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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.

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Hint: try $c e^{-2 t} t \cos(4 t)$. Do you know why is the more important question? Regards –  Amzoti Apr 24 '13 at 5:02
    
Are you asked to use a specific technique to find a particular solution? –  Mhenni Benghorbal Apr 24 '13 at 5:10
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Here is a technique you can use to find a particular solution. –  Mhenni Benghorbal Apr 24 '13 at 5:19
    
There is no specific technique that i have to use. When I go through using the $ce^{−2t}tcos(4t)$ I end up with $ce^{-2t}(-8sin(4t) - 4cos(4t)) = e^{-2t}sin(4t)$ which doesn't seem to work and when I try it with $ce^{−2t}tsin(4t)$ I end up with a similiar problem –  Bob Dale Apr 24 '13 at 5:44
    
Anyways, see my answer. I gave you your trial function. –  Mhenni Benghorbal Apr 24 '13 at 5:47

2 Answers 2

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

$$ y(t)=c_1\,{{\rm e}^{-2\,t}}\sin \left( 4\,t \right)+c_2\, {{\rm e}^{-2\,t}} \cos\left( 4\,t \right)-\frac{1}{8}t\,{{\rm e}^{-2\,t}}\cos \left( 4\,t \right).$$

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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

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