Trying to go from Euler to Heun's Method

We have done Euler's method, and now I need to solve this ODE: y' = 2x +y, y(0)=16 using Heun's method. Is this right?

h = 0.01;
X = 20;
N = round(X/h);
x = zeros(1,N+1);
y = zeros(1,N+1);
x(1) = 0;
y(1) = 16;
for n = 1:N
  x(n+1) = x(n) + h;
  w = y(n)+ h*(2*x(n)+y(n))
  y(n+1) = y(n) + h*0.5*((2*x(n)+y(n))+w);
  %yH(n+1) = y(n) +h*0.5*(yH(n)+ 2*x(n)+y(n));
end

Answer 1

[James Tursa]

No, but you are close. You basically need to use the average of y' at the current state with y' at the Euler estimated next step state. I.e., you take an Euler step as usual and calculate the y' at that point and average it with the y' at the starting point. E.g.,

x at the current state is x(n)

y at the current state is y(n)

y' at the current state is 2*x(n) + y(n)

x at the Euler estimated next state is x(n+1) = x(n) + h

y at the Euler estimated next state is w = y(n) + h*(2*x(n) + y(n))

y' at the Euler estimated next state is 2*x(n+1) + w

Taking the next step with an average of the two y' values:

y(n+1) = y(n) + h*0.5*((2*x(n) + y(n)) + (2*x(n+1) + w))