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?

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:
$$y_g=y_h+y_p=e^{-t}(A\cos2t+B\sin2t)+x^2-\frac{4}{5}x+\frac{2}{25}+\sin2x-4\cos 2x$$
