integration

Find y if y'' + y' + 1 = 0 by integration.

y'' + y' + 1 = 0
y'' + y' = - 1
∫(y'' + y') dx = ∫- 1 dx
∫y'' dx + ∫y' dx = ∫- 1 dx
y' + y = - x + C
(e^x)(y' + y) = (e^x)(- x + C)
(e^x)y' + (e^x)y = - xe^x + Ce^x
(ye^x)' = - xe^x + Ce^x
ye^x = ∫(- xe^x + Ce^x) dx
ye^x = ∫- xe^x dx + ∫Ce^x dx
ye^x = ∫- x d(e^x) + Ce^x + C_0
ye^x = - xe^x - ∫e^x d(- x) + Ce^x + C_0
ye^x = - xe^x + ∫e^x dx + Ce^x + C_0
ye^x = - xe^x + e^x + Ce^x + C_1
ye^x = - xe^x + (C_2)e^x + C_1
y = - x + (C_1)e^(- x) + C_2

From the given equation, you may let y = e^(rx) for some numbers r.
(Since only exponential function will be possible to arrive such an equation)

Then by succesive differentiation and substitution, you get r^2 + r + 1 = 0
so r = - 1/2 + ((root 3) / 2)i or - 1/2 - ((root 3) / 2)i

You can see r is a complex number, so use the euler's formula
e^(ix) = cosx + isinx, then u'll get the function u want

唔用integration
用其他方法可以嗎??
http://i299.photobucket.com/albums/mm309/lokwanshan/ScreenHunter_01Jun040121.gif