SystemsDifferentialEquationsForPrint

# SystemsDifferentialEquationsForPrint - Systems of...

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Systems of Differential Equations S. Bonnot S. Bonnot Systems of Differential Equations

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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 = 0 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 0 1 + E , so we deduce Ri 0 2 = - Li 00 1 + E 0 . Replace i 2 and i 0 2 in the second equation, and obtain ( - Li 00 1 + E 0 ) + 1 RC ( - Li 0 1 + E ) - 1 C = 0. Solve this equation to get i 1 and then use Ri 2 = - Li 0 1 + E to obtain i 2 . S. Bonnot Systems of Differential Equations

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General case: system of first order equ. with constant coefficients Given the system a 1 x 0 + b 1 y 0 + c 1 x + d 1 y = f 1 ( t ) a 2 x 0 + b 2 y 0 + c 2 x + d 2 y = f 2 ( t ) let’s try to eliminate y 0 by taking b 2 . ( Line 1 ) - b 1 ( Line 2 ) . we obtain ( a 1 b 2 - a 2 b 1 ) x 0 + ( 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 0 + Bx + Cy = F ( t ) . Case C = 0. Solve Ax 0 + Bx = F ( t ) then substitute in either equation to get y . Case C 6 = 0. We know y = ( - Ax 0 - Bx + F ( t )) / C , so we know y 0 and we can substitute y , y 0 in Equ.1 or Equ.2. S. Bonnot Systems of Differential Equations
Solving systems of first order equ. with constant coefficients Example : 3 x 0 + 2 y 0 + x + y = e t 6 x 0 + 4 y 0 + x = t We eliminate y 0 by taking Line 2 - 2 Line 1, to obtain - x - 2 y = t - 2 e t . Observe that by accident we also eliminated x 0 . So far we know y = 1 2 ( - x - t + 2 e t ) , so we also deduce y 0 = 1 2 ( - x 0 - 1 + 2 e t ) . Substitute y , y 0 in Equ.1 to get a first order equ. in x alone: 3 x 0 + ( - x 0 - 1 + 2 e t ) + x + ( 1 2 ( - x - t + 2 e t )) = e t . See textbook for an example where x 0 doesn’t disappear accidentally. S. Bonnot Systems of Differential Equations

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From higher order equ. to systems of first order equ.
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