Unformatted text preview: Matlab 4
Leonard Harpster A06802204 C03 Exercise 4.1 Exercise 4.2
1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 2.5 At x = 1, y = 0.2850 for h = .1 and y = 0.3284 for h = .01 At x = 2, y = 1.4700 for h = .1 and y = 1.6467 for h = .01 b) c) Euler's method gives better estimates as h decreases because the smaller the intervals over which the slope is taken, the more it approaches the actual slope. The farther from the initial point the graph goes, the more a small h differs from a large h. This is because Euler's method relies on adding each approximation to the one before it so the size of the error grows at a rate dependent on h. Exercise 4.3
a) For x = 5, y = 9.9978 for h = .1 and y = 9.9961 for h = .01 For x = 10, y = 10.0000 for h = .1 and y = 10.0000 for h = .01 b) dsolve('Dy = y*(2-y/5)', 'y(0)=1') ans = 10/(1+9*exp(-2*t)) g(5) ans = 9.9959 g(10) ans = 10.0000 c) The estimates get better as h decreases but only up to a certain point, y = 10 in this case, where they become equivalent. All solutions tend towards y = 10, so that no matter the size of h, f isn't changing, so all estimates are the same after a certain point. Exercise 4.4 The best estimate I got for y was with the h = .1, which predicted the value of y being about 2.5. This agrees with the estimate from dfield, which was about 3. Exercise 4.5
a) Except for slight periodic errors of 1/10000th the ode45 function mirrors the actual values very closely b) Euler's method is much less accurate here. ...
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