$ Which, from what I can see online, is correct. However, if I was going by what was in the book, I surely would assume a solution of $$cxe^{cx},$$ or $$2xe^{2x},$$ to match the power of the RHS. I'm clearly misunderstanding what the book is suggesting. I was just wondering if someone could explain where I can use those 3 assumed particular solutions/ why. Thanks in advance.

"From what I can see online". Of course I have no idea what you refer to.

Your particular solution $\frac14\,e^{2x}$ is correct. And it's exactly what the book tells you to do.

Had $2$ being a root of your characteristic equation, you would have needed $Cxe^{2x}$, or even with $x^2$ if $2$ is a double root.

• That's what I originally thought, what was confusing me however, cxe^{2x} is not of form cxe^{cx}, otherwise c would have to be 2 surely? and it'd be 2xe^{2x}, which i know is wrong – George Jan 5 '18 at 20:21
• Sorry, I am a complete idiot. I just realised it is Cxe^{cx} and C $\neq$ c. Thanks for your help :) – George Jan 5 '18 at 20:24