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Ordinary Differential Equations Using MATLAB, Third Edition by John Polking

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Similarly, to find ∂f/∂t, think of x as a constant and differentiate with respect to t to obtain ∂f/∂t = −1. 9. The solution of dx/dt = x 2 , x(0) = 1 is unique. 10. The solution “blows up” at t = 1. we find that 1 = 1/(C − 0), or C = 1, so the solution is x(t) = 1/(1 − t). From this equation, we see that limt→1− x(t) = +∞. 6 In this particular case, if the independent variable t represents time (in seconds), then the solution trajectory reaches positive infinity when one second of time elapses.

In this case x0 is called a stable equilibrium point. If f (x0 ) > 0, then every solution curve that has an initial value near x0 diverges away from x0 as t → ∞, and x0 is called an unstable equilibrium point. If f (x0 ) = 0, no conclusion can be drawn about the behavior of solution curves. In this case the equilibrium point may fail to be either stable or unstable. Apply this test to each of the equilibrium points. 16. x = cos(π x), x ∈ [−3, 3]. 17. x = x(x − 2), −∞ < x < ∞. 18. x = x(x − 2)2 , −∞ < x < ∞.

C) As the length of the t interval is increased, how close must y0 be to 0 in order to insure the same accuracy? To find out, repeat part b) with tf = 4, 6, 8, and 10. The results of the last problem show that the solutions can be extremely sensitive to changes in the initial conditions. This sensitivity allows chaos to occur in deterministic systems, which is the subject of much current research. ” For the ODEs in Exercises 29–33, use dfield6 to find approximately the value of x0 such that the solution x(t) to the initial value problem with initial condition x(0) = x0 satisfies x(t1 ) = x1 .

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