- #1

PsychonautQQ

- 784

- 10

## Homework Statement

consider y'' + 2y' - 3y = 1 + xe^x, find the particular solution

## The Attempt at a Solution

so

f(x) = 1 + xe^x

f'(x) =e^x + xe^x

f''(x) = 2e^x + xe^x

so it looks like my particular solution is going to have a constant term, an e^x term and an xe^x term,

so I can write

Particular Solution:

y(x) = A + Be^x + Cxe^x

and then differentiate this twice and plug into the original equation? Is this on the correct track? I ask because an online source says that I should have

y(x) = A + Bxe^x + C(x^2)e^x

can someone help me understand why I am wrong if I am wrong? Why have an (x^2)e^x but no e^x?