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

for Y(1) =1:1:numel(t)

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

Answer 1

[Torsten]

Y(1) is a scalar (single number), 1:numel(t) is a vector. You can't set a scalar to a vector.

Use an unused variable name instead, e.g.

for k = 1:numel(t)

Answer 2

[Image Analyst]

Try

for k = 1 : numel(t)
    this_t = t(k);
    Y(k) = 5 * this_t ^ 2 + 42;  % Replace with whatever your formula for Y is.
end

[Navaneetha Krishnan Murugadoss]

h=3600;
A0=1; 
B0=3; 
P0=0; 
K=5*10^-5; 
Yb=1; 
Yp=0.15;
tspan = [0 43200];
Fyt = @(t,y) [(-K*Y(1)*Y(2));
    (-Yb*(K*Y(1)*Y(2)));
    (Yp*(K*Y(1)*Y(2)))];
for k=1 : numel(t)
    k1 = fyt ((Y(1),Y(2));

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

Error in connector.internal.fevalMatlab

Error in connector.internal.fevalJSON

    k2 = fyt ((Y(1)+0.5*h,Y(2)+0.5*h*k1));
    k3 = fyt ((Y(1)+0.5*h,Y(2)+0.5*h*k2));
    k4 = fyt((Y(1)+h),(Y(2)+h*k3));
    Y(i+1) = Y(2) + (h/6)*(k1+2*k2+2*k3+k4);
    figure (1)
    plot(t,Y(:,1))
    figure (2)
    plot(t,Y(:,2))
    figure (3)
    plot(t,Y(:,3))
    
end

[Walter Roberson]

    k1 = fyt ((Y(1),Y(2));
             12 3 2  3 21

You have one more ( than you have )

[Torsten]

Fyt = @(t,y) [(-K*y(1)*y(2));
    (-Yb*(K*y(1)*y(2)));
    (Yp*(K*y(1)*y(2)))];

instead of

Fyt = @(t,y) [(-K*Y(1)*Y(2));
    (-Yb*(K*Y(1)*Y(2)));
    (Yp*(K*Y(1)*Y(2)))];

[Navaneetha Krishnan Murugadoss]

I rectified that but still getting error

[Jan]

Which error message do you get? Avoid overdoing the parentheses:

Fyt = @(t,Y) [-K * Y(1) * Y(2); ...
              -Yb * K * Y(1) * Y(2); ...
              Yp *  K * Y(1) * Y(2)];

This looks cleaner.

Note that you have defined "Fyt", but call "fyt" later.

This cannot work also:

k2 = fyt ((Y(1)+0.5*h, Y(2)+0.5*h*k1));
%         ^                         ^

Here the inner parentheses combine the two arguments, but the function Fyt (not fyt) should get 2 arguments. Better:

k2 = Fyt(Y(1) + 0.5 * h, Y(2) + 0.5 * h * k1);

Answer 3

[Walter Roberson]

The variable for a for loop must be a plain unindexed variable.

Answer 4

[Torsten]

h=3600;
A0=1; 
B0=3; 
P0=0; 
K=5*10^-5; 
Yb=1; 
Yp=0.15;
t = 0:0.1:10;   
fyt = @(t,y) [(-K*y(1)*y(2));
    (-Yb*(K*y(1)*y(2)));
    (Yp*(K*y(1)*y(2)))];
Y = zeros(3,numel(t))
Y(1,1) = 1.0;
Y(2,1) = 3.0;
Y(3,1) = 0;
for i=1 : numel(t)-1
    k1 = fyt(t(i),Y(:,i));
    k2 = fyt(t(i)+0.5*h,Y(:,i)+0.5*h*k1);
    k3 = fyt(t(i)+0.5*h,Y(:,i)+0.5*h*k2);
    k4 = fyt(t(i)+h,Y(:,i)+h*k3);
    Y(:,i+1) = Y(:,i) + (h/6)*(k1+2*k2+2*k3+k4);  
end
figure (1)
plot(t,Y(1,:))
figure (2)
plot(t,Y(2,:))
figure (3)
plot(t,Y(3,:))

[Navaneetha Krishnan Murugadoss]

My timestep is 3600

[Navaneetha Krishnan Murugadoss]

Can you explain this t = 0:0.1:10 ?

[Walter Roberson]

Change

t = 0:0.1:10;

to

t = 0:h:43200;

[Torsten]

@Navaneetha Krishnan Murugadoss

Look at the results. They won't change after t=10 (unless you change the function to be integrated).