The following initial value problem of ODE has an exact solution. To understand

Question

The following initial value problem of ODE has an exact solution. To understand how accurate different numeric methods can be used to approximate the exact solutions, 1) use respectively to find the numeric solutions to the given initial value problem of ODE. 2) Determine the absolute error of each numeric solution relative to the corresponding exact solution. Calculate the solutions with a step size of 0.1 and the nodes xi=0.1, x2-0.2 x, = 0.3 , x,-0.4 and x, = 0.5 . Keep 4 decimal digits in your calculations. sXy dx y(0) 1

Explanation / Answer

MATLAB CODE IS ATTCHED

clc;
clear all;

%% EXACT SOLUTION

x=[0:0.1:0.5];
disp(' Exact Solutions at x=0,x=0.1,x=0.2,x=0.3,x=0.4,x=0.5')
y_exact(1:length(x))=0.5+2*x(1:length(x))+0.5*exp(2*x(1:length(x)))

%% EXPLICIT EULER METHOD

x=[0:0.1:0.5];
y(1)=1; % initial value condition

for i=1:length(x)
der_y(i)=1-4*x(i)+2*y(i);
y(i+1)=y(i)+0.1*der_y(i);
end;

disp(' Solutions at x=0,x=0.1,x=0.2,x=0.3,x=0.4,x=0.5 in Euler Method ')
y(1:length(x)) % the solutions at x=0,x=0.1,x=0.2,x=0.3,x=0.4,x=0.5

disp(' Absolute Error at x=0,x=0.1,x=0.2,x=0.3,x=0.4,x=0.5 in Euler Method ')
abs(y(1:length(x))-y_exact(1:length(x)))

%% IMPROVED EULER METHOD

for i=1:length(x)
x(7)=0.6;
der1_y(i)=1-4*x(i)+2*y(i);
y(i+1)=y(i)+0.1*der1_y(i);
der2_y(i)=1-4*x(i+1)+2*y(i+1);
y(i+1)=y(i)+0.1*0.5*(der1_y(i)+der2_y(i));
end;

x=[0:0.1:0.5];
disp(' Solutions at x=0,x=0.1,x=0.2,x=0.3,x=0.4,x=0.5 in Improved Euler Method ')
y(1:length(x)) % the solutions at x=0,x=0.1,x=0.2,x=0.3,x=0.4,x=0.5

disp(' Absolute Error at x=0,x=0.1,x=0.2,x=0.3,x=0.4,x=0.5 in Improved Euler Method ')
abs(y(1:length(x))-y_exact(1:length(x)))  
  
%% 4TH ORDER RK METHOD

x=[0:0.1:0.5];
y(1)=1; % initial value condition

for i=1:length(x)
der_y(i)=1-4*x(i)+2*y(i);
k1=0.1*der_y(i);
k2=0.1*(1-4*(x(i)+0.1/2)+2*(y(i)+k1/2));
k3=0.1*(1-4*(x(i)+0.1/2)+2*(y(i)+k2/2));
k4=0.1*(1-4*(x(i)+0.1)+2*(y(i)+k3));
y(i+1)=y(i)+(1/6)*(k1+2*k2+2*k3+k4);
end;

disp(' Solutions at x=0,x=0.1,x=0.2,x=0.3,x=0.4,x=0.5 in RK-4 Method ')
y(1:length(x)) % the solutions at x=0,x=0.1,x=0.2,x=0.3,x=0.4,x=0.5

disp(' Absolute Error at x=0,x=0.1,x=0.2,x=0.3,x=0.4,x=0.5 in RK-4 Method ')
abs(y(1:length(x))-y_exact(1:length(x)))  

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