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$u'' + 0.125u' + u = 0$

I'm told to let $x_1 = u$ and $x_2 = u'$. Thus, $x_1' = x_2$ and $x_2' = u''$. Thus, we have:

$x_2' + 0.125x_2 + x_1 = 0$

Now here is where I get confused. The book says that $x_1$ and $x_2$ now satisfy the equations:

$x_1' = x_2$ and $x_2' = -x_1 - 0.125x_2$

Can someone elaborate on why this is the answer we're looking for? And how I can generalize this for differential equations of higher order?

Thank you!

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1 Answer

up vote 0 down vote accepted

$x_1'=x_2$ is a first-order equation; $x_2'=-x_1-(1/8)x_2$ is a first-order equation; together, they form a system of first-order equations; that's why it is the answer we are looking for.

For equations of higher order, introduce more new variables: $x_1=u,x_2=u',x_3=u'',\dots,x_n=u^{(n-1)}$. Then you'll get a system of first-order equations $x_1'=x_2,x_2'=x_3,\dots,x_{n-1}'=x_n$ and one equation for $x_n'$ involving some or all of the $x_i$.

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