Solve the differential equation $ \dfrac{ d^2x }{ dt^2 } + 6 \dfrac{ dx }{ dt } + 9x = 4t^2 + 5 $ using variation of parameters Would solving $ \dfrac{ d^2x }{ dt^2 } + 6 \dfrac{ dx }{ dt } + 9x = 4t^2 + 5 $ using variation of parameters require integration by parts or can I solve it without knowing integration by parts? 
I'm not sure if I'm just using the method wrong or if it requires integration by parts. I'm new to variation of parameters, and I haven't encountered integration by parts. Thanks.
The non homogeneous characteristic equation is $ r^2 + 6r + 9 = 4t^2 + 5 $
The characteristic homogeneous equation is $ r^2 + 6r + 9 = 0 $
$ \Rightarrow (r + 3)(r + 3) $
$ \Rightarrow r = -3 $
$ \therefore x(t) = C_1e^{-3t} + C_2te^{-3t}$ is the general solution for the homogeneous differential equation.
I now find the particular solution.
$-e^{-3t} \displaystyle\int \dfrac{te^{-3t}(4t^2 + 5)}{W(x_1, x_2)} dt + te^{-3t} \displaystyle\int \dfrac{e^{-3t}(4t^2 + 5)}{W(x_1, x_2)} dt$
The Wronskian $W(x_1, x_2) = e^{-6t} $
$-e^{-3t} \displaystyle\int \dfrac{te^{-3t}(4t^2 + 5)}{W(x_1, x_2)} dt + te^{-3t} \displaystyle\int \dfrac{e^{-3t}(4t^2 + 5)}{W(x_1, x_2)} dt $
$= -e^{-3t} \displaystyle\int \dfrac{te^{-3t}(4t^2 + 5)}{e^{-6t}} dt +
te^{-3t} \displaystyle\int \dfrac{e^{-3t}(4t^2 + 5)}{e^{-6t}} dt$
 A: Considering the general case of $$I_k=\int t^k e^{r t}\,dt$$ Using integration by parts $$u=t^k \implies du=k t^{k-1}\,dt$$ $$dv=e^{r t}\,dt\implies v=\frac{e^{r t}}{r}$$ makes $$I_k=\frac{t^k e^{r t}}{r}-\frac k r\int  t^{k-1} e^{r t}\,dt=\frac{t^k e^{r t}}{r}-\frac k r I_{k-1}$$ with $I_0=\frac{e^{r t}}{r}$.
If you already heard about the incomplete gamma function, almost from definition, you would have $$I_k=-\frac{t^{k+1}} {(-r t)^{k+1}} \Gamma (k+1,-r t)$$ which, in the case where $r<0$ reduces to $$I_k=-\frac{\Gamma (k+1,-r t)} {(-r )^{k+1}} $$
A: $$
\int e^{r t} \sum_{i=0}^na_it^i dt  = \frac{1}{D} e^{rt} \sum_{i=0}^na_it^i = e^{rt} \frac{1}{D+r}\sum_{i=0}^na_it^i = e^{rt}\sum_{i=0}^n \frac{a_i}{r}  \frac{1}{1+D/r}t^i =  
$$
$$
e^{rt}\sum_{i=0}^n \frac{a_i}{r} \sum_{j=0}^{\infty} \left(-\frac{D}{r}\right)^j t^i = e^{rt}\sum_{i=0}^n \frac{a_i}{r} \sum_{j=0}^{i} \left(-\frac{D}{r}\right)^j t^i = e^{rt}\sum_{i=0}^n \frac{a_i}{r} \sum_{j=0}^{i} \frac{(-1)^j}{r^j} \frac{i!}{(i-j)!} t^{i-j} 
$$
