Simple substitution will reveal that x(t) = π/2 is a solution of x′ = et cosx. Moreover, both f(t,x) = et cosx and df/dx = −et sinx are continuous everywhere. Therefore, solutions are unique and no two solutions can ever share a common point (solutions cannot intersect). Use Euler’s method with a step size h = 0.1 to plot the solution of x′ = et cosx with initial condition x(0) = 1 on the interval [0,2π] and hold the graph (hold on). Overlay the graph of x = π/2 with the command line ([0, 2*pi] , [pi/2, pi/2] ,1 'color', 'r') and note that the solutions intersect. Does reducing the step size help? If so, does this reduced step size hold up if you increase the interval to [0,10]? How do rk2 and rk4 perform on this same problem?
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