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systems_intro_real

# systems_intro_real - System of Eq Distinct and Real...

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System of Eq. – Distinct and Real eigenvalues April 16, 2011 1 Introduction/Motivation Suppose we’re given the follow system of differential equations and asked to solve it: Note: here x is the dependent variable, and t is the indepen- dent variable x 0 1 ( t ) = 1 x 1 ( t ) + 1 x 2 ( t ) (1) x 0 2 ( t ) = 4 x 1 ( t ) + 1 x 2 ( t ) (2) If x 2 ( t ) did not appear in the first equation, we’d have a simple first-order equation, and could certainly solve it. Therefore, it is only reasonable to try and eliminate x 2 ( t ) from Eq. ( 2 ). Using Eq. ( 1 ), we find x 2 ( t ) = x 0 1 ( t ) - x 1 ( t ) . (3) Plugging this expression into Eq. ( 2 ) we have ( x 0 1 ( t ) - x 1 ( t )) 0 = 4 x 1 ( t ) + 1 ( x 0 1 ( t ) - x 1 ( t )) . Expanding the left-hand-side: x 00 1 ( t ) - x 0 1 ( t ) = 4 x 1 ( t ) + x 0 1 ( t ) - x 1 ( t ) , or finally, x 00 1 - x 0 1 - 3 x 1 = 0 . (4) We can easily find the solution to above equation. The answer ends up being x 1 ( t ) = c 1 e 3 t + c 2 e - t . Using above expression and Eq. ( 3 ), we can easily solve for x 2 ( t ) . The result turns out to ! We won’t absorb the 2’s into the c 1 and c 2 here. be x 2 ( t ) = 2 c 1 e 3 t - 2 c 2 e - t . (5) To make a few observations, we will re-write our results in the following vector form: x 1 x 2 = c 1 e 3 t 2 c 1 e 3 t + c 2 e - t - 2 c 2 e - t = c 1 e 3 t 2 e 3 t + c 2 e - t - 2 e - t = c 1 1 2 e 3 t + c 2 1 - 2 e - t Key observations: 1 c HF, 2011

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System of Eq. – Distinct and Real eigenvalues April 16, 2011 1. We started with a system of first -order equations, and ended up with a second -order equation. 2. Our solution is in the following form: x 1 x 2 = ξ 1 ξ 2 e rt , where ξ 1 and ξ 2 and constants. In a more compact form, above equation may be written as x ( t ) = ξ e rt , (6) where x (t) (bold) denotes a vector-valued function. This should motivate us to seek two solutions having the form of above equation. In other words, we seek a general solution of the form x ( t ) = c 1 ξ (1) e r 1 ( t ) + c 2 ξ (2) e r 2 ( t ) .
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