004 - 320 CHAPTER 5 NONLINEAR SYSTEMS which at(0 0 is equal...

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Unformatted text preview: 320 CHAPTER 5 NONLINEAR SYSTEMS which at (0. 0) is equal to (‘3?) so the linearized system at (0, 0) is (we could also see this by “dropping the higher order terms“). (c) The eigenvalues of the linearized system at the origin are -l and 1. so the origin is a saddle. The linearized system decouples, so solutions approach the origin along the x-axis and tend away form the origin along the y-axis. 5. (a) Using separation of variables (or simple guessing). we have x0) = 10:" . (b) The equation dy _=_ 3 dr 4x +y is a ﬁrst-order. linear equation. We write the equation as Therefore. the integrating factor is e". Multiplying both sides of the equation by 9" yields ‘0 -r 3 -r ( d! y) e -— 41: e . Note that the left-hand side is just the derivative of ye" and the right-hand side is —4x3e"3’ e“ since x(t) = we" . Therefore we have i 9", dry -4r ‘3'e" = “use . = -4X3€ After integrating and simplifying, we have y(r) = x3e'3' + (yo — xgk‘. (c) The general solution of the system is x0) = x0e" y(t) = x3?” + (yo — x3)e‘. (d) For all solutions, x(r) —> 0 as t -> 00. For a solution to tend to the origin as r —> 00, we must have 370‘) —> 0, and this can happen only if yo - x3 = 0. (e) Since 1: = we", we see that a solution will tend toward the origin as r —> -00 only if x0 = 0. In that case, y(t) = yoe‘. and y(!) -v 0 as t -> —oo. ...
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• Spring '08
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