Please help me with this one:
$\displaystyle y''-9y'+20y=x^{2}e^{4x}$
Lets use Variation of Parameters. I have always preferred that method, but that's me.
From the auxiliary equation we get:
$\displaystyle m^{2}-9m+20=(m-5)(m-4), \;\ m=5, \;\ m=4$
$\displaystyle y_{c}=C_{1}e^{4x}+C_{2}e^{5x}$
Now use the determinants to find W, W1, W2:
$\displaystyle W=\begin{vmatrix}y_{1}&y_{2}\\y_{1}^{'}&y_{2}^{'}\ end{vmatrix}=\begin{vmatrix}e^{4x}&e^{5x}\\4e^{4x} &5e^{5x}\end{vmatrix}=e^{9x}$
$\displaystyle W_{1}=\begin{vmatrix}0&y_{2}\\f(x)&y_{2}^{'}\end{v matrix}=\begin{vmatrix}0&e^{5x}\\x^{2}e^{4x}&5e^{5 x}\end{vmatrix}=-x^{2}e^{9x}$
$\displaystyle W_{2}=\begin{vmatrix}y_{1}&0\\y_{1}^{'}&f(x)\end{v matrix}=\begin{vmatrix}e^{4x}&0\\4e^{4x}&x^{2}e^{4 x}\end{vmatrix}=x^{2}e^{8x}$
$\displaystyle u_{1}^{'}=\frac{W_{1}}{W}=-x^{2}$
$\displaystyle u_{2}^{'}=\frac{W_{2}}{W}=\frac{x^{2}}{e^{x}}$
Integrate to find u1 and u2:
$\displaystyle u_{1}=-\int x^{2}dx=\frac{-1}{3}x^{3}$
$\displaystyle u_{2}=\int\frac{x^{2}}{e^{x}}dx=\frac{-x^{2}-2x-2}{e^{x}}$
$\displaystyle y_{p}=u_{1}+u_{2}$
$\displaystyle \boxed{y=y_{c}+y_{p}=\left(\frac{-1}{3}x^{3}-x^{2}-2x-2\right)e^{4x}+C_{1}e^{4x}+C_{2}e^{5x}}$
There is a nice outline for future reference.