Exploring the Errors of RK4 Method for y' = -2y + x, y(0) = 1

[musicmar]

Homework Statement

Consider the initial value problem y' = -2y + x, y(0) = 1
The analytical solution is y(x) = (1/2)x²-(1/4) + (5/4)e^-2x

a. Approximate y(0.1) using one step and the RK4 method.
b. Find a bound for the local truncation error in y₁.
c. Compare the error in y₁ with the error bound.
d. Approximate y(0.1) using two steps and the RK4 method.
e. Verify that the global truncation error for the RK4 method is O(h⁴) by comparing the errors in parts (a) and (d).

The Attempt at a Solution

I think I know how to approximate y(0.1). I don't have time to write out my work now. I'll be back in about an hour and can do it then.

Can someone please help me understand the errors of the RK4 method? Thank you!

 

[mighty2000]

Thank you for your post. The RK4 method is a numerical method used to approximate solutions to initial value problems. It is important to understand the errors involved in this method in order to accurately interpret the results.

a. To approximate y(0.1) using one step and the RK4 method, we can use the formula y1 = y0 + (1/6)(k1 + 2k2 + 2k3 + k4), where k1 = hf(x0, y0), k2 = hf(x0 + h/2, y0 + k1/2), k3 = hf(x0 + h/2, y0 + k2/2), and k4 = hf(x0 + h, y0 + k3). In this case, y0 = 1, x0 = 0, and h = 0.1. Plugging these values into the formula, we get y1 = 0.905.

b. The local truncation error in y1 is given by the formula LTE = (1/720)(f''''(x0, y0)h^5), where f'''' is the fourth derivative of the function f(x,y). In this case, f'''' = 2, x0 = 0, y0 = 1, and h = 0.1. Therefore, the bound for the local truncation error is LTE = (1/720)(2)(0.1)^5 = 2.7778 x 10^-7.

c. To compare the error in y1 with the error bound, we can calculate the actual error using the analytical solution y(x) = (1/2)x^2 - (1/4) + (5/4)e^-2x. Plugging in x = 0.1, we get y(0.1) = 0.9048. Therefore, the error in y1 is 0.9048 - 0.905 = 0.0002, which is much larger than the local truncation error bound of 2.7778 x 10^-7. This indicates that the RK4 method has a relatively large error in this case.

d. To approximate y(0.1) using two steps and the RK4 method, we can use the formula y2 = y1 + (1/6)(k