An ODE is solved numerically three different ways, each using a time step of t=0.1  (Euler's method, the midpoint method and a 4th order Runge Kutta method), and then again using t=0.1 .  The solution is NOT shown!  (neither is the problem!)  However, the errors in the solution (measured at t = 10) using that time step are shown in the table below:   Euler Error Midpoint Error Runge-Kutta Error 1 1.3 0.20 0.05 0.1 0.13 0.0020 5 x 10^6           For each method, write an approximate function which describes how the error varies with time-step .

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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An ODE is solved numerically three different ways, each using a time step of t=0.1  (Euler's method, the midpoint method and a 4th order Runge Kutta method), and then again using t=0.1 .  The solution is NOT shown!  (neither is the problem!)  However, the errors in the solution (measured at t = 10) using that time step are shown in the table below:

  Euler Error Midpoint Error Runge-Kutta Error
1 1.3 0.20 0.05
0.1 0.13 0.0020 5 x 10^6
       

 

For each method, write an approximate function which describes how the error varies with time-step .

E_euler = ...

E_mp = ...

E_rk4 = ...

 

Expert Solution
Step 1

We know that the differential equation is solved using three different methods at two different time steps, t = 0.1, t = 1.

We are given the value of error obtained in solutions, from different methods at both the time steps.

The objective here is to form a relation between the time step used and the error obtained, for each method.

 

The data is given as follows:

time stepEuler ErrorMidpoint ErrorRange-Kutta Error0.10.130.00200.00000511.30.200.05

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