# Euler method and Graph

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Nasir Holliday el 22 de Feb. de 2020
Editada: Pravin Jagtap el 25 de Feb. de 2020
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
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Walter Roberson el 23 de Feb. de 2020
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

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### Respuestas (1)

Pravin Jagtap el 25 de Feb. de 2020
Editada: Pravin Jagtap el 25 de Feb. de 2020
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)
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