# How to solve a ode within a for loop?

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mathru on 10 Jul 2021
Commented: mathru on 10 Jul 2021
I am trying to solve a second order diferential function using ode45. I have defined a function for the differential function where in the expresion there is a constant, say, A. For single value, code is working good. I have written my code as follows:
global A
A = 3;
tspan = 0: 0.1: 1;
r = 0.7;
r0 = 2;
c = [r r0];
[t1 p1] = ode45('height',tspan,c);
where the function is as follows:
global A
function pdot = height(t,p)
pdot = zeros(size(p))
pdot(1)=p(2);
B=5*A*p(1);
pdot(2)=B/5*p(2)^2-3*p(2)+6;
end
Now I want to solve the ode for different values of A using a for loop as follows:
global A
A = 0: 5: 10;
p32 = zeros(length(tspan),length(A));
for k = 1: length(A)
tspan = 0: 0.1: 1;
r = 0.7;
r0 = 2;
c = [r r0];
[t1 p1] = ode45('height',tspan,c);
p32(k) = p(:,1);
end
I am getting the following errors:
"Unable to perform assignment because the left and right sides have a different number of elements."
"f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0"
How can I solve the problem?

Alan Stevens on 10 Jul 2021
Like this
A = 0:0.05:0.35;
tspan = 0: 0.1: 1;
p32 = zeros(length(tspan),length(A));
r = 0.7;
r0 = 2;
p0 = [r; r0];
for k = 1: length(A)
[t1, p1] = ode45(@(t,p) height(t,p,A(k)),tspan,p0);
p32(:,k) = p1(:,1);
end
plot(t1,p32),grid
xlabel('time'), ylabel('p32')
function pdot = height(~,p,A)
pdot = zeros(size(p));
pdot(1)=p(2);
B=5*A*p(1);
pdot(2)=B/5*p(2)^2-3*p(2)+6;
end
However, it fails with A larger than about 0.35. Have you got your equations right? (For example, should p(2)^2 be in the denominator?).
##### 1 CommentShowHide None
mathru on 10 Jul 2021
Hi Alan,
It works! Thanks for the code.
My original equation is too big. I just provided the first 3 terms.

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