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Use the variables $x_1 = x, x_2 = x', x_3 = x''$, etc.

$x_1' = x_2$

$x_2' = x_3$

$x_3' = x_4$

$x_4' = -3\cos (6t)+21x_1-9x_2+14x_3$

My $x_4'$ is wrong and I'm not sure why.

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Your signs are wrong, isolate $x^{(4)}$ in the equation above –  Jean-Sébastien Oct 26 '12 at 19:58
    
Thanks. Got it. –  user1038665 Oct 26 '12 at 20:00

3 Answers 3

up vote 2 down vote accepted

From my comment, all you did wrong was forgetting to reverse the signs when you isolated $x^{(4)}$.

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It's $x_4' = -3\cos (6t) -21x_1+9x_2-14x_3$

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You know that $$x_4'+14x_3-9x_2+21x_1=-3\cos(6t)$$ by substitution. You must solve this equation for $x_4'$.

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