Second Order Differential Equations e times sin particular solution The differential equation I am trying to solve is 
$$
\dfrac{d^2y}{dt^2} + 4\dfrac{dy}{dt} + 20y = e^{-2t}\sin(4t)
$$
I know how to start off. I have done the $s^2 + 4s + 20 = 0$ to get $s = -2-4i$ and $s = -2+4i$ And I know how to get the general solution from that using Euler's Formula. But I am having trouble guessing a solution for the particular. I just can't seem to get the form $e^{-2t}\sin(4t)$ I would very much appreciate any help you guys can give me. 
 A: A related problem. Here what you should assume for your particular solution
$$ y_p(t)= t e^{-2t}(A\sin(4t) + B \cos(4t)). $$
Note that, we multiplied by $t$ in the above equation, because the terms of the trial function coincide with the fundamental set of solutions (the solution of the homogeneous eq.). Here is your final solution
$$ y(t)=c_1\,{{\rm e}^{-2\,t}}\sin \left( 4\,t \right)+c_2\, {{\rm e}^{-2\,t}}
\cos\left( 4\,t \right)-\frac{1}{8}t\,{{\rm e}^{-2\,t}}\cos \left( 
4\,t \right).$$ 
A: You can solve this problem much easier by noticing the identity 
$$e^{-2t}\sin 4t = \textrm{Im}\left[e^{(-2 + 4i)t}\right].\ $$
Then you proceed to solve the ode in the real variable $t$ 
$$ z'' + 4z' + 20z = e^{(-2 + 4i)t}=1.e^{(-2 + 4i)t},$$
where $z:\mathbb R \to \mathbb C$. The solution that you are looking for is the imaginary part  of $z=z(t)$.
Now, let us solve the complex ode. We will find a polynomial $Q(t)$ such that 
$$z(t)=Q(t)e^{(-2 + 4i)t}$$
is a solution of the complex ode shown above. As you wrote, the characteristic polynomial is 
$$p(\lambda)= \lambda^2 + 4\lambda + 20= (\lambda + 2 - 4i)(\lambda + 2 + 4i).$$
It can be proved that [see O. R. B. de Oliveira ''An alternative to the undetermined coefficients and annihilator methods'' in arxiv.org/pdf/1110.4425‎, or ''A formula substituting the undetermined coefficients and the annihilator methods'' in Int. J. Math. Educ. Sci. Tech. 44-3, pp. 462-468, http://dx.doi.org/10.1080/0020739X.2012.714496] the polynomial $Q$ satisfies the simple ode
$$\frac{p''(-2 +4i)}{2!}Q'' + \frac{p'(-2+4i)}{1!}Q' + \frac{p(-2+4i)}{0!}Q = 1.$$
Since we have $p(-2+4i)=0$, $p''(t)=2$ for all $t$, and $p'(-2 +4i)= 8i$, this last ode boils down to
$$Q'' + 8iQ'=1.$$
Well, it is clear that $Q' =\frac{1}{8i}= -\frac{i}{8}$ is a solution of such equation. So, we can take
$$Q(t)= -\frac{t}{8}i.$$
Hence, we have
$$z(t)= -\frac{t}{8}ie^{-2t}(\cos 4t + i\sin 4t).$$
Finally, the particular solution that you are looking for is
$$y(t)=\textrm{Im}[z(t)]= -\frac{t}{8}e^{-2t}\cos 4t.$$
Best wishes,
O.R.B. de Oliveira
