MIT18_03S10_c32

MIT18_03S10_c32 - 18.03 Class 32 April 26 2010 Linear first...

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18.03 Class 32 , April 26 , 2010 Linear first order systems: Introduction 1. Elimination 2. Matrices 3. Anti-elimination: companion matrices 4. Shakespeare [1] There are two fields in which rabbits are breeding like rabbits, Farmer Jones's field contains x(t) rabbits, Farmer McGregor's field contains y(t) rabbits. Breeding like rabbits in this case means that Jones's show a net growth rate of .7 rabbits per rabbit per month, and McGregor's show .6. They can also hop over the hedge between the fields, and since the grass is greener in Farmer McGregor's field, Jones's rabbits jump over a the rate of .2 per month, MacGregor's at the rate of .1 per month. Flow diagram: .5 .5 ______ ______ / \ / \ / ------------ .2 ------------- \ \ | | -----------> | | / --->| Jones | | McGregor |<--- | | <----------- | | ------------ .1 ------------- So the equations are x' = .5x - .2x + .1y = .3x + .1y (1) y' = .5y - .1y + .2x = .2x + .4y (2) This is a linear SYSTEM of equations, homogeneous, and it seems to be impossible to solve, since you need to know y to solve for x and you need to know x to solve for y. We *can* solve, though: use (1) to express y in terms of x : y = 10 x' - 3 x and then plug this into (2): 10 x" - 3 x' = .2 x + 4x' - 1.2 x 10 x" - 7 x' + x = 0 or x" - .7 x' + .1 x = 0 This is a SECOND ORDER ODE , which we can solve: s^2 - .7s + .1 = 0

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has roots r1 = .5 and r2 = .2 . (Remember,
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MIT18_03S10_c32 - 18.03 Class 32 April 26 2010 Linear first...

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