How do I use Euler's forward difference method and the second order Runge-Kutta method to work out a third order ODE?

The given equation is y'''-3y'-3xy=0 with initial conditons y(0)=1.11983, y'(0)=0.83697, y''(0)=0.12356. h=0.1. The question asks to find the numerical solution y at x=0.1, x=0.2 and x=0.3 using the Euler's forward difference method and the second order Runge-Kutta method based on, k1=hf(xn,yn) k2=hf(xn+h/2, yn+k1/2) yn+1=yn+k2.

I first turned the given equation into three first oder differential equations such that dy/dx=v so y(0)=1.11983, dv/dx=w so v(0)=y'(0)=0.83697 and dw/dx=3v+3xy so w(0)=y''(0)=0.12356.

However, Im not sure on how to work out y1,y2 and y3 for both Eulers and Runge-Kutta methods.

[James Tursa]

Can you post the code you have written so far? Then we can help steer you in the right direction.

Answer 1

[Torsten]

Don't work with y1, y2 and y3, but with y = (y(1),y(2),y(3)).

Define the vector of derivatives as

 f = @(t,y) [y(2) y(3) 3*y(2)+3*t*y(1)];

and the vector of initial conditions as

 Y(1,:) = [1.11983 0.83697 0.12356];

Then, for Euler explicit, e.g., you can update all three ODE solutions as if it was a case for one unknown function:

 Y(i+1,:) = Y(i,:) + dt*f(t(i),Y(i,:))