Consider the differential equation C. What is the magnitude of the error in the

Question

Consider the differential equation

C. What is the magnitude of the error in the two Euler approximations you found?
Magnitude of error in Euler with 2 steps =
Magnitude of error in Euler with 4 steps =

D. By what factor should the error in these approximations change (that is, the error with two steps should be what number times the error with four)?
factor =
(How close to this is the result you obtained above?)

Consider the differential equation Y = C. What is the magnitude of the error in the two Euler approximations you found? Magnitude of error in Euler with 2 steps = Magnitude of error in Euler with 4 steps = D. By what factor should the error in these approximations change (that is, the error with two steps should be what number times the error with four)? factor = (How close to this is the result you obtained above?) y(1) (Be sure not to round your calculations at each step!) B. What is the solution to this differential equation (with the given initial condition)? (Be sure not to round your calculations at each step!) Now use four steps: : when . A. Use Euler's method with two steps to estimate with initial condition

Explanation / Answer

Do the problem for y(0) = 2 The Euler stepping method for y' = f(x) is y(x(n+1)) = y(x(n)) + dx f(x(n)) (A). dx = 0.25, for four steps. So x(1) = dx = 0.25, x(2) = 2dx = 0.5, x(3) = 0.75. y(0) = 1. y(0.25) = y(0) + dx * f(0) = 1 + 0.25*(4*0.) = 1. y(0.5) = y(0.25) + dx * f(0.25) = 1 + 0.25*(4.*0.25) = 1.25. y(0.75) = y(0.5) + dx * f(0.5) = 1.25 + 0.25*(4.*0.5) = 1.75. y(1.0) = y(0.75) + dx * f(0.75) = 1.75 + 0.25*(4.*0.75) = 2.50. The exact solution is y(x) = 2x^2 + 1. So y(0.25) = 1.125, y(0.5) = 1.5, y(0.75) = 2.125, and y(1) = 3. The error at x=1: E = |2.5-3| = 0.5. For (D), you should be able to conclude that the error is halved if the step dx is halved. Therefore, E = O(dx).

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