alright. Heres the problem:

1. $\displaystyle y''' + y'' = 8x^2$

Heres what I do:

$\displaystyle m^3 + m^2 = 0$

$\displaystyle m = 0, m = -1$

$\displaystyle D^3(D^3+D^2)y = D^38x^2$

$\displaystyle m^3(m^3 + m^2) = 0 auxillary equation$

$\displaystyle m^5(m+1) = 0$

$\displaystyle m = 0,0,0,0,0, -1$

$\displaystyle Yc = C1 + C2x + C3x^2 +C4+x^3 +C5x^4 + C6e^-^x$

The particular solution has the form:

$\displaystyle Yp = Ax^2 + Bx + C$

$\displaystyle Y'p = 2Ax + B$

$\displaystyle Y''p = 2A$

$\displaystyle Y'''p = 0$

Plugging in the particular solution into the DE gives:

$\displaystyle 2A = 8x^2$

The book says the answer is:

$\displaystyle y = C1 + C2x +C3e^-^x + \frac{2}{3}x^4 - \frac{8}{3}x^3 + 8x^2 $

What im lost on(yes its algebra) on how they get 8/3, 2/3, and 8. And how they knew where to put them.

Anyone willing to help me out? Thanks!!!