Systems of First Order Linear Equations

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Homework Statement



Systems of first order equations can sometimes be transformed into a single equation of
higher order. Consider the system

(1) x1' = -2x1 + x2
(2) x2' = x1 - 2x2

Solve the first equation for x2 and substitute into the second equation, thereby obtaining a second order equation for x1. Solve this equation for x1 and then determine x2 also.

Homework Equations



The Attempt at a Solution



Solving (1) for x2 yields:

x2 = x1' + 2x1

Substituting into (2) yields:

x2' = x1 - 2(x1' + 2x1)

Simplifying...

x2' = -3x1 - 2x1'

How am I supposed to solve this now with the x2' term there?
 
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You aren't. You haven't yet reduced it to a higher order equation in one variable.
"Solve the first equation for x2 and substitute into the second equation, thereby obtaining a second order equation for x1" obviously won't give you a second order differential equation. I think you have dropped the first 2/3 of the instruction!

Start by differentiating the x_1&#039;&#039;= -2x_1+ x_2&#039;. Now use the second equation to replace that x_2&#039;: x_1&#039;&#039;= -2x_1+ (x_1- 2x_2)= -x_1+ 2x_2. <br /> <br /> <b>NOW</b> solve the first equation for x_2 and substitute that into the equation x_1&amp;#039;&amp;#039;= -x_1+ 2x_2.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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