Determine the general solution of the differential equation $y''+2y'+5y=17 \sin(2x)+x^2$ As a starter: the auxiliary equation is given by $t^2+2t+5=0$ so $t=-1-2i$, $t=-1+2i$, so you can use this to determine the complementary function.
The next step is determining the particular integral. How would you approach this?
 A: We need to distinguish between private and homogeneous responses to the differential equation. The homogeneous answer can be found solving below algebraic equation:$$\lambda^2+2\lambda+5=0$$$$\to \lambda=-1\pm 2i$$ which gives us the following answer:$$y_h=e^{-t}(A\cos2t+B\sin2t)$$
The private answer has been obtained from response of the equation to two functions $x^2$ and $17\sin2t$. The response due to $x^2$ must be to the form $ax^2+bx+c$ that by substituting gives us:
$$a=1 \\ b=-\frac{4}{5} \\ c=-\frac{2}{25}$$ and the response due to $17\sin 2x$ must be of form $d\sin2x+e\cos2x$ which similarly gives us:
$$d=1 \\ e=-4$$
Then by summing both private and homogeneous responses we conclude that:
$$y_g=y_h+y_p=e^{-t}(A\cos2t+B\sin2t)+x^2-\frac{4}{5}x+\frac{2}{25}+\sin2x-4\cos 2x$$
A: Well, you already have the solution to the homogeneous associate equation:
$$y_h=c_1e^{-t}\cos 2t+c_2e^{-1}\sin2t$$
and now find by cases a particular solution to each of the following diff. eq.'s:
$$(I)\;\;\;y''+2y'+5y=17\sin2x\implies y(t)=A\sin2x+B\cos2x\implies$$
$$\begin{cases}y'=2A\cos2x-2B\sin2x\\{}\\y''=-4A\sin2x-4B\cos2x\end{cases}$$
so
$$y''+2y'+5y=-4A\sin2x-4B\cos2x+4A\cos2x-4B\sin2x+5A\sin2x+5B\cos2x$$
$$\implies (A-4B)\sin2x+(B+4A)\cos2x=17\sin2x\implies\begin{cases}A-4B=17\\{}\\4A+B=0\end{cases}\implies$$
$$17A=17\implies A=1\;,\;\;B=-4\implies y_{P_1}=\sin2x-4\cos2x$$
and now the dif. eq.:
$$(II)\;\;\;y''+2y'+5y=x^2\implies y(t)=ax^2+bx+c\implies\begin{cases}y'=2ax+b\\{}\\y''=2a\end{cases}\implies$$
$$y''+2y'+5y=2a+4ax+2b+5ax^2+5bx+5c\stackrel?=x^2\implies$$
$$5ax^2+(4a+5b)x+(2a+2b+5c)=x^2\implies\begin{cases}5a=1\implies a=\frac15\\{}\\4a+5b=0\implies5b=-\frac45\implies b=-\frac4{25}\\{}\\2a+2b+5c=0\implies5c=-\frac25+\frac5{25}=-\frac15\implies c=-\frac1{25}\end{cases}$$
and from here
$$y_{p_2}=\frac1{25}\left(5x^2-4x-1\right)$$
Finally, check that $\;y_{p_1}+y_{p_2}\;$ is a particular solution to the dif. eq. (the actual one, meaning: not the homogeneous case!)
And at the end you get the general solution $\;y_h+y_{p_1}+y_{p_2}\;$ 
Disclaimer: Check the above carefully; sign and/or other little mistakes could have sneaked. I'm not too careful with these annoying things...but students most be!). Anyway, the general way to the solution you have it above.
