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Suppose you are given:

$x'' = \left( \begin{array}{ccc} -50&10 \\ 30&-30 \\ \end{array} \right)\ x + \left( \begin{array}{ccc} 5 \\ 0 \\ \end{array} \right)\ \sin10t$

Find the particular solution of the system.

A separate part of the problem asked to find the natural frequencies of the masses, which I found to be $\omega_1^2 = 20$ and $\omega_2^2 = 60$. I tried solving the problem using $(A+\omega^2I)c = -F_0$ such as:

$\left( \begin{array}{ccc} \omega^2-50&10 \\ 30&\omega^2-30 \\ \end{array} \right)\ \left( \begin{array}{ccc} c_1 \\ c_2 \\ \end{array} \right)\ = \left( \begin{array}{ccc} -5 \\ 0 \\ \end{array} \right)\ $

But I don't know how to continue. I thought I should plug in $\omega^2 = 20$ (since the force is acting on the first mass), but that results in the system:

$\left( \begin{array}{ccc} -30&10 \\ 30&-10 \\ \end{array} \right)\ \left( \begin{array}{ccc} c_1 \\ c_2 \\ \end{array} \right)\ = \left( \begin{array}{ccc} -5 \\ 0 \\ \end{array} \right)\ $

Which doesn't have any solutions for $c_1$ and $c_2$ in which I would then be able to write the particular solution as $x(t) = c_1\sin20t + c_2\sin20t$, correct?

Any help would be great. Thanks.

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up vote 1 down vote accepted

Hint: Look for a particular solution of the form $\pmatrix{a\cr b\cr} \cos(10 t) + \pmatrix{c\cr d\cr} \sin(10 t)$.

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I'm sorry, I don't know how this helps. – user1038665 Nov 28 '12 at 23:36
Plug that in to the differential equation. You get two equations (one for each component) involving $\cos(10 t)$ and $\sin(10 t)$. For these to be true for all $t$, the coefficients of $\cos(10 t)$ one the left and right must be equal, and similarly for $\sin(10 t)$. So you should get four equations in the four unknowns $a,b,c,d$, which you can then solve. – Robert Israel Nov 28 '12 at 23:41
It's actually a bit easier if you can use complex numbers: find a solution of the form $u e^{10it}$ of the system with $\sin(10t)$ replaced by $e^{10it}$, and take the imaginary part. – Robert Israel Nov 28 '12 at 23:45
$a,b = 0$ and $c = \frac{-7}{64}$ and $d = \frac{3}{64}$. Does this seem correct? – user1038665 Nov 28 '12 at 23:46

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