Pharaoh's modified Euler method question from Yahoo Answers

[CaptainBlack]

Part 2 of Pharaoh's Taylor series and modified Euler question from Yahoo Answers

consider van der pol's equation
y" - 0.2(1-y^2)y' + y = 0 y(0)=0.1 y'(0)=0.1
2)
Write down the above problem as a system of first order differential equations.
Calculate the numerical solution at x = 0.2 using the Modified Euler method. Take the
step-length h = 0.1 and work to 6 decimal places accuracy. Compare with your solution
in part (i) and comment on your answers.

There is a standard method of converting higher order ODEs into first order systems, in this case it is to introduce the state vector:

\(Y(t)=\left[ \begin{array}{c} y(t) \\ y'(t) \end{array} \right] \)​

Then the ODE becomes:

\( Y'(t)= F(t,Y)=\left[ \begin{array}{c} y'(t) \\ y''(t) \end{array} \right]= \left[ \begin{array}{c} y'(t) \\ 0.2 \left(1-(y(t))^2 \right) y'(t)-y(t) \end{array} \right] \)​

Now you can use the standard form of the modified Euler method on this vector first order ODE.CB

 

[jvicens]

= 0.1
t = 0:CB:0.2
N = length(t)-1
Y = zeros(2,N+1)
Y(:,1) = [0.1;0.1] #initial conditions
for n = 1:N
Y(:,n+1) = Y(:,n)+CB*F(t(n),Y(:,n))
end
Y(:,end) #solution at t = 0.2The solution is [0.112494;0.109809]. Comparing this with the solution in part (i), [0.112503;0.110108], we can see that the solutions are very close. This is expected since the modified Euler method is a second order method and the problem is well-behaved. In general, the Modified Euler method is more accurate than the Euler method, but it requires more computations. Therefore, it is important to consider the trade-off between accuracy and efficiency when choosing a numerical method for solving ODEs.