An improved method that is similar to Euler's method is what is usually called the Improved Euler's method. It works like this: Consider an equation y' = f(x, y). From (xn, Yn), our approximation to the solution of the differential equation at the n-th stage, we find the next stage by computing the x-step Xn+1 = xn +h, and then k1, the slope at (xn, Yn). The predicted new value of the solution . İs Zn+1 = Yn + h · k₁. Then we find the slope at the predicted new point k₁ = f(xn+1, Zn+1) and get the corrected point by averaging slopes h Yn+1 = = Yn + 1½ ½ (k1 + k₂). Suppose that we use the Improved Euler's method to approximate the solution to the differential equation dy dx = x - 0.5y, y(0.5) = 9. We let xo = 0.5 and yo 9 and pick a step size h = 0.25. Complete the following table: n xn Yn k1 Zn+1 k₂ 0 0.59-48 -3.25 ♡ <+ help (numbers) The exact solution can also be found for the linear equation. Write the answer as a function of x. y(x) = = help (formulas) Thus the actual value of the function at the point x = : 1.5 is y(1.5)= | 6.795 help (numbers) While the approximate solution found above is y(1.5) ≈ help (numbers) Book: Section 1.7 of Notes on Diffy Qs

An improved method that is similar to Euler's method is what is usually called the Improved Euler's method. It works like this: Consider an equation y' = f(x, y). From (xn, Yn), our approximation to the solution of the differential equation at the n-th stage, we find the next stage by computing the x-step Xn+1 = xn +h, and then k1, the slope at (xn, Yn). The predicted new value of the solution . İs Zn+1 = Yn + h · k₁. Then we find the slope at the predicted new point k₁ = f(xn+1, Zn+1) and get the corrected point by averaging slopes h Yn+1 = = Yn + 1½ ½ (k1 + k₂). Suppose that we use the Improved Euler's method to approximate the solution to the differential equation dy dx = x - 0.5y, y(0.5) = 9. We let xo = 0.5 and yo 9 and pick a step size h = 0.25. Complete the following table: n xn Yn k1 Zn+1 k₂ 0 0.59-48 -3.25 ♡ <+ help (numbers) The exact solution can also be found for the linear equation. Write the answer as a function of x. y(x) = = help (formulas) Thus the actual value of the function at the point x = : 1.5 is y(1.5)= | 6.795 help (numbers) While the approximate solution found above is y(1.5) ≈ help (numbers) Book: Section 1.7 of Notes on Diffy Qs

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.6: The Inverse Trigonometric Functions

Problem 92E

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