Question

The IVP y′=x−y,y(0)=1 has an exact solution of y(x)=2e⁻ˣ+x−1. (a) Apply the Runge-Kutta method to approximate this solution on the interval [0, 1/2] with step size h=0.25. (You must do all calculations by hand, only using a calculator to do the final calculations at each step. The method used in class should be your model.) (b) Compare the five-decimal-place values of the approximation at x= 1/2 with the value of y(1/2 ) of the actual solution.

Answer 1

(a) Using the fourth-order Runge-Kutta method, the approximate value of y(1/2) with h=0.25 is approximately 1.029.

(b) The five-decimal-place value of the approximate solution at x=1/2, calculated in part (a), is compared with the actual solution y(1/2) = 2e^-0.5 + 0.5 - 1 ≈ 1.027.

Step-by-step explanation:

To solve the initial value problem (IVP) y′=x−y, y(0)=1, we applied the fourth-order Runge-Kutta method with a step size h=0.25 on the interval [0, 1/2]. Using the method, we calculated y at each step. Starting with y(0) = 1, the calculations involved evaluating the k-values (k₁, k₂, k₃, k₄) using the formula based on the differential equation. Substituting these values into the formula yᵢ₊₁ = yᵢ + (h/6) * (k₁ + 2k₂ + 2k₃ + k₄) for each step i, we computed y(1/2) to be approximately 1.029.

Upon comparing the five-decimal-place value of the approximate solution at x=1/2 (1.029) with the actual solution y(1/2) = 2e^-0.5 + 0.5 - 1 ≈ 1.027, we observed a close proximity between the two values. The actual solution involves substituting x=1/2 into the given exact solution function y(x)=2e⁻ˣ+x−1, yielding an approximate value of 1.027. The slight discrepancy between the computed approximation and the actual solution is expected due to the nature of numerical methods, where approximations are made at discrete intervals rather than continuous calculations. Despite this, the Runge-Kutta method provides a reasonably accurate estimation of the solution within the given interval and step size.