- #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?