Euler method: ODE with different initial conditions

Question

Atom 2023-1-21

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https://ww2.mathworks.cn/matlabcentral/answers/1898175-euler-method-ode-with-different-initial-conditions

编辑： Torsten 2023-1-21

采纳的回答： Torsten

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I want to solve an ODE by Euler method with different initial conditions.

I have used for y(1)=0.5:0.05:1.5; as the different initial conditions. But it given an error

for y(1)=0.5:0.05:1.5;

↑

Error: Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

How to modify the code so that I can able to run a loop for different values of y(1).

NB: Tried with

for v(1) = [0.05 0.01 0.15 0.2]
    
end

but got the error.

h=0.5;
x=0:h:4;
y=zeros(size(x));
for y(1)=0.5:0.05:1.5;
    n=numel(y);
    for i = 1:n-1
        dydx= -2*x(i).^3 +12*x(i).^2 -20*x(i)+8.5 ;
        y(i+1) = y(i)+dydx*h ;
        fprintf('="Y"\n\t   %0.01f',y(i));
    end
    figure;
    plot(x,y);
end