SystemsDifferentialEquationsForPrint.pdf

# SystemsDifferentialEquationsForPrint.pdf - Systems of...

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Unformatted text preview: Systems of Differential Equations S. Bonnot S. Bonnot Systems of Differential Equations Examples coming from physics Goal : Physical systems where separate parts interact on each other create automatically systems of differential equations. Example : a system with 2 consecutive springs. The motion of the first mass is described by m 1 x 00 1 =- k 1 x 1 + k 2 ( x 2- x 1 ) . and the motion of the second mass at the bottom is given by m 2 x 00 2 =- k 2 ( x 2- x 1 ) . System of two equations combined together : m 1 x 00 1 =- k 1 x 1 + k 2 ( x 2- x 1 ) m 2 x 00 2 =- k 2 ( x 2- x 1 ) S. Bonnot Systems of Differential Equations Examples, continued Electric Circuits: L di 1 dt + Ri 2 = E R di 2 dt + 1 C ( i 2- i 1 = We get a system of two first order linear equations with constant coefficients . How to solve it? Eliminate one variable (say i 2 ) to obtain a single differential equation in i 1 . We know Ri 2 =- Li 1 + E , so we deduce Ri 2 =- Li 00 1 + E . Replace i 2 and i 2 in the second equation, and obtain (- Li 00 1 + E ) + 1 RC (- Li 1 + E )- 1 C = 0. Solve this equation to get i 1 and then use Ri 2 =- Li 1 + E to obtain i 2 . S. Bonnot Systems of Differential Equations General case: system of first order equ. with constant coefficients Given the system a 1 x + b 1 y + c 1 x + d 1 y = f 1 ( t ) a 2 x + b 2 y + c 2 x + d 2 y = f 2 ( t ) let’s try to eliminate y by taking b 2 . ( Line 1 )- b 1 ( Line 2 ) . we obtain ( a 1 b 2- a 2 b 1 ) x + ( c 1 b 2- c 2 b 1 ) x + ( d 1 b 2- d 2 b 1 ) y = b 2 f 1 ( t )- b 1 f 2 ( t ) this equation is of type Ax + Bx + Cy = F ( t ) . Case C = 0. Solve Ax + Bx = F ( t ) then substitute in either equation to get y . Case C 6 = 0. We know y = (- Ax- Bx + F ( t )) / C , so we know y and we can substitute y , y in Equ.1 or Equ.2. S. Bonnot Systems of Differential Equations Solving systems of first order equ. with constant coefficients Example : 3 x + 2 y + x + y = e t 6 x + 4 y + x = t We eliminate y by taking Line 2- 2 Line 1, to obtain- x- 2 y = t- 2 e t . Observe that by accident we also eliminated x . So far we know y = 1 2 (- x- t + 2 e t ) , so we also deduce y = 1 2 (- x- 1 + 2 e t ) . Substitute y , y in Equ.1 to get a first order equ. in x alone: 3 x + (- x- 1 + 2 e t ) + x + ( 1 2 (- x- t + 2 e t )) = e t ....
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## This document was uploaded on 04/01/2011.

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SystemsDifferentialEquationsForPrint.pdf - Systems of...

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