Euler's Method check

Hello, I am trying to model a first order ODE using Euler's Method. I have shown the Euler's step in the code below and I wanted somebody to double-check it and see if it is written correctly. All the R,C,L, Ron values are given. Thankx for the help

 h = 0.0001;                         % Adjustable time-step
 t = 0:h:0.002;

    for i = 1:length(t)-1 
        switch (switch_state)
            case 1
               k11 = x2(i)*1/L - x1(i)*1/(R*C);
               k12 = x2(i)*-Ron/L - x1(i)*1/L + Vi*1/L;
            otherwise
               k11 = x2(i)*1/L - x1(i)*1/(R*C);
               k12 = x2(i)*-Ron/L - x1(i)*1/L;
        end

        x1(i+1) = x1(i) + h*k11;
        x2(i+1) = x2(i) + h*k12;  

    end
 plot(t,x1)
 plot(t,x2)

Answer 1

[Iain]

You've implemented euler's method correctly. Whether or not you have calculated k11 and k12 correctly, and chosen sensible values for h, and LCR is another question.

[Thierry Kayiranga]

I am retracing my steps and reviewing every single part. Thankx for the help

[Thierry Kayiranga]

Hello, I have an additional question. Instead of using equation form, would this matrix form of the Euler's be correct? Assuming A and B were calculated correctly. Thankx for your help again.

 for j=1:f
    switch (switch_state)
        case 1
            A = [ -Ron/L    -1/L
                     1/C    -1/(R*C)];
            B = [ 1/L
                  0  ];
        otherwise
            A = [ -Ron/L     -1/L
                     1/C     -1/(R*C)];
            B = [ 0
                  0 ];    
    end

    for i = 1:length(t)-1 
            k1 = A*x(:,i) + B*Vi;
            x(:,i+1) = x(:,i) + h*k1;  
    end

 end
 plot(x(1,:))
 plot(x(2,:))

[Iain]

They're not the same equations.

Remember, matrix multiplication is Row times column

But, yes, the euler implementation is correct providing that k1 is the differential of x.

[Thierry Kayiranga]

Right. Row times column. Correct me if I'm wrong. So basically, the loop would take each row of A and multiply it by the corresponding column of x and add it B*Vi. And k1 would be [k11 k12]' from the earlier code. right?