# Finding a particular solution to y'' + y' = 4x

So, I found the homogenous solution via solving the characteristic quad equation. It's $Ae^{0*x} + Be^{-x}$.

Now, I'm trying to find the particular solution to add onto that. Here's what I was thinking:

1. First, the particular solution should have the form $Qx + R$ (Q and R are constants)
2. Now, I differentiated this twice, to get $y'(x) = Q$ and $y''(x) = 0$
3. Now I just substituted these back into the original equation to get $0 + Q = 4x$...
4. ...wtf?

If anyone can help me out on where I'm messing up, I'd greatly appreciate it! I'll respond quickly!

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Point 1 is incorrect. It should be quadratic, not linear. –  anon Aug 4 '11 at 3:14
Whaaa? Really? So the particular solution should have a form of one degree higher than the f(x)? (in this case f(x) = 4x) Man, cuz I'm using the calclifesaver book, and his table says that if f(x) = 3x - 2, the particular y has the form ax+b –  user13327 Aug 4 '11 at 3:32
Yes, but I'm sure that book mentions some special cases which you are overlooking. –  Gerry Myerson Aug 4 '11 at 3:34
oooooo, there's a little section at the bottom of the table that says, if Yp (I don't know how to do subcase) conflicts with Yh, multiply the form by x or x^2 as appropriate. oooo Alright cool thanks a lot! –  user13327 Aug 4 '11 at 3:37

Since a constant is a solution to the homogeneous equation, it can't be part of the particular solution. As anon suggests in the comment, you have to try $Qx^2+Rx$.