1–4 EULER METHOD Do 10 steps. Solve exactly. Compute the error. Show details. 1. y' + 0.2y = 0, y(0) = 5, h = 0.2 2. y' = (1/2)??(1 - y^2), y(0) = 0, h = 0.1 3. y' = (y - x)^2, y(0) = 0, h = 0.1 4. y' = (y + x)^2, y(0) = 0, h = 0.1 5–10 IMPROVED EULER METHOD Do 10 steps. Solve exactly. Compute the error. Show details. 5. y' = y, y(0) = 1, h = 0.1 6. y' = 2(1 + y^2), y(0) = 0, h = 0.05 7. y' = -x y^2, y(0) = 1, h = 0.1 8. Logistic population model: y' = y - y^2, y(0) = 0.2, h = 0.1 9. Do Prob. 7 using Euler's method with h = 0.1 and compare the accuracy. 10. Do Prob. 7 using the improved Euler method, 20 steps with h = 0.05. Compare. 11–17 CLASSICAL RUNGE–KUTTA METHOD OF FOURTH ORDER Do 10 steps. Compare as indicated. Show details. 11. y' - x y^2 = 0, y(0) = 1, h = 0.1. Compare with Prob. 7. Apply the error estimate (10) to y10. 12. y' = y - y^2, y(0) = 0.2, h = 0.1. Compare with Prob. 8. 13. y' = 1 + y^2, y(0) = 0, h = 0.1 14. y' = (1 - x^{-1})y, y(1) = 1, h = 0.1 15. y' + y tan x = sin 2x, y(0) = 1, h = 0.1 16. Do Prob. 15 with h = 0.2, 5 steps, and compare the errors with those in Prob. 15.
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> 0.2y = 0.2(75), h = 0.2 This is a simple linear equation, y = 75. The exact solution is y = 75 for all x. Using Euler's method with h = 0.2, we start with y(0) = 0 and update y using the formula y(n+1) = y(n) + h*f(x(n), y(n)), where f(x, y) = 0.2y. After 10 Show more…
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Use Euler's method with each of the following step sizes to estimate the value of y(0.8), where y is the solution of the initial-value problem: y' = y, y(0) = 5. (i) h = 0.8 y(0.8) = 9 (ii) h = 0.4 y(0.8) = 9.8 (iii) h = 0.2 y(0.8) = ? The error in Euler's method is the difference between the exact value and the approximate value. Find the errors made in part (a) in using Euler's method to estimate the true value of y(0.8), namely 5e^0.8. (Round your answers to four decimal places.) h = 0.8 error = (exact value) - (approximate value) = 2.1277 h = 0.4 error = (exact value) - (approximate value) = 1.3277 h = 0.2 error = (exact value) - (approximate value) = ?
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(a) Use Euler's method with each of the following step sizes to estimate the value of $ y(0.4), $ where $ y $ is the solution of the initial-value problem $ y' = y, y(0) = 1. $ (i) $ h = 0.4 $ (ii) $ h = 0.2 $ (iii) $ h = 0.1 $ (b) We know that the exact solution of the initial-value problem in part (a) is $ y = e^x. $ Draw, as accurately as you can, the graph of $ y = e^x, 0 \le x \le 0.4, $ together with the Euler approximations using the step sizes in part (a). (Your sketches should resemble Figure 12, 13, and 14.) Use your sketches to decide whether your estimates in part (a) are underestimates or overestimates. (c) The error in Euler's method is the difference between the exact value and the approximate value. Find the errors made in part (a) in using Euler's method to estimate the true value of $ y(0.4), $ namely $ e^{0.4}. $ What happens to the errors each time the steps size is halved?
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Direction Fields and Euler's Method
Modify the M-file euler.m to implement the algorithm for Improved Euler. Call the new file impeuler.m (include the file in your report). Test your code for the IVP y' = 1.5 y, y(0) = 1, in the interval [0, 1.4], using N = 7 steps. First enter the function f(t,y) = 1.5 y as anonymous function and then enter the following: >> [t7,y7] = impeuler(f,[0,1.4],1,7); % use @f if defined in separate m-file >> [t7,y7] ans = 0 1.0000 0.2000 1.3450 0.4000 1.8090 0.6000 2.4331 0.8000 3.2726 1.0000 4.4016 1.2000 5.9202 1.4000 7.9626 Compare your output with the one above. You should obtain the same values. Consider the IVP: y' = 1.5 y, y(0) = 1, 0 ≤ t ≤ 1.4 (a) Determine the Improved Euler's approximation for N = 70, N = 700 and N = 7000. Fill in the following table with the values of the approximations, errors and ratios of consecutive errors at t = 1.4 . Two values have already been entered to help you check your results. Recall that the exact solution to the ODE is y(t) = e^{1.5t}. Include the table in your report, as well as the MATLAB commands used to find the entries. N | approximation yN(end) | error eN | ratio eN/eNprev 7 | 7.9626 | | N/A 70 | | | 80.9417 700 | | | 7000 | | | (b) Examine the last column. How does the ratio of the errors relate to the number of steps used? Your answer to this question should confirm the fact that Improved Euler's method is a ‐second-order‐ method. That is, every time the step size is decreased by a factor k, the error is reduced (approximately) by a factor of k^2. Note: Since Euler's method is only of the 1st order, the Improved Euler's method is more efficient (hence the ‐improved‐).
Sri K.
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