Euler method and Graph

Hi, I have to solve a an ODE with the Euler method for two separate step sizes and have to place them on same graph. I'm very confused on how to do this... Please help me...

y'=1/y; Initial Condition y(0) = 1;

I have to use Euler method to solve for y(1) for step size deltat = 0.1 and also deltat = 0.01

This is as far as I got and I'm just completely stuck.....

t0=0; t1=1; dt=0.1;  %define time range and step size
y0=1; %initial condition
t=t0:dt:t1; 
y=zeros(size(t,1), size(t,2))
for i=1: ((t1-t0)/dt)
    y(i+1)=y(i)+1/y(i)*dt
end

[Walter Roberson]

Note: your line

for i=1: ((t1-t0)/dt)

risks round-off errors in the calculation of the number of elements. You would be better using

for i = 1 : length(t)

Also note that your y matrix is created as 2D but you are only using a single index of it. In this particular case, using length(t) would work, but more generally you would use

y = zeros(size(t));
for i = 1 : numel(y)

I would also recommend you rethink your termination condition: is your output intended to be one element for each time, or is the output intended to be one element for each time, plus one final element that is the prediction for the next time? Because you are assigning to y(i+1)...

[Jim Riggs]

The solution vector is sized to be y=zeros(size(t)), and the looping computes y(i+1), therefore the loop index should be

for i=1:length(t) -1

or

for i=1:numel(y)-1

Keeping vectors y and t the same length is a good idea.

[Nasir Holliday]

I'm sorry but I'm not really understanding

[Walter Roberson]

t0=0; t1=1; dt=0.1;  %define time range and step size
y0=1; %initial condition
t=t0:dt:t1; 
y=zeros(size(t));
num_t = length(t);
for i = 1: num_t - 1
    y(i+1)=y(i)+1/y(i)*dt
end

Answer 1

[Pravin Jagtap]

Hello Nasir,

Refer to the following function which takes ‘dt’ and ‘y0’ as an input argument and returns the ‘y’ as solution for time vector from 0 to 1 with ‘dt’ as timestep (This is just a demonstrative example. You can modify as per your requirements).

function y = EulerMethod(dt, y0)
    % Preparing the Time vector from tStart to tEnd
    t_Start = 0; 
    t_End = 1;  
    t = t_Start:dt:t_End;
  
    % Initialize the solution vector
    y = zeros(size(t)); 
    % Impose Initial Condition
    y(1) = y0;
    
    % Compute the Solution vector using the Eulers Method
    num_t = length(t);
    for i = 1: num_t - 1
        y(i+1)=y(i)+((1/y(i))*dt);
    end
end

Call the above function for different values of ‘dt’ as follows and plot it

% call the function for dt 0.1
y1 = EulerMethod(0.1,1);
% call the function for dt 0.01
y2 = EulerMethod(0.01,1);
% Plot the corresponding solutions 
t1 = 0:0.1:1;
plot(t1,y1)
hold on
t2  = 0:0.01:1;
plot(t2,y2)